Using operators and finding expectation value

[A9876]

Homework Statement

The expectation value of the time derivative of an arbitrary quantum operator $\hat{O}$ is given by the expression:

d$\langle$$\hat{O}$$\rangle$/dt$\equiv$$\langle$d$\hat{O}$/dt$\rangle$=$\langle$∂$\hat{O}$/∂t$\rangle$+i/hbar$\langle$$[$$\hat{H}$,$\hat{O}$$]$$\rangle$​

Obtain an expression for $\langle$d$\hat{L}$_x/dt+d$\hat{L}$_y/dt$\rangle$ where $\hat{H}$=$\hat{H}$₀+μ₀B_z$\hat{L}$_z/hbar

Homework Equations

$[$$\hat{L}$_x,$\hat{L}$_y$]$=i*hbar$\hat{L}$_z
$[$$\hat{L}$_y,$\hat{L}$_z$]$=i*hbar$\hat{L}$_x
$[$$\hat{L}$_z,$\hat{L}$_x$]$=i*hbar$\hat{L}$_y

$[$A,B$]$=AB-BA

The Attempt at a Solution

$\langle$d$\hat{L}$_x/dt+d$\hat{L}$_y/dt$\rangle$=d$\langle$$\hat{L}$_x+$\hat{L}$_y$\rangle$
=$\frac{1}{ih}$ d$\langle$$[$$\hat{L}$_y,$\hat{L}$_z$]$+$[$$\hat{L}$_z,$\hat{L}$_x$]$$\rangle$/dt
=$\frac{1}{ih}$$\langle$ ∂$[$$\hat{L}$_y,$\hat{L}$_z$]$+$[$$\hat{L}$_z,$\hat{L}$_x$]$/∂t$\rangle$+$\frac{i}{hbar}$$\langle$$[$$\hat{H}$,$[$$\hat{L}$_y,$\hat{L}$_z$]$+$[$$\hat{L}$_z,$\hat{L}$_x$]$$]$$\rangle$

I'm not sure how to continue on from this

 

[clamtrox]

Just calculate the commutators $[L_x, H]$ and $[L_y, H]$? You know that H₀ should commute with all L's (why?), and the commutator with the extra term is easy to calculate.

 

[A9876]

Just calculate the commutators $[L_x, H]$ and $[L_y, H]$? You know that H₀ should commute with all L's (why?), and the commutator with the extra term is easy to calculate.

I calculated the commutators $[L_x, H]$ and $[L_y, H]$ but I don't see how this helps me answer the question. I also can't figured out why H₀ should commute with all L's.

Also $\hat{H}$₀=-hbar²/2mr² $\{$$\frac{∂}{∂r}$(r²$\frac{∂}{∂r}$)+$\frac{1}{sinθ}$$\frac{∂}{∂θ}$(sinθ$\frac{∂}{∂θ}$)+$\frac{1}{sin squared θ}$$\frac{∂ squared}{∂\phi squared}$$\}$+V(r)

where the angular-dependent part of the Hamiltonian corresponds to the total angular momentum operator $\hat{L}$²

 

[vela]

I calculated the commutators $[L_x, H]$ and $[L_y, H]$ but I don't see how this helps me answer the question.

You were given that
$$\bigg\langle \frac{d\hat{O}}{dt} \bigg\rangle = \bigg\langle \frac{\partial \hat{O}}{\partial t} \bigg\rangle + \frac{i}{\hbar} \langle [\hat{H},\hat{O}] \rangle.$$ What do you get if you let $\hat{O} = \hat{L}_x$?

I also can't figured out why H₀ should commute with all L's.

Also
$$\hat{H}_0 = -\frac{\hbar^2}{2mr^2}\left[
\frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r}\right) +
\frac{1}{\sin\theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial}{\partial \theta}
\right) +
\frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial\phi^2}\right]+V(r)$$ where the angular-dependent part of the Hamiltonian corresponds to the total angular momentum operator $\hat{L}^2$

You should be able to prove that $\hat{L}_i$ commutes with $\hat{L}^2$ using the property [AB,C]=A[B,C]+[A,C]B and the commutation relations you listed above.

 

[A9876]

You were given that
$$\bigg\langle \frac{d\hat{O}}{dt} \bigg\rangle = \bigg\langle \frac{\partial \hat{O}}{\partial t} \bigg\rangle + \frac{i}{\hbar} \langle [\hat{H},\hat{O}] \rangle.$$ What do you get if you let $\hat{O} = \hat{L}_x$?

But shouldn't I substitute $\hat{O}$=$\hat{L}$_x+$\hat{L}$_y instead?

You should be able to prove that $\hat{L}_i$ commutes with $\hat{L}^2$ using the property [AB,C]=A[B,C]+[A,C]B and the commutation relations you listed above.

Oh I understand now. Thanks

 

[vela]

You asked why you want to calculate $[\hat{H},\hat{L}_i]$. Do you see why?

 

[A9876]

Yh I do. My final answer is

$\langle$d$\hat{L}$_x/dt+d$\hat{L}$_y/dt$\rangle$=$\langle$∂($\hat{L}$_x+$\hat{L}$_y)/∂t$\rangle$+$\frac{i}{hbar}$$\langle$iμ₀B_z($\hat{L}$_y - $\hat{L}$_x)$\rangle$

Could I further simplify this?