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A free particle paradox?

  1. Feb 26, 2013 #1
    For a free particle stationary state, one would expect the expectation values are constants, for example, <x> (t) = 0. From its one dimensional wave function psi(x,t)= exp(ipx/h-iEt/h)/L^1/2, <x> is undefinied (or zero since x has odd parity).

    How does one reconcile the above with Ehrenfest's theorem which states that <x> is a fucntion of time:

    <x>(t) = <x>(0) + pt/m?

    Thanks in advance for your help!

    rays
     
  2. jcsd
  3. Feb 26, 2013 #2

    Bill_K

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    This applies only to operators that commute with the Hamiltonian.
     
  4. Feb 26, 2013 #3
    for a stationary state (energy eigenstate)

    psi(x,t)=u(x)exp(-iEt/h)

    <A> = integral of psi* A psi dx. The time dependence exponentials cancel each other. <A> has to be time-independent regardless of its commutation properties with the Hamiltonian (there is a proof of it in Sakurai, 2nd ed., Eq. 2.1.45).

    From the free particle wave function of momentum eigen state with eigen value of p, psi(x,t)= exp(ipx/h-iEt/h)/L^1/2, how does one prove

    <x>(t) = <x>(0) + pt/m?

    Thanks.

    rays
     
  5. Feb 26, 2013 #4

    Demystifier

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    Ehrenfest theorem makes sense for states for which <x> is well defined. For your state proportional to exp(ipx/h) this is not the case.
     
  6. Feb 26, 2013 #5
    Thanks, Demystifier.
    I can understand the difficulty with the free particle wave function. But even qualitatively, the Ehrenfest theorem (a time-dependent <x> makes sense to me) contradicts what the free particle wave function says (<x> = 0 at all times).

    if Ehrenfest theorem does not apply to the free particle wavefunction, which statements on time-dependency is correct?

    rays
     
  7. Feb 26, 2013 #6

    Bill_K

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    No, of course not.

    In complete generality, iħ d/dt <A> = <[A,H]> + iħ <∂A/∂t>. Quoting Messiah,
     
  8. Feb 26, 2013 #7
    I've just found out that there is a proof of

    x(t) = x(0) + p(0)t/m

    for a stationary state free particle in Sakurai (2nd ed. Eq. 2.2.27).

    But for any stationary state <x> = const at all times.

    rays
     
  9. Feb 26, 2013 #8
    Bill_K,

    Agree with your post. But the ih<dA/dt> = <[A,H]> is where the Ehrenfest theorem comes from.

    In Sakurai 2nd ed, Page 73, it was proved that for a stationary state, expectation value of any observavle (regardless of whether it commutes with H) is independent of time. The proof is given in the following:

    |a(t)> = U(t) |a(0)>

    for an arbitrary operator B (which may not commute with H)

    <B> = <a(0)|exp(iEt/h)Bexp(-iEt/h)|a(0)> = <a(0)|B|a(0)>, independent of time.

    If we set B = x, <x> = const.

    Using the equation you quoted, if we set A = x, it says <x>(t) = <x>(0) + pt/m, which is the Ehrenfest theorem for a free particle.

    Which statement on <x>(t) is correct?

    rays
     
  10. Feb 27, 2013 #9

    Demystifier

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    It does not apply to a plane-wave free particle wavefunction. But it may apply to some SUPERPOSITION of plane-wave free particle wavefunctions, which is a free particle wavefunction itself. The superposition must be such that the superposed wave function has a shape of a localized wave packet.
     
  11. Feb 27, 2013 #10
    Demystifier,

    Thank you. I think I get it this time.
    Please correct me if I am wrong: Based on what you said and some research I believe that the free particle wave function (exp(ipx)) is fundamentally flawed since it says <x>(t) = const. With a Gaussian wave packet (which is not an energy eigen state of H= p^2/2m), <x>(t) = <x>(0) + pt/m, regardless of the width of the gaussian function. I guess that the free particle wave function can be considered as a gaussian wave packet with very large width. Then, it no longer contradicts the Ehrenfest theorem as long as one does not really take the width to inifinity.

    rays
     
  12. Feb 27, 2013 #11

    Demystifier

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    Yes, now you get it. :approve:
     
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