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Equation of Motion in Heisenberg Picture

  1. Dec 10, 2013 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m is in a harmonic oscillator potential with spring constant k. An observable quantity is given in the Schrodinger picture by the operator:

    [itex] Z = a^{\dagger}a a^{\dagger} a [/itex]

    a) Determine the equation of motion of the operator in the Heisenberg picture

    b) Solve the equation of motion to calculate the form of Z as a function of time.

    2. Relevant equations

    [itex]i\hbar \frac{dZ(t)}{dt} = [Z(t),H] [/itex]

    [itex] Z(t) = U^{\dagger}ZU [/itex]

    [itex] U = e^{-iHt/\hbar} [/itex]

    [itex] <Z>_{t} = <\psi(x,0) | Z(t) |\psi(x,0)> [/itex]


    3. The attempt at a solution

    a)

    From a textbook, it said the equation of motion for a time-independent operator (schrodinger picture) in the Heisenberg picture is:
    [itex]i\hbar \frac{dZ(t)}{dt} = [Z(t),H] [/itex]

    Where I am assuming [itex] H = \frac{p^2}{2m} + 1/2kx^2 [/itex] because it's a harmonic oscillator.

    Is that it for this part of the problem? Just write it down? Seems a bit silly to me.

    b)

    This part is confusing to me. It asks for me to solve the equation of motion to get Z as a function of time. However, can't I just use

    [itex] Z(t) = U^{\dagger}ZU [/itex]

    [itex] U = e^{-iHt/\hbar} [/itex]

    [itex] <Z>_{t} = <\psi(x,0) | Z(t) |\psi(x,0)> [/itex]

    to obtain Z as a function of time? I assume this because I know with the creation and annihilation operators, I can act on [itex] <\psi(x,0)| [/itex] to (if I assume the ground state) eliminate some operators in Z(t).

    Okay so this is the method I think I should go with??

    writing out the last equation above, I get:

    [itex] <Z>_{t} = <\psi(x,0) | U^{\dagger}ZU|\psi(x,0)> [/itex]

    [itex] <Z>_{t} = <\psi(x,0) | U^{\dagger}a^{\dagger}a a^{\dagger} aU|\psi(x,0)> [/itex]

    now I am stuck and not convinced i am going about this problem correctly.
     
  2. jcsd
  3. Dec 10, 2013 #2

    George Jones

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    Staff Emeritus
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    Gold Member

    Can you write the Hamiltonian in terms of raising and lowering operators?
     
  4. Dec 11, 2013 #3
    Ah! yes! I completely forgot about that.

    [itex] H = (a^{\dagger}a + 1/2)\hbar \omega [/itex]
    I know what to do now. I compute the commutator using Z, and the U's will go away because they commute with H.

    Thank you!!

    in the end I got that Z(t) = Z because

    dz(t)/dt = 0 which means that z(t) = some constant.
     
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