Understanding Projectors in Quantum Mechanics: A Mathematical Approach

[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 anyone 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 into 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
$$( \int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq , \int_{ - \infty }^\infty d(q \prime ) ^\rightarrow{q \prime } dq \prime ) =$$
$$( \int_{ - \infty }^\infty \overline{d(q \prime)} c ( q \prime) dq \prime$$[/tex]

 

[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 anyone 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]

Responses to posts #84,86,88,89

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, P_I defined by

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

or equivalently,

P_I ≡ ∫_I |q><q| dq ,

does satisfy P_I² = P_I.
______________

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|P_I|ψ> .

The last expression corresponds to (ψ, P_Iψ) 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" I_n corresponding to each "point" q_n, say

q_n = n∙∆q , and I_n = (q_n-∆q/2 , q_n+∆q/2] .

Then, our "discrete" position observable, call it Q_∆q, will be a degenerate observable. It will have eigenvalues q_n and corresponding eigenprojectors (not kets!) P_{I_n}. That is,

Q_∆q = ∑_n q_nP_{I_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 c_n|φ_n> ,

[2'] c_n = <φ_n|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|c_n|² < ∞ .

Everything is the 'same' in both cases, except for the fact that <q|q> = ∞ , whereas <φ_n|φ_n> = 1. That is, each |q> is not a member of H, whereas each |φ_n> 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(q_o)|²dq_o = 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]

Response to post #93

(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 |φ_n>, what is the φ_n-representation of |ψ>? ... It is just <φ_n|ψ>. And if we write |ψ> = ∑_n c_n|φ_n>, we then have <φ_n|ψ> = c_n. So,

c_n is just the n-component of |ψ> in the |φ_n>-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 b_i be a basis. Then, (using the "summation convention" for repeated indices) any vector v can be written as

v = v_ib_i .

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

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

[1] Lb_j = L_ijb_i .

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

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

w_ib_i

= w

= Lv

= L(v_jb_j)

= v_j(Lb_j)

= v_j(L_ijb_i) ... (from [1])

= (L_ijv_j)b_i .

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

[2] w_i = L_ijv_j ,

where, L_ij 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 b_i.
_____

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

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

(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:

b_i = |b_i>

v = |v>

v_i = <b_i|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]

Response to post #91

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 anyone 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 into 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)> ≡ e^iø(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
$$( \int_{ - \infty }^\infty c(q) ^\rightarrow{q} dq , \int_{ - \infty }^\infty d(q \prime ) ^\rightarrow{q \prime } dq \prime ) =$$
$$( \int_{ - \infty }^\infty \overline{d(q \prime)} c ( q \prime) dq \prime$$[/tex][tex][/tex]

$$Again, this looks fine (except for a typo in a couple of indices).$$

 

[Eye_in_the_Sky]

Response to post #94

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 |q_o>. It is

∫ δ(q - q_o) e^ik_o|q> dq = e^ik_o|q_o> .

Thus, |q_o> = e^ik_o|q_o>; so e^ik_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 anyone 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]

Response to post #97

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 P_I onto an interval I = (q, q+∆q) , ∆q > 0 , it would be defined as

lim_{∆q → 0+} P_I/∆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" I_n corresponding to each "point" q_n, say

q_n = n∙∆q , and I_n = (q_n-∆q/2 , q_n+∆q/2] .

Then, our "discrete" position observable, call it Q_∆q, will be a degenerate observable. It will have eigenvalues q_n and corresponding eigenprojectors (not kets!) P_{I_n}. That is,

Q_∆q = ∑_n q_nP_{I_n} .

Each eigenvalue q_n of Q_∆q is infinitely-degenerate. If you think about it, then you will realize that each degenerate eigensubspace E_n , corresponding to q_n , is nothing but the set of square-integrable functions on the interval (q_n- ∆q/2 , q_n+ ∆q/2). The 'magic' of it all is that ... in the limit ∆q → 0+ ... each ∞-dimensional eigensubspace E_n '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:

δ_n(q) ≡
n , q Є I_n
0 , otherwise ,

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

Its norm, however, is

sqrt{ ∫ |δ_n(q)|² dq } = √n .

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

 

[Eye_in_the_Sky]

Response to post #105

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 "|ψ_q>" is what I have referred to as "belonging to a class of generalized vector", and it satisfies the relation d|ψ_q> = |q>dq, giving d|ψ_q> the interpretation of an "infinitesimal vector".

But ... the object "|ψ_q>" 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 "ψ_q'"? And remember what I said above:

The object "|ψ_q>" 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

|ψ_q'> = ∫ |q><q|ψ_q'> 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|ψ_q> = |q>dq, and therefore a suitable definition for |ψ_q> is

|ψ_q> = ∫_-∞^q |q'> dq' .

If we introduce the "step function"

Θ(x) ≡
1 , x > 0
0 , x < 0 ,

then we can write

|ψ_q> = ∫ Θ(q - q') |q'> dq' .

As you can see, the "representation" of |ψ_{q_o}> in "q-space" is just
Θ(q_o - 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, <q_o|q_o> = δ(0) = ∞. ... What about <ψ_{q_o}|ψ_{q_o}> ?

Well, the "q-space" representation of |ψ_{q_o}> is just Θ(q_o - q). So,

<ψ_{q_o}|ψ_{q_o}> = ∫ |Θ(q_o - q)|² dq

= ∫ Θ(q_o - q) dq

= ∫_-∞^q_o dq

= ∞ .

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

However, if, instead, we define an object

|ψ_q',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 ∂|ψ_q',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> = e^{i[θ(p) - Φ(q)]} 1/sqrt{2π} e^ipq/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 continuous 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 continuous 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 into make it work.

Thanks

 

[Eye_in_the_Sky]

Response to posts #98-102

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 was an arbitrary linear operator:

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

[1] Lb_j = L_ijb_i .

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 L_ij – this is not a typo! The summation needs to be defined that way in order for L_ij to obey the "component transformation rule":

(Lv)_i = L_ijv_j .

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 |b_i> the "components" relative to this basis are given by

<b_i|v> ... vector ,

<b_i|L|b_j> ... 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 |b_j><b_j| = 1 (the identity on H) ,

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

<b_i|L|v> = <b_i|L (∑_j |b_j><b_j|) |v>

= ∑_j <b_i|L|b_j> <b_j||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]

Another Postulate

Recall the previous postulates:

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

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

P2: To a physical quantity A measurable on (the quantum system) S, there corresponds a self-adjoint linear operator A acting in H_S. 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 P_a , then the probability of obtaining the result a is given by

P(a) = <ψ|P_a|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 S_p(A). The continuous part is referred to as "the continuous spectrum of A", denoted S_c(A). The (overall) spectrum of A is given by
S(A) = S_p(A) U S_c(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 S_p(A), let P_a be the corresponding eigenprojector. For a Є S_c(A), let |a> be the corresponding generalized eigenket. Then,

A = ∑_{a Є S_p(A)} a P_a + ∫_{S_c(A)} a |a><a| da ,

and

∑_{a Є S_p(A)} P_a + ∫_{S_c(A)} |a><a| da = 1 ,

where 1 is the identity on H_S . 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 S_p(A) or S_c(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 a_n Є S_p(A) be a nondegenerate eigenvalue with corresponding eigenket |a_n>. Let the quantum system be in the pure state |ψ>.

(a) Verify that P(a_n) = |<a_n|ψ>|².

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

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

(a) Verify that P(a_n) = ∑_k=1^g(n) |<a_n^k|ψ>|².

(b) Define ρ = |ψ><ψ|. Verify that P(a_n) = ∑_k=1^g(n) <a_n^k|ρ|a_n^k>.

E.3.5) Verify the analogous expression for (b) of E.3.3, in the case of a (nondegenerate) eigenvalue a Є S_c(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 S_p(A), and p(a) is the probability density of obtaining a Є S_c(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=1^g(n) <a_n^k|ρA|a_n^k> + ∫_{S_c(A)} <a|ρA|a>da .

 

[Sammywu]

E 3.1)

The one I can think of that mixes discrete and continuous 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)
$$< \psi | A | \psi > =$$
$$< \psi | \ \sum_{S{p(a)}} a P_a + \int_{S{c(a)}} a |a> <a| da \ | \psi > =$$
$$< \psi | \ \sum_{S{p(a)}} a P_a \ | \psi > \ + \ < \psi | \int_{S{c(a)}} a |a> <a| da \ | \psi > =$$
$$\sum_{S{p(a)}} a < \psi | P_a | \psi > \ + \ \int_{S{c(a)}} a < \psi |a> <a| \psi > da =$$
$$\sum_{S{p(a)}} a P(a) + \ \int_{S{c(a)}} a P(a) da =$$
$$< A >$$