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

Obtain an expression for \langled\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

\langled\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

\langled\hat{L}_x/dt+d\hat{L}_y/dt\rangle=\langle∂(\hat{L}_x+\hat{L}_y)/∂t\rangle+\frac{i}{hbar}\langleiμ₀B_z(\hat{L}_y - \hat{L}_x)\rangle

Could I further simplify this?