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Time Evolution of Operators

  • #1

Homework Statement

[/B]

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
$$ \frac{d\hat{A_{mn}}}{dt} = i \frac{\hat{A_{mn}}}{\hbar} (E_{n}-E_{m}) $$

Homework Equations




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!
 

Answers and Replies

  • #2
blue_leaf77
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$$
d\frac{[\hat{A},\hat{H}]}{dt} =(E_{n}-E_{m}) exp\bigg[-i\frac{(E_{n}-E_{m})t}{\hbar}\bigg] \bigg( -\frac{i \hat{A_{mn}} (E_{n}-E_{m})}{\hbar} + \frac{d \hat{A_{mn}}}{dt}\bigg)
$$
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
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
blue_leaf77
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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
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.
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
blue_leaf77
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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
nrqed
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Homework Statement

[/B]

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 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
blue_leaf77
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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).
I also asked the same question and the OP said he/she got (or probably infer) it from another question (post #5).
 

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