I am trying to understand a line in the quantum mechanics book by Merzbacher, specifically the second line of equation 14.106.

The problem is a forced quantum harmonic oscillator. The Hamiltonian operator in the Schrodinger picture is

$H_S(t) = \hbar \omega_0 (a^{\dagger}a + 1/2) + \hat{x}f_x(t) +\hat{y}f_y(t)$

Here $\hat{x}$ and $\hat{y}$ are the quadratures of motion of the oscillator, ie. dimensionless position and momentum. In other words $a = (1/\sqrt{2})(\hat{x}+i\hat{y})$

If we define $f(t)\equiv -f_x(t)+if_y(y)$ we can rewrite the Hamiltonian as

$H_S(t) = \hbar \omega_0 (a^{\dagger}a + 1/2) + af(t) + a^{\dagger}f^*(t)$

The thing that I don't understand is that Merzbacher goes from the previous equation to

$\bar{H}(t) = \hbar \omega_0 (\bar{a}^{\dagger}(t)\bar{a}(t) + 1/2) + f(t)\bar{a}(t) + f^*(t)\bar{a}^{\dagger}(t)$

where the overbars indicate Heisenberg picture operators. The question is, how does he do this? Since the Schrodinger picture Hamiltonian is time dependent and doesn't commute with itself at different times, the conversion from Schrodinger to Heisenberg pictures isn't simple. Can someone explain how this is done?

My approach is to say that the propogator in the Schrodinger picture is some operator $T(t)$, ie. $T(t)$ propagates the Schrodinger picture state vector forward from time 0 to time $t$. Then I define the Heisenberg picture operators by

$\bar{A}(t) \equiv T^{\dagger}(t)A_S(t)T(t)$

Using $i\hbar \dot{T}(t) = H_S(t)T(t)$ you can show

$i\hbar \frac{d\bar{A}(t)}{dt} = \left[ \bar{A}(t), H_S(t) \right] + T^{\dagger}(t)(\partial A_S(t) / \partial t)T(t)$

Suppose we want to solve for $\bar{a}(t)$. The Schrodinger picture operator $a$ has no time dependence so the equation of motion is

$i\hbar \frac{d\bar{a}(t)}{dt} = \left[ \bar{a}(t), H_S(t) \right]$

If the Schrodinger picture Hamiltonian commuted with itself at different times this would be simple to work with, because in that case $H_S(t)=\bar{H}(t)$. We would then be able to use the fact that

$\left[\bar{a}(t), \bar{a}^{\dagger}(t) \right] = 1$

to evaluate the commutator on the right hand side of the equation of motion. However, in the present problem this is not the case so I don't know how to compute the commutator.

  • $\begingroup$ I don't understand the problem. The choice of picture can not modify the commutation relations because the time evolution operator is always unitary. $\endgroup$
    – Bubble
    Sep 10, 2014 at 21:05
  • $\begingroup$ @Bubble: You seem to be onto something. Care to be more explicit? $\endgroup$
    – DanielSank
    Sep 10, 2014 at 23:15
  • $\begingroup$ I've posted what I think as an answer because it's too long for a comment. If that doesn't clear up your confusion I'll delete it. $\endgroup$
    – Bubble
    Sep 10, 2014 at 23:17

1 Answer 1


The choice of picture can not modify the commutation relations.

The time evolution operator, $U(t)$ is unitary regardless of how complicated it is to calculate it.

Now observe,

$$AB-BA=C$$ Use the fact that $U(t)^\dagger U(t)=U(t) U(t)^\dagger=1$ to get, $$AU(t) U(t)^\dagger B-BU(t) U(t)^\dagger A=C$$ and then multiply by the conjugate transpose from the left and $U(t)$ from the right, $$ U(t)^\dagger AU(t) U(t)^\dagger B U(t)^\dagger- U(t)^\dagger BU(t) U(t)^\dagger AU(t)^\dagger= U(t)^\dagger C U(t)=\\ A(t)B(t)-B(t)A(t)=C(t)$$


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