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Show that expecting value of energy is independent of time

  1. Oct 8, 2014 #1
    Sorry I failed to make Latex work, don't know why...

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

    We consider
    for a given potential. Psi is also normalized...

    h%29%5Cpsi_%7Bn%7D%28x%29%24%24.gif


    2. Relevant equations
    Show that expectation value of energy is independent of time.

    3. The attempt at a solution

    Well, I'm use to expecting values, but not of the energy...

    I started with <H>psi=<E>psi
    then if I compute <H>=<psi|H|psi> = E|Psi> = E

    Then I told myself H=p^2/2m + V(x)

    So I guess if I compute dH/dt and I find it equals to 0 I can say E is independent of time?

    I feel like this is wrong tough... It looks too simple

    Thanks a lot for checking that out guys!
     
    Last edited: Oct 8, 2014
  2. jcsd
  3. Oct 8, 2014 #2

    BvU

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    Hello, and welcome to PF :-)
    They aren't called expecting values, but expectation values...

    the thing in the middle doesn't look right . What exactly do you compute ?
     
  4. Oct 8, 2014 #3
    Thanks about that!

    Well, I'm really not quite sure...found that somewhere and I tought it made sens but the more I look at it the more this doesn't help at all...
     
  5. Oct 8, 2014 #4

    Matterwave

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    Are you given that ##V=V(x)\neq V(x,t)## is true, so that your Hamiltonian is explicitly time-independent? Otherwise, I don't think the statement in this problem is true.

    I'm guessing this is assumed because you have written:

    $$\Psi(x,t)=\sum_n c_n(t=0)e^{-iE_n t/\hbar}\psi_n(x)$$

    If so, all you have to prove is that ##\left<H\right>## is independent of time, since that IS the expectation value of the energy. In other words, prove:

    $$\frac{d\left<H\right>}{dt}=0$$
     
  6. Oct 8, 2014 #5

    BvU

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    If ##H## is independent of time, the wave functions ##\Psi({\bf r},t)## in ##i\hbar {\partial \Psi\over\partial t} = H\Psi ## can be written as ##f(t) \;\psi({\bf r})## and the equation can be re-written as ##i\hbar {\partial f \over\partial t} \psi({\bf r}) = f(t) \;H\psi (\bf r)## or $$i\hbar {\partial f \over\partial t} /f(t) = H\psi (\bf r)/\psi({\bf r})$$ Now left hand side is a function of t only, right hand of r only. Can only be if both sides are constant. name the constant E and you get the time-independent Schroedinger eqn ##H\psi ({\bf r}) = E\;\psi({\bf r})##.
    For the complete wave function there still is the time part, for which ##i\hbar {\partial f \over\partial t} = E \; f(t)## with solution ## f(t) = e^{{-i E\over \hbar} t }##. That's where the ##\Psi(x,t)## in your exercise come from.

    The full wording of your exercise probably says the ##\psi_n## are normalized ? And they satisfy ##H|\psi_n> = E_n|\psi_n>## ?

    Now your <H>=<psi|H|psi> can be written out in full (by you) as $$\int_{-\infty}^{\infty}\; \Psi^*(x,t) \; H \; \Psi^*(x,t) \; dx $$to give the expectation value of E. Fill in the given ##\Psi##, see that the exp (and hence the time dependence) disappears
     
  7. Oct 8, 2014 #6
    I had to read that a couple times but thanks a lot! that really helped!
     
  8. Oct 8, 2014 #7
    Oh, and while I'm there,

    I also need to prove that ##<H> <= E_{1}##, where ##E_{1}## is the fundamental energy. How will I proceed to do that?

    Do I need to compute ##E_{1}##? Cause if I do, I'll end up with terms in ##\psi_{1}## that I'm not quite sure how to compare with ##\psi_{n}##
     
  9. Oct 9, 2014 #8

    Matterwave

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    I think you mean ##\left<H\right> \geq E_1## right? ##E_1## is the ground state energy, so you shouldn't be getting energies lower than it.
     
  10. Oct 9, 2014 #9
    Yes sorry your right! I know it sound really obvious, but I need to prove it...That's what bugging me :S
     
  11. Oct 9, 2014 #10

    Matterwave

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    Well start with the definition: ##\left<H\right>=\left<\Psi|H|\Psi\right>## and then break down the ##\left|\Psi\right>## into the superposition of eigenvectors of the Hamiltonian to see how this will work out.
     
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