# Average of any operator with Hamiltonian

1. Sep 16, 2014

### andre220

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

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

3. 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 dont know where that will get me.

2. Sep 16, 2014

### Matterwave

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

3. Sep 16, 2014

### andre220

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

4. Sep 16, 2014

### Orodruin

Staff Emeritus
And how would you compute the expectation value?

5. Sep 16, 2014

### 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 $\langle\psi\mid\hat{A}\mid\psi\rangle =\int\psi^* \hat{A}\psi dx$

6. Sep 16, 2014

### Orodruin

Staff Emeritus
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?

7. Sep 16, 2014

### 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?

8. Sep 16, 2014

### Orodruin

Staff Emeritus
Are you working in Heisenberg or Schrödinger representation?

9. Sep 16, 2014

### andre220

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

10. Sep 16, 2014

### nrqed

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.

11. Sep 16, 2014

### Orodruin

Staff Emeritus
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.

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

12. Sep 16, 2014

### 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 Schrodinger picture the operators cannot be time-dependent (a major bluff).

13. Sep 16, 2014

### 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:
$\langle\psi\left|[\hat{A},\hat{H}]\right|\psi\rangle = \langle\psi\left|\hat{A}\hat{H}-\hat{H}\hat{A}\right|\psi\rangle =\langle\psi\left|\hat{A}\hat{H}\right|\psi\rangle-\langle\psi\left|\hat{H}\hat{A}\right|\psi\rangle = \int\psi^*\hat{A}\hat{H}\psi\,dx -\int\psi^*\hat{H}\hat{A}\psi\,dx$.

14. Sep 16, 2014

### Orodruin

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

15. Sep 16, 2014

### 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, $E_i$.

16. Sep 16, 2014

### Orodruin

Staff Emeritus
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?

17. Sep 16, 2014

### andre220

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

18. Sep 16, 2014

### nrqed

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.

19. Sep 16, 2014

### 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?

20. Sep 16, 2014

### andre220

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