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## Homework Statement

Let

[tex]H= \frac{1}{2}m(V_x^2+V_y^2+V_z^2)+u(\vec{Q})[/tex]

be the hamiltonian operator for a particle wich has mass m>0 with

[tex]u(\vec{Q})=\lambda_0 (Q_x^2+ Q_y^2)[/tex].

Knowing that

[tex][Q_\alpha, m V_\beta]=i \delta_{\alpha \beta}[/tex].

Show that If

[tex]\displaystyle A_\alpha= \frac{d V_\alpha^{(t)}}{d t}[/tex]

then

[tex]\displaystyle mA_\alpha = -\frac{\partial u(\vec{Q})}{\partial x_\alpha}[/tex]

## Homework Equations

Note that

Q is position operator, V is velocity operator, A is acceleration operator

[tex][A_1, A_2]= A_1 A_2-A_2 A_1[/tex]

## The Attempt at a Solution

The problem is equivalent to show that [tex]A_\alpha= -\frac{\partial u(\vec{Q})}{m\partial x_\alpha}[/tex], but it is difficult to me, because of time dependent operator [tex]V_{\alpha}^{(t)}[/tex].

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

[tex][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}} [/tex]

So

[tex]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[/tex].

Now I'm stuck. Any helps will be appreciated. Thank you.

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

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