How Does Time Dependence Affect Quantum Operators?

[Mathitalian]

Homework Statement

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 which 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
$$\displaystyle mA_\alpha = -\frac{\partial u(\vec{Q})}{\partial x_\alpha}$$

Homework Equations

Note that
Q is position operator, V is velocity operator, A is acceleration operator
$$[A_1, A_2]= A_1 A_2-A_2 A_1$$

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.