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Is the expectation value of this commutator zero?

  1. Sep 4, 2010 #1
    If I have [tex]H=p^2/2m+V(x)[/tex], [tex]|a'>[/tex] are energy eigenkets with eigenvalue [tex]E_{a'}[/tex], isn't the expectation value of [tex][H,x][/tex] wrt [tex]|a'>[/tex] not always 0? Don't I have that
    [tex]<a'|[H,x]|a'> = <a'|(Hx-xH)|a'> = <a'|Hx|a'> - <a'|xH|a'> = 0 [/tex]?
    But if I calculate the commutator, I get:
    [tex]<a'|[H,x]|a'> = <a'|-i p \hbar / m|a'> \neq 0 [/tex]

    I suddenly can't see where I did wrong...
  2. jcsd
  3. Sep 4, 2010 #2
    And what is for you the value of [tex]<a'|p|a'>[/tex]?
    Last edited: Sep 4, 2010
  4. Sep 5, 2010 #3
    I don't know, but do we know it is zero?
  5. Sep 5, 2010 #4
    Take two simple, exactly solvable, examples in one dimension:

    1) free particle
    2) harmonic oscillator

    Calculate both values in each case. One which you think should be zero, and one which you think should not be zero.

    Very easy and entertaining exercise.
    Last edited: Sep 5, 2010
  6. Sep 5, 2010 #5
    Now suppose I did two examples and they both give zero. So now I know it is true for all hamiltonians? nice proof
  7. Sep 5, 2010 #6
    I do not understand. So you calculated all four values (two for each of the cases) for each energy eigenstate, and all gave zero, 0=0?. So where is the problem? Can you be more precise?
  8. Sep 5, 2010 #7
    Hmm okay you do not understand, I hope some other on this forum will understand and answer.
  9. Sep 5, 2010 #8
    I do understand. For the harmonic oscillator you will get 0=0. No problem.

    For free particle you will not get reasonable answer at all, except 0xinfinity, which can be any number, because energy eigenstates are not normalized there. But let's see, maybe someone will fall into your trap.
  10. Sep 10, 2010 #9
    I think your math constitutes a proof that [tex]\left \langle a \right | p \left | a \right \rangle = 0[/tex] when "a" is a normalizable eigenstate of the Hamiltonian (I guess as arkajad points out you have to be careful about non-normalizable states).

    This makes sense intuitively. Eigenstates of H are "stationary states": the probably density does not move around with time. For a bound (normalizable) state, this means that the expectation of the position of the particle is not changing, so we expect (hah) the expectation of momentum to be 0.
  11. Sep 10, 2010 #10
    Well, even for normalizable states you may have a surprise if you find that your energy eigenstate is not in the the dense domain on which the momentum operator is self-adjoint. These subtleties (anomalies?) are discussed in the old good von Neumann's book "Mathematical Foundations of Quantum Mechanics" (be careful because of some old-fashioned terminology, not in use today), but not too often in typical physics books.
  12. Sep 10, 2010 #11

    ppx-pxp=h'/i p
    pxp-xpp=h'/i p
    ppx-xpp=2h'/i p
    [H,x]=[p^2,x]/2m=h'/i p/m

    Energy eigenstate is decomposed to symmetric and antisymmetric states
    [H,x] is antisymmetric, i.e. [H(x),x]=-[H(-x),-x]

    = <s|[H,x]|s> + <a|[H,x]|a> whose integrand is odd function of x
    + <a|[H,x]|s> + <s|[H,x]|a> whose integrand is even function of x

    So <e|[H,x]|e>=0 in case |s>=0 or |a>=0. Otherwise <e|[H,x]|e> can be different from zero and is a pure imaginary number.
    It depends on how you choose V(x) in the problem. If V(x) is symmetric i.e.V(x)=V(-x), |s>=0 or |a>=0 so <e|[H,x]|e>=0.

    Last edited: Sep 11, 2010
  13. Sep 11, 2010 #12
    I feel like IFNT's simple manipulations show that for "nice" eigenstates of H the expectation of p must be 0 no matter whether the state is even or odd or neither. Can you find a bound eigenstate |e> of any Hamiltonian with a nonzero value for <e|[H,x]|e> ?
  14. Sep 11, 2010 #13
    I here show another condition for <p>=0. By partial integaration,
    <p>=h'/i ∫Psi*(x) Psi'(x)dx =h'/i [Psi*(x) Psi(x)] at both the infinity - h'/i ∫Psi*'(x)Psi(x)dx
    If wave function Psi(x) is real and Psi(x)=0 at x=plus and minus infinity,
    <p>=h'/2i [Psi(x) Psi(x)] at both the infinity = 0.
  15. Sep 12, 2010 #14
    I don't understand why you think [tex] [H,x] = 0 [/tex] in general. Why should this be the case, position and momentum operators don't commute. If I remember correctly this means that you can't simultaneously have a basis |a'> which obeys the eigenvalue rules for both p and x.
  16. Sep 12, 2010 #15
    Actually I do not have examples to show you with me. In return, can you show me proof that <a|[H,x]|s> + <s|[H,x]|a> = 0 or <a|p|s> + <s|p|a> = 2Re<a|p|s> = 0 where |e>=|a>+|s> ?

    |e>=|a>+|s> may hold in case of non symmetric potential e.g. V(x)=ax+bx^2.
    Last edited: Sep 13, 2010
  17. Sep 14, 2010 #16
    For a famous particle-in-a-box which is on the train of speed v, <p>=mv. Can it be the case that <p>≠0 ?
  18. Sep 14, 2010 #17
    The canonical commutation relation


    does not really hold for a particle in a box. A whole line is needed.
  19. Sep 14, 2010 #18


    User Avatar
    Science Advisor

    Adding to all the other answers in this thread...

    This seems like a variation on Dirac's famous "0 = 1" paradox: if Q,P are the
    usual canonically-conjugate operators satisfying [Q,P] = ih, we can (naively) write:

    \langle q|\,[Q,P]\,|q\rangle ~=~ i\hbar \langla q|I|q\rangle ~=~ i\hbar \langle q|q\rangle

    \frac{1}{i \hbar} \Big( \langle q|QP|q\rangle - \langle q|PQ|q\rangle \Big)
    ~=~ \langle q|q \rangle ~.

    In the first term, let Q act on the bra; in the second, let Q act on the ket:

    \Big( q \langle q|P|q\rangle - q \langle q|P|q\rangle \!\!\Big) ~=~ \langle q|q \rangle
    ~~,~~\Longrightarrow 0 = 1 ~(!)

    The resolution of this paradox in standard quantum theory is that the
    spectra of both operators is continuous, so the inner products in the above
    are meaningless. You can't validly manipulate adjoints as above for this case.
    This is a consequence of another theorem that for canonically conjugate
    operators as above, at least one of the operators must be unbounded.
    (Proof can be found in Reed & Simon.) Another theorem (Hellinger-Toeplitz)
    then says this means that Q and P cannot both have the whole Hilbert space
    as their domain.

    This is (one of) the primary motivations why people move to rigged Hilbert spaces.
    (Anyone using the Dirac bra-ket formalism in this context is secretly using
    rigged Hilbert spaces whether they realize it or not.)

    For a gentle introduction, try Ballentine section 1.4

    Also see quant-ph/0502053 by Rafael de la Madrid. It also gives a gentle
    introduction, and also covers the example of the 1D rectangular barrier
    potential -- which might shed some light more directly relevant to the original
    post in this thread.


    PS: I wish there was a sticky thread about this... :-)
  20. Sep 15, 2010 #19
    Please see wiki "Particle in a box" where QM is applied. I see it holds there.

    Last edited: Sep 15, 2010
  21. Sep 15, 2010 #20
    Wikipedia is not to be considered as a Bible in such questions involving rather subtle tricky mathematical concepts (unbounded, symmetric but not self-adjoint operators).

    You should rather consult Reed and Simon, "Methods of Mathematical Physics", Vol 2, pp. 141-143.

    You are dealing with unbounded operators and trying to formally manipulate them. This can easily lead to paradoxes. From [p,x]=-ih' you easily

    exp(ipa/h') x exp(-ipa/h')= x+a

    and this will lead you out of the box for a big enough.

    See also the discussion in the chapter "C. Driven particle in a box. Quantum particle" in http://eprints.whiterose.ac.uk/1374/1/weigerts28.pdf".
    Last edited by a moderator: Apr 25, 2017
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