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Paradox or not ?

  1. Dec 20, 2006 #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.
     
    Last edited: Dec 20, 2006
  2. jcsd
  3. Dec 20, 2006 #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.
     
  4. Dec 20, 2006 #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...
     
  5. Dec 21, 2006 #4

    dextercioby

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    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.
     
    Last edited: Dec 21, 2006
  6. Dec 21, 2006 #5
  7. Dec 21, 2006 #6

    dextercioby

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    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.
     
    Last edited: Dec 21, 2006
  8. Dec 21, 2006 #7

    dextercioby

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    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.
     
    Last edited: Dec 21, 2006
  9. Dec 21, 2006 #8

    vanesch

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    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...).
     
  10. Dec 21, 2006 #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.
     
    Last edited: Dec 21, 2006
  11. Dec 21, 2006 #10
    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.
     
  12. Dec 21, 2006 #11

    dextercioby

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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.
     
  13. Dec 21, 2006 #12
    :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.
     
    Last edited: Dec 21, 2006
  14. Dec 21, 2006 #13

    vanesch

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    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.
     
  15. Dec 21, 2006 #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.
     
  16. Dec 21, 2006 #15
    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.

    :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
     
  17. Dec 21, 2006 #16

    vanesch

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    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...)
     
  18. Dec 21, 2006 #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.
     
  19. Dec 21, 2006 #18

    Demystifier

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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.
     
  20. Dec 21, 2006 #19
    :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

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

    Careful
     
  21. Dec 21, 2006 #20

    dextercioby

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    I admit, i went too far with that assertion. Sorry.

    Daniel.
     
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