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Peculiar feature of a commutator, can anyone explain?

  1. Feb 9, 2009 #1
    http://img209.imageshack.us/img209/4922/14662031eo8.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 9, 2009 #2
    1. Note that we are only interested in wavefunctions such that [tex]f(\phi) = f(\phi+2\pi)[/tex].

    2. When the operator [tex]-i\hbar \frac{\partial}{\partial\phi}[/tex] is applied to a [tex]2\pi[/tex] periodic function, the resulting function is also [tex]2\pi[/tex] periodic. -> No problem

    3. Yet, the operator [tex] \phi [/tex] applied to a [tex]2\pi[/tex] periodic function doesn't give a periodic function. -> This is problematic. [tex] \phi [/tex] is not a legitimate operator in the Hilbert space considered.

    3-2. Instead, one should understand "[tex] \phi [/tex]" as a periodic function which has period [tex]2\pi[/tex] and coincide with [tex] \phi [/tex] in the interval [tex][0,2\pi][/tex] (Just imagine the sawtooth wave)

    4. Now, if one evaluates the commutator with more carefully defined "[tex]\phi[/tex]", additional delta function term follows, and everything should agree with the other method(utilizing Bra-Ket notations).

    5. I think my explanation isn't rigorous in mathematicians' sense but still suffices to give the right answer. Hope it helped.
     
    Last edited: Feb 9, 2009
  4. Feb 9, 2009 #3
    What you proved is that the expectation value of the commutator with respect to some spherical symmetric eigenstate of L_z is zero. But this is always the case, namely, suppose if:
    [itex] [A,B] = C[/itex]
    For some hermitian operators A, B and a third operator C. Now take an eigenstate of the operator B, let's name it [itex]|b\rangle[/itex], such that [itex]B|b\rangle = b|b\rangle[/itex]. Then:

    [itex]\langle b|[A,B]|b\rangle = \langle b|AB|b\rangle - \langle b|BA|b\rangle
    =b\langle b|A|b\rangle - b\langle b|A|b\rangle = 0 [/itex]

    Which is precisely what you showed, and I haven't even defined the operators yet. The fact is that we cannot conclude that the commutator is zero, because we have only considered a very particular set of expecation values. For instance, consider the following (b and b' correspond to different states):
    [itex]\langle b'|[A,B]|b\rangle = b'\langle b'|A|b\rangle - b\langle b'|A|b\rangle [/itex]
    In general, this will not be identical to zero.
     
    Last edited: Feb 9, 2009
  5. Feb 9, 2009 #4
    Hey weejee! You are really astute! I think You pinpointed the problem that has been bothering me for a month!

    I believe you got the point, just to clarify

    How is the wavefunction in spherical coordinate defined? Is it defined with phi belonging to [0,2pi) or is it defined by imposing the condition f(phi+2pi)=f(phi)? And please give me a book (where it is to refered to also... I did not find a definition even in the dictionary like Cohen-Tannoudji book)

    What I am wondering is that if the wavefunction is indeed defined with phi belonging to [0,2pi) (which in my humble opinion sounds far more natural to associate each point in space with a single point in the function) then the periodity problem of phi operator really dosen't matter as we would never get out of the 2pi domain...
     
    Last edited: Feb 9, 2009
  6. Feb 9, 2009 #5
    Hi,xepma I do not agree with you.

    If the operator is indeed as the coordinate representation calculated, i.e. ih(bar), then no matter what state we use, we should always get the expectation value is ih(bar), since the commutator is essentially the operator ih(bar)I.
    So if we can show that in a PARTICULAR state, the expectation value is NOT ih(bar), then we have shown that the representation is wrong (Proof by finding a counterexample)

    So I am not saying that the operator is 0, but I am say the operator is definitely not ih(bar) as calculated in the coordinate representation
     
  7. Feb 9, 2009 #6
    Hi, I think I can add a reference, although it is not a textbook. There is a nice old article by Carruthers et al. in Reviews of Modern Physics, 40, 411 (1968) that discusses this issue. If you do not have access to PROLA just google "phase and angle variables in quantum mechanics" and you should be able to find a downloadable version around 6th place (I don't want to give direct link due to forum policy)


    The base manifold (space on which function is defined) is a circle, i.e. the points phi and phi+2pi are the same points, so it does not really make sense to say that the wave-function is defined outside [0,2pi). The topological constraint imposes the periodicity of the wave-functions, and any operator on this (reduced)Hilbert space must transform a periodic function to another periodic function. This is not satisfied by phi as pointed out by weejee, so it must be made periodic.

    The paradox in the commutation relation calculation you showed is a direct consequence of the fact that phi*f(phi) is not periodic. It is easy to see that in fact L_z=id/dphi is not self adjoint unless it acts on periodic functions (check it).

    I am not completely sure this answered your question, I realize I mostly repeated what weejee said.. anyway hope it helped in some way.
     
  8. Feb 9, 2009 #7
    Well, thanks. Actually I had pondered over this problem for months and reached this conclusion after debates with my friends.


    I think the condition ( [tex]f(\phi)=f(\phi+2\pi) [/tex]) should be imposed, as [tex]\phi[/tex] and [tex]\phi+2\pi[/tex] corresponds to the same physical point.

    To fit the function [tex]f(\phi)=\phi[/tex] in such scheme, the sudden [tex]2\pi[/tex] drop between [tex]2\pi-\epsilon[/tex] and [tex]0+\epsilon[/tex] should be accounted for.
    Therefore, its derivative is not 1 but additional delta function term follows.
     
    Last edited: Feb 9, 2009
  9. Feb 9, 2009 #8
    Yes, you definitely answered my question. And your reference makes excellent reading. Thanks a lot. Although I am still reading the article, I am eager to clarify a bit more on what u just said,

    So do u mean by what u said in red that the wavefunction is NOT defined outside [0,2pi)? Or from what u said in blue u seems to be defining the wavefunction outside [0,2pi) by imposing periodicity?
    So after all is the wavefunction defined outside [0,2pi) ?
     
  10. Feb 9, 2009 #9
    You can also ask what is the operator conjugate to the number operator of the harmonic oscillator. Supposedly this is how Dirac found the creation and annihilation operators (though I'm not sure). In any event, you might want to look at the Weyl representation of the canonical commutation relations.
     
  11. Feb 9, 2009 #10
    I was afraid that I was not very clear in what I said. The reason for this is that the issue involves topology which I am only vaguely familiar with. But I will try to explain what I meant.

    As I mentioned before, the space that the wavefunction is defined on is the circle. The problem is to find a proper coordinate system for this space. If we simply use the angle [itex]\phi[/itex] then we run into the problem that the points [itex]\phi=0, \phi=2\pi[/itex] describe the same point but have different "labels". Hence it does not correctly represent the non-trivial topology of the cirlce.

    So how do we go about dealing with this difficulty? In topology one makes use of several coordinate systems on different "patches" of the manifold and "glue them togehter" (via a map/relation between the two coordinate systems). In most cases it is much simpler to do the following:

    Define the angle to be in the interval [0,2pi), i.e. excluding the point 2pi, and then impose the limiting behaviour [itex]\psi(\phi\rightarrow 2\pi)=\psi(0)[/itex]. This makes sure the wave-function knows about the topological properties of the space it is defined on (this is what I mean when I say impose periodic boundary conditions). Expectation values and such can then be obtained by integrating over [0,R] and take the limit R goes to 2pi. Effectively this last step just means to integrate over [0,2pi]. The important thing is just to make sure the wave-functions obey the topological constraints..

    Does this answer your question?
     
    Last edited: Feb 9, 2009
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