[QM] Time dependent operator

[Mathitalian]

1. The problem statement, all variables and given/known data
Let
H= \frac{1}{2}m(V_x^2+V_y^2+V_z^2)+u(\vec{Q})
be the hamiltonian operator for a particle wich has mass m>0 with
u(\vec{Q})=\lambda_0 (Q_x^2+ Q_y^2).
Knowing that
[Q_\alpha, m V_\beta]=i \delta_{\alpha \beta}.
Show that If
\displaystyle A_\alpha= \frac{d V_\alpha^{(t)}}{d t}
then
\displaystyle mA_\alpha = -\frac{\partial u(\vec{Q})}{\partial x_\alpha}
2. Relevant equations
Note that
Q is position operator, V is velocity operator, A is acceleration operator
[A_1, A_2]= A_1 A_2-A_2 A_1

3. The attempt at a solution

The problem is equivalent to show that A_\alpha= -\frac{\partial u(\vec{Q})}{m\partial x_\alpha}, but it is difficult to me, because of time dependent operator V_{\alpha}^{(t)}.
I think that the relation V_{\alpha}^{(t)}= i U_{t}^{-1}[H, V_{\alpha}]U_t is useful. Now
[H, V_\alpha]= [u(\vec{Q}), V_\alpha]= [\lambda_0 (Q_x^2+ Q_y^2), V_\alpha]=\frac{\lambda_0}{m}(2 i (\delta_{\alpha, x}Q_x+ \delta_{\alpha, y}Q_y))= i \frac{\partial u(\vec{Q})}{m x_{\alpha}}

So
V_\alpha^{(t)}=i U_t^{-1}[H, V_\alpha]U_t = - U_t^{-1} \frac{\partial u(\vec{Q})}{m x_\alpha}U_t.
Now I'm stuck. Any helps will be appreciated. Thank you.

Sorry for mistakes in english language. I'm italian :)

[G01]

I think that the relation V_{\alpha}^{(t)}= i U_{t}^{-1}[H, V_{\alpha}]U_t is useful.

This is not correct. I think you are mixing together two different relations. One is the time dependence of an operator in the Heisenberg picture:

V_a(t)=U^{-1}V_aU

where U is the time translation operator.

The other relation is the Heisenberg Equation of Motion for operators in the Heisenberg picture:

\frac{dV_a(t)}{dt}=\frac{1}{i\hbar}[H,V_a(t)]

from which we can find an "expectation value equation of motion:"

\frac{d<V_a(t)>}{dt}=\frac{1}{i\hbar}<[H,V_a(t)]>

You are essentially being asked to show that the quantum expectation values obey equations similar in form to their classical variable counterparts.

You should be able to get to the result directly if you start from the third equation in this post.

[Mathitalian]

Oh, i'm so sorry for this terrible mistake. You're right. To tell the truth, when I posted the question, it was night in Italy.

My intent was to use

\frac{dV_\alpha^{(t)}}{d t}=i U_t^{-1}[H, V_\alpha]U_t = - U_t^{-1} \frac{\partial u(\vec{Q})}{m\partial x_\alpha}U_t


Now, from
i U_t^{-1}[H, V_\alpha]U_t = - U_t^{-1} \frac{\partial u(\vec{Q})}{m \partial x_\alpha}U_t

How can i complete the proof? Thank you for your help.

[G01]

Bring the U operators inside the derivative (they do not depend on x). Then remember that the Hamiltonian, and thus the potential energy operator, is time independent.

[Mathitalian]

Bring the U operators inside the derivative (they do not depend on x). Then remember that the Hamiltonian, and thus the potential energy operator, is time independent.

I'm not so good in physics, so forgive me if i'm going to say something stupid..

From your word, can i conclude that
\displaystyle\left[U_t, \frac{\partial u(\vec{Q})}{m \partial x_\alpha}\right]=O
i.e. the operators commute?

[G01]

Yes. And what does that then imply?

[Mathitalian]

\displaystyle\left[U_t, \frac{\partial u(\vec{Q})}{m \partial x_\alpha}\right]=O\implies U_t\frac{\partial u(\vec{Q})}{m \partial x_\alpha}=\frac{\partial u(\vec{Q})}{m \partial x_\alpha} U_t

then
- U_t^{-1} \frac{\partial u(\vec{Q})}{m\partial x_\alpha}U_t = - U_t^{-1} U_t\frac{\partial u(\vec{Q})}{m\partial x_\alpha}= -\frac{\partial u(\vec{Q})}{m\partial x_\alpha}
as we want.

I think it is ok now, right?

Thanks a ton

[G01]

Yes, you got it. You're welcome. :smile: