1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    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!

Operator in quantum mechanics

  1. Jul 31, 2011 #1
    1. The problem statement, all variables and given/known data
    i know that
    [itex] H |n> = \varepsilon_n |n>[/itex]
    [itex] H |i> = \varepsilon_i |i>[/itex]

    and i want to estimate
    [itex] < n | e^{i H t / \hbar} V(t) e^{-i H t / \hbar} |i> [/itex]

    2. Relevant equations

    [itex] <\psi | \varphi> = \int \psi^*(q) \varphi(q) dq [/itex]


    3. The attempt at a solution

    i don't understand well how the operator interacts with the wave function.... i'm pretty sure that the right solution is

    [itex] < n | e^{i H t / \hbar} V(t) e^{-i H t / \hbar} |i> = \int{ \left(n e^{- i H t / \hbar}\right)^* V e^{-i \varepsilon_i t / \hbar}} i = \int n^* e^{i \varepsilon_n t / \hbar} V e^{-i \varepsilon_f t / \hbar} i= <n | V(t) | i> e^{i \frac{\varepsilon_n - \varepsilon_i}{\hbar}t} [/itex]

    but i'm not sure about "why is this the right way to do this" ... for example i dont know why [itex] H |n> = \varepsilon_n |n> [/itex] implies [itex] e^{iHt / \hbar} |n> = e^{i \varepsilon_n t / \hbar } |n> [/itex]... or why this way to do is wrong:

    [itex] < n | e^{i H t / \hbar} V(t) e^{-i H t / \hbar} |i> = \int \left(n e^{i H t / \hbar}\right)^* e^{i H t / \hbar} V(t) i = \int n^* V(t) i = <n|V(t)|i> [/itex]

    (i didn't put the "dq" in the integral because i dont know in wich variable i should integrate)
    if someone can explain me why this is the right way to do this expression i'll be happy.
     
    Last edited: Jul 31, 2011
  2. jcsd
  3. Jul 31, 2011 #2
    Expand the exponential into Taylor series, then operate term-by-term, then recombine.

    If [itex]V(t)[/itex] is just a c-number, you can just pull it out, which then makes the problem trivial. So I am guessing that [itex]V(t)[/itex] is an operator, in which case, you cannot commute it with the exponential on the right.
     
  4. Jul 31, 2011 #3
    ...thanks mathfeel!
     
  5. Aug 1, 2011 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    When you calculate an amplitude, the integral comes from inserting a complete set[tex]\int dq\,\lvert q \rangle\langle q \rvert = \mathbf{1}[/tex] to get
    \begin{align*}
    \langle \phi \vert \psi \rangle &= \langle \phi \vert \left(\int dq\,\lvert q \rangle\langle q \rvert\right) \vert \psi \rangle \\
    &= \int dq\,\langle \phi \vert q \rangle\langle q | \psi \rangle
    \end{align*}
    Then using the fact that [itex]\langle\phi\vert q\rangle = \phi^*(q)[/itex] and [itex]\langle q\vert\psi\rangle = \psi(q)[/itex], you obtain[tex]\langle \phi \vert \psi \rangle = \int dq\,\phi^*(q) \psi(q)[/tex]
    Note that the bra and ket are related to but are not exactly the same thing as the wave functions. The wave function is the representation of ket relative to some basis.

    In this problem, you never need to introduce a complete set and get an integral. Just work with the kets. There's no need to use the wave functions.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Operator in quantum mechanics
Loading...