# Time Evolution of Operators

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1. Jun 1, 2016

### bananabandana

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

For a general operator $\hat{O}$, let $\hat{O}_{mn}(t)$ be defined as:
$$\hat{O_{mn}}(t) = \int u^{*}_{m}(x,t) \hat{O} u_{n}(x,t)$$
and
$$\hat{O_{mn}} = \int u^{*}_{m}(x) \hat{O} u_{n}(x)$$
$u_{m}$ and $u_{n}$ are energy eigenstates with corresponding energy $E_{n}$ and $E_{m}$.
From this it can be shown that: (sticking in $\hat{O} = [\hat{A},\hat{H}]$
$$d\frac{[\hat{A},\hat{H}]}{dt} =(E_{n}-E_{m}) exp\bigg[-i\frac{(E_{n}-E_{m})t}{\hbar}\bigg] \bigg( -\frac{i \hat{A_{mn}} (E_{n}-E_{m})}{\hbar} + \frac{d \hat{A_{mn}}}{dt}\bigg)$$
But simultaneously, it is also possible to prove that: (as we'd expect from Ehrenfest anyhow)
$$\frac{d\hat{A_{mn}}}{dt} = -\frac{i}{\hbar} [\hat{A_{mn}},\hat{H}]$$
Where $\hat{H}$ is the Hamiltonian, and $\hat{A}$ some general Hermitian operator.

This is an issue, since if $[\hat{A},\hat{H}] =0$ we have simultaneously:
$$\frac{d \hat{A_{mn}}}{dt} = 0$$
and
$$\frac{d\hat{A_{mn}}}{dt} = i \frac{\hat{A_{mn}}}{\hbar} (E_{n}-E_{m})$$
2. Relevant equations

3. The attempt at a solution
This is a problem I made up and got stuck with! I think the resolution should be fairly trivial, but I can't see it - I'd bet that there is probably a simpler way of looking at this using Dirac notation, but I don't really know that very well...

Thanks!

2. Jun 1, 2016

### blue_leaf77

The LHS is an operator while the RHS is a scalar function, is such an equation possible? I would like to know how you obtained it in the first place?

3. Jun 2, 2016

### bananabandana

Ah, $[\hat{A},\hat{H}]$ should in fact read $[\hat{A}, \hat{H}]_{mn}$ But otherwise:

$$[\hat{A},\hat{H}]_{mn} = exp\bigg( i \frac{(E_{m} -E_{n})t}{\hbar} \bigg) \int u^{*}_{m}(x) [\hat{A},\hat{H}] u_{n}(x) \ dx$$
$$= exp\bigg( i \frac{(E_{m} -E_{n})t}{\hbar} \bigg) (E_{n}-E_{m}) \int u^{*}_{m}\hat{A}u_{n}$$
Where the second line follows from that Hermitian nature of the operators, ( $\int \phi^{*} \hat{H}\hat{A} \psi \ dx = \int (\hat{A}\hat{H}\phi)^{*}\psi dx$)

Differentiating that under the integral sign ( right phrase? I mean - $\frac{d}{dt}\int \rightarrow \int \frac{\partial}{\partial t}$) with respect to $t$ gets the result shown.

About the $[\hat{A},\hat{H}]$ vs. $[\hat{A},\hat{H}]_{mn}$ I think, given that any arbitrary wave function can be expanded out in terms of the eigenstates of energy, we can still get the result in general for the expectation commutator? I mean, if we do:

$$\int \psi^{*}(x,t) [\hat{A},\hat{H} ] \psi(x,t) dx$$

and write $$\psi = \sum c_{n}u_{n}(x)exp(-\frac{E_{n}t}{\hbar})$$
Then, if the commutator $[\hat{A},\hat{H}] = 0$ then the integral must be zero, meaning that each term in $n$ individually yields zero, which gives us a sum of exponentials [Setting third eq'n in first posting = 0]...

Does that make sense?

Thanks!

4. Jun 3, 2016

### blue_leaf77

This equation
$$\frac{d\hat{A_{mn}}}{dt} = -\frac{i}{\hbar} [\hat{A_{mn}},\hat{H}]$$
doesn't look right to me. $A_{mn}$ which appears in the derivative depends on $t$ internally, e.g. oscillating electric field, hence it does not need to obey the Heisenberg equation of motion. Given Heisenberg equation of motion for a time-independent operator $O$
$$\frac{d}{dt}O_H(t) = -\frac{i}{\hbar} [O,\hat{H}]$$
where $O_H(t) = U^\dagger O U$. On the other hand, $A_{mn}(t)$ in the first equation clearly is not defined this way because the dependence on time is already inherent in the physical quantity represented by the operator itself.

5. Jun 3, 2016

### bananabandana

Unfortunately, it turns out I got that result as a given from another question- I'm not sure where they derived it from exactly! I've double checked, and they are the same formula, so I'm not sure what to do.

6. Jun 3, 2016

### blue_leaf77

One more strange thing, no matter how you look at it, as it stands now $A_{mn}$ is a scalar, therefore $[A_{mn},H]=0$. Are you sure the other question you are talking about address exactly the same issue?

7. Jun 3, 2016

### nrqed

Where did you get this last equation from? Ehrenfest's theorem applies to operators, not matrix elements (and there is in addition a partial derivative $\partial \hat{A}/\partial t$ in general).

8. Jun 3, 2016

### blue_leaf77

I also asked the same question and the OP said he/she got (or probably infer) it from another question (post #5).