Equation of Motion in Heisenberg Picture

[QuarksAbove]

Homework Statement

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.

Homework 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]

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.

 

[George Jones]

Can you write the Hamiltonian in terms of raising and lowering operators?

 

[QuarksAbove]

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.