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Time Evolution of Operators

  1. Jun 1, 2016 #1
    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. jcsd
  3. Jun 1, 2016 #2

    blue_leaf77

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    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?
     
  4. Jun 2, 2016 #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!
     
  5. Jun 3, 2016 #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.
     
  6. Jun 3, 2016 #5
    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.
     
  7. Jun 3, 2016 #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?
     
  8. Jun 3, 2016 #7

    nrqed

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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).
     
  9. Jun 3, 2016 #8

    blue_leaf77

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    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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