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!

Showing energy is expectation of the Hamiltonian

  1. Jan 10, 2014 #1
    1. The problem statement, all variables and given/known data

    The vector [itex]\psi =\psi_{n}[/itex] is a normalized eigenvector for the energy level [itex]E=E_{n}=(n+\frac{1}{2})\hbar\omega [/itex] of the harmonic oscillator with Hamiltonian [itex]H=\frac{P^{2}}{2m}+\frac{1}{2}m\omega^{2}X^{2}[/itex]. Show that:

    [itex]E=\frac{\mathbb{E}_{\psi}[P^{2}]}{2m}+\frac{1}{2}m\omega^{2}\mathbb{E}_{\psi}[X^{2}][/itex]

    2. Relevant equations

    Time independent Schrodinger: [itex]H\psi=E\psi[/itex]


    3. The attempt at a solution

    Am I wrong in thinking it's as simple as taking the expectation of both sides? I feel like I must be as that gives [itex]\mathbb{E}_{\psi}[E]=\frac{\mathbb{E}_{\psi}[P^{2}]}{2m}+\frac{1}{2}m\omega^{2}\mathbb{E}_{\psi}[X^{2}][/itex] which isn't quite right.

    But I can't see how else I'd do it?

    Thanks for your help in advance!

    EDIT: I believe the extra information is needed to answer the rest of the question (this is just part 1).
     
  2. jcsd
  3. Jan 10, 2014 #2

    DrClaude

    User Avatar

    Staff: Mentor

    I guess you mean (using your notation)
    $$
    \mathbb{E}_{\psi}[H]=\frac{\mathbb{E}_{\psi}[P^{2}]}{2m}+\frac{1}{2}m\omega^{2}\mathbb{E}_{\psi}[X^{2}]
    $$

    I would start from ##E = \mathbb{E}_{\psi}[H]## and develop the right-hand-side until the desired result (the equation just above) is obtained.
     
  4. Jan 10, 2014 #3

    maajdl

    User Avatar
    Gold Member

    Indeed, taking the EV of both sides is the solution.
    Disturbingly simple ;>) .
     
  5. Jan 10, 2014 #4
    We're then asked to show by considering [itex]<\psi\mid (P\pm im\omega X)^{k}\psi>[/itex] for k=1,2 and using orthogonality properties of eigenvectors that:

    [itex]\mathbb{E}_{\psi}(P)=0=\mathbb{E}_{\psi}(X)[/itex] and [itex]\mathbb{E}_{\psi}(P^{2})=m^{2}\omega^{2}\mathbb{E}_{\psi}(X^{2})=mE[/itex]

    I've shown that [itex](P-im\omega X)*=P+im\omega X[/itex] and I can see that [itex]<\psi\mid (P- im\omega X)\psi>=\mathbb{E}_{\psi}(P)-im\omega\mathbb{E}_{\psi}(X)[/itex] etc.

    But I really can't see how to get to the answer?

    Am I supposed to show that [itex]<\psi\mid (P- im\omega X)\psi>=0=<\psi\mid (P+ im\omega X)\psi>[/itex]? Because I can't seem to do that...
     
  6. Jan 10, 2014 #5

    DrClaude

    User Avatar

    Staff: Mentor

    Have you seen ladder operators?
     
  7. Jan 10, 2014 #6
    Not yet. But they do come in the next chapter of the notes?
     
  8. Jan 10, 2014 #7

    DrClaude

    User Avatar

    Staff: Mentor

    Then I don't get how you're supposed to solve the problem. Which textbook are you using?
     
  9. Jan 10, 2014 #8
    It's my university notes - we could well have to use ladder operators to solve this. In the past we've occasionally had questions that can only be covered with material from the next chapter. What in specific should I be looking for with regard to ladder operators?
     
  10. Jan 10, 2014 #9

    DrClaude

    User Avatar

    Staff: Mentor

    You should be able to express ##P \pm i m \omega X## in terms of the ladder operators. Once this is done, you can figure out what ##(P \pm i m \omega X) \left| \psi \right\rangle## does, keeping in mind that ##\psi## is actually ##\psi_n##.
     
  11. Jan 10, 2014 #10
    I've got it! Thanks very much.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted