# 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}$00Bz$\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

## Answers and Replies

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.

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

Last edited:
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.

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

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

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$iμ0Bz($\hat{L}$y - $\hat{L}$x)$\rangle$

Could I further simplify this?