Understanding Projectors in Quantum Mechanics: A Mathematical Approach

[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]

Two Beginning Postulates of QM

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 H_S.

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

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 anyone 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.


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

 

[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]

Another Postulate

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".
___________

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 H_S).

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 a_n|u><u_n| .

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

A = ∑_n a_nP_n .

(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 P_n = |u_n><u_n|). For the general case of (2), the a_n are the eigenvalues of A and the P_n are the corresponding eigenprojectors.

Examples of case (3) are any of the components (Q_j or P_k) 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 a_nP_n ,

it follows that

P_jP_k = δ_jkP_k .

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

∑_nP_n = 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 a_n|u_n><u_n| .

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 anyone of the three observables, said Q_x.
To be continued ...
7). For

 

[Eye_in_the_Sky]

Back in post #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 |φ_k'> = a_k|φ_k> (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 a_k. 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 a_n|u_n><u_n| .

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 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 H_S).

Define E_n = { |ψ> Є H_S │ A|ψ> = a_n|ψ> } . From (b), E_{n ┴} E_n' , for n ≠ n'; so,

(b') E_n^┴ כ E_n' , for n ≠ n' .

From (c), it follows that

(c') the vectors of U_nE_n span H_S ,

Now, from (b') and (c'), it follows that for any |ψ> Є H_S there exist unique |ψ_n> Є E_n such that

|ψ> = ∑_n|ψ_n> .

From the uniqueness of the |ψ_n>, we can define the "eigenprojectors" P_n by

P_n|ψ> = |ψ_n> .

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

P_n = ∑_k=1^g(n) |u_n^k><u_n^k| .

You can then convince yourself that this definition of P_n 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, "c_l" → "d_l". 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 "c_l" → "d_l", 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 anyone of the three observables, said Q_x.
To be continued ...
7). For

1) 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 ≠ a_k", 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 = k_o, where k_o 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, P_Iψ) ,

where P_I is the "projector" defined by

[P_Iψ](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 P_I already contains in it the idea of |q><q|, since they are connected by the simple relation

P_(a,b) = ∫_a^b |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 or Wikipedia, 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

ih_bar ∂_t|ψ(t)> = H|ψ(t)> .

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

|ψ(t)> = ∑_n exp{ -iE_nt / h_bar } c_n |φ_n> .

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|ψ_q> = q|ψ_q>, 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 a_n|u_n><u_n|

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 a_n|n><n| .

Finally, the analogue of

[3'] A = ∫ a |a><a| da

is

[3] A = ∑_{a_n} a_n|a_n><a_n| .
_______________

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 P_I which I mentioned before; i.e.

[P_Iψ](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, P_Iψ) ,
-----------------------------------------

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".

 

[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]

Response to posts #79-81

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 H_S.

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

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 |φ_i> with probability p_i. 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.)