The time-derivative of the Hamiltonian for a 1D harmonic potential

Question

I do not undersand how to take the time derivative of the following Hamiltonian $\hat{H}(t) = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2(\hat{x}-a(t))^2$ where $a(t) = v_0t$. For instance how does $\hat{p}$ change when taking the time derivative of it? And the same for $\hat{x}$.

Answer 1

The way you have written it, without time arguments in $\hat p$ and $\hat x$, these two operators are presumed to be in the Schrödinger representation, i.e., time independent. Thus $$ \frac{d}{dt} \hat H(t)=\partial_t \hat H(t)=v_0(v_0 t-\hat x). $$

Note this is not the Galilean transform of the oscillator hamiltonian, but, instead, a Hamiltonian canonically equivalent to the $v_0=0$ case, and with the same spectrum as it. (Under a quantum canonical transformation, the commutation relations are preserved.)

It would be perverse & unfriendly to assume your expression is in the Heisenberg picture while omitting the (t), as the Schrödinger operator above would have to be converted into the corresponding Heisenberg one, $U(t)^\dagger v_0(v_0 t-\hat x) U(t)$, liable to confuse you, if you are asking this question.