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Quantum time corr: expectation value of particle motion in Schro. pic

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data
    The expectation value of motion of a particle over a time interval t-to is

    C(t,to) = <0|x(t)x(to)|0>
    (product of position operators in Heisenberg representation for ground state harmonic oscillator)

    2. Relevant equations
    Schrodinger picture:
    <ψ(t)|Ω|ψ(t)> = <ψ(to)|U+(t,to) Ω U(t,to)|ψ(to)>

    I believe:
    U(t,to) = exp[-iωt/h]
    U+(t,to) = exp[iωt/h]

    Heisenberg rep: time-indep states, time-dep operators
    Schrodinger rep: time-dep states, time-indep operators

    3. The attempt at a solution
    I'm confused about how to convert x(t) in the Heisenberg representation to the Schrodinger representation and how to incorporate it into C(t,to).

    Here is what I scribbled:
    <x(t)> = <ψ(t)|x(to)|ψ(t)> = <ψ(to)|exp[iωt/h] x(to) exp[-iωt/h]|ψ(to)>
    = <0|exp[iωt/h] x(to) exp[-iωt/h]|0>
    Is the first equality even true?

    The question seems like it should be simple, but I am sincerely confused. Thanks for your time and for any help!
     
  2. jcsd
  3. Mar 10, 2013 #2
    In the Heisenberg picture all time dependence is attributed to the operators, the states carry no time dependence. In the Schrödinger picture this is reversed, the operators are constants but the states change in time.

    The connection between the pictures is established by requiring physics to be unchanged. That is expectations values must give the same numbers in both cases.

    This is a little wrong, why do you write ωt/h?
    In the Schrödinger picture you can formally solve the Schrödinger equation by means of the evolution operator U(t,t0) = exp[-iHt/h], where H is the Hamiltonian of the system. So the states will be given by ψ(t)> = U(t,to)|ψ(to)>. To go to the Heisenberg picture you simply have to get rid of the time dependence so you can write |ψ(Heisen.)> = U+(t,to) U(t,to)|ψ(to)> = ψ(to, Schr.)> due to the unitarity of the evolution operator.

    Now you can simply apply the same thing in any of your expectation value equations and find how your operators must change in order for the expectation values to be invariant.
     
  4. Mar 10, 2013 #3
    Hello and thanks for responding. The omega represents energy and the h is supposed to be h bar. I could see how that would look strange actually, so I won't use omega anymore.

    Anyway, here is another attempt:

    <0|x(t)x(to)|0> = <0|exp[iHt/h]exp[-iHt/h] x(to)x(to)exp [-iHt/h]exp[iHt/h]|0>
    = <0|x2|0> = h/(2mω)

    This doesn't match my answer for solving it another way :/. I think I understand what to do conceptually, but not mathematically. Are the exponentials supposed to cancel? Aren't those what make the states time-dependent?

    I hope I am making some amount of sense. Thanks again.
     
  5. Mar 10, 2013 #4
    How did you get exp[iHt/h]exp[-iHt/h] on either side?

    The correct way should be:

    In the Heisenberg picture you have <0|x(t)x(to)|0>
    So using the evolution operator to switch the states to the Schrödinger picture you get:
    <0|exp[iHt/h] x(t) x(to) exp[-iHt/h]|0> = <0|U+ x(t) x(to) U |0>

    So in order for the expectation value to be invariant you must write
    <0|U+ U x(t) U+ U x(to) U+ U |0> = <0|x(t)x(to)|0> due to unitarity of U.

    So the operators must change like U x(t) U+

    Hope this helps. Note the ambiguity when i write x(t) even in the Schrödinger picture.
     
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