Average of any operator with Hamiltonian

[andre220]

Homework Statement

Prove that for any stationary state the average of the commutator of any operator with the Hamiltonian is zero: $$\langle\left[\hat{A},\hat{H}\right]\rangle = 0.$$

Substitute for $\hat{A}$ the (virial) operator:$$\hat{A} = \frac{1}{2}\sum\limits_i\left(\hat{p}_ix_i +x_i\hat{p}_i\right)$$
and prove the virial theorem.

Homework Equations

$[\hat{A},\hat{H}] =\hat{A}\hat{H}-\hat{H}\hat{A}$

The Attempt at a Solution

So for stationary states we have that $H(q_i,\dot{q}_i,...)$, namely $H$ is not a function of $t$, and we know that the commutator $[\hat{A},\hat{H}] =\hat{A}\hat{H}-\hat{H}\hat{A}$. For any operator $\hat{A}$ the expectation value $\langle\hat{A}\rangle = \langle\psi\mid\hat{A}\psi\mid\rangle$, where $\psi$ is some given state. So I'm stuck as to where to start. Initially my thought is to take $\frac{\partial \hat{A}}{\partial t}$, but I don't know where that will get me.

 

[Matterwave]

If a state $\left|\psi\right>$ is a stationary state, what does $\hat{H}\left|\psi\right>$ look like?

 

[andre220]

Then it would be: $\hat{H}\mid\psi_n\rangle = E_n\psi_n$

 

[Orodruin]

And how would you compute the expectation value?

 

[andre220]

The expectation value of $\hat{H}$ would then be $\langle\psi\mid\hat{H}\mid\psi\rangle =\int\psi^* \hat{H}\psi dx$, and similarly the expectation value $\hat{H}$ would be $\langle\psi\mid\hat{A}\mid\psi\rangle =\int\psi^* \hat{A}\psi dx$

 

[Orodruin]

So how do you compute the expectation of the commutator? (It is also an operator...) How can you use the information that $|\psi\rangle$ is a stationary state?

 

[andre220]

Well if it is a stationary state that implies that $\frac{\partial \hat{H}}{\partial t} = 0$, however, that does not mean that $\frac{\partial \hat{A}}{\partial t} = 0$, correct?

 

[Orodruin]

Are you working in Heisenberg or Schrödinger representation?

 

[andre220]

No specification, but in class we haven't covered the Heisenberg representation yet, so assume schordinger.

 

[nrqed]

No specification, but in class we haven't covered the Heisenberg representation yet, so assume schordinger.

You do need to do anything fancy. Just sandwich the commutator between the wave functions and use the fact that they are eigenstates of the hamiltonian. You will be done in two lines.

 

[Orodruin]

The reason that I am asking is that you seem to want to take time derivatives of the operators and in Schrödinger representation the operators are time-independent while the time-dependence is all in the states, but in Heisenberg representation the states are time-independent while the operators are time-dependent.

[edit] nrqed essentially already told you how to do it ... I was trying to lead you to this realisation from your own reasoning.

 

[andre220]

Right, right, that makes sense, I have seen the Heisenberg representation before, but it has been a while. I guess I forgot that in the Schrödinger picture the operators cannot be time-dependent (a major bluff).

 

[andre220]

Okay can someone please check my work, I think I am close, but it is still not quite clicking here is what I have:
$\langle\psi\left|[\hat{A},\hat{H}]\right|\psi\rangle = \langle\psi\left|\hat{A}\hat{H}-\hat{H}\hat{A}\right|\psi\rangle =\langle\psi\left|\hat{A}\hat{H}\right|\psi\rangle-\langle\psi\left|\hat{H}\hat{A}\right|\psi\rangle = \int\psi^*\hat{A}\hat{H}\psi\,dx -\int\psi^*\hat{H}\hat{A}\psi\,dx$.

 

[Orodruin]

Hint: You should never have to go to the integral representation of the expectation values. Use the defining property of the stationary state.

 

[andre220]

Okay so then perhaps this $\langle\psi\left|\hat{A}\hat{H}\right|\psi\rangle - \langle\psi\left|\hat{A}\hat{H}\right|\psi\rangle \implies \langle\psi\left|\hat{A}\hat{H}\right|\psi\rangle - \langle\psi\left|\hat{A}\hat{H}\right|\psi\rangle = 0$ because $\langle\psi\left|\hat{A}\hat{H}\right|\psi\rangle$ and $\langle\psi\left|\hat{H}\hat{A}\right|\psi\rangle$ produce the same eigenvalue of the state $|\psi\rangle$ namely, $E_i$.

 

[Orodruin]

Almost there, I would like to see some intermediate steps to be sure you got it for the right reasons. Note that $E$ is not an eigenvalue of the commutator but it does hold that $AH|\psi\rangle = AE |\psi\rangle = EA |\psi\rangle$, since $E$ is just a number which commutes with anything. Can you fill in the steps for the expectation value of $AH$ and for that of the other term?

 

[andre220]

Ok, slowly getting there, $\langle[A,H]\rangle = \langle\psi|AH|\psi\rangle-\langle\psi|HA|\psi\rangle =\langle E_i|AH|E_i\rangle-\langle E_i|HA|E_i\rangle = E_i\langle E_i|A|E_i\rangle-E_i\langle E_i|A|E_i\rangle =0$.

 

[nrqed]

Ok, slowly getting there, $\langle[A,H]\rangle = \langle\psi|AH|\psi\rangle-\langle\psi|HA|\psi\rangle =\langle E_i|AH|E_i\rangle-\langle E_i|HA|E_i\rangle = E_i\langle E_i|A|E_i\rangle-E_i\langle E_i|A|E_i\rangle =0$.

Right.
And the key point is to understand why we may say $\langle \psi_i | H = \langle \psi_i | E_i$. If this is clear to you then you understand completely the solution.

 

[andre220]

Yes this makes sense now. And for part two, I won't go through the whole thing on here, but the main thing I want to clear up is that for the given virial operator I use that to prove the Ehrenfest (virial theorem), correct?

 

[andre220]

Thank you all for helping me understand all of this by the way.

 

[Orodruin]

Yes this makes sense now. And for part two, I won't go through the whole thing on here, but the main thing I want to clear up is that for the given virial operator I use that to prove the Ehrenfest (virial theorem), correct?

You will need to compute the commutator of the Hamiltonian with the virial operator, which will be based on the canonical commutation relations. Other than that, it is just a matter of inserting the result into what you already showed.

 

[andre220]

Okay that makes sense. Thanks.