Paradox or not ?

  • #1
dextercioby
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Here's something interesting i found:

If one introduces polar coordinates in the plane or spherical coordinates in space, then the polar angle [itex] \phi [/itex] and the component [itex] L_{z} [/itex] of angular momentum are canonically conjugate
variables in classical mechanics. In quantum theory, the variable [itex] \phi [/itex] becomes the operator of "multiplication of the wave function [itex] \psi(\phi) [/itex] by [itex] \phi [/itex]" and

[tex] L_{z}=-\frac{\hbar}{i}\frac{\partial}{\partial \phi} [/tex]

, which implies the commutation relation

[tex] [L_{z}, \phi] =\frac{\hbar}{i} \hat{1} [/tex]

These operators acting on periodic wave functions (i.e. [itex] \psi(0) = \psi(2\pi)[/itex]) are Hermitian. Furthermore, [itex] L_{z} [/itex] admits a complete system of orthonormal eigenfunctions [itex] \psi_{m} [/itex]

[tex] L_{z}\psi_{m}(\phi) = m\hbar\psi_{m}(\phi) [/tex]

, with [itex] \psi_{m}(\phi) =\frac{1}{\sqrt{2\pi}} e^{im\phi}, m\in\mathbb{Z} [/itex].

By evaluating the average value of the operator [itex] [L_{z}, \phi] [/itex] in the state [itex] \psi_{m} [/itex] and by taking into account the fact that Lz is Hermitian, one finds that

[tex] \frac{\hbar}{i} =\langle \psi_{m}, \frac{\hbar}{i}\hat{1}\psi_{m}\rangle = \langle \psi_{m} , L_{z}\phi\psi_{m}\rangle -\langle \psi_{m},\phi L_{z}\psi_{m}\rangle [/tex]

[tex] =\langle \L_{z}^{\dagger}\psi_{m} , \phi\psi_{m}\rangle -m\hbar\langle \psi_{m},\phi \psi_{m}\rangle
=(m\hbar-m\hbar)\langle \psi_{m},\phi\psi_{m} \rangle = 0 !! [/tex]

Hmm, paradoxical or not ? Or is it really necessary that [itex] \hbar=0 [/itex] for consistency ?

Daniel.
 
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  • #2
Demystifier
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Good one!
The catch is that the state [tex]\phi|\psi_m>[/tex] is not in the Hilbert space because it does not satisfy the periodic bondary condition. Therefore the last bracket is ill defined, so one does not need to set h=0, of course.
 
  • #3
vanesch
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This looks like the compact version of:
[tex] i \hbar = \langle \psi | x p - p x | \rangle [/tex]

Now take an eigenstate of p:

[tex] i \hbar = \langle \psi_p |x p | \psi_p \rangle - \langle \psi_p |p x | \psi_p \rangle = (p - p) \langle \psi_p |x | \psi_p \rangle = 0 [/tex]

Then there was a discussion of when we limited ourselves to a finite interval of x, and what boundary conditions to use etc...
 
  • #4
dextercioby
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Good one!
The catch is that the state [tex]\phi|\psi_m>[/tex] is not in the Hilbert space because it does not satisfy the periodic bondary condition. Therefore the last bracket is ill defined, so one does not need to set h=0, of course.
In case you didn't notice, the Hilbert space is simply

[tex] \mathcal{H}=L^{2}\left([0,2\pi], d\phi\right) [/tex]

in which the eigenfunctions of [itex] L_{z} [/itex] pictured above determine an orthonormal basis wrt the scalar product:

[tex] \langle \varphi,\psi \rangle =\int_{0}^{2\pi} \overline{\varphi(\phi)}{}\psi(\phi){} d\phi [/tex]

The [tex] \hat{\phi} [/tex] operator appearing in the commutation relation is bounded (hence by continuity everywhere defined) and self-adjoint on the Hilbert space [itex] \mathcal{H}=L^{2}\left([0,2\pi], d\phi\right) [/itex].

So to claim that [tex] \hat{\phi}\psi \notin \mathcal{H} [/tex] is simply erroneous.

Daniel.
 
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  • #6
dextercioby
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This looks like the compact version of:
[tex] i \hbar = \langle \psi | x p - p x |\psi \rangle [/tex] (1)

Now take an eigenstate of p:

[tex] i \hbar = \langle \psi_p |x p | \psi_p \rangle - \langle \psi_p |p x | \psi_p \rangle = (p - p) \langle \psi_p |x | \psi_p \rangle = 0 [/tex]

Then there was a discussion of when we limited ourselves to a finite interval of x, and what boundary conditions to use etc...
Unfortunately your argument is flawed and the last equality is wrong. And for this claim to be made, the spectrum of the position operator, whether compact or not, semibounded or the entire real axis, is irrelevant. That's why i argue using the free nonrelativistic particle, for it's the simplest mathematical problem.

The formula (1) you wrote is correct, if [itex] \psi \in \mathcal{S}(\mathbb{R}) [/itex] which is known to be a dense everywhere common domain for essential self-adjointness of the position and momentum operators in [itex] L^{2}\left(R^{3}, dx\right) [/itex]. All this is done using the norm topology in [itex] L^{2}\left(R^{3}, dx\right) [/itex].

Unfortunately, you cannot apply it, since an eigenstate of [itex] \hat{p} [/itex] is not an element of the Hilbert space in question and neither of the Schwartz space. It's rather an element of the antidual of the Schwartz space wrt the nuclear topology on the Schwartz space.

Incidentally

[tex] (p-p) \langle p|Q|p\rangle = 0\cdot \left(i\hbar \frac{d}{dp} \langle p|p\rangle \right) =0\cdot \left(i\hbar \frac{d}{dp} \delta (0)\right) [/tex]

which doesn't make any sense mathematically.

Daniel.
 
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  • #7
dextercioby
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I suppose you already know the answer because you took the "mathematical surprise" from here:

"Mathematical surprises and Dirac's formalism in quantum mechanics", F. Gieres
http://arxiv.org/abs/quant-ph/9907069
http://www.iop.org/EJ/abstract/0034-4885/63/12/201

http://arxiv.org/abs/quant-ph/9907070
(French version)
It's true. I had the article. I chose the simplest problem which didn't get a correct response until your posting the solution. My (and the author's) aim was to demonstrate that formal computations in quantum mechanics could lead to erroneus results. It's actually the same motivation that von Neumann found for rejecting formal computations introduced by Dirac in his 1930 book.

This is also an example against low-level books on QM which abound in formal computations and present the bra ^ ket approch as simply a tool that will alway work. As Vanesch's post proved, it doesn't always work.

Daniel.
 
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  • #8
vanesch
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So to claim that [tex] \hat{\phi}\psi \notin \mathcal{H} [/tex] is simply erroneous.

Isn't there a requirement that the elements be periodic and continuous, so that we should have that [tex]\psi(0) = \psi(2 \pi)[/tex] ? But then we don't have that [tex]\hat{\phi}\psi(0) = \hat{\phi}\psi(2 \pi)[/tex] because we have [tex] 0 \psi(0) = 2 \pi \psi(2 \pi)[/tex] which can only be satisfied if we also have the extra condition that [tex] \psi(0) = 0[/tex]. But if that is the case, then [tex]\psi_m[/tex] is not an element of the space in question... exactly as you point out in my "argument" (which I know is flawed! that was the point...).
 
  • #9
dextercioby
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There are no boundary conditions on the vectors in the Hilbert space. Boundary conditions are relevant only when trying to render the L_{z} operator self-adjoint. The whole problem is with the domains of definition and that is because the L_{z} operator is unbounded. You can quite easily follow the whole discussion and the resolution of the apparent paradox in the quoted article on page 39 and 40.

Daniel.
 
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  • #10
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The whole problem is with the domains of definition and that is because the L_{z} operator is unbounded. You can quite easily follow the whole discussion and the resolution of the apparent paradox in the quoted article on page 39 and 40.
Daniel.
Euh, the point Demystifier as well as Vanesch made is completely correct. The [tex] \frac{\hbar}{i} [/tex] comes from the nonzero boundary term you simply forget in the partial integration in the last step. You do not even need to consider domain issues here, since all operations involved are well defined.
 
  • #11
dextercioby
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Euh, the point Demystifier as well as Vanesch made is completely correct. The [tex] \frac{\hbar}{i} [/tex] comes from the nonzero boundary term you simply forget in the partial integration in the last step. You do not even need to consider domain issues here, since all operations involved are well defined.
I don't follow. There are many [itex] \frac{\hbar}{i} [/itex] involved here. The whole point of the problem is that the domain of the commutator doesn't contain the eigenvectors of the L_{z} operaor.

Daniel.
 
  • #12
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I don't follow. There are many [itex] \frac{\hbar}{i} [/itex] involved here. The whole point of the problem is that the domain of the commutator doesn't contain the eigenvectors of the L_{z} operaor.

Daniel.
:bugeye: DO the partial integration yourself !! Then you will see that the mistake is in the step :
[tex] < \psi_m , L_z \phi \psi_m > = < L_z \psi_m , \phi \psi_m > [/tex]
actually, you wrote the conjugate of [tex] L_z [/tex], but use [tex]L_z[/tex] anyway.
 
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  • #13
vanesch
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There are no boundary conditions on the vectors in the Hilbert space. Boundary conditions are relevant only when trying to render the L_{z} operator self-adjoint. The whole problem is with the domains of definition and that is because the L_{z} operator is unbounded. You can quite easily follow the whole discussion and the resolution of the apparent paradox in the quoted article on page 39 and 40.

Daniel.
You can consider 3 different spaces: you can consider the square-integrable functions on [ 0,2 pi ]. You can consider the square-integrable functions on [0 , 2pi ] which are periodic [tex] \psi(0) = \psi(2 \pi) [/tex]. Or you can consider the square-integrable functions on [ 0, 2pi ] which are 0 on the boundaries. Let's call these spaces respectively case A, B and C.

Now, in order for [tex]L_z = \frac{\hbar}{i}\frac{\partial}{\partial \phi}[/tex] to be a self-adjoint operator, we need to have that:
[tex](L_z u,v) = (u,L_z v)[/tex] or that:

[tex] \int d\phi \frac{\partial u^*}{\partial \phi} v = - \int d\phi \frac{\partial v}{\partial \phi}u^*[/tex] which can be shown by partial integration, on the condition that [tex]u^*(0)v(0) = u^*(2\pi) v(2 \pi)[/tex], because otherwise there is a boundary term that is added.

Clearly, the above condition only holds in the cases B and C.

However, we also need other things. We need the eigenfunctions of [tex]L_z[/tex], which are of the form [tex]\psi_m(\phi) = exp(i m \phi)[/tex]. Clearly, this eliminates the case C (the eigenfunctions are not 0 on the boundary) and we're left with the case B: periodic, non-zero boundary.

Well, in this space, if [tex]\psi[/tex] is an element of the space, then [tex]\phi \psi [/tex] isn't, because the boundary is not periodic.
 
  • #14
dextercioby
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Yes, Vanesch your argument is shown to be equivalent to the one written by the author of the article. As for Careful, he argues against the neglection of the boundary term in the movement of L_{z} from the right of the scalar product to the left of the scalar product. It should be said that this move, even after putting in the [itex] \frac{\hbar}{i} [/itex] correction makes no sense, simply because the vector

[tex] L_{z}\phi \psi_{m} [/tex] does not exist !!

It looks like Careful sees the resolution in the fact that [itex] \frac{i}{\hbar} \neq 0 [/itex] and the paradox is solved by reaching the uninteresting tautology [itex] \frac{i}{\hbar} = \frac{i}{\hbar} [/itex] , while the true resolution is simply reaching the conclusion that in the derivation, right from the very first "=" sign one's making illegal computations with vectors. I might be wrong though, maybe Careful meant the right thing.

Yes, Vanesch made me understand that Demystifier got the idea right, however mistook the Hilbert space with the domain of L_{z}.

Either way, i guess the article is one of the best i've seen on hep lately.

Daniel.
 
  • #15
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As for Careful, he argues against the neglection of the boundary term in the movement of L_{z} from the right of the scalar product to the left of the scalar product. It should be said that this move, even after putting in the [itex] \frac{\hbar}{i} [/itex] correction makes no sense, simply because the vector

[tex] L_{z}\phi \psi_{m} [/tex] does not exist !!
I advise you not to come up with ridiculous arguments :
(a) I said in post 10 that I agree with Vanesch and Demystifier. The latter who explicitely said to you in post 2 that the vector does not exist in the Hilbert space at hand.
(b) My point about partial integration was a consideration *independent* of any Hilbert space embedding, merely integral calculus.

It looks like Careful sees the resolution in the fact that [itex] \frac{i}{\hbar} \neq 0 [/itex] and the paradox is solved by reaching the uninteresting tautology [itex] \frac{i}{\hbar} = \frac{i}{\hbar} [/itex] , while the true resolution is simply reaching the conclusion that in the derivation, right from the very first "=" sign one's making illegal computations with vectors. I might be wrong though, maybe Careful meant the right thing.
:grumpy: :grumpy: Clearly the two points are more or less equivalent : the partial integration only works out as you use it if and only if the function under the integral sign is periodic. And yes, if you do things correctly you arrive at a tautology; so the only lesson here is that people should learn to use math properly.

Careful
 
  • #16
vanesch
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Yes, Vanesch your argument is shown to be equivalent to the one written by the author of the article. As for Careful, he argues against the neglection of the boundary term in the movement of L_{z} from the right of the scalar product to the left of the scalar product.
I had the impression that Careful was thinking of exactly the same argument as I did (and as demystifier did): the trick has to do with the boundary conditions. You can impose them from the start (which makes then phi psi non-existent in that space), or you can NOT impose them, which makes L_z non-self-adjoint because of the non-zero boundary term in the partial integration, or you can require them to be 0, in which case the eigenfunctions are not in the space. I only spelled out all the different possibilities more explicitly, while these other posters took them implicitly. Then again, they can speak for themselves...
(and please no silly flame wars of the kind "I was right" - "no, you didn't understand, *I* was right...)
 
  • #17
dextercioby
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Working with non self-adjoint L_{z} is only mathematics for the sake of mathematics. Choosing/imposing L_{z} self-adjoint is mathematics for the sake of physics. That's for sure.

Daniel.
 
  • #18
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Note also the following! The quantity [tex]<\psi_m|\phi|\psi_m>[/tex] should be the average value of [tex]\phi[/tex] in the state [tex]|\psi_m>[/tex]. However, if you try to calculate that average value of [tex]\phi[/tex], you will find out that this value is not well defined. Similarly, in the noncompact case, you will find out that the average value of x is not well defined in an momentum eigenstate. This ambiguity can be viewed as a physical interpretation of the mathematical ambiguity discussed so far.
 
  • #19
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Working with non self-adjoint L_{z} is only mathematics for the sake of mathematics. Choosing/imposing L_{z} self-adjoint is mathematics for the sake of physics. That's for sure.

Daniel.
:rofl: Really, then you must say that physicists who are studying non self adjoint Hamiltonians with a *real* spectrum, aren't doing physics. Neverthess, complex Hamiltonians are studied and for very good reasons. Here is a fairly recent paper by Patrick Dorey on the subject for example http://www.arxiv.org/PS_cache/hep-th/pdf/0309/0309209.pdf [Broken]

Again, I advise you not to play such games with me.

Careful
 
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  • #20
dextercioby
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I admit, i went too far with that assertion. Sorry.

Daniel.
 
  • #21
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Dextercioby:” Here's something interesting i found:
If one introduces polar coordinates in the plane or spherical coordinates in space, then the polar angle and the component of angular momentum are canonically conjugate variables in classical mechanics.”

The detailed introduction in that and the related questions may be found also in P.Carruthers, M. Nieto, Rev. Mod. Phys., 40, 411 (1968).

"Nothing is more practical than a good theory"

Compare with Reilly “tends to make one practical”. Perhaps above questions will convince Reilly that we (math-ph) also “moving lead bricks around”.
 
  • #22
reilly
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Here's something interesting i found:

If one introduces polar coordinates in the plane or spherical coordinates in space, then the polar angle [itex] \phi [/itex] and the component [itex] L_{z} [/itex] of angular momentum are canonically conjugate
variables in classical mechanics. In quantum theory, the variable [itex] \phi [/itex] becomes the operator of "multiplication of the wave function [itex] \psi(\phi) [/itex] by [itex] \phi [/itex]" and

[tex] L_{z}=-\frac{\hbar}{i}\frac{\partial}{\partial \phi} [/tex]

, which implies the commutation relation

[tex] [L_{z}, \phi] =\frac{\hbar}{i} \hat{1} [/tex]

These operators acting on periodic wave functions (i.e. [itex] \psi(0) = \psi(2\pi)[/itex]) are Hermitian. Furthermore, [itex] L_{z} [/itex] admits a complete system of orthonormal eigenfunctions [itex] \psi_{m} [/itex]

[tex] L_{z}\psi_{m}(\phi) = m\hbar\psi_{m}(\phi) [/tex]

, with [itex] \psi_{m}(\phi) =\frac{1}{\sqrt{2\pi}} e^{im\phi}, m\in\mathbb{Z} [/itex].

By evaluating the average value of the operator [itex] [L_{z}, \phi] [/itex] in the state [itex] \psi_{m} [/itex] and by taking into account the fact that Lz is Hermitian, one finds that

[tex] \frac{\hbar}{i} =\langle \psi_{m}, \frac{\hbar}{i}\hat{1}\psi_{m}\rangle = \langle \psi_{m} , L_{z}\phi\psi_{m}\rangle -\langle \psi_{m},\phi L_{z}\psi_{m}\rangle [/tex]

[tex] =\langle \L_{z}^{\dagger}\psi_{m} , \phi\psi_{m}\rangle -m\hbar\langle \psi_{m},\phi \psi_{m}\rangle
=(m\hbar-m\hbar)\langle \psi_{m},\phi\psi_{m} \rangle = 0 !! [/tex]

Hmm, paradoxical or not ? Or is it really necessary that [itex] \hbar=0 [/itex] for consistency ?

Daniel.
I haven't looked at the posts here -- wanted to figure out the puzzle without any clues.
First of all, dexterc's result must be wrong -- with all due respect. When doing the step of taking Lz ->Lz adjoint, integration by parts is at issue, as is commonly done in discussions of Sterm-Liouville equations.
See the first step in the last row of equations, the key term is dropped.

(Sorry 'bout my lack of clue about math notation)
If you forget the abstract notation and work directly with integrals over phi, then you get: Necessarily


INT(W*{-i d/d(phi)} phi W d(phi) = INT (W*-i { W +(phi) d/(phi) W)

This can be shown to be true, as it must be, when the adjoint approach is used via integration by parts. No problem -- everything is done on a finite interval.

In the space of SL ordinary differential equations, the whole idea of an adjoint stems from properties of integration by parts.

With L2 functions on a finite interval, the adjoint of Lz is well defined, and all the integrals are well defined. Therefore, dexterc's result is necessarily incorrect. And, the sticky point is the relationship between adjoint and integration by parts
My apologies if I've repeated an earlier post.

We'll take W to be periodic in 2pi, by the way.
Regards,
Reilly
 
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  • #23
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I admit, i went too far with that assertion. Sorry.

Daniel.
:approve: You went already too far before that, but let's not talk about it anymore. Let me tell you something : you might have noticed that I have rather ``obstinate'' ideas (I don't know if it is the right word) about many things. This is not a comfortable way of life, so one time I decided to take it easy for myself and simply defend some standard point of view in a discussion about a very delicate subject in quantum mechanics (oh yes, here she is again). The other party was kindly presenting something what many (textbooks) would label as heretic or even worse, simply impossible... Anyway, the upshot of what occured is that I felt silly afterwards you see, because what this person told was simply correct and I did not notice it myself previously (even while studying ``the books'').

We all miss something or don't notice possible objections or ways to do things differently. That is why one learns not to use the word ``correct'', ``to be'' or alike when one gets older especially in a context where the assumptions underlying the conclusions are not as firm as it might seem to be at first sight.
 
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  • #24
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Dextercioby:” Here's something interesting i found:
If one introduces polar coordinates in the plane or spherical coordinates in space, then the polar angle and the component of angular momentum are canonically conjugate variables in classical mechanics.”

Edgardo:” I suppose you already know the answer because you took the "mathematical surprise" from here:

"Mathematical surprises and Dirac's formalism in quantum mechanics", F. Gieres http://arxiv.org/abs/quant-ph/9907069” [Broken]

Dextercioby:“It's true. I had the article. I chose the simplest problem which didn't get a correct response until your posting the solution. My (and the author's) aim was to demonstrate that formal computations in quantum mechanics could lead to erroneus results.”
“This is also an example against low-level books on QM which abound in formal computations and present the bra ^ ket approch as simply a tool that will alway work. As Vanesch's post proved, it doesn't always work.”


I do not understand why sometimes you use garbage to demonstrate good point.The writer have no idea what he is talking about. On p.4 (Example 1) he introduce Tr operation without definition. Then he wrote Tr(i) not =0. This is the same as to state that the Pauli matrices are not traceless. On p.10 he wrote:”Clearly, two non equivalent experimental set-ups for the measurement of a given physical observable generally lead to different experimental results”. Perfect idiot. I was not able to proceed reading after that.

The key notion in your example is “canonically conjugate variables in classical mechanics”. The solution should satisfy the minimum uncertainty requirement.

“Philosophically” he consider the mathematical physics to be responsible for formal mathematical rigour of the physical theory. It is only less significant aspect of it. The purpose of math-ph is to write mathematical relations that describe adequately the underlined physical phenomena. Sometimes it requires the invention of unknown mathematical tools ( for example:I. Newton, J.C.Maxwell, D. Hilbert, W. Heisenberg, P.A.M. Dirac, R.P. Feynman).
 
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  • #25
Demystifier
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It is ironic that physics attained the best sussess in areas where it was not mathematically rigorous (e.g. quantum field theory with renormalization of infinities).
 

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