Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question Regarding Harmonic Oscillator Eigenkets

  1. Oct 26, 2011 #1
    Hi everyone!

    Given that a harmonic oscillator has eigenkstates |n> where n = 1,2,3,..., how can we calculate <X>, <P>, <X^2>, etc. Is there a need to define a wavefunction in the |n> basis?

    Thanks!
     
  2. jcsd
  3. Oct 26, 2011 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Essentially no, because |n>'s are eigenkets of the number operator/Hamiltonian and X and P, though unbounded & with purely continuous spectrum, can be expressed as linear combinations of the raising & lowering ladder operators whose action on the eigenket's space becomes known once you establish that |n>'s are eigenkets of H and N.
     
  4. Oct 26, 2011 #3
    So that means just express the |n> kets as linear combinations of the ladder operators, and then use them as ψ in the formula <X> = <ψ|X|ψ>? But how would you deal with the infinite dimensionality? Will the answer be finite in that case?

    Thank you by the way for the idea!
     
  5. Oct 26, 2011 #4
    Consider that
    [tex]
    \hat{a}=\frac{1}{\sqrt{2}}(\hat{x}+i\hat{p})
    [/tex]
    and
    [tex]
    \hat{a}^{\dagger}=\frac{1}{\sqrt{2}}(\hat{x}-i\hat{p})
    [/tex]
    You can use these to write [itex]\hat{x}[/itex] and [itex]\hat{p}[/itex] in terms of [itex]\hat{a}[/itex] and [itex]\hat{a}^{\dagger}[/itex]. Then you know
    [tex]
    \hat{a}|n\rangle = \sqrt{n}|n-1\rangle
    [/tex]
    and
    [tex]
    \hat{a}^{\dagger}|n\rangle=\sqrt{n+1}|n+1\rangle
    [/tex]
    You have the tools to take the expectation value.
     
  6. Oct 27, 2011 #5

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    You don't express the kets as ladder operators acting on the vacuum, you express x and p as ladder operators.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook