Equation of Motion in Heisenberg Picture

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SUMMARY

The discussion centers on determining the equation of motion for the operator Z in the Heisenberg picture for a harmonic oscillator potential. The operator is defined as Z = a^{\dagger}a a^{\dagger} a, and the equation of motion is derived using the relation i\hbar dZ(t)/dt = [Z(t),H], where H is the Hamiltonian H = (a^{\dagger}a + 1/2)\hbar \omega. The solution indicates that Z(t) remains constant over time, leading to the conclusion that dZ(t)/dt = 0.

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  • Understanding of quantum mechanics principles, specifically the Heisenberg and Schrödinger pictures.
  • Familiarity with harmonic oscillators and their Hamiltonians.
  • Knowledge of operator algebra, including commutation relations.
  • Experience with creation and annihilation operators in quantum mechanics.
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Homework Statement



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

Z = a^{\dagger}a a^{\dagger} a

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



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

Z(t) = U^{\dagger}ZU

U = e^{-iHt/\hbar}

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


The Attempt at a Solution



a)

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

Where I am assuming H = \frac{p^2}{2m} + 1/2kx^2 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

Z(t) = U^{\dagger}ZU

U = e^{-iHt/\hbar}

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

to obtain Z as a function of time? I assume this because I know with the creation and annihilation operators, I can act on <\psi(x,0)| 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:

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

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

now I am stuck and not convinced i am going about this problem correctly.
 
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Can you write the Hamiltonian in terms of raising and lowering operators?
 
Ah! yes! I completely forgot about that.

H = (a^{\dagger}a + 1/2)\hbar \omega
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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