Projector in QM

[Sammywu]

I saw some discussions about projectors in some threads. Also, the projector is used in this book to define pure state but did not provide what is a projector.

http://www.math.sunysb.edu/~leontak/book.pdf

Thru some math. check by assuming TR AP_\psi = (A (\psi) , \psi ) ( In this book, that is the expectation vale for AP_\psi ) , I got the answer by
P_\psi ( e_n ) = \sum_i c_i \overline{c_n } e_i
if
\psi= \sum_i c_i e_i
for an orthonormal basis { e_n } .

That sounds to be a good one for it.

Also, if
\psi_1 = c_1 \psi + c_2 \bot \psi
then
P_\psi ( \Psi_1) = c_1 \psi , where
( \bot\psi, . \psi ) = 0 .

Is this right?

[Eye_in_the_Sky]

... did not provide what is a projector.Definition: P is a "projector" if (and only if):

(i) Pt = P ,

and

(ii) P has eigenvalues 0 and/or 1 .

It then follows that P is a "projector" if, and only if, Pt = P and P2 = P.

--------------------

Thru some math. check by assuming TR AP_\psi = (A (\psi) , \psi ) ( In this book, that is the expectation vale for AP_\psi ) , I got the answer by
P_\psi ( e_n ) = \sum_i c_i \overline{c_n } e_i
if
\psi= \sum_i c_i e_i
for an orthonormal basis { e_n } .

That sounds to be a good one for it.I'm not quite sure what you are asking here. But all of the equations are correct. However, the "Trace" equation is said to be "the expectation value for A when the system is in the state ψ", not "the expectation value for APψ".

--------------------

Also, if
\psi_1 = c_1 \psi + c_2 \bot \psi
then
P_\psi ( \Psi_1) = c_1 \psi , where
( \bot\psi, . \psi ) = 0 .

Is this right?Yes.

[Sammywu]

Eye,

Thanks.

The error you pointed out was my typo. I will think about the link between the EQ. I derived and your defintion.

[Sammywu]

Eye,

Shall it be \overline{P^t} = P ?

Thanks

[Sammywu]

By the way, this condition is also needed in my EQ:

\sum_i c_i \overline{c_i } = 1

Otherwise, P_\psi ( \psi ) = \sum_i c_i \overline{c_i } \psi

Or, in general, the EQ shall be:

P_\psi ( e_n ) = ( 1 / \sum_j c_j \overline{c_j } ) \sum_i c_i \overline{c_n } e_i

[Sammywu]

Notation conversion between inner product and bra/ket for C and D belonging to the Hilbert Space.

( C , D ) = < D | C > = \sum_j c_j \overline{d_j }

if C = \sum_j c_j e_j and D = \sum_j d_j e_j

[Eye_in_the_Sky]

Eye,

Shall it be \overline{P^t} = P ?

ThanksIf by "t" you don't mean the "Hermitian transpose" (called the "adjoint"), then it shall.

------------------------

By the way, this condition is also needed in my EQ:

\sum_i c_i \overline{c_i } = 1 Oh, I was assuming that. It means that the state ψ is "normalized" (i.e. to unity).


Otherwise, P_\psi ( \psi ) = \sum_i c_i \overline{c_i } \psi Yes. BUT, then, for nontrivial Pψ (i.e. Pψ ≠ 0), it is not a projector (because the eigenvalue 1 has become Σi|ci|2).


Or, in general, the EQ shall be:

P_\psi ( e_n ) = ( 1 / \sum_j c_j \overline{c_j } ) \sum_i c_i \overline{c_n } e_i The factor in front should still be a 1 (not 1/Σj|cj|2). (BUT remember: this more general case is no longer a projector!)

------------------------

Notation conversion between inner product and bra/ket for C and D belonging to the Hilbert Space.

( C , D ) = < D | C > = \sum_j c_j \overline{d_j }

if C = \sum_j c_j e_j and D = \sum_j d_j e_j Looks fine.

[turin]

Does a projector have an inverse. I never thought of it before, but now I'm wondering ... (it doesn't seem to have an inverse)

[Eye_in_the_Sky]

P2 = P

implies

P-1P2 = P-1P

P = 1 ;

the only one with an inverse is the identity

(alternatively, you can say that on account of a 0 eigenvalue det = 0, and therefore, there is no inverse, except when P = 1)

(alternatively, you can say that when P ≠ 1 (and visualizing it geometrically) the mapping is MANY-to-ONE, and therefore has no inverse)

[Sammywu]

I think I derived a few things that seem important to me:

1). For \psi_1 and \psi_2 \in H ,

In order for
( P_\psi_1 + P_\psi_2 ) = P_{\psi_1+\psi_2}
then
\psi_1 \bot \psi_2 i.e. ( \psi_1 , \psi_2 ) = 0

2). From that,

\sum_n P_{e_n} = P_{\sum_n e_n}

[Sammywu]

I verified that:

For \psi and \psi_n \in H ,

where \psi_n are eigenbasis of A,

this holds true:
< A | P_\psi > = \sum_n c_n \overline{c_n} E_n

where
\psi = \sum_n c_n \psi_n .

[Sammywu]

A question is now whether P_{e_n} can serve as a basis ( or generator ) of GL(H) or A(H)?

[Sammywu]

If
\sum_n a_n P_{e_n} = P_{\sum_n a_n e_n}
is true, then P_{e_n} can not serve as a basis, because all its linear combinations are also projectors then.

But 2 P_{e_n} does not seem to be the projector for
2 e_n ; so in general, there seems to be possibility.

[Eye_in_the_Sky]

I think I derived a few things that seem important to me:

1). For \psi_1 and \psi_2 \in H ,

In order for
(a) ( P_\psi_1 + P_\psi_2 ) = P_{\psi_1+\psi_2}
then
(b) \psi_1 \bot \psi_2 i.e. ( \psi_1 , \psi_2 ) = 0
Do you mean "(a) implies (b)", "(b) implies (a)", or both ?


2). From that,

\sum_n P_{e_n} = P_{\sum_n e_n}
Yup.

--------------------------

I verified that:

For \psi and \psi_n \in H ,

where \psi_n are eigenbasis of A,

this holds true:
< A | P_\psi > = \sum_n c_n \overline{c_n} E_n

where
\psi = \sum_n c_n \psi_n .
You didn't mention that the En are the eigenvalues of A. (... I assume you assumed <ψ|ψ> = 1.)

--------------------------

... All of this will become much, much simpler once you start using Dirac notation.

--------------------------------------------

[Sammywu]

Eye,

Thanks.

I have a question for the lemma 1.1 at page 40 in that book.

I don't think there is anythings that says the eigenbasis of a self-adjoint operator can always span the entire Hilbert space.

If that's the case, I can write a

\psi = \sum_i a_i \Psi_i + b \psi_\bot

where

\Psi_\bot \bot \Psi_n , for all n.

( M (\psi) , \psi )
might not equal to
( \sum_i a_i P_\psi_n ( \psi ) , \psi ) .

[Sammywu]

Eye,

What I found is below is the sufficient condition for this Lemma to be true:

( M (\psi) , \psi_\bot ) = 0

Any Way, it there is a countable basis, this seems to be true.

I aslo found I can easily set up a self-adjoint operator that only has limited number of eigen values and map the rest of basis to zero.

[vanesch]

Does a projector have an inverse. I never thought of it before, but now I'm wondering ... (it doesn't seem to have an inverse)

No it doesn't, except for the trivial projector on the whole space. The "feel it in the bones" argument is that when you project something, you don't know what the orthogonal component was (the one that has been filtered out).
The mathematical argument is of course that a non-trivial projector has 0 eigenvalues (of the orthogonal directions that have been filtered out!).

In fact, this property is at the heart of the "measurement problem" in QM: a projection (such as happens in von Neuman's view) can never be the result of a unitary evolution.

cheers,
Patrick.

[Eye_in_the_Sky]

A question is now whether P_{e_n} can serve as a basis ( or generator ) of GL(H) or A(H)?
GL(n,C) is a group with respect to matrix multiplication. It is not "closed" under matrix addition, and therefore, does not have the vector space structure you are assuming in your question. On the other hand, the full set of n x n matrices with entries from C, M(n,C), is "closed" under matrix addition (as well as, multiplication, of course). So, you might want to pose your question with respect to M(n,C).

In that case, for n > 1, the answer is "no". M(n,C) has dimension n2, whereas, the Pi will span a subspace with dimension no larger than n (of course, the Pi are in fact linearly independent, so they will span a subspace of dimension equal to n).

[Eye_in_the_Sky]

If

[1] \sum_n a_n P_{e_n} = P_{\sum_n a_n e_n} ...
The object on the left-hand-side is not (in general) a projector. That object has eigenvalues an , whereas a projector has eigenvalues
0 and / or 1.

--------------------------

At this juncture, it is instructive to consider ordinary 3-D Euclidean space. Pick any unit vector n. Then, the projector corresponding to this unit vector is given by

[2] Pn(v) = (v∙n) n , for any vector v .

The description of [2] is "the projection of v along n". Do you remember what this means geometrically? (see figure (http://www.physicsforums.com/attachment.php?attachmentid=1248&stc=1)).

--------

NOTE:

In Dirac notation, [2] becomes

Pn|v> = <n|v> |n> = |n> <n|v> = (|n><n|) |v> , for any |v> .

We therefore write:

Pn = |n><n| .

If you think of each ket as a column matrix and the corresponding bra as its Hermitian transpose (a row matrix) then this notation can be taken "literally".

--------------------------

I suggest you reserve the symbol "P", in the above type of context, only for a "projector" proper. Also, I suggest you invoke the rule that the "subscript" of P is always a "unit" vector. These two prescriptions would then disqualify the "legitimacy" of the right-hand-side of [1] on both counts.

At the same time, if you want to consider generalizations for which a relation like [1] holds, then use a symbol "R" (or whatever else) instead of "P". The generalization of [2] which gives a relation like [1] is then simply:

[2'] Ru(v) = [ v ∙ (u/|u|) ] u , for any vector v .

But what is the motivation for reserving a special "symbol" for this operation? Its description is "project the vector v into the direction of u and then multiply by the magnitude of u". The meaningful aspects of this operation are much better expressed by writing the corresponding operator as |u|P(u/|u|).

--------------------------

Now, let's go back the definition I gave in post #2.


Definition: P is a "projector" if (and only if):

(i) Pt = P ,

and

(ii) P has eigenvalues 0 and/or 1 .

It then follows that P is a "projector" if, and only if, Pt = P and P2 = P.
I am now strongly suggesting that we, instead, use the following as our "official" definition:

*************************
* 1a) Given any unit vector e, definite the "projector onto e" by:
*
* Pe(v) = (v,e) e , for any vector v .
*
* Such a projector is said to be "1-dimensional".
*
* 1b) An operator P is said to be a "projector" if (and only if)
* it can be written as a sum of 1-dimensional projectors
* which project onto mutually orthogonal unit vectors.
*
*************************

This definition [1a) and 1b) taken together] is equivalent to the original one I gave. But I think it makes the meaning of "projector" much clearer.

------------------------------------------------------

All that I have said above should clarify matters like:



But 2 P_{e_n} does not seem to be the projector for
2 e_n ...
-----------

[Eye_in_the_Sky]

I don't think there is anythings that says the eigenbasis of a self-adjoint operator can always span the entire Hilbert space.
[Note: you wrote "eigenbasis", when you meant "eigenvectors".]

The answer to your query is given in that book by Theorem 1.1 (called "The Spectral Theorem"), on p. 38. In simple language, it is saying that the answer is: "Yes, a self-adjoint operator will always have eigenvectors (or "generalized" eigenvectors) spanning the entire Hilbert space."

HOWEVER, you must NOTE that the definition of "self-adjoint" (in the case of an infinite-dimensional Hilbert space) is nontrivial (... in that book, the appropriate definition is given at the top of p. 36, in 1.1.1 Notations).

------------------------

For the sake of giving you (at least) something, here is some basic information. Let A be a linear operator acting in the Hilbert space.

Definition: A is "symmetric" iff <g|Af> = <Ag|f> for all f,g Є Domain(A).

Definition: A is "self-adjoint" iff: (i) A is symmetric; (ii) the "adjoint" At exists; and (iii) Domain(A) = Domain(At).

Lemma: At, the "adjoint" of A, exists iff Domain(A) is dense in the Hilbert space.

All that is missing in the above is a definition of "adjoint" (which I have omitted for the sake of brevity and simplicity). That definition would then give us a specification of Domain(At) and thereby complete the definition of "self-adjoint".

------------------------

Now, you might ask: How can it be that there is a linear operator A with a domain "smaller" than the whole Hilbert space, yet, at the same time, A has eigenvectors which span the entire space?

Well, first of all, this can only happen in an infinite-dimensional Hilbert space. Suppose A has eigenfunctions φn(x) with corresponding eigenvalues an. So,

[1] Aφn(x) = anφn(x) .

Since the φn(x) span the entire space, an arbitrary element ψ(x) of the space can be written as

[2] ψ(x) = Σn cnφn(x) .

The right-hand-side of [2] is an infinite sum, and, therefore, involves a limit. While every finite subsum is necessarily in the domain of A, it is possible that in the limit of the infinite sum, the resulting vector is no longer in that domain. ... As you can see, this sort of phenomenon can only occur when the Hilbert space is infinite-dimensional.

But what do we get if we, nevertheless, attempt to "apply" A to ψ(x) by linearity and use [1]? Let's try it:

Aψ(x) = Σn cnAφn(x)

= Σn ancnφn(x) [3] .

As you may have guessed, when ψ(x) is not in Domain(A), the following occurs in [3]: while every finite subsum is necessarily an element of the Hilbert space, in the limit of the infinite sum the "result" is no longer in the Hilbert space.

--------

[Sammywu]

Eye,

I wanted to print your response in oder to read it clearer. Unfortunatelly, I got some troubles with my printer. Hopefully I can print it tomorrow.

One of the issue I saw in your last response is that you seem to extend your defintion of self-adjoint to "non-trivial" one, my guess is you want to extend that to the one with eigenvectors that can span the entire Hilbert space.

Is it necessary?

Because if you do so, then P_{e_n} is not self-adjoint then. Note it has only e_n as its eigen vector.

I did overlook that theorem, I shall look closer into it.

Any way, I found these facts that help me to see it clearer:
Below, I was assuming a relaxed self-adjoint definition.

1).
\sum_n a_n P_{e_n}
can be simply represented as the diagonal matrix as diag( a_1, a_2, a_3 ... ) .

So, it actually spans the group ( or ring ) formed by the matrix with only diagonal elements with values.

2). In general, they are not self-adjoint unless all a_i are real.

3). If all a_i are real, for any a_i not zero, e_i is one of its eigenvectors.

4). In particular, if
\sum_n a_n = 1 and a_n >= 0
then it's a "state". If more than one a_n > 0 , then it's a mixed state.

5). This also tells me that a set of { P_{e_n} } is defintely not enough to span ( or generate ) all states, even though the { e_n } can span the Hilbert Space.

[Sammywu]

Eye,

By the way, are you Leon?

In the page 37, the mapping defining projection valued measure is not necessary 1-1 & onto, right?

[Eye_in_the_Sky]

Because if you do so, then P_{e_n} is not self-adjoint then. Note it has only e_n as its eigen vector.Every vector orthogonal to en is an eigenvector of Pn with eigenvalue zero. Clearly, Pn has a complete set of eigenvectors.


1).
\sum_n a_n P_{e_n}
can be simply represented as the diagonal matrix as diag( a_1, a_2, a_3 ... ) .

So, it actually spans the group ( or ring ) formed by the matrix with only diagonal elements with values.

2). In general, they are not self-adjoint unless all a_i are real.

3). If all a_i are real, for any a_i not zero, e_i is one of its eigenvectors.

4). In particular, if
\sum_n a_n = 1 and a_n >= 0
then it's a "state". If more than one a_n > 0 , then it's a mixed state.

5). This also tells me that a set of { P_{e_n} } is defintely not enough to span ( or generate ) all states, even though the { e_n } can span the Hilbert Space.1) The only problematic part here is the expression "spans the group (or ring)". True, the said objects form a group with respect to +, and monoid with respect to ∙ , and that defines a ring. But when you talk about spanning, you are thinking of the of the group aspect (with the + operation) over the field C. This gives a vector space ... and if you want to acknowledge the monoid aspect with respect to ∙ , then it's called an (associative) algebra. In short, the simplest correct thing to say is:

So, it actually spans the vector space formed by the matrices ... over C.

(... if you have a "thing" for such terminologies try mathworld (http://mathworld.wolfram.com/))

2) True.

3) This is the same error as the one identified at the beginning of this post ... eigenvalues can be 0 (it's the eigenvectors! which can't).

4) True.

5) True, the Pn are not enough. But what is the "this" that tells you?

[Sammywu]

Eye,

Got you. My error was assuming zero can not be eigenvalue.

[Sammywu]

Eye,

About (5), what it means to me is that I will have no gurantee that I can generate a "position" eigenstate by linear combination of "energy"'s eigenstates even though I can generate its eigenfunction by "energy"'s eigenfunction. Or, a mixed state by "energy"'s eigenfunction won't equal to any combination of "postion"s' pure states.

I followed through the rest to page 45. Operators P and Q should be unbounded. What does it exactly mean by that?

Thanks

[Eye_in_the_Sky]

By the way, are you Leon?No. (... I think he is too busy writing difficult eBooks to be in the forum.)

-----------------------

In the page 37, the mapping defining projection valued measure is not necessary 1-1 & onto, right?It appears to me that such a mapping can never be one-to-one. It is certainly never onto.

-----------------------

Operators P and Q should be unbounded. What does it exactly mean by that?Definition: A linear operator L is bounded iff there exists a constant C such that

(Lψ, Lψ) < C (ψ, ψ) , for all ψ Є H .

Can you see that there is no such constant C for Q or for P?

[Sammywu]

Eye,

I take definition of inner product in L2(R, dq) as
( \varphi, \psi ) = \int \overline{\psi} \varphi dq
.

(q \psi , q \psi ) = \int \overline{q \psi } q \psi dq
= \int \overline{q} \overline{\psi} q \psi dq
= \int q^2 \overline{\psi } \psi dq

because q is real.

If it's bounded, then

\int ( q^2 - C ) \overline{\psi } \psi dq < 0
.

Now all I need to do is proving this is not possible for a C exits.

Am I on right track?

Thanks

[Eye_in_the_Sky]

I think that this track will lead you to a proof that Q is an unbounded operator.

[Sammywu]

Eye,

Now to keep it easy, I pick a
\psi = ( | q - q_0 | e^{-(q-q_0)^2 })^{1/2}
.

It's easy to prove that
( \psi , \psi ) = 1
.

Considering in the interval
[ q_0-1 , q_0 +1 ] ,
\int_{q_0-1}^{q_0+1) q^2 * | \psi |^2 >= ( q_0 - 1)^2 * 2 * 1/2 * 1/e
.

I can prove ( q \psi , q \psi ) > ( q_0 - 1 )^2 / e
.

This is definitely unbounded, because all I need to do is moving q_0 to one end of R line, this value will grow with it without bound.

Does this look fine?

Thanks

[Sammywu]

OK. If that shows how Q is an unbounded operator.

What I did not show is how the commutator of two bounded operators can not be I. At this point, all I find is
( (AB-BA) \psi , (AB-BA) \psi ) = ( AB \psi, AB \psi) + (BA \psi, BA \psi )
.

So if AB-BA = I, then 1/2 <= C_1 * C_2
. Not further.

Any way, I guess originally I was bothered is because the definition of < Q | M > = TR QM and only bounded operators' trace was defined.

So, here the problem is actually QM need to be bounded even if Q is not bounded.

[Sammywu]

So, I just try to see whether QM is always of trace class.

Let M = \sum_n a_n P_{\psi_n}
where
\sum_n a_n = 1

TR QM = \sum_n ( Q \sum_i a_i P_{\psi_i} \psi_n , \psi_n )
= \sum_n ( Q a_n \psi_n , \psi_n )
= \sum_n a_n ( Q \psi_n, \psi_n )

Assuming
b_n = | ( Q \psi_n, \psi_n ) |
, whether QM is of trace class will depends on whether
\sum_n a_n b_n converges.

So, if I can find a set of \psi_n such that b_n = 2/a_n , then I have a QM not of trace class.

[Eye_in_the_Sky]

Now to keep it easy, I pick a
\psi = ( | q - q_0 | e^{-(q-q_0)^2 })^{1/2} This is not easy.


( \psi , \psi ) = 1 True.


Considering in the interval
[ q_0-1 , q_0 +1 ] ,
q^2 * | \psi |^2 >= ( q_0 - 1)^2 * 2 * 1/2 * 1/e How so? For q = qo the LHS is 0.

------
... The idea behind the proof will work. But for technical reasons the proof has failed. Why not really keep it easy and choose ψ like below?

ψ(q)

= 1 , q Є (qo, qo + 1)
= 0 , otherwise

Then (Qψ, Qψ) > qo2 .

------------------------------------

Define: ║φ║ = √(φ, φ)

Then: ║φ + ξ║ ≤ ║φ║ + ║ξ║

............. :surprise:

------------------------------------

So, I just try to see whether QM is always of trace class.In the most general case, QM may or may not be.

In the following, you carry out the inquiry well:

So, I just try to see whether QM is always of trace class.

Let M = \sum_n a_n P_{\psi_n}
where
\sum_n a_n = 1

TR QM = \sum_n ( Q \sum_i a_i P_{\psi_i} \psi_n , \psi_n )
= \sum_n ( Q a_n \psi_n , \psi_n )
= \sum_n a_n ( Q \psi_n, \psi_n )

Assuming
b_n = | ( Q \psi_n, \psi_n ) |
, whether QM is of trace class will depends on whether
\sum_n a_n b_n converges.Remember also that M is a state. So we also have

0≤an≤1 as well as ∑nan=1 .

But still, this is not enough. In general the series can still diverge.

This tells us that our current definition for a state M is still too general. While all physical states do satisfy the definition of M, not all M's are physical states.

[Sammywu]

Eye,

Note I just added an integral sign in front.

Considering in the interval
[ q_0-1 , q_0 +1 ] ,
\int_{q_0-1}^{q_0+1} q^2 * | \psi |^2 >= ( q_0 - 1)^2 * 2 * 1/2 * 1/e

Any way, I agree your proof is much easier and quicker than mine.

I need to read your response more thoroughly.

It seems you used the extended triangular inequality; that was what I was thinking yesterday but can't get it proved and working.

I did branch out to think the issue of mixed state:

< Q , P_{\sum_n a_n \psi_n} > = TR \ QP_{\sum_n a_n \psi_n} =
\sum_n ( QP_{\sum_i a_i \psi_i} \psi_n , \psi_n ) = \sum_n ( Q \sum_j a_j \overline{a_n} \psi_j , \psi_n ) =
\sum_n \overline{a_n} ( \sum_j a_j Q \psi_j , \psi_n ) = \sum_n \overline{a_n} \sum_j a_j ( Q \psi_j , \psi_n )

<Q | M > = < Q , \sum_n a_n P_{\psi_n } > = TR \ Q \sum_n a_n P_{\psi_n} =
\sum_n a_n ( Q P_{\psi_n} \psi_n , \psi_n ) = \sum_n a_n ( Q \psi_n , \psi_n )

This shows that even though P_{\sum_n a_n \psi_n} \ and \ \sum_n a_n P_{\psi_n } , but they do have the same expectation value.

Now, for P_\sum_n a_n \psi_n to be a state, (\sum_n a_n \psi_n , \sum_n a_n \psi_n ) = \sum a_n \overline\a_n = \sum a_n^2 \ needs \ to \ be \ one . This condition is different from the condition for M to be a state in that TR \ M = \sum_n (\sum_i a_i P_\psi_i \psi_n , \psi_n ) = \sum a_n needs \ to \ be \ one .

I thought I might be able to show some Q wll have the same expectation values for the mixed state and the pure state. Apparently this is a little tedious than I thought.

[Sammywu]

Eye,

The previous one is a little messy. I was trying to see whether the '//' will give me a new line.

So, I redo it here.

I did branch out to think the issue of mixed state:

< Q , P_{\sum_n a_n \psi_n} > = TR \ QP_{\sum_n a_n \psi_n} =
\sum_n ( QP_{\sum_i a_i \psi_i} \psi_n , \psi_n ) =
\sum_n ( Q \sum_j a_j \overline{a_n} \psi_j , \psi_n ) =
\sum_n \overline{a_n} ( \sum_j a_j Q \psi_j , \psi_n ) =
\sum_n \overline{a_n} \sum_j a_j ( Q \psi_j , \psi_n )

<Q | M > = \sum_n a_n ( Q \psi_n , \psi_n )

as we alreday showed in earlier one.


Now, for P_\sum_n a_n \psi_n to be a state,
(\sum_n a_n \psi_n , \sum_n a_n \psi_n ) =
\sum a_n \overline{a_n} = \sum a_n^2 [\tex]
needs to be one.
This condition is different from the condition for M to be a state in that
[tex] TR \ M = \sum_n (\sum_i a_i P_\psi_i \psi_n , \psi_n ) =
\sum a_n
needs to be one.

I thought I might be able to show some Qs wll have the same expectation values for the mixed state and the pure state. Apparently this is a little tedious than I thought.

Any way, why do you say I is unbounded in a infinte dimensional space?

( I \psi , I \psi ) = ( \psi , \psi ) < 2 ( \psi , \psi )

whether the space is infinte or finite dimensional.

Thanks

[Sammywu]

Eye,

Changing my tactic, I gathered some facts:

Let
\psi = \sum b_i \psi_i
where
b_i \overline{b_i} = a_i
and
\sum b_i \overline{b_i} = 1
.

Now, this can be changed to:

< Q , P_{\sum_n b_n \psi_n} > = TR \ QP_{\sum_n b_n \psi_n} =
\sum_n ( QP_{\sum_i b_i \psi_i} \psi_n , \psi_n ) =
\sum_n ( Q \sum_j b_j \overline{b_n} \psi_j , \psi_n ) =
\sum_n \overline{b_n} ( \sum_j b_j Q \psi_j , \psi_n ) =
\sum_n \overline{b_n} \sum_j b_j ( Q \psi_j , \psi_n )

Compare that to:

<Q | M > = \sum_n a_n ( Q \psi_n , \psi_n )

If I set
Q = \sum c_n P_{\psi_n}
, then
< Q | P_\psi > = \sum_n \overline{b_n} b_n c_n

and

< Q | M > = \sum_n a_n c_n

They are the same.

So for any Qs as \sum c_n P_{\psi_n} , we will have the same expectation value for the mixed state and the pure state.

[Sammywu]

Eye,

You know sometimes this latex thing is strange.

My question is:

Any way, why do you say I is unbounded in a infinte dimensional space?

I mean,

( I \psi , I \psi ) = ( \psi , \psi ) < 2 ( \psi , \psi )

whether the space is infinte or finite dimensional.

Thanks

[Sammywu]

Any way, back to the issue between the "mixed" state and the "pure" states, further questions shall be:

1). Can we relax the conditions for the "pure" states and the observables A ( I wrote Q earlier , since this is a general observable not the "position", I think A is better)?

2). Even though they have the same expection value, shall they have different distribution such like < A | M > might have a multi-nodal distribution and < A | P_\psi > has a central normal distribution?

[Eye_in_the_Sky]

Any way, why do you say I is unbounded in a infinte dimensional space?

I mean,

( I \psi , I \psi ) = ( \psi , \psi ) < 2 ( \psi , \psi ) I forgot to mention the following:

Metatheorem: A necessary condition for I to be unbounded (in the ∞-dimensional case) is serious confusion. :surprise:


I'll have to fix that.

[Sammywu]

Eye,

Now, let me take an example.

Assuming two free particles,

\psi_1 = \int \int \delta ( k - k_1 , w - w_1 ) e^{i(k(x-x_1)+w(t-t_1))} dk dw

and

\psi_2 = \int \int \delta ( k - k_2 , w - w_2 ) e^{i(k(x-x_2)+w(t-t_2))} dk dw

represent their wave functions; their states shall be
P_{\psi_1} \\ and \\ P_{\psi_2} .

In combination, this mixed state M = 1/2 ( P_{\psi_1} + P_{\psi_2} ) shall represent their combined state, or the ensemble.

Correct?

[Eye_in_the_Sky]

I see you are still using that notation where P is not a projector and its subscript isn't a unit vector:

P_{\sum_n a_n \psi_n} .

According to that notation, is it not true that the preceding expression equals

\sum_n a_n P_{\psi_n } ?

So, why don't you just use the second one?

-------------------------------

The step below has an error:


\sum_n ( QP_{\sum_i a_i \psi_i} \psi_n , \psi_n ) = \sum_n ( Q \sum_j a_j \overline{a_n} \psi_j , \psi_n ) I think this error results from what I just pointed out above, that you are treating this P as a projector when it is not! I also explained this same point in post #19:

[1] \sum_n a_n P_{e_n} = P_{\sum_n a_n e_n}

The object on the left-hand-side is not (in general) a projector. That object has eigenvalues an , whereas a projector has eigenvalues 0 and / or 1.

---------------

[2] Pn(v) = (v∙n) n , for any vector v

----------------

I suggest you reserve the symbol "P", in the above type of context, only for a "projector" proper. Also, I suggest you invoke the rule that the "subscript" of P is always a "unit" vector. These two prescriptions would then disqualify the "legitimacy" of the right-hand-side of [1] on both counts.

At the same time, if you want to consider generalizations for which a relation like [1] holds, then use a symbol "R" (or whatever else) instead of "P". The generalization of [2] which gives a relation like [1] is then simply:

[2'] Ru(v) = [ v ∙ (u/|u|) ] u , for any vector v .

But what is the motivation for reserving a special "symbol" for this operation? Its description is "project the vector v into the direction of u and then multiply by the magnitude of u". The meaningful aspects of this operation are much better expressed by writing the corresponding operator as |u|P(u/|u|).Do you follow what I am saying?

------

All of this means there is no second condition ∑nan2=1. (In which case, your post #35 no longer stands (I think :confused: ).)

[Sammywu]

Eye,

Yes. I followed what you say. The correct defintion of projector here has a necessary condition in that the norm of the ket needs to be one or in other words it's a normal vector.

I correct that in the next post. \sum b_i \overline{b_i} = 1 gurantees that \psi is a normal vector.

Thanks

[Sammywu]

Eye,

You are right.

I shall have used the second one to differentiate it from P.

OK.

[Eye_in_the_Sky]

Any way, back to the issue between the "mixed" state and the "pure" states, further questions shall be:

1). Can we relax the conditions for the "pure" states and the observables A ( I wrote Q earlier , since this is a general observable not the "position", I think A is better)?

2). Even though they have the same expection value, shall they have different distribution such like < A | M > might have a multi-nodal distribution and < A | P_\psi > has a central normal distribution?In 1): relax how? ... or is that what 2) is explaining?

I don't understand 2).

[Sammywu]

Eye,

1) Relax.. means I could even find simpler conditions such as maybe all vectors in H_M can be treated and found a role in this issue. Or maybe A as any function of P_{\psi_n} can satisfy similar property.

My guess to the question is probably NO. The conditions I found pretty describe the situations of indistinguishable "mixed" and "pure" states.

2). What I am saying is even though their expectation values are the same. The statistics will show us two different distributions. So now it's time to investigate their probability decomposition of AM and AP_\psi .

Thoughts came out from me sometimes just are not described immediately in correct math. language but a pragmatic thought to begin with. So, I wrote distribution of < A | M >, things like that.

Multi-nodal distribution is my words for the distribution that you see multiple distingushiable points like two clear different frequency of light pulse will leave two lines in the light spectrum.

Central normal distribution is my words to emphasize that a normal probability distribution has a central node with an inverse exponetial shape of distribution.

I am sorry I do not remember what are correct math. terms for them.

Regards

[Sammywu]

The example I show is two particles shot rightward at different times and at time T we shine rays of lights from top downward. Now will we predict from this math. the light detector at the bottom will show two spots instead one spot.

[Eye_in_the_Sky]

A pure state is represented by a unit vector |φ>, or equivalently, by a density operator ρ = |φ><φ|. In that case, ρ2 = ρ.

Suppose we are unsure whether or not the state is |φ1> or |φ2>, but know enough to say that the state is |φi> with probability pi. Then the corresponding density operator is given by

ρ = p1|φ1><φ1| + p2|φ2><φ2| .

In that case ρ2 ≠ ρ, and the state is said to be mixed. Note that the two states |φ1> and |φ2> need not be orthogonal (however, if they are parallel (i.e. differ only by a phase factor), then we don't have mixed case but rather a pure case).

This is the real place to begin. And Dirac's notation is the superior way of representing these objects.

--------------------

Tell me, Sammy, have you learned the postulates of Quantum Mechanics in terms of simple, basic statements like the one below?

The probability of obtaining the result an in a measurement of the nondegenerate observable A on the system in the state |φ> is given by

P(an) = |<ψn|φ>|2 ,

where |ψn> is the eigenvector corresponding to the eigenvalue an.

[Sammywu]

Eye,

I have seen that postulate but not fully convinced in that and tried to understand that and see whether my interpretation is correct.

So, I try to translate that here:

P(a_n) = ( \int \overline{\psi_n} \varphi )^2

Then
a_n P(a_n) = a_n ( \int \overline{\psi_n} \varphi )^2 =
( \int \overline{\psi_n} (a_n)^{1/2} \varphi )^2

Is there something wrong with my translation?

It seems that this is more likely.
P(a_n) = \int \overline{\psi_n} \varphi

[Sammywu]

I thought :

< A | M > = < \varphi | A | \varphi > =
\int \overline{\varphi} (A) \varphi

In order to lead us there.

If
P(a_n) = \int \overline{\psi_n} \varphi

Then
\sum_n a_n P(a_n) =
\int \sum_n ( a_n \overline{\psi_n} ) \varphi

This might lead us there because a_n \psi_n = A \psi_n .

[Sammywu]

Eye,

If I use Dirac notation,

< \varphi | A | \varphi > =

< \varphi | A \sum_n | \psi_n > < \psi_n | \varphi> =

\sum_n < \varphi | A | \psi_n > c_n =
\sum_n \overline{c_n} c_n < \psi_n | A | \psi_n > =
\sum_n \overline{c_n} c_n a_n =
\sum_n | < \psi_n | \varphi > |^2 a_n


Assuming < \psi_n | \varphi > = c_n
, so
| \varphi > = \sum_n c_n | \psi_n >
.
That do agree with your formula.

Does that look good to you?

But I do have some troubles to show that in an integral of Hermitian OP.

[Sammywu]

Eye,

Back to what you said, the mixed state does not seem to apply to my proposed case.

I was confused about how we can use this mixed state. Now I have better idea but i still want to think further about it.

Any way, I thought the translation between the three notation is :

( u , v ) = < v | u > = \int \overline{v} u

Is this correct?

[Sammywu]

Eye,

Is this where I did wrong?

I think I shall start with this.

P(a_n) =
( \int \overline{\psi_n(x)} \varphi (x) dx ) \overline{ ( \int \overline{\psi_n(y)} \varphi(y) dy ) }

\sum_n a_n c_n \overline{c_n} =
\sum_n a_n P(a_n) =
\sum_n a_n \int \int \overline{\psi_n(x)} \varphi(x) \psi_n(y) \overline{\varphi(y)} dx dy =
\sum_n \int \int \overline{\psi_n(x)} \varphi(x) A \psi_n(y) \overline{\varphi(y)} dx dy =

I still need to see how this can be :

\int \overline{\varphi(x)} A \varphi(x) dx



Thanks

[Sammywu]

Eye,

Based on the two assumptions

\varphi = \sum_n c_n \psi_n
and
\int \overline{\psi_i} \psi_j = \delta_{i,j}
,

\int \overline{\varphi} A \varphi =
\int \overline{\varphi} A \sum_n c_n \psi_n =
\int \overline{\varphi} \sum_n c_n a_n \psi_n =
\sum c_n a_n \int \overline{\varphi} \psi_n =
\sum c_n a_n \overline{ \int \varphi overline{\psi_n} =
\sum c_n a_n \overline{c_n}

So, I verified all of these three approaches show the same expectation value.

Now, back to your question, why shall
P(a_n) = c_n \overline{c_n}
?

Any good argument about it?

[Sammywu]

Eye,

Now I see where you are leading.

The probability decomposition for AM will be

P(a) =
\sum_{a_n<=a} P(a_n) =
\sum_{a_n<=a} a_n c_n \overline{c_n}
.

Is it?

[Sammywu]

Now, if I use my original model of
M = 1/2 ( P_{\psi_1} + P_{\psi_2} )
or
M = 1/2 | \psi_1> < \psi_1 | + 1/2 | \psi_2> < \psi_2 |
even though
< \psi_2 | \psi_1 > not = 0
and let
P_{XM}(a)
be its probability decomposition,
this does seem to lead to a two spots measurement.

A trouble I might need to take care of is the two wavefunctions are not orthogonal.

While I recall that a multiple particle model in a book using a model
\psi = \psi_1 \psi_2
, I wonder how these two different models will turn out in this case.

[Sammywu]

Any way, back to the basic postulate, it does now seem very reasonable for probability for eigenvalue

P(a_n) = < \psi_n | \varphi > < \varphi | \psi_n > = ( \psi_n , \varphi) ( \varphi , \psi_n ) =
\int \overline{\psi_n} \varphi \int \psi_n \overline{\varphi}

[Eye_in_the_Sky]

If I use Dirac notation,

< \varphi | A | \varphi > =

\sum_n | < \psi_n | \varphi > |^2 a_n

Assuming < \psi_n | \varphi > = c_n
, so
| \varphi > = \sum_n c_n | \psi_n >

Does that look good to you?It is correct.
_________

Any way, I thought the translation between the three notation is :

( u , v ) = < v | u > = \int \overline{v} u

Is this correct?Looks fine.
_________
_________

... back to the basic postulate, it does now seem very reasonable for probability for eigenvalue

P(a_n) = < \psi_n | \varphi > < \varphi | \psi_n > = ( \psi_n , \varphi) ( \varphi , \psi_n ) =
\int \overline{\psi_n} \varphi \int \psi_n \overline{\varphi} The basic postulates are the "true" starting point of Quantum Mechanics. From those postulates, one can then build the more complex versions which talk about "mixed states" and "expectation values" of observables ("nondegenerate" and "degenerate" cases) ... like on page 37 of that eBook.

Try to find a book in which those basic postulates are written in clear, concise terms. Those postulates should talk about a nondegenerate observable with a discrete set of eigenvalues, where the quantum system is in a pure state. There should also be a statement concerning the time evolution of a pure state in terms of the Schrödinger equation.

Once you understand and accept those basic postulates, then from them you will be able to derive - in a most transparent way - all of the more complex versions.

From that perspective, everything will be much clearer. I am quite sure of this.

[Sammywu]

Eye,

I think I have a book that shall have a discussion of that postulate. I shall have read it but just going over it without too much deeper thought.

I always wanted to try thinking things in my logical reasoning and then compare that to what is there any way. So I tried to analyze what it is here.

What I have here is a state \varphi representing a probability and all I know is
( \varphi , \varphi ) =
< \varphi , \varphi > =
\int \overline{\varphi} \varphi = 1
.
Now with an observable A, I might have different observated value a_n and I can associate with each one of them with a state \psi_n .

By the assumption of { \psi_n } being a orthonornal basis,

\varphi = \sum_i c_i \psi_n
.

The total probability as unity shall be decomposed into components representing probability for each a_n .
1 = (\varphi, \varphi ) = ( \sum_n c_n \psi_n , \varphi ) = \sum_n c_n ( \psi_n , \varphi )
.

Since this decomposition of unity has a coresponding number to each a_n , I can assume that the probaility for each a_n will be
c_n ( \psi_n , \varphi ) = ( \varphi , \psi_n ) (\psi_n , \varphi )
.

While analyzing this, there seems to be a condition that shall be in effect, i. e all of different possible outcomes as a_n need to be independent to each other and that's what orthogonality of eigenfunctions guranteed.

If the a_n are not independent to each other, i.e. { \psi_n } is not orthonormal, then there will be some interference can be derived here.

Any way, this is my own way to try to comprehend this postulate, but it does seem to make it clearer how this postulate was formulated and also indicate how the wavefunction interference might have come into play.

[Sammywu]

Another important fact I noticed is that c \psi_n is also an eigenfunction and regarded the same as \psi_n as logon as | c | = 1; this is related to what is stationary state.

So \varphi can be decomposed into a way such that all c_i are real. Still \sum_i c_i does not equal to one but \sum_i c_i^2 = 1 so that it will show me why [/tex] c_i^2 [/tex] not c_i is the probability.

My intuition is c_i shall be the probability but not c_i^2, this makes it clearer to me why my intuition is wrong.

[Sammywu]

I think I have a book that shall have a discussion of that postulate. I shall have read it but just going over it without too much deeper thought.

I always wanted to try thinking things in my logical reasoning and then compare that to what is there any way. So I tried to analyze what it is here.

What I have here is a state \varphi representing a probability and all I know is
( \varphi , \varphi ) =
< \varphi , \varphi > =
\int \overline{\varphi} \varphi = 1
.
Now with an observable A, I might have different observated value a_n and I can associate with each one of them with a state \psi_n .

By the assumption of { \psi_n } being a orthonornal basis,

\varphi = \sum_i c_i \psi_n
.

The total probability as unity shall be decomposed into components representing probability for each a_n .
1 = (\varphi, \varphi ) = ( \sum_n c_n \psi_n , \varphi ) = \sum_n c_n ( \psi_n , \varphi )
.

Another way to look into this:

An abstract state can be represented by different orthonormal basis. By an unitary transformtion, we can translate a state to different representation in another basis, but its norm remians as one. So the square of its coefficiens can be conviniently used for probability representation. The basis is used as the "coordinate" of the measurable.

For example, Fourier transformation as an unitary transformation can transform a "position" represtation to a "momentum" representation.

An operator as an obserable can be only measured as certain real values; these are the eigenvalues and they work just like the "coordinate" of the measurement. Only certain probability distribution ( i.e. states ) can be exacted to one of these real values: they are the eigenfunctions and pure states.

For example, the eigenfunction of "position" observable of x_0 can be seen as the \delta ( x - x_0) .

When measuring other states, only eigenvalues will appear, but since it has non-zero coefficients on different eigenvalues so it will show a dstribution among these eigen values.

An degenerate observable has more than two orthogonal functions measured the same value, so its probability measured for this value shall be the sum of the square of their ciefficients.

[Eye_in_the_Sky]

... there seems to be a condition that shall be in effect, i. e all of different possible outcomes as a_n need to be independent to each other and that's what orthogonality of eigenfunctions guranteed.I am not quite sure what you mean by "independent of each other". If the |ψn> are not all mutually orthogonal, then the whole postulate 'falls apart'! ... we will no longer have ∑nP(an) = 1.

-----

Another important fact I noticed is that c \psi_n is also an eigenfunction and regarded the same as \psi_n as logon as | c | = 1; this is related to what is stationary state.Yes, there is a connection.

------------------------------
------------------------------

Sammy, looking at all of the different points you are making, I get the sense that it might be instructive for us to go through each one of the postulates in a clear and concise way, one by one, step by step, ... . What do you say?

[Sammywu]

Eye,

What I was saying about independency is that it came to my mind that interference of two wave functions will be zero when they are orthogonal. Their interference seems to be represented by their inner product.

Of course, I know the necessary condition in here is that these are all orthonormal basises. I am saying is we can ignore their interference because they are orthogonal, but it would be interesting to check the relationship between interference and the inner product of two arbitrary wave functions.

I think your suggestion of going over all postulations is good.

Thanks

[Eye_in_the_Sky]

What I was saying about independency is that it came to my mind that interference of two wave functions will be zero when they are orthogonal. Their interference seems to be represented by their inner product.Yes, I see.

If

|φ> = c1|φ1> + c2|φ2> ,

then

<φ|φ> = |c1|2<φ1|φ1> + |c2|2<φ2|φ2> + 2 Re{c1*c2<φ1|φ2>} .

The last term is the "interference" term, and it vanishes if <φ1|φ2> = 0 .

------------------

Is the above what you meant?

... it would be interesting to check the relationship between interference and the inner product of two arbitrary wave functions.------------------
------------------

I think your suggestion of going over all postulations is good.I will post something soon.

[Eye_in_the_Sky]

ATTENTION: Anyone following this thread ... if you find any points (or 'near' points) of error on my part, please do point them out.
___________

P0: To a quantum system S there corresponds an associated Hilbert space HS.

P1: A pure state of S is represented by a ray (i.e. a one-dimensional subspace) of HS.
___________

Notes:

N.0.1) Regarding postulate P0, in certain contexts it is possible to associate a Hilbert space with a particular dynamical 'aspect' of the quantum system (e.g. a Hilbert space corresponding to "spin", decoupled from, say, "position").

N.1.1) In postulate P1, a "ray" is understood to represent the "(pure) state" of a single quantum system. Some physicists prefer to let those terms designate an ensemble of "identically prepared" quantum systems. Such a distinction becomes relevant only in cases where one considers possible interpretations of the theory.

N.1.2) A ray is determined by any one of its (non-zero) vectors. Our convention is to use a "normalized" (i.e. to unity) ket |ψ> to designate the corresponding ray, and hence, the corresponding pure state. This means that two normalized kets |ψ> and |ψ'> which differ by only a phase factor (i.e. |ψ'> = α|ψ>, where |α| = 1) will represent the same "ray", and hence, the same "state".
___________

Exercise:

E.1.1) What, if anything, is wrong with the following?

Suppose that the kets |φ1> and |φ1'> represent the same state. Then, the kets |ψ> and |ψ'> given below will also represent the same state:

|ψ> = c1|φ1> + c2|φ2> ,

|ψ'> = c1|φ1'> + c2|φ2> .
___________

[Sammywu]

Eye,

About the interference of wave functions: Yes. You got what I meant. I will have to ruminate over your simple answer, though.

Answer to the exercise:

If
| \psi \prime > = a | \psi >
| \varphi_1 \prime > = a_1 | \varphi_1 >
| \varphi_2 \prime > = a_1 | \varphi_2 >
then
a c_1 | \varphi_1 > + a c_2 | \varphi_2 > =
a_1 c_1 | \varphi_1 > + a_2 c_2 | \varphi_2 >
.

One solution to it is:
a = a_1 = a_2
i. e. they are mutiplied by the same phase factor.

If
a not = a_1
, then
| \varphi_1 > = ( a_2 -a ) c_2 / ( a - a_1 ) c_1 | \varphi_2 >
; i. e. | \varphi_1 > and | \varphi_2 > have to be the same state.

This shows that statement is in general incorrect except in the two solutions I show above.

[Sammywu]

Actually, for the second solution of them being the same state, there are other troubles to take care, | (a_1 - a ) / (a - a_2 ) \ needs to be one.

It does not seem to be easy to get solution for this. I have to think about how to resolve it. The possibility is this solution will not even work.

[Sammywu]

| ( a_1 - a ) c_1 / ( a_2 -a ) c_2 | =1 is equivalent to | a_1 - a | / | a_2 -a | = | c_2 | / | c_1 | ; knowing | a | = | a_1 | = | a_ 2 | = 1, we can view a, a_1 and a_2 as three normal vectors in the unit circle on the Complex plane; a shall be chosen as a point on the unit circle such that ratio of the cord length between a_1 and a over the cord length a_2 and a is | c_2 | / | c_1| .

[Eye_in_the_Sky]

First off, you have attempted to answer a question slightly more general than the one I posed. In my question, there was no |φ2'>. But that doesn't really matter. ... We'll use your version of the question.

You start off well.If
| \psi \prime > = a | \psi >
| \varphi_1 \prime > = a_1 | \varphi_1 >
| \varphi_2 \prime > = a_1 | \varphi_2 >
then
a c_1 | \varphi_1 > + a c_2 | \varphi_2 > =
a_1 c_1 | \varphi_1 > + a_2 c_2 | \varphi_2 > But your final conclusion suggests to me that the main point has been missed.This shows that statement is in general incorrect except in the two solutions ...Let's look at your second "solution". You write:If
a not = a_1
, then
| \varphi_1 > = ( a_2 -a ) c_2 / ( a - a_1 ) c_1 | \varphi_2 >
; i. e. | \varphi_1 > and | \varphi_2 > have to be the same state.In this case, you are right. If |φ1> and |φ2> themselves represent the same state, then so too will |ψ> and |ψ'>.

Now, let's look at your first "solution".One solution to it is:
a = a_1 = a_2
i. e. they are mutiplied by the same phase factor.Can we say that the statement is correct in this case? For the statement to be correct, we must have a1 and a2 as arbitrary free parameters, except for the constraint |a1| = |a2| = 1; but this solution produces an additional constraint over and above that.

... So, your conclusion should have been:

The statement is incorrect in all cases, except when |φ1> and |φ2> represent the same state.

------------------------

NOW ...
Just to make sure that the main point hasn't been lost in all of this abstraction, let's look at a concrete example.

Given the state

c1|φ1> + c2|φ2> ,

then

i(c1|φ1> + c2|φ2>)

represents the same state,

but

ic1|φ1> + c2|φ2>

does not (unless, of course, |φ1> and |φ2> represent the same state).

In the first case, we have inserted a "global phase-factor", and that is OK. In the second case, however, we have inserted a "relative phase-factor", and that is not OK.

------------------------

Finally, for the sake of completeness, I offer a solution of my own along the same lines as the one you gave above.
____

First, we assume:

(i) c1, c2 ≠ 0 ,

and

(ii) |φ1> and |φ2> are linearly independent .

For, otherwise, it follows trivially that |ψ> and |ψ'> will define the same ray (i.e. represent the same state).


Next, by considerations similar to yours above, we note that

|ψ> and |ψ'> define the same ray if, and only if

[1] c1(a - a1)|φ1> + c2(a - a2)|φ2> = 0 ,

where (and this is the important part!) the parameters a1 and a2 are completely arbitrary except for the constraint |a1| = |a2| = 1.

We now reach our conclusion. We say: but from assumptions (i) and (ii) we see that relation [1] holds iff a = a1 = a2, implying that a1 and a2 cannot be completely arbitrary (except for the constraint) as required; therefore, |ψ> and |ψ'> do not define the same ray.

:smile:
------------------------

[Sammywu]

Eye,

I am sorry. I did not notice there is no prime on the second \varphi_2.

I agree to your answer.

One part of my calculation was wrong. Somehow | c_1 | / | c_2| = 1 slipped into my mind, which is not true, so I corrected my answer in case somebody else were reading this post. Any way, that was showing even if they are all "same" state, "a" need to be chosen in a correct math. way.

[Eye_in_the_Sky]

P2: To a physical quantity A measurable on (the quantum system) S, there corresponds a self-adjoint linear operator A acting in HS. Such an operator is said to be an "observable".
___________

Notes:

N.2.1) From "The Spectral Theorem" for self-adjoint operators, it follows that:

(a) A has real eigenvalues;

(b) the eigenvectors corresponding to distinct eigenvalues are orthogonal;

(c) the eigenvectors of A are complete (i.e. they span HS).

The set of eigenvalues of A is called the "spectrum" of A. If A has a continuous spectrum, then the eigenvectors of A are said to be "generalized" and A is said to satisfy a "generalized" eigenvalue equation.

N.2.2) We now enumerate three special cases for A:

(1) A has a discrete, nondegenerate spectrum; then

A = ∑n an|u><un| .

(2) A has a discrete (possibly degenerate) spectrum; then

A = ∑n anPn .

(3) A has a continuous, nondegenerate spectrum; then

A = ∫a |a><a| da .

In each case, the RHS of each of the above relations is referred to as the "spectral decomposition" for A.

Case (1) is a particularization of case (2) (with Pn = |un><un|). For the general case of (2), the an are the eigenvalues of A and the Pn are the corresponding eigenprojectors.

Examples of case (3) are any of the components (Qj or Pk) of the position or momentum observables.
___________

Exercises:

E.2.1) From N.2.1) (b) (i.e. the eigenvectors corresponding to distinct eigenvalues are orthogonal), show that for a spectral decomposition of the type in N.2.2) (2), i.e.

A = ∑n anPn ,

it follows that

PjPk = δjkPk .

From N.2.1) (c) (i.e. the eigenvectors of A are complete), show that

∑nPn = I .

E.2.2) Use, as an example, the observable Q (for the position of a spinless particle moving in 1-dimension) to explain why the eigenkets |q> are said to be "generalized" and, therefore, why Q is said to satisfy a "generalized" eigenvalue equation.
___________

[Sammywu]

Eye,

I).
I think you have a typo here. Did you miss a subscriptor n here in this paragraph.

N.2.2) We now enumerate three special cases for A:

(1) A has a discrete, nondegenerate spectrum; then

A = ∑n an|un><un| .

II)
Also, For acontinuous spectrum, why are the eigenvectors called " generalized"? Does this have anythings to do with the fact that most likely they will be "generalized" function such as dirac-delta function isntead of regular functions?

III). Answer to the Exercise. It's actually a little difficult because you did not define eigenprojectors.

So, I just gave a guess on this:

P_n ( \sum_i c_i u_n_i + b u_\bot ) = \sum_i c_i u_n_i
where
u_n_i are eigenvectors of a_n and u_\bot \bot all u_n .

Under that assumption,
P_n^2 ( \sum_i c_i u_n_i + b u_\bot ) = P_n ( \sum_i c_i u_n_i ) =
\sum_i c_i u_n_i

so, P_n^2 = P_n .

If i does not equal to j, then
P_i P_J ( \sum_k c_k u_i_k + \sum_l c_l u_j_l + b u_\bot ) =
P_i ( \sum_l c_l u_j_l ) = 0
; so P_i P_j = 0 .

That shall take care of the first part of E.2.1).

The question could be asked is how I can prove all u can be decomposed to
\sum_k c_k u_i_k + \sum_l c_l u_j_l + b u_\bot
.

I think setting
c_k = < u_i_k | u >
c_l = < u_j_l | u >
and
b u_\bot = u - \sum_k c_k u_i_k - \sum_l c_l u_j_l
could take care of that.

I might verify this part later.

I am still working on the other two exercises.

[Sammywu]

Because { u_n_i } spans the Hilbert space, any u can be expressed as:
\sum_n \sum_i c_n_i u_n_i

\sum_n P_n ( \sum_j \sum_i c_j_i u_j_i ) =
\sum_n \sum_j \sum_i c_j_i P_n ( u_j_i ) =
( because
P_n ( u_j_i ) = u_n_i
when n =j,
P_n ( u_j_i ) = 0
when n , j not equal,
)
\sum_n \sum_i c_n_i u_n_i

That takes care of
\sum P_n = I
.

[Sammywu]

About Q:
1) Our universe can be regarded as a system being observed for years.
2). We have noted every "position" in this system can be described by three real values by setting a a 3-dim reference "coordinate" or frame.
3) These three real values (q_x, q_y, q_z ) have to be regarded as eigenvalues of 3 different observables ( i. e. Operators Q_x, Q_y, Q_z ) if this approach of Hilbert Space and states is to be used to investigate the system.
4). These real values (q_x, q_y, q_z ) have been observed to form three continuous real lines ( three continuous spetra ).
5). By the arguments above and the expansion postulate, the base of our system shall be constituted of a Hilbert Space spanned by these three sets of eigenvectors, which has a minimum one-to-one relationship with the eigenvalues by assumptions.
6) To Simplify our analysis, we can just look at any one of the three observables, said Q_x.
To be continued ...
7). For

[Eye_in_the_Sky]

Back in post #67 ( http://www.physicsforums.com/showpost.php?p=288658&postcount=67), I "messed up"! There I wrote:

Now, let's look at your first "solution".One solution to it is:
a = a_1 = a_2
i. e. they are mutiplied by the same phase factor.Can we say that the statement is correct in this case? ...
-----

The continuation of what I wrote there is wrong! The kets |φ1'> and |φ2'> are specific kets, not a general "family" of kets. This means that as long as a1 = a2, then there exists a value of a (namely, a = a1 (= a2)) such that |ψ'> = a|ψ>. Thus, the overall conclusion should have been:

The statement is incorrect in all cases, except: (i) when |φ1> and |φ2> represent the same state, or (ii) a1 = a2 in the relations |φk'> = ak|φk> (k = 1,2).

This means that your first "solution" is OK. I would only have added another sentence something like this:

So, we see that if a1 = a2 , then |ψ'> and |ψ> represent the same state.

... As you can see, the general "structure" of our question is like this: given "a1" and "a2", does there exist an "a"?

--------------

Now, however, this means that there is still a slight difficulty with your second "solution":If
a not = a_1
, then
| \varphi_1 > = ( a_2 -a ) c_2 / ( a - a_1 ) c_1 | \varphi_2 >
; i. e. | \varphi_1 > and | \varphi_2 > have to be the same state.In the first "solution", you have covered the case a1 = a2. So, now you need to cover the case a1 ≠ a2. This must be the starting point for the second "solution". ... How can we continue?

Well ... suppose there exists an a. Since, a1 ≠ a2, then a must be different from one of the ak. Suppose for definiteness – without loss of generality – that a ≠ a1. Then, ... [the rest of your "solution" is fine].

(Do you understand the meaning of the expression "suppose for definiteness – without loss of generality"?)

-----
... Sorry about the confusion on my part.

Sometime soon I hope to fix that post (#67).
-----------
But now I see that "editing privileges" have changed, so it will just have to stay that way!
______________

[Eye_in_the_Sky]

I).
I think you have a typo here. Did you miss a subscriptor n here in this paragraph.

N.2.2) We now enumerate three special cases for A:

(1) A has a discrete, nondegenerate spectrum; then

A = ∑n an|un><un| .Yes, I missed a subscript. (I would go put that in now, but there's no more "edit" for me for that post!)
__________________II)
Also, For acontinuous spectrum, why are the eigenvectors called " generalized"? Does this have anythings to do with the fact that most likely they will be "generalized" function such as dirac-delta function isntead of regular functions?Yes, that is the basic idea. More generally, we can say that the resulting "eigenvectors" will not be square-integrable; we will have <a|a> = ∞, and so, technically the "function" a(x) ≡ <x|a> will not belong to the Hilbert space of "square-integrable" functions. Nevertheless, this difficulty is not in any way serious. In a self-consistent way, we find that we are able write <a|a'> = δ(a - a') .
__________________III). Answer to the Exercise. It's actually a little difficult because you did not define eigenprojectors.Yes, some details have been omitted.

Here are some (but not all) of those details (for the discrete case). [Note: By assumption, A is a self-adjoint operator according to a technical definition which has not been stated in full (however, in post #20 (http://www.physicsforums.com/showpost.php?p=280841&postcount=20) of this thread, part of such a definition was given). Thus, in the following, certain related subtleties will not be given full explicit attention.]

Recall (b) and (c) from N.2.1): (b) the eigenvectors corresponding to distinct eigenvalues are orthogonal;

(c) the eigenvectors of A are complete (i.e. they span HS).Define En = { |ψ> Є HS │ A|ψ> = an|ψ> } . From (b), En ┴ En' , for n ≠ n'; so,

(b') En┴ כ En' , for n ≠ n' .

From (c), it follows that

(c') the vectors of UnEn span HS ,

Now, from (b') and (c'), it follows that for any |ψ> Є HS there exist unique |ψn> Є En such that

|ψ> = ∑n|ψn> .

From the uniqueness of the |ψn>, we can define the "eigenprojectors" Pn by

Pn|ψ> = |ψn> .

... Alternatively, we can do it this way. Each En itself is a (closed (in the sense of "limits")) linear subspace of HS, and, therefore, has a basis, which we can set up to be orthonormal. Let |unk> be such a basis, where k = 1, ... , g(n) (where g(n) is the degeneracy (possibly infinite) of the eigenvalue an). We then define

Pn = ∑k=1g(n) |unk><unk| .

You can then convince yourself that this definition of Pn is independent of choice of basis.
__________________

So, let's continue.So, I just gave a guess on this:

P_n ( \sum_i c_i u_n_i + b u_\bot ) = \sum_i c_i u_n_i
where
u_n_i are eigenvectors of a_n and u_\bot \bot all u_n .This "guess" was fine.

Next:If i does not equal to j, then
P_i P_J ( \sum_k c_k u_i_k + \sum_l c_l u_j_l + b u_\bot ) =
P_i ( \sum_l c_l u_j_l ) = 0
; so P_i P_j = 0 .Just one small problem: you already used "c" in the first sum over "k"; in the sum over "l" you need to use a different letter, say, "cl" → "dl". Then, your answer is fine.

Next:The question could be asked is how I can prove all u can be decomposed to
\sum_k c_k u_i_k + \sum_l c_l u_j_l + b u_\bot
.

I think setting
c_k = < u_i_k | u >
c_l = < u_j_l | u >
and
b u_\bot = u - \sum_k c_k u_i_k - \sum_l c_l u_j_l
could take care of that.Yes. Your idea works just fine. (Note: You need to make the change "cl" → "dl", or something like it.)
__________________

And now for the next part.Because { u_n_i } spans the Hilbert space, any u can be expressed as:
\sum_n \sum_i c_n_i u_n_i

\sum_n P_n ( \sum_j \sum_i c_j_i u_j_i ) =
\sum_n \sum_j \sum_i c_j_i P_n ( u_j_i ) =
( because
P_n ( u_j_i ) = u_n_i
when n =j,
P_n ( u_j_i ) = 0
when n , j not equal,
)
\sum_n \sum_i c_n_i u_n_i

That takes care of
\sum P_n = I
.Yes!
__________________

[Eye_in_the_Sky]

About Q:
1) Our universe can be regarded as a system being observed for years.
2). We have noted every "position" in this system can be described by three real values by setting a a 3-dim reference "coordinate" or frame.
3) These three real values (q_x, q_y, q_z ) have to be regarded as eigenvalues of 3 different observables ( i. e. Operators Q_x, Q_y, Q_z ) if this approach of Hilbert Space and states is to be used to investigate the system.
4). These real values (q_x, q_y, q_z ) have been observed to form three continuous real lines ( three continuous spetra ).
5). By the arguments above and the expansion postulate, the base of our system shall be constituted of a Hilbert Space spanned by these three sets of eigenvectors, which has a minimum one-to-one relationship with the eigenvalues by assumptions.
6) To Simplify our analysis, we can just look at any one of the three observables, said Q_x.
To be continued ...
7). For1) All we really care about (for the moment, at least) is whether or not our "model" will "explain" the "observed phenomena".

2) This is an essential part of our "model". It doesn't necessarily have to apply to the universe as a whole, but only some part of it (e.g. the "laboratory").

3) Yes. But we still have to define the Hilbert space and set up an "eigenvalue equation".

4) Yes ... at least to some (very good) approximation. We will use this hypothesis in our "model".

5) This is curious. I was expecting to define the Hilbert space first, and then define Q on it afterwards. I was expecting something like: let H be the space of all square-integrable functions R3 → C ... and then we set up an "eigenvalue equation"

Q f(q) = q f(q) .

Afterwards, we would then "find out" (or "show") that the components of Q are is self-adjoint, etc ... .

6) This is actually the "starting point" of E.2.2). You have taken some time to consider its 'justification' – that is good.

--------------------------

I think I'm just going to give the answer I had in mind. (I basically said it already in the previous post.) When we set up the eigenvalue equation for Q, we find that there are no "square-integrable" functions f such that [Qf](q) =qf(q) . So, strictly speaking, with respect to our Hilbert space, Q has no eigenfunctions. But then again, we can still make 'sense' of the "eigenvalue equation" ... and so, we say it has been "generalized", and that the solutions f are "generalized" eigenfunctions.

[Sammywu]

Eye,

Yes, it's strange. We can no longer edit our previous responses.

I agree that you caught me. I shall use a different letter for it.

Actually, I have no idea what you mean by this --
"suppose for definiteness – without loss of generality"?
Would you mind elaborating it?

About Q, I guess I actually went back to justify the Hilbert space. There could be more I can explore it a little latter; you have already noticed that. Basically, what will the Hilbert space look if I build it with the eigenvectors of the Q operator? Since we know its eigenvectors are not square integrable in the real line, This hilbert space might be bigger than the space of square integrable functions. Some thorough knowledge of functional analysis and probability theory might be needed here.

Actually, that also can lead to another question. That justification brings me a Hilbert space with probability decomposition but not necessarily Complex value coefficiened. I think the need of Complex value coefficient seems to be explanable by the scattering of electron diffraction or its interference.

My first response to this question was like this:
I did not submit it because I am seeing some holes there, but it's wotrthy to show a different perspective in approaching this question.
I was not sure the scope of your question, so I decided to go back to discuss what are "positions" observed ( as real value eigenvaues as we can see ) and the assumed Q operator associate with it. .

--->
In order to define Q | \psi > , you have to have a background manifold, then you can say Q | \psi > = q | \psi > , where q is not a constant.

In order to look for an answer of
Q | \psi_n > = q | \psi_n > = q_n | \psi_n >
where q_n needs to be a constant.

If | \psi_n > is a function f(q) of the coordinate q of the manifold M, then q*f(q) = q_n*f(q) ; | \psi_n > will have to be a function f(q) that is one when q equals to q_n and zero elsewhere.

---> Then this will lead to f(q) is not a square integrable function.

[Sammywu]

This is what I was going to continue on the Hilbert space built by the Q eigenvectors:

7). We can see two approches to expand the space now.
a. Discreet approach : | | u > = \sum_n a_n | u_n > or
A = \sum_n a_n | u_n > < u_n |
b. Continuous approach : | u > = \int a_n | u_n > or
| A = \int a | u > < u| du

Note I used a for the coefficient instead of u. It sounds better to me in that it shows that's a function of u but dependent on A. It's more comparable to the discreet notation too. Hope you agree.

8). Of course, if by the postulation N.2.2) (3), we only need to continue with 7.b.

I have to pause here before I can continue.

[Sammywu]

9). I paused to ponder about what this integration means. I think, it's a path integral of a single parameter of a family of operators | q > < q | ( u )( I now use q for the ket instead of u, and use u as the parameter to denote this family of operator ) and a(u) is a certain coeffient of A for the subcomponent of the operator | q > < q |.

Now, we will have to think what is the integration and differential of operators.

In a completely abstract setup with infinite and possibly uncountable basis, to define an operator's integration will have to deal with something like examing the change | q > < q | ( u ) | \psi > of the faimily of operator | q > < q | ( u ) for all \psi in H. I will assume I do not need to go so far.

10). Bottom line here is that we will face a problem that the eigenvector | q_0 > of the eigenvalue q_0 is unable to be represented by such an integral except when setting the a(u) as a function such that its value is \infty when u = q_0 and zero elsewhere, but its integration will be one.

11). In a way, if we already assume only square integrable functions are legitimate coefficients, then this is basically admitting there is no true eigenvalues or eigenvectors for operator Q. The eigenvalue we have observed as a point q_0 might be actually q_0+\triangle q .

12). Note I am able to separate q and u in the integration. The equation of with equating q and u actually is basically a special case when the parameter u was set to be the coordinate itself.

13). Now back to 7.a, I will try to show whether there is possibility that we can set up a Hilbert space that includes the square integrable functions and the generalized functions in a different way.

Pause...

[Sammywu]

14). Back to 9), I have said that using
A = \int a(u) | q > < q | ( u ) du
, we can represent a general representation of "mixed" states.
We will also find out if we trsanform the parameter to another parameter v. then
A = \int a(v) | q > < q | ( v ) (du/dv) dv .
So the coefficient for a new parameter v will be a(v)*(du/dv).
The coefficient will change with the introduction of a different parameter.
If we intend to standardize the coefficient, the easiest choice will be using q as the standard parameter, and so a(q) could be used to represent a state, and
A = \int a(q) | q > < q | ( q ) dq
or
A = \int a(v) | q > < q | ( v ) (dq/dv) dq .

15. If we compare this to a discreet case in that
A = \sum_n a_n | q_n > < q_n |
, we can note it's like we place q_n on a real line and
\int_\infty^{q_0} a(q) | q > < q | ( q ) dq \cong
\sum_{q_n <= q_0} a_n | q_n > < q_n |

16. Back to 7.a, if we want to build a Hilbert space in which the state can be a discreet sum of eigenvectors of the "position" eigenvalues, we will write
\sum_n a_n | q_n > < q_n | .
Looking into 15, is there a way we can have both forms of summation and integration coexist. I believe I have seen this in probability theory, you can set a probability distribution like this P([- \infty, a ]) where P([- \infty, - \infty ]) = 0 and P([- \infty, \infty ]) = 1 ; also it shall be a incresing function. This probability distribution is not necessary continuous or differentiable at everywhere, where it's differentiable the derivative will be square integrable and where it's not dfferentiable it will have "jump" points whose "generalized" derivatives just work similar to delta function.

Pause

[Sammywu]

17) So, we can see the derivative of P([ - \infty, q]) , denoted as f(q), is related to the a(q). To clarify their relationship, I need to add the conditions for a(q) that you might forget. For a mixed state, in a discreet case, \sum_n a_n = 1 ; so for a continuous case, I would say \int a(q) dq = 1 is needed.
By that, we can see a(q) is f(q).
Note, a(q) is not the wavefunction \psi(q) then, beacuse
\int \overline{\psi(q)} \psi(q) dq = 1
If we want to relate them, then
\overline{\psi(q)} \psi(q) = a(q)
seems to be a possible solution.
Actually, there is an issue here, which is related to the exercise you show as c_1 \psi_1 + c_2 \psi_2 | = c_1 \psi_1 \prime + c_2 \psi_2 \prime .

[Sammywu]

19). To illustrate this, I need to differntiate | q_n \prime > from | q_n > in that | q_n \prime > = a_n | q_n > where | a_n | = 1 but a_n |= 1 .
First, | q_n \prime > < q_n \prime | = | q_n > < q_n | .
so
\int a(q \prime ) | q \prime > < q \prime | dq \prime = \int a(q) | q > < q | dq .
No way to distinguish two "mixed" states from this point of view.
If we look from the perspective of a ket, compare
\int c(q \prime ) | q \prime > dq \prime
to
\int c(q) | q > dq
, even if c is the same function, they could be two different kets by the exercise we have shown in that even if | q \prime > and |q > are the "same", their complex linear combinations are not the "same", and the integration here can be viewed as a continuous linear combination of infinitely many "same" kets. Note these kets are assocaied with a "pure" state though.

[Eye_in_the_Sky]

Yes, it's strange. We can no longer edit our previous responses.For me, it is not only "strange", but also, "too bad". This means that (apart from any 'embarrassment' that incorrect posts will remain "permanently" on line) the data base, as a whole, as a 'resource' for someone who just "surfs-in" (looking for information) will no longer be as reliable as it could have been. This is unfortunate. Someone "surfing" the net may arrive at a post in some thread and think that what is written there is correct without realizing that several posts later on a comment has been made explaining how that post was in fact incorrect.

I was envisioning that this website would become a real reliable "source" of accurate information. Now, I see that as far my own posting is concerned, this will only be possible with additional 'care', over and above the usual amount, to make sure that posts are placed "correctly" at the onset (or shortly thereafter). Given my own limits of "time" and "knowledge", such a constraint may prove to be too demanding.
_______________Actually, I have no idea what you mean by this --
"suppose for definiteness – without loss of generality"?
Would you mind elaborating it?Sometimes, in the midst of a mathematical proof, one reaches a stage where a certain proposition P(k) will hold for at least one value of k. This particular value of k, however, is 'unknown' but nevertheless 'definite'. (For example, in the case of your "solution" above, the proposition P(k) was simply "a ≠ ak", and this had to be true for at least one of k =1 or k =2.)

Moreover, it is sometimes the case that the continuation of the proof proceeds in 'identical' fashion regardless of the particular value of k for which P(k) is true. (This was indeed the case for your "solution".) So, instead of saying that P(k) is true for some 'definite' value of k, say k = ko, where ko is 'unspecified', one says "suppose for definiteness that P(1) is true", and since the proof is the 'same' for any other 'choice' of k, one adds the remark "... without loss of generality".

The statement is, therefore, a sort of 'shorthand' which allows one to bypass certain 'mechanical' details and go straight to the essential idea behind the proof.
_______________Basically, what will the Hilbert space look if I build it with the eigenvectors of the Q operator? Since we know its eigenvectors are not square integrable in the real line, This hilbert space might be bigger than the space of square integrable functions. Some thorough knowledge of functional analysis and probability theory might be needed here.In the "functional analysis" approach, one begins with a Hilbert space of square-integrable functions R → C. The 'justification' for this comes about from the Schrödinger equation (in "x-space") coupled with the Born probability rule that ψ*(x)ψ(x) is the "probability density", where the latter of these implies that the (physical) wavefunctions are all square-integrable. Thus, the probability P(I) of finding the particle in the (non-infinitesimal) interval I is given by

P(I) = (ψ, PIψ) ,

where PI is the "projector" defined by

[PIψ](x) ≡
ψ(x) , x Є I
0 , otherwise

and we have defined an "inner product"

(φ, ψ) = ∫φ*(x)ψ(x) dx .

This 'family' of projectors PI already contains in it the idea of |q><q|, since they are connected by the simple relation

P(a,b) = ∫ab |q><q| dq .

... Now, let's look more closely at what you say:This hilbert space might be bigger than the space of square integrable functions.Here, you are suggesting the idea of "building" a space from the |q>'s in such a way that those objects themselves are included in the space. I have never thought abut such a proposition in any detail. Nevertheless, the original Hilbert space would then be seen as "embedded" in a larger 'extended' vector space which would include the |q>'s (and whatever else).

For the record, you may want to know the 'technical' definition of a "Hilbert space" H:

(i) H is a "vector space";

(ii) H has an "inner product" ( , );

(iii) H is "complete" in the "induced norm" ║ ║ ≡ √( , );

(iv) H is "separable".

The last of these is usually not included in the definition. I have put it in here, since the Hilbert spaces of QM are always "separable". You may want to 'Google' some these terms or check at mathworld (http://mathworld.wolfram.com/) or Wikipedia (http://en.wikipedia.org/wiki/Main_Page), or the like.

Note that such a notion of an "extended" space is used in what is called a "rigged" Hilbert space. I do not know much about such a construction and, in particular, I am unsure as to what its 'utility' is from a 'practical' point of view.

There is also the "Theory of Distributions" (or "Distribution Theory"), which deals with this idea of "generalized" functions (i.e. "distributions") in a formally rigorous way.
_______________Actually, that also can lead to another question. That justification brings me a Hilbert space with probability decomposition but not necessarily Complex value coefficiened. I think the need of Complex value coefficient seems to be explanable by the scattering of electron diffraction or its interference.So far, we have been viewing the situation from a "static" perspective. As soon as we admit "motion" into the picture, then complex-valued coefficients come into play by way of necessity.

Think of a (time-independent) Hamiltonian, and the Schrödinger equation

ihbar ∂t|ψ(t)> = H|ψ(t)> .

With |φn> a basis of eigenkets such that H|φn> = En|φn> , we then have general solutions of the form

|ψ(t)> = ∑n exp{ -iEnt / hbar } cn |φn> .

There is no way 'around' this. The coefficients must be complex-valued.

Your example of "diffraction" or "interference" appears (to me) to be a special case of this general fact. On the other hand, we know that such problems can be 'treated' by the formalism of "classical optics", in which case the use of complex-valued coefficients is merely a matter of 'convenience', and not one of 'necessity' (so, I'm not so sure that this is in fact a 'good' example).
_______________In order to define Q | \psi > , you have to have a background manifold, then you can say Q | \psi > = q | \psi > , where q is not a constant.You mean: Q|ψq> = q|ψq>, where q is not a constant.
---> Then this will lead to f(q) is not a square integrable function.Yes. ... And as I mentioned above, "Distribution Theory" handles this 'difficulty' in a perfectly rigorous way.
_______________

[Eye_in_the_Sky]

7). We can see two approches to expand the space now.
a. Discreet approach : | | u > = \sum_n a_n | u_n > or
A = \sum_n a_n | u_n > < u_n |
b. Continuous approach : | u > = \int a_n | u_n > or
| A = \int a | u > < u| du

Note I used a for the coefficient instead of u. It sounds better to me in that it shows that's a function of u but dependent on A. It's more comparable to the discreet notation too. Hope you agree.Here are some "notational" details:

The 'continuous analogue' of the notation for the 'discrete case'

[1] A = ∑n an|un><un|

is

[1'] A = ∫ a(s)|u(s)><u(s)| ds .

What you wrote, i.e. (note: I have put "a" → "a(u)")

[2'] A = ∫ a(u)|u><u| du ,

is the analogue of

[2] A = ∑n an|n><n| .

Finally, the analogue of

[3'] A = ∫ a |a><a| da

is

[3] A = ∑a_n an|an><an| .
_______________9). I paused to ponder about what this integration means. I think, it's a path integral of a single parameter of a family of operators | q > < q | ( u )( I now use q for the ket instead of u, and use u as the parameter to denote this family of operator ) and a(u) is a certain coeffient of A for the subcomponent of the operator | q > < q |.I don't see how it can be construed as a "path integral". In a path-integral formulation of the problem for a particle moving in one-dimension, the single parameter q is construed a function of time, i.e. q(t), where that function is varied over all 'possible' functions on t Є [t1, t2] subject to the constraint δq(t1) = δq(t2) = 0. We would then have

<q(t2)|q(t1)> = a path integral .

But here, the 'closest' thing I can see is

<q'|q> = δ(q' - q) .

In short, a "path integral" can come into play once we consider the "time evolution" of the quantum system. Right now, we are only concerned with the situation at a single 'given' time.
_______________Now, we will have to think what is the integration and differential of operators.

In a completely abstract setup with infinite and possibly uncountable basis, to define an operator's integration will have to deal with something like examing the change | q > < q | ( u ) | \psi > of the faimily of operator | q > < q | ( u ) for all \psi in H. I will assume I do not need to go so far.Now, the "family" of operators you are considering, what I will call |q><q|, is very much like a 'derivative' of the projector PI which I mentioned before; i.e.

[PIψ](x) ≡
ψ(x) , x Є I
0 , otherwise .

Let us define E(q) ≡ P(-∞,q). Then, 'formally' we have

dE(q) = |q><q| dq .

The LHS is the 'formal' expression for the "differential of the spectral family" in the context of "functional analysis"; the RHS is the "Dirac" equivalent.
_______________10). Bottom line here is that we will face a problem that the eigenvector | q_0 > of the eigenvalue q_0 is unable to be represented by such an integral except when setting the a(u) as a function such that its value is \infty when u = q_0 and zero elsewhere, but its integration will be one.

11). In a way, if we already assume only square integrable functions are legitimate coefficients, then this is basically admitting there is no true eigenvalues or eigenvectors for operator Q.Yes. And this is where "Distribution Theory" comes in.


... The eigenvalue we have observed as a point q_0 might be actually q_0+\triangle q .I don't see this (... unless we take the limit Δq → 0).
_______________
_______________

... As it turns out, unfortunately, starting this week and continuing on for the next several months(!), I will become very busy. Consequently, I will have little time for any significant activity in the Forum here. I have already reduced my posting to only this thread alone (over the last few weeks).

This week, however, I still do hope to at least get to the next two postulates and connect them to the original issue which was of concern – "expectation values", "mixed states", and the "Trace" operation. If you recall, it was matters of this kind which caused me to ask you if you had gone over the postulates in a clear, concise way.

... After that, there will be only one more postulate, that of "time evolution". If we deal with that here, I must tell you in advance that my input into this thread will 'evolve' only very slowly.
_______________

[Sammywu]

Eye,

Thanks for your reply.

I think the conecpt of projector of interval is more straightforward and better than my approach, even though I think the way I approach it can be proved the same eventually. There is just a misunderstnding here. maybe I shall not use the word "path integral"; I did not mean to associate that integration with any time parameter. The parameter is just any real line in this case. Of course, in this case, I will have to be able to define how to integrate a operator function of a real line. The idea of projector of interval and so the point projector being its derivative takes care of the issue of what is the integration here.

[Sammywu]

Eye,

Sorry about this stupid question.

But what does LHS and RHS stand for? I can't find it in mathworld.

Thanks

[Sammywu]

Actually, I did a little bit verification here to see how this is derived.

The probability P(I) of finding the particle in the (non-infinitesimal) interval I is given by

P(I) = (ψ, PIψ) ,
-----------------------------------------

First, in a discreet case,
P(I) = \sum_n (\psi, q_n) (q_n , \psi)
Take P_I \psi = \sum_n (\psi, q_n) \q_n ,
(\psi, P_I \psi ) = \sum_n \overline{( \psi, q_n )} ( \psi , q_n) =
\sum_n ( q_n , \psi ) ( \psi , q_n) =

Then, translate into continuous case,
P(I) = \int_a^b (\psi, q) (q ,\psi) dq
Take P_I \psi = \int_a^b (\psi, q) | q > dq ,
(\psi, P_I \psi ) = ( \psi, \int_a^b ( \psi, q ) | q > dq ) =
\int_a^b \overline{( \psi, q )} ( \psi , q) dq =
\int_a^b ( q , \psi ) ( \psi , q) dq =

Now, this looks better.

[Eye_in_the_Sky]

Eye,

Sorry about this stupid question.

But what does LHS and RHS stand for? I can't find it in mathworld.

Thanks"LHS" stands for "left-hand-side"; "RHS" stands for "right-hand-side". :smile:

[Sammywu]

Eye,

Thanks. I actually thought they could stand for some special Hilbert spaces.

Any way, your mention of "rigged" Hilbert space probably is what I was led to do with a ket defined as a function series { f[SUB}n[/SUB] } and
lim_{ n \rightarrow \infty } \int_{-\infty}^\infty f_n = 1
. So all kets can be treated as a funcion series. Just as you said, it might not be of any practical use. I guess there is no need to continue.

Any way, I have gone thru an exercise showing me that I can construct a "wavefunction" space with any observed continuous eigenvalues.

Note the arguments applied is not specific to "position" but applicable to any continuous eigenvalues.

[Sammywu]

I am not sure whether this is too much, but I found I can go even further; something is interesting here.

21). I can represent a ket in such a way:
\int \psi(q) | q > dq
This shows that the wavefunction is actually a abbreviated way of this ket.

The eigenvactor of an eigenvalue q_0 can be then written as
\int \delta(q_0) | q > dq .

Or in general, I can extend this into a sample such as a function series { f_n }
in that:

lim_{ n \rightarrow \infty } lim_{ q \rightarrow \q_1 , q_2 } f_n ( q) = \infty
and
lim_{ n \rightarrow \infty } \int_{ - \infty }^q_1 f_n ( q) dq = a_1
lim_{ n \rightarrow \infty } \int_{ - \infty }^q_2 f_n ( q) dq = 1

22). I can even check what shall the inner products of two kets without clear prior definition of inner products:

< \psi_1 | \psi_2 > =
< \int \psi_1(q) |q > dq | \int \psi_2(q \prime) | q \prime > dq \prime > =
\int \overline{\psi_1(q)} < q | \int \psi_2(q \prime) | q \prime> dq \prime > dq =
\int \overline{\psi_1(q)} \int \psi_2(q \prime) < q | q \prime > dq \prime dq =

[Eye_in_the_Sky]

14). ... I have said that using
A = \int a(u) | q > < q | ( u ) du
, we can represent a general representation of "mixed" states.Now, wait just a moment! How did we get onto the subject of "states" in a decomposition like that of above? Up until now, we have been talking about "observables". ... "Mixed states" will come soon.
______________... If we intend to standardize the coefficient, the easiest choice will be using q as the standard parameter ... and
A = \int a(q) | q > < q | ( q ) dq Yes, the easiest choice of "notation" is

A = ∫ a(q) |q><q| dq .

Note, however, that such an operator is merely a function of Q. Specifically, A = a(Q). In other words, the matrix elements of A, in the "generalized" |q>-basis, are given by

<q|A|q'> = a(q) δ(q – q') .

(It turns out that: any linear operator L is a 'function' of Q iff [L,Q] = 0. (This, of course, applies to a spinless particle moving in one dimension.))

BUT ...

In all of this, I am getting the feeling that each of us is misunderstanding what the other means. In the above, if you 'meant' that A is some self-adjoint operator whose spectrum is (simple) continuous, then 'automatically' we can write

[1] A = ∫ a |a><a| da

with no difficulty whatsoever. There is no reason to write it any other way, because by 'hypothesis'

[2] A|a> = a|a> .

The exact analogue of these expressions in the corresponding (nondegenerate) discrete case is

[1'] A = ∑a a |a><a| ,

and

[2'] A|a> = a|a> .

In the discrete case, however, we modify the notation by introducing an index like "n" because somehow 'it is more pleasing to the eye'. But to do an analogous thing in the continuous case is completely uncalled for, since doing so will introduce a new element of "complexity" which provides no advantage whatsoever. ... Why should we write "a" as a function of some parameter "s", say a = w(s), and then have da = w'(s)ds? ... We will get nothing in return for this action except additional "complexity"! (Note that changing the "label" for the generalized ket |a> → |u(a)> introduces no such difficulties.)
______________16. ... we will write
\sum_n a_n | q_n > < q_n | .
Looking into 15, is there a way we can have both forms of summation and integration coexist.Yes. Given a self-adjoint operator A, then "the spectrum of A" (i.e. "the set of all eigenvalues ('generalized' or otherwise) of A") can have both discrete and continuous parts. A simple example of such an observable is the Hamiltonian for a finite square-well potential. The "bound states" are discrete (i.e. "quantized" energy levels), whereas the "unbound states" are continuous (i.e. a "continuum" of possible energies).
______________17) ... For a mixed state, in a discreet case, \sum_n a_n = 1 ; so for a continuous case, I would say \int a(q) dq = 1 is needed.Hopefully, soon we will be able to talk 'sensibly' about "mixed states". Once we do that, you will see that a 'state' like

ρ = ∫p(q)|q><q|dq (with, of course, p(q) ≥0 (for all q) and ∫ p(q) dq = 1) ,

is not 'physically reasonable'.

So far, we have explained only "pure states", as given by our postulate P1. Recall:P0: To a quantum system S there corresponds an associated Hilbert space HS.

P1: A pure state of S is represented a ray (i.e. a one-dimensional subspace) of HS.When we get to discussing "mixed states", we will not explain them in terms of a "postulate", but rather, those objects will be introduced by way of a 'construction' in terms of "pure states". I have already alluded to such a "construction" in post #46 of this thread. There I wrote:A pure state is represented by a unit vector |φ>, or equivalently, by a density operator ρ = |φ><φ|. In that case, ρ2 = ρ.

Suppose we are unsure whether or not the state is |φ1> or |φ2>, but know enough to say that the state is |φi> with probability pi. Then the corresponding density operator is given by

ρ = p1|φ1><φ1| + p2|φ2><φ2| .

In that case ρ2 ≠ ρ, and the state is said to be mixed. Note that the two states |φ1> and |φ2> need not be orthogonal (however, if they are parallel (i.e. differ only by a phase factor), then we don't have mixed case but rather a pure case).______________
______________

Sammy, I am hoping to post a response your posts #84,86,88,89 by Monday. After that I hope to get at least one more postulate out. (There are also (at least) two items from our previous exchanges which I wanted to address.)

[Sammywu]

23). In trying to evaluate 22), I found I need something clearer about how to represent all vectors. Let me put all eigenvalues in one real line; for each q in this real line, we associate an eigenvector ^\rightarrow{q} with it. I want to avoid using | q > for now, because | q > is actually a ray. also, remember there are many vectors as c * ^\rightarrow{q} where | c | =1 can be placed here; let's just pick any one of them.

So, now with a function c(q), we can do a vector integration over the q real line as:
\int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq
Note q in c(q) and dq is just a parameter and ^\rightarrow{q} is a vector, and also viewed a vector function paramterized by q.

24).Refering back to 21), all vectors can be represented now by:
lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty c_n(q) ^\rightarrow{q} dq

25). In particular, let
\delta_n( q - q_0 ) = 1/n for q_0 - 1/2n <= q <= q_0 +1/2n and 0 elsewhere,
the eigenvector for q_0 can be represented as:
lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty \delta_n(q - q_0) ^\rightarrow{q} dq

26). And, for other vectors, c_n(q) can be set to a constant function c(q);
we can verify its consistency with the normal representation:

lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty c_n(q) ^\rightarrow{q} dq =
\int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq =
\int_{ - \infty }^\infty c(q) lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty \delta_n(q \prime -q) ^\rightarrow{q \prime } dq \prime dq =
lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty \int_{ - \infty }^\infty c(q) \delta_n(q \prime - q) ^\rightarrow{q \prime} dq dq \prime =
lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty ( \int_{ - \infty }^\infty c(q) \delta_n(q \prime - q) dq ) ^\rightarrow{q \prime} dq \prime =
lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty c(q \prime) ^\rightarrow{q \prime} dq \prime

27). For inner products of c and d,
( \int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq , \int_{ - \infty }^\infty d(q \prime ) ^\rightarrow{q \prime } dq \prime ) =
\int_{ - \infty }^\infty \overline{d(q \prime)} ( \int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq , ^\rightarrow{q \prime} ) dq \prime

28). Now, I have to discuss what shall it be for
( \int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq , ^\rightarrow{q \prime } )

First, if we look into the inner products of two eigenvectors \rightarrow{q \prime } and \rightarrow{q } , we can first think about what shall be the innerproduct betwen
lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty \delta_n(u - q) ^\rightarrow{u} du
and
lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty \delta_n(v - q \prime) ^\rightarrow{v} dv
.

Comparing it to a discreet case, I guess this could be
| ( q , q \prime) | = lim_{ i \rightarrow \infty } lim_{ j \rightarrow \infty } \int_{ - \infty }^\infty \delta_i( u - q \prime) \delta_i( u - q ) du

So, in general,
( q , q \prime) = e^{ia} lim_{ i \rightarrow \infty } lim_{ j \rightarrow \infty } \int_{ - \infty }^\infty \delta_n( u - q \prime) \delta_n( u - q ) du

The phase factor is put in to show the possibility of two out-of-phase eiegnvector. For now, we can assume our standard basis are in-phase vectors.

With this, we can further translate the inside part of 27) to.

( \int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq , ^\rightarrow{q \prime} ) =
\int_{ - \infty }^\infty c(q) ( ^\rightarrow{q} , ^\rightarrow{q \prime} ) dq =
\int_{ - \infty }^\infty c(q) lim_{ i \rightarrow \infty } lim_{ j \rightarrow \infty } \int_{ - \infty }^\infty \delta_n( u - q \prime) \delta_n( u - q ) du dq =
lim_{ i \rightarrow \infty } lim_{ j \rightarrow \infty } \int_{ - \infty }^\infty \delta_i( u - q \prime) \int_{ - \infty }^\infty c(q) \delta_j( u - q ) dq du =
lim_{ i \rightarrow \infty } \int_{ - \infty }^\infty \delta_i( u - q \prime) c(u) du =
[tex] c(q \prime)

Placing that into 27), I got
( \int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq , \int_{ - \infty }^\infty d(q \prime ) ^\rightarrow{q \prime } dq \prime ) =
( \int_{ - \infty }^\infty \overline{d(q \prime)} c ( q \prime) dq \prime

[Sammywu]

Eye,

Thanks for the reply. It defintely caught my misunderstandings and stimulated my thoughts too.

[Sammywu]

Eye,

Now I really know what you were showing me. I definitely went on a different direction. You are showing me that a self-adjoint operator can be decompsed into an integration of its eigenvalues multiplied by its eigenprojectors.

So, Q = \int q | q> < q | dq . Defintely correct.

And can I do this?
Q | \psi > = ( \int q | q> < q | dq ) | \int \psi(q) |q > dq = \int q \psi(q)| q > dq
By the above EQ., if we see \psi(q) representing | \psi > , then
Q | \psi > = q * \psi ( q ) = q * | \psi > .

This is of course due to the \psi ( q ) is the coeffient when choosing the eigenvectors of Q as basis.

If the energy eigenvectors are chosen as the basis, then we can write
H | \psi > = E * | \psi >
, because Hamiltonian's eigenvalues are energies.

While I use
| \psi > = \int c(q) |q > dq
because I treat q as a paramter here, but I think I saw another notation in this way
| \psi > = \int c(q) d |q >
Do you have any comments on that?

[Sammywu]

Just make some conclusions on my deduction:

I. After including the phase factor consideration, | q_0 > as the normal eigenvector of the eigenvalue q_0 can be denoted as:

lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty \delta_n(q - q_0) e^{ik(q)} | q> dq
, or
\int_{ - \infty }^\infty \delta(q - q_0) e^{ik_0} | q> dq

This can be checked against that it shall be representable as
\sum_n c_n \psi_n =
\sum_n c_n \int_{ - \infty }^\infty \psi_n( q ) | q> dq
where \psi_n is the normal eigenvector of Hamiltonian, because this infinite summation can be viewed the limit of a function sequence ( To use correct math. term, I think I probably shall say sequence instead of series , series is reserved for infinite summation. correct ? ) as well.

II. If we do not simplify the 3-dim eigenvalues observed into the discussion of any one of them, then we will find out the three measurables form a 3-dim vector, and we will need to think about what is a "vector operator" or a "vector observable".

We will have more to explore such as, what does the rotation of the vector operator mean, what shall two vector operators' inner or scalor product and their outer product be.

[Eye_in_the_Sky]

About the object "|q><q|" which you referred to as a "point projector". Do you realize that "|q><q|" is not a "projector"? ... A necessary condition for an object P to be a projector is P2 = P. But

(|q><q|)2 = |q><q| δ(0) = ∞ .

On the other hand, PI defined by

[PIψ](x) ≡
ψ(x) , x Є I
0 , otherwise ,

or equivalently,

PI ≡ ∫I |q><q| dq ,

does satisfy PI2 = PI.
______________

According to the "Born rule", the probability for finding the particle in the interval I is given by

P(I) = ∫I ψ*(q)ψ(q) dq .

But ψ(q) ≡ <q|ψ>, so that

P(I) = ∫I <ψ|q><q|ψ> dq

= <ψ| { ∫I|q><q|dq } |ψ>

= <ψ|PI|ψ> .

The last expression corresponds to (ψ, PIψ) in the notation of "functional analysis".

Apart from some 'notational difficulties', you have performed this verification correctly:P(I) = \int_a^b (\psi, q) (q ,\psi) dq
Take P_I \psi = \int_a^b (\psi, q) | q > dq ,
(\psi, P_I \psi ) = ( \psi, \int_a^b ( \psi, q ) | q > dq ) =
\int_a^b \overline{( \psi, q )} ( \psi , q) dq =
\int_a^b ( q , \psi ) ( \psi , q) dq ____

As for what you say regarding a discrete case:P(I) = \sum_n (\psi, q_n) (q_n , \psi)
Take P_I \psi = \sum_n (\psi, q_n) \q_n ,
(\psi, P_I \psi ) = \sum_n \overline{( \psi, q_n )} ( \psi , q_n) =
\sum_n ( q_n , \psi ) ( \psi , q_n) ... the picture you are presenting of a "discrete" position observable in terms of "discrete" eigenkets does not make.

To make the position observable "discrete" we want to have a small "interval" In corresponding to each "point" qn, say

qn = n∙∆q , and In = (qn-∆q/2 , qn+∆q/2] .

Then, our "discrete" position observable, call it Q∆q, will be a degenerate observable. It will have eigenvalues qn and corresponding eigenprojectors (not kets!) PI_n. That is,

Q∆q = ∑n qnPI_n .
______________21). I can represent a ket in such a way:
\int \psi(q) | q > dq
This shows that the wavefunction is actually a abbreviated way of this ket.Yes. In general,

[1] |ψ> = ∫ ψ(q) |q> dq ,

which is equivalent to

[2] ψ(q) = <q|ψ> .

Thus, ψ(q) is the "component" of |ψ> in the ("generalized") |q>-basis.

Relative to an 'ordinary' basis we have two similar expressions:

[1'] |ψ> = ∑n cn|φn> ,

[2'] cn = <φn|ψ> .

The requirement that |ψ> Є H, i.e. <ψ|ψ> < ∞ , in terms of [1] (and [2]) means

∫ |ψ(q)|2dq < ∞ ,

whereas in terms of [1'] (and [2']) it means

∑n|cn|2 < ∞ .

Everything is the 'same' in both cases, except for the fact that <q|q> = ∞ , whereas <φn|φn> = 1. That is, each |q> is not a member of H, whereas each |φn> is. Indeed, as you say:The eigenvactor of an eigenvalue q_0 can be then written as
\int \delta(q_0) | q > dq .... and <q|q> = ∫|δ(qo)|2dqo = δ(0) = ∞ .
______________Any way, I have gone thru an exercise showing me that I can construct a "wavefunction" space with any observed continuous eigenvalues. Yes. The position observable Q is used the most frequently for this purpose. The next most frequently used is the momentum observable P.
______________22). I can even check what shall the inner products of two kets without clear prior definition of inner products:

< \psi_1 | \psi_2 > =
< \int \psi_1(q) |q > dq | \int \psi_2(q \prime) | q \prime > dq \prime > =
\int \overline{\psi_1(q)} < q | \int \psi_2(q \prime) | q \prime> dq \prime > dq =
\int \overline{\psi_1(q)} \int \psi_2(q \prime) < q | q \prime > dq \prime dq = In the above, the internal consistency of our formulation is brought out as soon as we write

<q|q'> = δ(q - q') .

Then, the last integral becomes

∫ψ1*(q) ∫ψ2(q') δ(q - q') dq' dq

= ∫ ψ1*(q) ψ2(q) dq ,

which is just what we want for <ψ1|ψ2>.
______________

[Eye_in_the_Sky]

(Note: I am deferring a response to post #91 until later.)



Now I really know what you were showing me. I definitely went on a different direction. You are showing me that a self-adjoint operator can be decompsed into an integration of its eigenvalues multiplied by its eigenprojectors.

So, Q = \int q | q> < q | dq . Defintely correct.Yes ... when Q is a self-adjoint operator with pure continuous (nondegenerate) spectrum.
____________And can I do this?
Q | \psi > = ( \int q | q> < q | dq ) | \int \psi(q) |q > dq = \int q \psi(q)| q > dq Yes, but use distinct integration variables in each of the integrals, say q in the first and q' in the second, so you can then show the 'computation' explicitly, like this:

Q|ψ>

= (∫ q|q><q| dq) (∫ ψ(q')|q'> dq')

= ∫dq q|q> ∫ψ(q')<q|q'> dq'

= ∫dq q|q> ∫ψ(q') δ(q - q') dq'

= ∫ qψ(q)|q> dq [E1] .
____________By the above EQ., if we see \psi(q) representing | \psi > , then
Q | \psi > = q * \psi ( q ) = q * | \psi > .No. The relation Q|ψ> = q|ψ> would mean that |ψ> is an eigenket of Q, something you do not wish imply. In words, what you want to express is this: "the action of Q on |ψ> when depicted in the q-space of functions is multiplication by q".

That is easy to do. Given any ket |φ>, then it's q-space representation is just <q|φ>, which we write as φ(q). Now, we want |φ> = Q|ψ> in q-representation, which is therefore just

<q|(Q|ψ>) = <q| (∫ q'ψ(q')|q'> dq') , using [E1] above

= ∫ q'ψ(q') <q |q'> dq'

= ∫ q'ψ(q') δ(q - q') dq'

= qψ(q) .

Alternatively, from Q|q> = q|q>, we have (Q|q>)† = (q|q>)†, which becomes <q|Q† = <q|q*. But Q† = Q and q* = q, so <q|Q = q<q|. Therefore,

<q|Q|ψ> = q<q|ψ> = qψ(q) .

This is of course due to the \psi ( q ) is the coeffient when choosing the eigenvectors of Q as basis.Yes,

ψ(q) is just the q-component of |ψ> in the generalized |q>-basis.
____________

Compare this last statement with the case of a (non-"generalized") discrete basis.

In a discrete basis |φn>, what is the φn-representation of |ψ>? ... It is just <φn|ψ>. And if we write |ψ> = ∑n cn|φn>, we then have <φn|ψ> = cn. So,

cn is just the n-component of |ψ> in the |φn>-basis.

... In the discrete case, this is 'obvious'. The continuous case should now be 'obvious' too.

Perhaps a 'connection' to "matrices" may offer further insight. So here we go!
____________

Note that, in what follows, no assumption is made concerning the existence of an "inner product" on the vector space in question. It is therefore quite general. (Note: I am just 'cutting and pasting' from an old post.)
_____

Let bi be a basis. Then, (using the "summation convention" for repeated indices) any vector v can be written as

v = vibi .

In this way, we can think the of vi as the components of a column matrix v which represents v in the bi basis. For example, in particular, the vector bk relative to its own basis is represented by a column matrix which has a 1 in the kth position and 0's everywhere else.

Now, let L be a linear operator. Let L act on one of the basis vectors bj; the result is another vector in the space which itself is a linear combination of the bi's. That is, for each bj, we have

[1] Lbj = Lijbi .

In a moment, we shall see that this definition of the "components" Lij is precisely what we need to define the matrix L corresponding to L in the bi basis.

Let us apply L to an arbitrary vector v = vjbj, and let the result be
w = wibi. We then have

wibi

= w

= Lv

= L(vjbj)

= vj(Lbj)

= vj(Lijbi) ... (from [1])

= (Lijvj)bi .

If we compare the first and last lines of this sequence of equalities, we are forced to conclude that

[2] wi = Lijvj ,

where, Lij was, of course, given by [1].

Now, relation [2] is precisely what we want for the component form of a matrix equation

w = L v .

We, therefore, conclude that [1] is the correct "rule" for giving us the matrix representation of a linear operator L relative to a basis bi.
_____

The above description of components is quite general. It relies on the following two "facts" concerning a "basis" bi:

(i) any vector can be written as a linear combination of the bi,

(ii) the coefficients in such a linear combination are unique.

Now, here is an exercise:

Draw the 'connection' between what was just described above to that of our Hilbert space.

Your answer should be short and straight 'to the point'. To show you what I mean, I will get you started:

bi = |bi>

v = |v>

vi = <bi|v>

etc ...
___

... What about a continuous basis, say |q>?
____________

Now getting back to your post:If the energy eigenvectors are chosen as the basis, then we can write
H | \psi > = E * | \psi >
, because Hamiltonian's eigenvalues are energies.This is the same mistake you made above with "Q|ψ> = q|ψ>", which you now know is wrong ... right?
____________While I use
| \psi > = \int c(q) |q > dq
because I treat q as a paramter here, but I think I saw another notation in this way
| \psi > = \int c(q) d |q >
Do you have any comments on that?The object "|q>" in each formula is obviously not the same.

In the first formula, we have "|q>dq" in an integral which produces a vector of the Hilbert space (provided that ∫|c(q)|2dq < ∞). The interpretation of "|q>" is, therefore, that of a "vector density" in the Hilbert space, while "dq" is the associated "measure". Their product, "|q>dq", then has the interpretation of an "infinitesimal vector".

In the second formula, we see "d|q>". Its interpretation is that of an "infinitesimal vector". I will change the notation to avoid confusion and write "d|q>" as "d|q)". An appropriate definition of "|q)" in terms of the usual "|q>" is then

|q) = ∫-∞q |q'> dq' .

Thus, |q) is also in a class of "generalized vector". If we now take |q) as the "given", then from it we can define |q> ≡ d|q)/dq.

From the perspective of any calculation I have ever performed in quantum mechanics, the "|q>" notation of Dirac is superior.

[Sammywu]

Eye,

I am still digesting your response. So it's going to take me a while to answer that exercise.

Just respond to some points you made:

1) I did not know | q > < q | is not a projector. I have to think about that.

2). I did hesiate to write Q | \psi > = q | \psi > in the same reasons you mentioned, but in both Leon's Ebook and another place I did see their mentioning about the Q's defintion is Q | \psi > = q | \psi > . Just as I mentioned, the only reason I could see this "make sense" is by either
Q | \psi > = \int q |q > < q> dq or
Q | \psi > = q \psi ( q ) in the form of wavefunvtions.

3). I think your defining that \psi ( q ) = < q \ \psi > actually will make many calculations I did in showing in general
< \psi \prime | \psi > = \int \overline{\psi \prime ( q ) } \psi ( q ) dq
much more straighforward thamn my cumbersome calculations.

But one problem is then what is < q | q >. The discrete answer will be it needs to be one. Which of course will lead to some kind of conflicts in a general requirement of
\int \overline{\psi ( q ) } \psi ( q ) dq = 1 .

This is probably related to the eigenprojector P_{I_n} you mentioned.

4). Actually, I noticed my deduction has a contradition unresolved.

There is an issue to be resolved in my eigenfunction for "position" q_0 as
lim_n { \rightarrow \infty } \int \delta_n(q - q_0) | q> dq
. The problem here is whether the norm of \delta_n( q - q_0 ) shall be one or its direct integration shall be one.
If the norm shall be one, then it shall be altered to be its square root then.

[Sammywu]

Eye,

Answer to the exercise:

What you show here is a vector space with a basis, and the Hilbert space is a vector space with inner product, so I think what behind here is how to establish the relationship between an arbitrary basis and an inner product.

I). Discrete case:

I.1) From an arbitray basis to an inner product:

For two vector v and w written as [
tex] v = \sum_i b_i [/tex] and w = \sum_i b_j
with any basis b_i , we can define an inner product as ( v , b_i ) = v_i and we can deduct from there that
( v , w ) = \sum_i v_i \overline{w_i} .
This inner product will satisfy all condition required for an inner product and { b_i } becomes an orthonormal basis automatically.

If
b_i \prime = L b_i = \sum_j L_{ij} b_j
transforms a basis b_i to b_i \prime and b_i \prime happens to be orthonormal in the inner product we defined, L shall be an unitary transformation. ( I haven't proved this yet, but I think this shall be right ).

I.2) From an inner product to a basis:

Let any two \psi_1 , \psi_2 \in H , set
b_1 = \psi_1 \div ( \psi_1, \psi_1 ) .
Set
\psi_2 \prime = \psi_2 - ( \psi_2 , b_1 ) b_1
.

If \psi_2 \prime is not zero, then set
b_2 = \psi_2 \prime \div ( \psi_2 \prime , \psi_2 \prime ) .
I can establish an orthonormal basis { b_1 , b_2 } for the space spanned by \psi_1 , \psi_2 .

Taking in a \psi_3 with
\psi_3 \prime = \psi_3 - ( \psi_3 , b_1 ) b_1 - ( \psi_3 , b_2 ) b_2
not zero, we can set
b_3 = \psi_3 \prime \div ( \psi_3 \prime , \psi_3 \prime )
and span the space even larger.

Continuing this process, we can establish an orthonormal basis as long as the Hilbert space has finite or infinite but countable dimension.

Does separability contribute to ensure its countability?

[Sammywu]

II) For a continuous spectrum:

II.1) From any basis to an inner product:

For two vector v and w written as v = \int v(q) | q> dq and w = \int w(q) | q > dq with any continuous vector density basis | q > , we can define an inner product as ( v , w ) = \int v(q) \overline{w(q)} dq . This inner product will satisfy all condition required for an inner product and { | q > } shall be a generalized orthonormal basis automatically.
( I need to work out the detail of ( v , |q> ) later ).

If | p > = L | q > = \int L(p,q) | q > dq transforms a basis | q > to | p > and | p > happens to be orthonormal in the inner product we defined, L shall be an unitary transformation. ( Again, pending detail proof. )

If L is unitary, then
| q > = \overline{L^T} | p > = \int \overline{L(q,p)} | p > dp
So for v = \int v(q) | q> dq , v can be transformed to
v = \int v(q) \int \overline{L(q,p)} | p > dp dq =
\int \int v(q) \overline{L(q,p)} dq | p > dp

So
\int v(q) \overline{L(q,p)} dq
become the coefficient representing in |p> .

I.2) From an inner product to a basis:

The process of this part is almost exactly the same as the discrete one.
I need to figure out how separability contribute to ensure its countability.

[Sammywu]

Addentum to II.1).

When dealing with ( \psi , q) = < q | \psi > , there shall be an extra care because |q> could be representing two different things here.

Inside the integral
\int \psi(q) | q > dq
, it's a " vector density".

And we aslo use it to denote the eigenvector or eigenket of "position", in this case it's a normal vector not a "vector density".

Strictly speaking, for
\psi (q) = ( \psi, q ) = < q | \psi >
, if |q> is a "vector density" here, then it's not an inner product but rather an "inner product density".

But with this in mind, I am able to write the eigenket as
| q > = \int \delta(q \prime - q ) |q \prime > dq \prime
, or more precisely if considering phase factors,
| q > = \int \delta(q \prime - q ) e^{ik(q \prime)} | q \prime >d q \prime =

lim_{n \rightarrow \infty} \int \frac{e^{- \frac{ n^2 ( q \prime - q)^2 }{2} }}{ \pi^{ \frac{1}{4} } \frac{1}{n} } e^{ik(q \prime)} | q \prime > d q \prime
.

Here I think using Gausian wave function as the approximate function sequence could be the best. And I have chosen a factor in such a way that
their norms could be one always.

Any way, with this we can say that the "inner product density" of an eigenket |q> and the vector density | q \prime > of the position operator is,
< q \prime | q > = \delta ( q \prime - q ) e^{ik(q)} e^{-ik(q \prime)}

And the "inner product" of two eigenkets |q> and | q \prime > shall be
\int \delta ( q \prime \prime - q \prime ) \delta ( q \prime \prime - q ) e^{ik ( q \prime \prime)} e^{-ik \prime (q \prime \prime)} dq \prime \prime

I will see whether I prove that they will be the same value any way?

And the "inner product" of an eigenket |q> and any ket \psi shall be
< \psi | q > = \int \delta ( q \prime - q ) \overline{\psi( q \prime )} e^{ik(q \prime )} dq \prime

[Sammywu]

Eye,

Actually after I read through your response, I already understand why
< q | Q | \psi > = q \psi(q)
.

Now I figured out that you expect me to explore more in that A as a self-adjoint linear operator here and about this EQ.

First, in discrete case.
We know A = \sum a_n P_{\psi_n} .

For
A \psi_n = a_n \psi_n
, So
A_{ij} = a_i \delta_ij

Fo any vector v = \sum_i v_i \psi_n ,
< \psi_n | A | v > = < \psi_n | A | \sum_i v_i \psi_n > =
\sum_i v_i < \psi_n | A | \psi_i > =
v_n a_n

[Sammywu]

Now, in the continuous case.
We know Q = \int q | q> < q | dq .

In analogous to A_{ij} = < b_j | A | b_i > = a_i \delta_{ij} ,
The component of Q as < q | Q | q \prime > is
q \prime \delta( q \prime - q)

Fo any vector
v = \int v(q \prime ) | q \prime > dq \prime ,
< q | Q | v > = < q | Q | \int v(q \prime ) | q \prime > dq \prime > =
\int v(q \prime ) < q | Q | q \prime > dq \prime =
\int v(q \prime ) q \prime \delta ( q \prime - q ) dq \prime =
v( q ) q

[Eye_in_the_Sky]

23). In trying to evaluate 22), I found I need something clearer about how to represent all vectors. Let me put all eigenvalues in one real line; for each q in this real line, we associate an eigenvector ^\rightarrow{q} with it. I want to avoid using | q > for now, because | q > is actually a ray. also, remember there are many vectors as c * ^\rightarrow{q} where | c | =1 can be placed here; let's just pick any one of them.

So, now with a function c(q), we can do a vector integration over the q real line as:
\int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq
Note q in c(q) and dq is just a parameter and ^\rightarrow{q} is a vector, and also viewed a vector function paramterized by q.But |q> is not a ray. (From what you have written in your later posts, it appears to me that you now realize this.) I see no difference at all between "→q" and "|q>".
____________24).Refering back to 21), all vectors can be represented now by:
lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty c_n(q) ^\rightarrow{q} dq Yes, even the generalized ones like →q. As you point out:25). In particular, let
\delta_n( q - q_0 ) = 1/n for q_0 - 1/2n <= q <= q_0 +1/2n and 0 elsewhere,
the eigenvector for q_0 can be represented as:
lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty \delta_n(q - q_0) ^\rightarrow{q} dq ____________26). And, for other vectors, c_n(q) can be set to a constant function c(q);
we can verify its consistency with the normal representation:

lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty c_n(q) ^\rightarrow{q} dq =
\int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq =
\int_{ - \infty }^\infty c(q) lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty \delta_n(q \prime -q) ^\rightarrow{q \prime } dq \prime dq =
lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty \int_{ - \infty }^\infty c(q) \delta_n(q \prime - q) ^\rightarrow{q \prime} dq dq \prime =
lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty ( \int_{ - \infty }^\infty c(q) \delta_n(q \prime - q) dq ) ^\rightarrow{q \prime} dq \prime =
lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty c(q \prime) ^\rightarrow{q \prime} dq \prime

27). For inner products of c and d,
( \int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq , \int_{ - \infty }^\infty d(q \prime ) ^\rightarrow{q \prime } dq \prime ) =
\int_{ - \infty }^\infty \overline{d(q \prime)} ( \int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq , ^\rightarrow{q \prime} ) dq \prime Apart from a typo in a couple of indices, these relations look fine. (I must, however, point out that I have never seen the use of such "limits" in expressions which also involve objects like "|q>" (or, as you are writing, "→q"). Usually, these limits are used only in the "function-space" representation of the Hilbert space in order 'justify' (or 'explain') the use of "distributions". Once that has been accomplished, then there is no longer any need to bring those limits into the picture when dealing with the "formal" space of "bras" and "kets", because the meanings of these objects are defined by 'correspondence' with the (now "generalized") function-space representation.)
____________28). Now, I have to discuss what shall it be for
( \int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq , ^\rightarrow{q \prime } )

First, if we look into the inner products of two eigenvectors \rightarrow{q \prime } and \rightarrow{q } , we can first think about what shall be the innerproduct betwen
lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty \delta_n(u - q) ^\rightarrow{u} du
and
lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty \delta_n(v - q \prime) ^\rightarrow{v} dv
.

Comparing it to a discreet case, I guess this could be
| ( q , q \prime) | = lim_{ i \rightarrow \infty } lim_{ j \rightarrow \infty } \int_{ - \infty }^\infty \delta_i( u - q \prime) \delta_i( u - q ) du

So, in general,
( q , q \prime) = e^{ia} lim_{ i \rightarrow \infty } lim_{ j \rightarrow \infty } \int_{ - \infty }^\infty \delta_n( u - q \prime) \delta_n( u - q ) du

The phase factor is put in to show the possibility of two out-of-phase eiegnvector. For now, we can assume our standard basis are in-phase vectors.I do not see why you are bringing phase factors into the picture here. The objects |q> are just the generalized eigenkets of Q. This means

[1] Q|q> = q|q> , and <q|q'> = δ(q - q') .

The second relation tells us that: (i) <q|q> = ∞; and (ii) for q ≠ q', <q|q'> = 0. There is no 'room' here for phase factors.

On the other hand, once we have designated one such "family" |q>, we can then talk about another such family, say |u(q)> ≡ eiø(q)|q>. Clearly, relations [1] will also be satisfied for |u(q)>, i.e.

[1'] Q|u(q)> = q|u(q)> , and <u(q)|u(q')> = δ(q - q') .

But, which "family" is the 'real' |q> ... "|q>" or "|u(q)>"? From this perspective, the answer is: Whichever one we want! It is much like the situation with imaginary numbers: Which is the 'real' i ... "i" or "-i"?
____________With this, we can further translate the inside part of 27) to.

( \int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq , ^\rightarrow{q \prime} ) =
\int_{ - \infty }^\infty c(q) ( ^\rightarrow{q} , ^\rightarrow{q \prime} ) dq =
\int_{ - \infty }^\infty c(q) lim_{ i \rightarrow \infty } lim_{ j \rightarrow \infty } \int_{ - \infty }^\infty \delta_n( u - q \prime) \delta_n( u - q ) du dq =
lim_{ i \rightarrow \infty } lim_{ j \rightarrow \infty } \int_{ - \infty }^\infty \delta_i( u - q \prime) \int_{ - \infty }^\infty c(q) \delta_j( u - q ) dq du =
lim_{ i \rightarrow \infty } \int_{ - \infty }^\infty \delta_i( u - q \prime) c(u) du =
[tex] c(q \prime)

Placing that into 27), I got
( \int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq , \int_{ - \infty }^\infty d(q \prime ) ^\rightarrow{q \prime } dq \prime ) =
( \int_{ - \infty }^\infty \overline{d(q \prime)} c ( q \prime) dq \prime Again, this looks fine (except for a typo in a couple of indices).

[Eye_in_the_Sky]

Just make some conclusions on my deduction:

I. After including the phase factor consideration, | q_0 > as the normal eigenvector of the eigenvalue q_0 can be denoted as:

lim_{ n \rightarrow \infty } \int_{ - \infty }^\infty \delta_n(q - q_0) e^{ik(q)} | q> dq
, or
\int_{ - \infty }^\infty \delta(q - q_0) e^{ik_0} | q> dq As I said in the previous post, there is no 'room' here for phase factors. Let's look at your last expression for |qo>. It is

∫ δ(q - qo) eik_o|q> dq = eik_o|qo> .

Thus, |qo> = eik_o|qo>; so eik_o = 1.
______________This can be checked against that it shall be representable as
\sum_n c_n \psi_n =
\sum_n c_n \int_{ - \infty }^\infty \psi_n( q ) | q> dq
where \psi_n is the normal eigenvector of Hamiltonian, because this infinite summation can be viewed the limit of a function sequence ( To use correct math. term, I think I probably shall say sequence instead of series , series is reserved for infinite summation. correct ? ) as well.This is an infinite sum. Nevertheless, each successive additional term in a "series" gives rise to a "sequence".
______________II. If we do not simplify the 3-dim eigenvalues observed into the discussion of any one of them, then we will find out the three measurables form a 3-dim vector, and we will need to think about what is a "vector operator" or a "vector observable".
The simplification involves the idea of a "tensor product" of Hilbert spaces.
____We will have more to explore such as, what does the rotation of the vector operator mean, what shall two vector operators' inner or scalor product and their outer product be.Yes. These are good questions.

[Sammywu]

Eye,

You know that after you pointed out the objects as "vector density" and "infinitesmal vector", I have made some corrections on my late posts.

Actually, that was what initially confussed me. In my mind, I have this \psi_q as the generalized eigen vector of eigen value q for the position operator Q. I have it confussed with |q>, which is the "vector density" for it.

Now I organized them, so I realized d \psi_q is the "infinitismal vector" as "d|q)" in your writing.

The "|q>" as the "vector density" is d \psi_q/dq \prime in my thought and "d|q)/dq" in your writing.

Now in that, this falls into place:

\psi = \int \psi(q) |q> dq = \int < q | \psi > < q | dq = \int | q > < q | \psi > dq
for any \psi .

In particular,
\psi_q \prime = \int | q > < q | \psi_q \prime > dq
for the generlized eigenvector of eigenvalue q \prime .

In order for the above EQ. to be true, My first impression was
< q | \psi_q \prime > = \delta ( q \prime - q ) dq
, but actually it turns out wrong, it will give an object more like |q \prime> in the sense of
\int \delta ( q \prime - q ) O(q) dq = O ( q \prime )
.

Actually, if I use
d \psi_q \prime /dq \prime \prime = | q \prime >
, I can get
\psi_q \prime = \int | q \prime > dq \prime \prime
.

I can put this into the previous formula
\psi_q \prime = \int | q > < q | \int | q \prime > dq \prime \prime > dq =
\int \int | q > < q | q \prime > dq dq \prime \prime
If
< q | q \prime > = \delta ( q - q \prime )
then this does make back to
\psi_q \prime = \int | q \prime > dq \prime \prime
.

This is basically my self-verification on the relationship between
<q|q \prime>
and
< q \prime | \psi_q > .

Because in my mind,
< \psi_q | \psi_q > =1
so
< q | q > = < d \psi_q \div dq \prime | d \psi_q \div dq \prime > =1
, but apparently it does not come out so.

II) I brought in the phase factor, because I thought e^ia \psi_q are also an eigenvector.
Even if we take | q > = e^ia d \psi_q/dq , we still have a phase factor left there.
I did mentioned that we can set aside a standard set of |q> as a basis, so we do not need the phase factor in this |q> but the phase factors will then appear as an "explicit" part in <q|\psi> .
As I am saying in my mind, there shall be multiple \psi_q by the difference of e^ia . For example, if we use the "momentum" representation to represent this "position" eigenvector, you can always multiply a phase facor such as e^{iwt} to it.

[Eye_in_the_Sky]

1) I did not know | q > < q | is not a projector. I have to think about that.It would more properly be called a "projector density".

In terms of the projector PI onto an interval I = (q, q+∆q) , ∆q > 0 , it would be defined as

lim∆q → 0+ PI/∆q .

As you can see, this is a "density", and as such when it is squared, the same thing does not come back ... instead, something infinite comes back.
____________2). I did hesiate to write Q | \psi > = q | \psi > in the same reasons you mentioned, but in both Leon's Ebook and another place I did see their mentioning about the Q's defintion is Q | \psi > = q | \psi > . Just as I mentioned, the only reason I could see this "make sense" is by either
Q | \psi > = \int q |q > < q> dq or
Q | \psi > = q \psi ( q ) in the form of wavefunvtions.I think what you saw in Leon's book and any other place is

[1] Qψ(q) = qψ(q) .

This is not the same as Q|ψ> = q|ψ>.

Strictly speaking, we should not be using the same "Q" in both cases. They are different 'objects'. The first Q acts on a "function space" (so, now I will use Q to denote it). The second Q acts on a "ket space". Although the two spaces are "isomorphic", they are nevertheless formally distinct.

Now, what is the difference then between [1], which I now write as

[1] Qψ(q) = qψ(q) ,

and Q|ψ> = q|ψ> ? Well ... let's write the Q|ψ> = q|ψ> in "q-space". But(!) wait ... we've already used the "q" ... so, let's write it in "q'-space". We then have

<q'|Q|ψ> = <q'|(Q|ψ>) = <q'|(q|ψ>) = q<q'|ψ> .

The LHS we write as <q'|Q|ψ> = Q'ψ(q'), and the RHS we write as q<q'|ψ> = qψ(q'). Since LHS = RHS, we then have

[2] Q'ψ(q') = qψ(q') .

Do you see the difference between [1] and [2]? ... In [1], Q takes the function ψ(q) to a new function qψ(q), for which the two 'instances' of q are the same variable. But in [2], Q' takes ψ(q') to qψ(q'), for which q' is the variable and q is a constant ... and the fact that q is a CONSTANT here is what makes [2] an "eigenvalue" equation. The q in [1] is not a constant – equation [1] is not an "eigenvalue" equation. Equation [1] is what defines the action of Q in "q-space"!
____________3). I think your defining that \psi ( q ) = < q \ \psi > actually will make many calculations I did in showing in general
< \psi \prime | \psi > = \int \overline{\psi \prime ( q ) } \psi ( q ) dq
much more straighforward thamn my cumbersome calculations.

But one problem is then what is < q | q >. The discrete answer will be it needs to be one. Which of course will lead to some kind of conflicts in a general requirement of
\int \overline{\psi ( q ) } \psi ( q ) dq = 1 .

This is probably related to the eigenprojector P_{I_n} you mentioned.<q|q> = ∞. That is what it means to say that these kets do not "belong" to the Hilbert space – they do not have finite norm ... and, on account of that, we say that these kets are "generalized". Of course, when the two q's are distinct, say q ≠ q', then <q|q'> = 0. But we can say more than just "∞" and "0" ... we can say more precisely

<q|q'> = δ(q - q') , for all q,q' Є R .

Now, going back to what I said about a "discrete" position observable:
To make the position observable "discrete" we want to have a small "interval" In corresponding to each "point" qn, say

qn = n∙∆q , and In = (qn-∆q/2 , qn+∆q/2] .

Then, our "discrete" position observable, call it Q∆q, will be a degenerate observable. It will have eigenvalues qn and corresponding eigenprojectors (not kets!) PI_n. That is,

Q∆q = ∑n qnPI_n .Each eigenvalue qn of Q∆q is infinitely-degenerate. If you think about it, then you will realize that each degenerate eigensubspace En , corresponding to qn , is nothing but the set of square-integrable functions on the interval (qn- ∆q/2 , qn+ ∆q/2). The 'magic' of it all is that ... in the limit ∆q → 0+ ... each ∞-dimensional eigensubspace En 'collapses'(!) into SOMETHING which can be characterized by a single object |q> whose interpretation is that of a "vector density", called a "generalized vector", and which has an infinite norm ... but nevertheless ... stands in the relation Q|q> = q|q>. ... abracadabra ... and so, you get a "generalized ket".
____________4). Actually, I noticed my deduction has a contradition unresolved.

There is an issue to be resolved in my eigenfunction for "position" q_0 as
lim_n { \rightarrow \infty } \int \delta_n(q - q_0) | q> dq
. The problem here is whether the norm of \delta_n( q - q_0 ) shall be one or its direct integration shall be one.
If the norm shall be one, then it shall be altered to be its square root then.Its direct integration shall be one. Look at the definition:

δn(q) ≡
n , q Є In
0 , otherwise ,

where In = (-[2n]-1 , [2n]-1) .

Its norm, however, is

sqrt{ ∫ |δn(q)|2 dq } = √n .

In the limit, this is ∞ ... as required.

[Eye_in_the_Sky]

I am tentatively bypassing a response to your posts #98-102 in order to address your current concerns.
______________You know that after you pointed out the objects as "vector density" and "infinitesmal vector", I have made some corrections on my late posts.I will keep this in mind when I get to them.
______________Actually, that was what initially confussed me. In my mind, I have this \psi_q as the generalized eigen vector of eigen value q for the position operator Q. I have it confussed with |q>, which is the "vector density" for it.Now, let's make sure we have gotten this straight. The object "|q>" is the "generalized eigenket for Q" ... and(!) it is also what I have referred to as a "vector density".

All "generalized eigenkets" are "vector densities".

The object which you are now referring to as "|ψq>" is what I have referred to as "belonging to a class of generalized vector", and it satisfies the relation d|ψq> = |q>dq, giving d|ψq> the interpretation of an "infinitesimal vector".

But ... the object "|ψq>" itself is neither(!) a "generalized eigenket" nor a "vector density"!
______________Now I organized them, so I realized d \psi_q is the "infinitismal vector" as "d|q)" in your writing.

The "|q>" as the "vector density" is d \psi_q/dq \prime in my thought and "d|q)/dq" in your writing.

Now in that, this falls into place:

\psi = \int \psi(q) |q> dq = \int < q | \psi > < q | dq = \int | q > < q | \psi > dq
for any \psi .

In particular,
\psi_q \prime = \int | q > < q | \psi_q \prime > dq
for the generlized eigenvector of eigenvalue q \prime .Where is the ket notation "| >" on your "ψ" and "ψq'"? And remember what I said above:

The object "|ψq>" is not(!) a "generalized eigenket" (... of Q).
______________In order for the above EQ. to be true, My first impression was
< q | \psi_q \prime > = \delta ( q \prime - q ) dq
, but actually it turns out wrong, it will give an object more like |q \prime> in the sense of
\int \delta ( q \prime - q ) O(q) dq = O ( q \prime ) The condition which makes

|ψq'> = ∫ |q><q|ψq'> dq

true is simply

∫ |q><q| dq = 1 ,

where the "1" here is the identity operator on the Hilbert space. Thus, there are three important properties to note about the |q>-family.

[1] Q|q> = q|q> ... "eigenkets" ,

[2] <q|q'> = δ(q - q') ... "generalized" (ortho-'normal') ,

[3] ∫ |q><q| dq = 1 ... "complete" .
______________Actually, if I use
d \psi_q \prime /dq \prime \prime = | q \prime >
, I can get
\psi_q \prime = \int | q \prime > dq \prime \prime
.This notation does not make sense. What you really mean to say is d|ψq> = |q>dq, and therefore a suitable definition for |ψq> is

|ψq> = ∫-∞q |q'> dq' .

If we introduce the "step function"

Θ(x) ≡
1 , x > 0
0 , x < 0 ,

then we can write

|ψq> = ∫ Θ(q - q') |q'> dq' .

As you can see, the "representation" of |ψq_o> in "q-space" is just
Θ(qo - q).
______________I can put this into the previous formula
\psi_q \prime = \int | q > < q | \int | q \prime > dq \prime \prime > dq =
\int \int | q > < q | q \prime > dq dq \prime \prime
If
< q | q \prime > = \delta ( q - q \prime )
then this does make back to
\psi_q \prime = \int | q \prime > dq \prime \prime And it must "take you BACK", because ∫ |q><q| dq = 1.
______________This is basically my self-verification on the relationship between
<q|q \prime>
and
< q \prime | \psi_q > .

Because in my mind,
< \psi_q | \psi_q > =1
so
< q | q > = < d \psi_q \div dq \prime | d \psi_q \div dq \prime > =1
, but apparently it does not come out so.Indeed, <qo|qo> = δ(0) = ∞. ... What about <ψq_o|ψq_o> ?

Well, the "q-space" representation of |ψq_o> is just Θ(qo - q). So,

<ψq_o|ψq_o> = ∫ |Θ(qo - q)|2 dq

= ∫ Θ(qo - q) dq

= ∫-∞q_o dq

= ∞ .

This "infinite norm" is the reason why I originally said that |ψq_o> is in a class of "generalized vector".

However, if, instead, we define an object

|ψq',q> = ∫q'q |q"> dq" , for q' < q ,

then this object would be an ORDINARY vector of the Hilbert space, and moreover, we would also have ∂|ψq',q>/∂q = |q>. But as I have already said:From the perspective of any calculation I have ever performed in quantum mechanics, the "|q>" notation of Dirac is superior.... And now I would like to suggest the following as well:

The only additional thing which bringing such considerations into a calculation of any kind can offer is a headache!

On the other hand, I do appreciate that 'playing around' with these objects can offer some measure of clarification of what is going on, and moreover, that this is what you are in fact accomplishing through such exercises.
______________II) ... For example, if we use the "momentum" representation to represent this "position" eigenvector, you can always multiply a phase facor such as e^{iwt} to it.No. You didn't write what you meant. The correct statement is:

The most general relationship between a pair of families of eigenkets |q> of Q and |p> of P is

<q|p> = ei[θ(p) - Φ(q)] 1/sqrt{2π} eipq/h_bar .
______________

P.S. On Sunday, regardless, of whether or not I am able to respond to any of your other posts, I will post something on the next "Postulate".

[Sammywu]

Eye,

I roughly got you, still reading it, glad that you clarify many points here .

I definitely agree that you can just bypass #98-102.

About my \psi_q and |q>, I tried some more in clarifing what's going on:

I). Using an self-adjoint operator as example, starting from a discrete case to a continuous case:
A = \sum a_n | \psi_n > < \psi_n |

Let's take
\triangle_n A = a_n | \psi_n > < \psi_n |

\triangle_n a = a_n - a _{n-1}

Now A can be written:
A = \sum_n ( \triangle_n A / \triangle_n a ) \triangle_n a =
\sum_n ( a_n | \psi_n > < \psi_n | / \triangle_n a ) \triangle_n a

Converting it to continous spectrum, that means
da = d_n a = \triangle_n a \rightarrow 0
for all n, and of course n \rightarrow \infty
so
A = \int d ( a | \psi_a > < \psi_a | / da ) da

Compare that to
A = \int a | a > < a | da

So, we know
d ( | \psi_a > < \psi_a | ) / da = | a> < a |

II) Look from a perspective of any nomal vector or ket, let
| \psi > = \sum_n | \psi_n > < \psi_n | \psi >

Let's take
\triangle_n \psi = | \psi_n > < \psi_n | \psi >
and again
\triangle_n a = a_n - a _{n-1} .

Now,
| \psi > = \sum_n ( \triangle_n \psi / \triangle_n a ) \triangle_n a =
\sum_n ( | \psi_n > < \psi_n | \psi > / \triangle_n a ) \triangle_n a

Converting it to continous spectrum, that means
da = d_n a = \triangle_n a \rightarrow 0
for all n, and of course n \rightarrow \infty
so
| \psi > = \int ( d ( | \psi_a > < \psi_a | \psi > ) / da ) da

Compare that to
| \psi > = \int | a > < a | \psi > da

Again we see
d ( | \psi_a > < \psi_a | ) / da = | a> < a |

III) If I want to make this formula in general:

| \psi > = \int | a > \psi(a) da =
\int | a > < a| \psi > da =

so, I expect to write
| \psi_q > = \int | a > \psi_q(a) da =
\int | a > \delta ( a - q) da =
\int | a > < a| q > da =

and also
| \psi_q > = \int | a > < a| \psi_q > da

while
| q > = \int | a > < a| q > da =
\int | a > < a| \psi_q > da = | \psi_q >

.

Take the conclusion of I) , II) and III), I found all I need is set
| \psi_q > = \ q>
and
d ( | \psi_q > < \psi_q | ) / dq = | q> < q |
.

Replacing | \psi_q > with |q> into the second EQ. I got

d ( | \psi_q > < \psi_q | ) / dq = d ( | q > < \psi_q | ) / dq= | q> < q |

All I need to do is
< q | = d < \psi_q | /dq

This can also well explain
< q| q > = \infty
and
| q > < q | is not a projector.


Thanks

[Sammywu]

Eye,

My previous post regarding the object \psi_q seems to make sense, but I just got into troube when trying to verify that with the inner products or norms.

So, I guess it's not working any way.

You can just disregard that and just move on.

To me, I really appreciate what you showed me. At least I have some ideas why this is not working, and what was brought in to make it work.

Thanks

[Eye_in_the_Sky]

Post #98

Overall, this section is handled very well.

What you say about the operator L which you construe as a "change of basis" is correct: L shall be unitary.

Concerning the Gram-Schmidt orthonormaliztion procedure which you outline in I.2), as you point out, there is a need for a countable set which "spans" the entire Hilbert space. You are quite right in identifying "separability" as the characteristic which ensures the existence of such a set. The formal definition is as follows:

Let H be a vector space with an "inner product" ( , ) , "complete" in the "induced norm" ║ ║ ≡ √( , ) . Then, H is said to be separable iff:

H has a countable dense subset.
___

This property is then equivalent to:

H has a countable orthonormal basis.
______________

Post #99

Again, overall, this section is handled quite well.

... And, yes, the operator L which transforms from one continuous "generalized" basis |q> to another one |p> given by

[1] |p> = L|q>

shall be unitary. Thus, we also have

[2] |q> = L†|p> .

If we write [1] and [2] in terms of "kernels", these two relations become:

[1'] |p> = ∫ L(p,q) |q> dq ,

[2'] |q> = ∫ M(q,p) |p> dp ,

where M(q,p) = L(p,q)*. I am pointing this out, because there is an ambiguity with your notation in the "kernel" where you have "switched" the order of p and q: | q > = \overline{L^T} | p > = \int \overline{L(q,p)} | p > dp Regarding what you say next:I.2) From an inner product to a basis:

The process of this part is almost exactly the same as the discrete one.
I need to figure out how separability contribute to ensure its countability.The families |q> and |p> are not countable. By assumption these are "generalized vectors" whose parameters q and p vary continuously over R such that

<q|q'> = δ(q - q') , ∫ |q><q| dq = 1 , and similarly for |p> .
______________

Post #100
When dealing with ( \psi , q) = < q | \psi > , there shall be an extra care because |q> could be representing two different things here.

Inside the integral
\int \psi(q) | q > dq
, it's a " vector density".

And we aslo use it to denote the eigenvector or eigenket of "position", in this case it's a normal vector not a "vector density".No! The object "|q>" is the same in both cases! This appears to be an essential point of confusion in some of your other posts. To repeat what I said in post #107, there are three important properties to note about the |q>-family (and others like it):

Q|q> = q|q> ... "eigenkets" ,

<q|q'> = δ(q - q') ... "generalized" (ortho-'normal') ,

∫ |q><q| dq = 1 ... "complete" .

Also:

All "generalized eigenkets" are "vector densities".
______________

Note that in your posts #98 and #99, you have missed my point concerning L. The operator L which I originally defined in post #96 (http://www.physicsforums.com/showpost.php?p=298408&postcount=96) was an arbitrary linear operator:Now, let L be a linear operator. Let L act on one of the basis vectors bj; the result is another vector in the space which itself is a linear combination of the bi's. That is, for each bj, we have

[1] Lbj = Lijbi .This is not (in general) a transformation from "one basis to another". L does not even have to have an inverse. Furthermore, notice that the summation is on the first index in Lij – this is not a typo! The summation needs to be defined that way in order for Lij to obey the "component transformation rule":

(Lv)i = Lijvj .

That this relation results from [1] above was shown explicitly in post #96.

... So, to answer the question I asked there, the 'connection' is:

For any orthonormal basis |bi> the "components" relative to this basis are given by

<bi|v> ... vector ,

<bi|L|bj> ... operator .

Note that with Dirac notation, all of this is, in a certain sense, 'trivialized' by the relation (I am now writing the summation explicitly)

∑j |bj><bj| = 1 (the identity on H) ,

because we can merely "insert" this relation into the appropriate spot so that

<bi|L|v> = <bi|L (∑j |bj><bj|) |v>

= ∑j <bi|L|bj> <bj||v> .
_____

Similarly, in the continuous case, we have, for the "components" relative to any "family" |a>,

<a|v> ... vector ,

<a|L|a'> ... operator ,

and this is "verified" by

<a|L|v> = <a|L ( ∫ |a'><a'| da' ) |v>

= ∫ <a|L|a'> <a'|v> da' .

_____

In your posts #101 and #102, you came close to this idea (... except your operators there were special).
______________

[Eye_in_the_Sky]

Recall the previous postulates:

P0: To a quantum system S there corresponds an associated Hilbert space HS.

P1: A pure state of S is represented a ray (i.e. a one-dimensional subspace) of HS.

P2: To a physical quantity A measurable on (the quantum system) S, there corresponds a self-adjoint linear operator A acting in HS. Such an operator is said to be an "observable".
____

And now here is the next postulate:

P3: The only possible result of a measurement of a physical quantity A is one of the eigenvalues a of the corresponding observable A. In the following, let the quantum system be in a pure state represented by |ψ>:

(i) If the eigenvalue a belongs to a discrete part of the spectrum of A with corresponding eigenprojector Pa , then the probability of obtaining the result a is given by

P(a) = <ψ|Pa|ψ> .

(ii) If the eigenvalue a belongs to a continuous part of the spectrum of A with corresponding generalized eigenket |a>, then the probability of obtaining a result in the infinitesimal interval (a, a + da) is given by

p(a) da = |<a|ψ>|2 da .
___________

Notes:

N.3.1) It is possible for an observable A to have a "mixed" spectrum – i.e. parts which are discrete as well as parts which are continuous. The discrete part of the spectrum is referred to as "the point spectrum of A", denoted Sp(A). The continuous part is referred to as "the continuous spectrum of A", denoted Sc(A). The (overall) spectrum of A is given by
S(A) = Sp(A) U Sc(A).

N.3.2) In (ii) of P3, an implicit assumption is being made, namely, that the continuous part of the spectrum of A is nondegenerate. In the examples which we encounter in quantum mechanics, whenever such an assumption does not hold, we will find that the Hilbert space in question admits a decomposition into a "tensor product" of Hilbert spaces, and that the assumption will then hold with regard to one of the Hilbert spaces in that decomposition.

N.3.3) Let A be an observable whose spectrum may have a continuous (but nondegenerate) part. For a Є Sp(A), let Pa be the corresponding eigenprojector. For a Є Sc(A), let |a> be the corresponding generalized eigenket. Then,

A = ∑a Є S_p(A) a Pa + ∫S_c(A) a |a><a| da ,

and

∑a Є S_p(A) Pa + ∫S_c(A) |a><a| da = 1 ,

where 1 is the identity on HS . This last relation is called the "closure relation" – it expresses the idea that A has a complete set of eigenkets ("generalized" or otherwise). In case the set Sp(A) or Sc(A) is empty, then the associated sum or integral is understood to be zero.
___________

Exercises:

E.3.1) Regarding N.3.1, give one or two specific examples.

In the following exercises, use Dirac notation.

E.3.2) Let A be an observable whose spectrum may have a continuous (but nondegenerate) part, and let the quantum system be in a pure state |ψ>. Use the postulate P3 and the "closure relation" of N.3.3 to verify that

∑a Є S_p(A) P(a) + ∫S_c(A) p(a)da = 1 .

E.3.3) Let an Є Sp(A) be a nondegenerate eigenvalue with corresponding eigenket |an>. Let the quantum system be in the pure state |ψ>.

(a) Verify that P(an) = |<an|ψ>|2.

(b) Define ρ = |ψ><ψ|. Verify that P(an) = <an|ρ|an>.

E.3.4) Let an Є Sp(A) be an eigenvalue (possibly degenerate) and let |ank>, k = 1, ... , g(n) , be an orthonormal basis of the eigensubspace corresponding to an. Let the quantum system be in the pure state |ψ>.

(a) Verify that P(an) = ∑k=1g(n) |<ank|ψ>|2.

(b) Define ρ = |ψ><ψ|. Verify that P(an) = ∑k=1g(n) <ank|ρ|ank>.

E.3.5) Verify the analogous expression for (b) of E.3.3, in the case of a (nondegenerate) eigenvalue a Є Sc(A).

E.3.6) Let A be an observable whose spectrum may have a continuous (but nondegenerate) part, and let the quantum system be in a pure state |ψ>. Define "the expectation value of A", denoted <A>, by

<A> ≡ ∑a Є S_p(A) a P(a) + ∫S_c(A) a p(a) da ,

where P(a) is the probability of obtaining a Є Sp(A), and p(a) is the probability density of obtaining a Є Sc(A).

(a) Explain why this definition is correct.

(b) From the definition of <A> and the postulate P3, show that <A> = <ψ|A|ψ>.

(c) Define ρ = |ψ><ψ|. In the notation of E.3.4 for the discrete part, verify that

<A> = ∑n∑k=1g(n) <ank|ρA|ank> + ∫S_c(A) <a|ρA|a>da .

[Sammywu]

E 3.1)

The one I can think of that mixes discrete and continous spetrum seems to be the simple one in that the discrete eigenvalues are basically the degenerate case of [ q , q + \triangle q ] .

By basically taking a function that maps the underlining continuous interval to discrete values, such as
f ( x ) = 2 \ when \ x \in ( 1 , 3 )
f ( x ) = 6 \ when \ x \in ( 5 , 7 )
f ( x ) = x \ when \ x \ not \ \in ( 1 , 3 ) \cup ( 5 , 7 )
, this function f of the continuous observables Q will have a mixed set of continuous spectrum and discrete spectrum.

S_{c(A)} = R - \ ( 1, 3) - \ ( 5 , 7 )

P_2 = \int_1^3 | q > < q | dq

P_6 = \int_5^7 | q > < q | dq

A = f(Q) = \int_{S_{c(A)}} q | q > < q | dq + 2 P_2 + 6 P_6

Is there any other more interesting ones?

[Eye_in_the_Sky]

Good. Your response shows that you have understood the concept.

Here are two other examples:

(i) the Hamiltonian for a particle moving in one dimension subject to a finite square-well potential;

(ii) the Hamiltonian for a particle subject to a Coulomb potential.

In each of these cases, the eigenstates of the Hamiltonian will involve bound and unbound states of the particle. The bound states are "quantized" – i.e. discrete – whereas, the unbound states are continuous. This is true in general.
____

By the way, here is another basic exercise which I forgot to include:

E.3.1) (b) Regarding N.3.2, give one or two specific examples.

(But ... if you haven't learned "tensor products" yet, this question will have to wait.)

[Sammywu]

I will do E 3.3) first.

(a)
Using
P_{a_{n}} | \psi > = | a_n > < a_n | \psi >
for nondegerate eigenvalue a_n because the projector will map | \psi > to its subcomponent of | a_n > , we get
P ( a_n ) = < \psi | P_{a_{n}} | \psi > =
< \psi ( | a_n > < a_n | \psi > ) =
< \psi | a_n > < a_n | \psi > =
| < a_n | \psi > |^2
, knowing that
< \psi | a_n > = \overline { <a_n | \psi > }
.

(b)
Using
\rho | a_n > = | \psi > < \psi | a_n >
, we get
< a_n | \rho | a_n > =
<a_n ( | \psi > <\psi | a_n > ) =
< a_n | \psi > < \psi | a_n > =
| < a_n | \psi > |^2 = P ( a_n )

[Sammywu]

E 3.4) is very similar to E 3.3).

(a)
Using
P_{a_{n}} | \psi > = \sum_{k=1}^{g(n)} | a_n^k > < a_n^k | \psi >
, we get
P ( a_n ) = < \psi | P_{a_{n}} | \psi > =
< \psi | ( \sum_{k=1}^{g(n)} | a_n^k > < a_n^k | \psi > ) =
\sum_{k=1}^{g(n)} < \psi | a_n^k > < a_n^k | \psi > =
\sum_{k=1}^{g(n)} | < a_n^k | \psi > |^2
, knowing that
< \psi | a_n^k > = \overline { <a_n^k | \psi > }
.

(b)
Using
\rho | a_n^k > = | \psi > < \psi | a_n^k >
, we get
\sum_{k=1}^{g(n)} < a_n^k | \rho | a_n^k > =
\sum_{k=1}^{g(n)} <a_n^k | ( | \psi > <\psi | a_n^k > ) =
\sum_{k=1}^{g(n)} < a_n^k | \psi > < \psi | a_n^k > =
\sum_{k=1}^{g(n)} | < a_n^k | \psi > |^2 = P ( a_n )

[Sammywu]

E 3.2)

\forall \psi \in H \ ,
< \psi | \ ( \sum_{S_{P(a)}} P_{a_{n}} + \int_{S_{c(a)}} | a > < a | da ) \ | \psi > =
< \psi | 1 | \psi > =
< \psi | \psi > = 1

< \psi | \ ( \ \sum_{S_{P(a)}} P_{a_{n}} + \int_{S_{c(a)}} | a > < a |da \ ) \ | \psi > =
< \psi | \ ( \ \sum_{S_{P(a)}} P_{a_{n}} \ ) \ | \psi > + < \psi | \ ( \ \int_{S_{c(a)}} | a > < a |da \ ) \ | \psi > =
\sum_{S_{P(a)}} < \psi | P_{a_{n}} | \psi > + < \psi | \ ( \ \int_{S_{c(a)}} | a > < a | da \ ) \ | \psi > =
\sum_{S_{P(a)}} P( a_n ) + < \psi | ( \ \int_{S_{c(a)}} | a > < a | da \ ) | \psi > =

Now all I need to prove is
< \psi | ( \int_{S_{c(a)}} | a > < a | da ) | \psi > =
\int_{S_{c(a)}} P(a) da

< \psi | ( \ \int_{S_{c(a)}} | a > < a | da \ ) | \psi > =
< \psi | ( \ \int_{S_{c(a)}} | a > < a | \psi > da \ ) > =
( By N 3.3 Completeness , it will be translated into below :)
< (( \ \sum_{S_{P(a)}} P_{a_{n}} | \psi > \ ) + ( \ \int_{S_{c(a)}} | a \prime > < a \prime | \psi > da \prime \ ) )| ( \int_{S_{c(a)}} | a > < a | \psi > da ) > =
( By orthogonality of eigenkets, we can reduce it to )
< ( \ \int_{S_{c(a)}} | a \prime > < a \prime | \psi > da \prime \ ) | ( \int_{S_{c(a)}} | a > < a | \psi > da ) > =
\int_{S_{c(a)}} \overline{< a \prime | \psi >} < ( | a \prime > ) | ( \int_{S_{c(a)}} | a > < a | \psi > da ) > da \prime =
\int_{S_{c(a)}} \overline{< a \prime | \psi >} \int_{S_{c(a)}} < a \prime | a > < a | \psi > da da \prime =
\int_{S_{c(a)}} \overline{< a \prime | \psi >} ( \int_{S_{c(a)}} \delta ( a - a \prime ) < a | \psi > da ) da \prime =
\int_{S_{c(a)}} \overline{< a \prime | \psi >} < a \prime | \psi > da \prime =
\int_{S_{c(a)}} | < a \prime | \psi > |^2 da \prime =
\int_{S_{c(a)}} P(a) da

[Sammywu]

E 3.5)
By P3,
P(a) = | < a | \psi > |^2 da =
< a | \psi > \overline{< a | \psi >} da
.

< a | \rho | a > = < a | ( \psi > < \psi | a > ) =
< a | \psi > < \psi | a >

We already know what is < a | \psi > , but unclear about what < \psi | a > could be.

By comparing the two EQs above, the only way we can equate them is
< \psi | a > = \overline{< a | \psi >} da
.

In a way this make sense, because a is a "generalized vector" and " vector density".

So , < \psi | a > shall be an infinitesmal amount proportional to da and also it shall be propotional to \overline{< a | \psi >} by the general rule of inner product.

I think maybe I can use similar mechanism of \triangle a to get a better analogous proof for it. Later I will give it a try.

[Sammywu]

Just try to summarize what I learn here before continue on E 3.5)

1) Let me start with
\sum_{S_{P(a)}} P_{a_{n}} + \int_{S_{c(a)}} | a > < a | da = I
, so
\int_{S_{c(a)}} | a > < a | da = I - \sum_{S_{P(a)}} P_{a_{n}}
.

2). Take derivative of a to it at the continuous part, then
\frac{d I}{ da } da = | a > < a | da
or in other words,
\int_{S_{c(a)}} \frac{d I}{ da } da = \int_{S_{c(a)}} | a > < a | da

3). Further,
\frac{d I}{ da } da | \psi > = | a > < a | \psi > da

4) Consider
P ( a ) =
< \psi | \frac{d I}{ da } da | \psi > =
< \psi | \frac{d I}{ da } | \psi > da

[Sammywu]

E 3.5)

By taking the generlized EQ. you posted at #110,

< \psi | L | \varphi > = \int < \psi | L | a> < a | \varphi > da
or
< a \prime | L \varphi > = \int < a \prime | L | a> < a | \varphi > da
, this is pretty easy.

Taking
< \psi | I | \varphi > = \int < \psi | a> < a | \varphi > da =
\int \overline{< \psi | a>} < a | \varphi > da
, we know
< \psi | a > = \overline{< a | \psi >}
.

Actually, it was my fault, somehow I missed the da in the LHS, the actual EQ. in your post is :
P(a) da = | < a | \psi > |^2 da
.

Verifying this, I take
D ( a \prime ) = \int_{ - \infty}^{ a \prime } \overline{< a | \psi >} < a | \psi > da
, then
P(a) = dD /da = \overline{< a | \psi >} < a | \psi >
, in that P(a) is a "probability density" and then
P(a) da = | < a | \psi > |^2 da can be an "infinitismal probability".

Other than E 3.5), this EQ also shows other facts you have posted:

In order for this to be true,
< q | \psi > = < q | I | q > = \int < q | q \prime > < q \prime | \psi > dq \prime
, we need
< q | q \prime > = \delta ( q \prime - q )
.

Also,
< q | q > = < q | I | q > = \int < q | q \prime > < q \prime | q > dq \prime =
\int \delta ( q - q \prime) ( q \prime - q ) dq \prime = \infty
.

[Sammywu]

E 3.6 )

A). If the probability of an obeservable showing a is 1, of course we will expect it always shows a.

If the observable has multiple possible outcomes, we really can not say which one it will definiitely show, we basically make a math. average of them and saying this is its average value, which is defined as " expectation value".

So, in a discrete case, it's of course:
\sum a P(a)

In a continuous case, we will take the approximate average as
\sum a \ \ Probability([a - \triangle a, a + \triangle a ])
when \triangle a \rightarrow 0 , we get
\int a P(a) da

If we have both discrete and continuous, we will take the average as
\sum_{S{p(a)}} a P( a | S_{p(a)} ) P( S_{p(a)} ) + \int_{S{c(a)}} a P( a | S_{c(a)} ) P( S_{c(a)} ) da =
\sum_{S{p(a)}} a P( a ) + \int_{S{c(a)}} a P( a ) da

b)
< \psi | A | \psi > =
< \psi | \ \sum_{S{p(a)}} a P_a + \int_{S{c(a)}} a |a> <a| da \ | \psi > =
< \psi | \ \sum_{S{p(a)}} a P_a \ | \psi > \ + \ < \psi | \int_{S{c(a)}} a |a> <a| da \ | \psi > =
\sum_{S{p(a)}} a < \psi | P_a | \psi > \ + \ \int_{S{c(a)}} a < \psi |a> <a| \psi > da =
\sum_{S{p(a)}} a P(a) + \ \int_{S{c(a)}} a P(a) da =
< A >

[Sammywu]

< A > =
\sum_{S{p(a)}} a P(a) \ + \ \int_{S{c(a)}} a P(a) da =

( By E 3.4 and E 3.5 , we derive below: )

\sum_n \sum_{k=1}^{g(n)} a < a_n^k | \rho | a_n^k > \ + \ \int_{S{c(a)}} a < a | \rho | a > da =

( Using the definition of eigenkets as,
A a_n^k = a a_n^k
A | a > = a | a >
we can derive below: )

\sum_n \sum_{k=1}^{g(n)} < a_n^k | \rho A | a_n^k > \ + \ \int_{S{c(a)}} < a | \rho A | a > da

[Sammywu]

E 3.1) a)

Even though my knowledge of "tensor product" of vector spaces might not be enough, I did give a thought on how to handle this exercise.

1). How do we know the spectrum of Q is degenerate?

Unless, we have another set of eigenbasis and self-adjoint operator, said E, and we found we have problems reconciling them with Q. For example,
< E | \psi > = \int < E | q > < q | \psi > dq
does not hold steady or meet the expectation for a state | \psi > .

2). We will then believe that we need to expand the Hilbert space by adding another one in. Why don't we choose "direct sum" instead of "tensor product"?

If we we choose "direct sum", the basis will be extended as
v_1, v_2, ... v_n, w_1, ... w_n
; we are basically just adding more eigenvalues and eigenvectors in doing so.

We need something like
< x | \psi > = \int < xy | \psi > dy
or
P (xy) dxdy = | < xy | \psi > |^2 dxdy
.

So we need to define a "product" of vector spaces that fit our needs.

I may continue later.

================================================

For E 3.1) a), I have "roughly" read a chapter about the unbounded and bounded solutions of square well potential problem.

I found the difference that led to the discrete bounded solution is basically that the solutions outside of the well are in the forms of A e^{-p_1 x/h} at the LHS and D e^{p_1 x/h} at the RHS of the well, because the condition of E < 0 took out the imaginary part.

It's very interesting.

[Sammywu]

E 3.1) B)

3) In looking what we already have,

| \psi_x > = | x> < x | \psi_x >
,
P(y) dy = < \psi | y> < y | \psi > dy
,
| \psi_y > = | y> <y | \psi_y >
,
P(x) dy = < \psi | x> < x | \psi > dx
and our belive that
| \psi > = | x, y > < x,y | \psi >
and
P(x,y) dx dy = < \psi | x,y> <x,y | \psi > dx dy
, our quickiest approach will be:
< x, y | \psi > = < y | \psi_y > < x | \psi_x >
and
< \psi | x,y > = \overline{< y | \psi_y > < x | \psi_x >} =
< \psi_y | y > < \psi_x | x >
.

This turned out it will satisfy both needs in representation of ket and probability.

Not only that, we also see
P(x,y) dx dy = P(x) p(y) dx dy
.

This satisfies the general probability rule:
P ( x, y ) = P(x)P(y|x)
; the issue here seems to be that
P(y) = P(y|x), which means x and y need to be independent to each other.

So, if we do find the eigenvalues of x and y independent to each other, we will try to expand our Hilbert space in the above way.

Our curent ket space of | x,y,z,s > is of course an example.

4). If x and y are not independent, what will we get?

For example, I can easily produce a degenerate continuous spetrum by using function f(x) = |x|.

[Sammywu]

E 3.1) B).

5). In 3), we actually have used the two property of "tensor product":

( | x_1 > + | x_2 > ) \otimes | y > = | x_1 , y > + | x_2 , y >
| x > \otimes ( | y_1 > + | y_2> ) = | x , y_1 > + | x , y_2 >

The last property
\alpha | x, y > = | \alpha x> \otimes | y> = | x> \otimes | \alpha y >
will be used in
< A > = < x,y | A | x,y >
.

[Sammywu]

Checked with a book, I found my answer to E3.1) B) will basically lead to "looking for the set of maximum commutable observables".

That is not what Eye asked any way. Eye's question already assumed there is a Hilbert space and we found multiple "generalized kets" for one eigenvalues of a continuous spectrum.

Any way, to look for that answer, I found I need to clarify there are only true discrete eigenkets in our assumption.

So I went sideway to find a proof why "separability" ensures "countable discrete eigenkets".

The proof is actually quite straightforward, after a few day's rumination:
1). A Hilbert space is a vector space with an inner product.
2). An inner product can define the "norm". The "norm" can define the distance between two vectors.
3). With the "distance", we can define the open sets and so the topology.
4). Definition of "separability" says if we have an open covering for the Hilbet space, then we have a "counatble" open covering as its subset.
5). For any two vectors belong to an orthonormal basis, their distance will be 2^{1/2} .
6). If we define an open coverings as all open balls with a radius 0.25, we will have a countable subset that covers the Hilbert space.
7). Because any two orthonormal vectors has a distance greater than 0.25, so no two orthonormal vectors could be in one open ball; this implys the number of elements in the countable open coverings is greater than the number of all orthonormal vectors, so the basis is definitely countable.

Does this look good?
==================================================

Any way, if we assume the Hilbert space only has countable orthonormal basis, then we know there is no true eigenvector for a continuous spectrum because any continuous real interval has uncountable numbers in it.

Of course, I think there must be a proof that shows the space of square-integrable functions has a countable orthonormal basis and is separable. So, if the "space" has an isomorphich structure with the "functoion" space, it has only countable orthonormal basises and is separable.

[Eye_in_the_Sky]

The limiting procedure which you allude to in post #108 is not at all "well-defined". You have given no "structure" which tells how A changes with each incremental step in the supposed "limiting procedure".
____

P.S. I had to change this post and cut out all of the quotes because the LaTeX was doing really strange things!

P.P.S. I also wanted to respond to your posts #114-124, but LaTex is malfunctioning. I may not have another opportunity to post more until the beginning of next week.

[Eye_in_the_Sky]

This is what I originally wanted to post.

Here is what you said in post #109 regarding post #108:My previous post regarding the object \psi_q seems to make sense, but I just got into troube when trying to verify that with the inner products or norms.

So, I guess it's not working any way.

You can just disregard that and just move on. ...To what you have said here, I will only add that the limiting procedure which you allude to in post #108 is not at all "well-defined". For example, when you say:Let's take
\triangle_n A = a_n | \psi_n > < \psi_n |

\triangle_n a = a_n - a _{n-1} ... you have given no "structure" which tells how A changes with each incremental step in the supposed "limiting procedure".

[Eye_in_the_Sky]

Posts #114, 115

Your answers to E.3.3 and E.3.4 look fine.
______________

Post #116

In your answer to E.3.2, you began with: \forall \psi \in H \ , ...I guess at that moment you forgot that H includes all of the vectors; i.e. even those with norm different from 1.

Later on you reach:Now all I need to prove is
< \psi | ( \int_{S_{c(a)}} | a > < a | da ) | \psi > =
\int_{S_{c(a)}} P(a) da The next step should have been to 'push' both the bra "<ψ|" and the ket "|ψ>" through and underneath the integral to get

∫S_c(A) <ψ|a><a|ψ> da .

For some reason, you were inclined to do this only with regard to the ket (with a slight 'abuse' of notation): < \psi | ( \ \int_{S_{c(a)}} | a > < a | da \ ) | \psi > =
< \psi | ( \ \int_{S_{c(a)}} | a > < a | \psi > da \ ) > = But you got around this by invoking an "inner product" (again, with a slight 'abuse' of notation), and then you convinced yourself (quite correctly) that <ψ|a> = <a|ψ>* ... which is what you needed to take the final step.

Let's put an end to this 'abuse' of Dirac notation. Here's what you wrote: < (( \ \sum_{S_{P(a)}} P_{a_{n}} | \psi > \ ) + ( \ \int_{S_{c(a)}} | a \prime > < a \prime | \psi > da \prime \ ) )| ( \int_{S_{c(a)}} | a > < a | \psi > da ) > = What you had there was the ket

∑a Є S_p(A) Pa|ψ> + ∫S_c(A) |a><a|ψ> da ,

which you needed to 'turn around' into a bra. That's easy to do: kets go to bras, bras go to kets, numbers go to their complex conjugates, and operators go to their adjoints. In this case, we get

∑a Є S_p(A) <ψ|Pa + ∫S_c(A) <ψ|a><a| da .

... And that's all there is to it. Now, you just need "slam" this expression on it's right side with the expression

∫S_c(A) |a'><a'|ψ> da'

and you will get the desired result. This is how to 'use' Dirac notation without 'abuse'.

______________

Post #117

This answer for E.3.5 is fine ... except for the "dangling" da: < \psi | a > = \overline{< a | \psi >} da But, later on, in post #119 you correct yourself.
______________

Posts #118, 119

In post #118 you say:1) Let me start with
\sum_{S_{P(a)}} P_{a_{n}} + \int_{S_{c(a)}} | a > < a | da = I
, so
\int_{S_{c(a)}} | a > < a | da = I - \sum_{S_{P(a)}} P_{a_{n}}
.

2). Take derivative of a to it at the continuous part, then
\frac{d I}{ da } da = | a > < a | da
or in other words,
\int_{S_{c(a)}} \frac{d I}{ da } da = \int_{S_{c(a)}} | a > < a | da The parameter "a" is the variable of integration. It is not "free" to take a derivative with respect to it.

What you do in post #119, along similar lines, however, is correct: D ( a \prime ) = \int_{ - \infty}^{ a \prime } \overline{< a | \psi >} < a | \psi > da
, then
P(a) = dD /da = \overline{< a | \psi >} < a | \psi > Indeed, you have a "free" parameter here.

Everything else looks fine (modulo a couple of minor typos).
______________

Posts #120, 121

These look fine.
______________

Post #122E 3.1) a)

Even though my knowledge of "tensor product" of vector spaces might not be enough, I did give a thought on how to handle this exercise.

1). How do we know the spectrum of Q is degenerate?

Unless, we have another set of eigenbasis and self-adjoint operator, said E, and we found we have problems reconciling them with Q. For example,
< E | \psi > = \int < E | q > < q | \psi > dq
does not hold steady or meet the expectation for a state | \psi > .I don't understand what you meant in the above.

Next:2). We will then believe that we need to expand the Hilbert space by adding another one in. Why don't we choose "direct sum" instead of "tensor product"?

If we we choose "direct sum", the basis will be extended as
v_1, v_2, ... v_n, w_1, ... w_n
; we are basically just adding more eigenvalues and eigenvectors in doing so.... Right. So, that's not what we want.We need something like
< x | \psi > = \int < xy | \psi > dy
or
P (xy) dxdy = | < xy | \psi > |^2 dxdy
.

So we need to define a "product" of vector spaces that fit our needs.Yes, this is the idea.
______________

Post #123

Looks good (... I do see one small typo, though).

However, about:, our quickiest approach will be:
< x, y | \psi > = < y | \psi_y > < x | \psi_x > This is not true in general; i.e. it is true only when the state is such that x and y are "independent".
___Our curent ket space of | x,y,z,s > is of course an example.... where "s", I assume, refers to spin. This is precisely the example I had in mind (except that I split it up into two examples: |x,y,z> and |x,s>).
___4). If x and y are not independent, what will we get?We will get "correlations".

Next:For example, I can easily produce a degenerate continuous spetrum by using function f(x) = |x|.Yes, |Q| has a continuous, doubly degenerate spectrum.
______________

Post #124

E 3.1) B).

5). In 3), we actually have used the two property of "tensor product":

( | x_1 > + | x_2 > ) \otimes | y > = | x_1 , y > + | x_2 , y >
| x > \otimes ( | y_1 > + | y_2> ) = | x , y_1 > + | x , y_2 >

The last property
\alpha | x, y > = | \alpha x> \otimes | y> = | x> \otimes | \alpha y >
will be used in
< A > = < x,y | A | x,y >
.Yes ... except, the last relation should read

<A> = ∫∫ <x,y|A|x,y> dx dy .
____________________

P.S. I may not be able to post again for another 3 weeks.