1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum physics problem: SE and operators

  1. Jun 29, 2012 #1
    I have this problem at hand:

    1. The problem statement, all variables and given/known data
    A quantum mechanical system has a hamilton operator [itex]\hat{H}[/itex] and another, time independent operator [itex]\hat{A}_{0}[/itex].
    Construct a time dependent operator [itex]\hat{A}(t)[/itex] so that:
    <ψ(t)|[itex]\hat{A}_{0}[/itex]|ψ(t)> = <ψ(0)|[itex]\hat{A}(t)[/itex]|ψ(0)>
    for all states ψ(t) that develop in time according to the SE.

    3. The attempt at a solution

    In the derivation of the Schrödinger equation, we use the unitary operator [itex]\hat{U}(t)[/itex] to calculate the effect of time on the state ψ(0)...
    ψ(t) = [itex]\hat{U}(t)[/itex] ψ(0) = exp(-i/[itex]\hbar \hat{H}[/itex] t) ψ(0).

    In other words:
    <ψ(t)|[itex]\hat{A}_{0}[/itex]|ψ(t)> = <[itex]\hat{U}(t)[/itex] ψ(0)|[itex]\hat{A}_{0}[/itex]|[itex]\hat{U}(t)[/itex] ψ(0)>
    =<ψ(0) |[itex]\hat{U}(t)^{+} \hat{A}_{0} \hat{U}(t)[/itex] | ψ(0)>.

    so my "solution" is that
    [itex]\hat{A}(t)[/itex] = [itex]\hat{U}(t)^{+} \hat{A}_{0} \hat{U}(t)[/itex]...

    But this is way too simple to be correct...

    So what am I missing?


  2. jcsd
  3. Jun 29, 2012 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Looks fine to me. I suppose you could write explicitly what ##U(t)^\dagger## is equal to.
  4. Jun 29, 2012 #3
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Quantum physics problem Date
Concurrence and three tangle Oct 21, 2016
How to calculate negativity measurement of quantum state? Aug 30, 2016
A problem based on Quantum Physics 2 Dec 22, 2013
A problem based on Quantum Physics. Dec 22, 2013