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Harmonic oscillator eigenvector/eigenvalue spectrum

  1. Jun 5, 2012 #1
    the problem is attached as an image.

    im having troubles with the question. i'm assuming this is an induction question?
    i can prove it for the basis step n=0.

    but im having trouble as to what i have to do for n+1 (inductive step).

    any help or hints would be great!thanks
     

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  2. jcsd
  3. Jun 5, 2012 #2

    George Jones

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    What is the commutator of [itex]H[/itex] and [itex]a^\dagger[/itex]?
     
  4. Jun 5, 2012 #3
    N=a†a

    [N,a†]=a†
    [N,a]=-a
     
  5. Jun 5, 2012 #4

    George Jones

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    What about [itex]H[/itex]?
     
  6. Jun 6, 2012 #5
    [H,a†] = [itex]\hbar[/itex][itex]\omega[/itex]a†

    [H,a] = -[itex]\hbar[/itex][itex]\omega[/itex]a
     
  7. Jun 6, 2012 #6

    George Jones

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    [itex]\phi_{n+1} = \left( n + 1 \right)^{-\frac{1}{2}} a^\dagger \phi_n[/itex] gives [itex]H \phi_{n+1} = \left( n + 1 \right)^{-\frac{1}{2}} H a^\dagger \phi_n[/itex]. Now use the commutator to reorder [itex]H a^\dagger[/itex].
     
  8. Jun 6, 2012 #7
    re-order how?

    by using: Ha†=[itex]\hbar\omega[/itex]a† + a†H
    ?
     
  9. Jun 6, 2012 #8

    George Jones

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  10. Jun 6, 2012 #9
    i then get:

    =(n+1)-1/2(a†[itex]\hbar\omega[/itex]+a†H)[itex]\phi[/itex]n
    =[itex]\hbar\omega[/itex][itex]\phi[/itex]n+1 + (n+1)-1/2a†H[itex]\phi[/itex]n

    not sure what to do from here..
     
  11. Jun 6, 2012 #10

    George Jones

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    The inductive step assumes what about [itex]H\phi_n[/itex]?
     
  12. Jun 6, 2012 #11
    it assumes that Hϕn = nϕn

    i.e. that ϕn is an eigenvector of H. n being the eigenvalue


    => = ([itex]\hbar\omega[/itex]+n)ϕn+1

    ?
     
  13. Jun 6, 2012 #12

    George Jones

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    The inductive step assumes that [itex]\phi_n[/itex] is an eigenvector of [itex]H[/itex], but it doesn't assume that the associated eigenvalue is [itex]n[/itex]. Energies are eigenvalues of the Hamiltonian, so call the the eigenvalue [itex]E_n[/itex]. Maybe [itex]E_n = n[/itex], but maybe it doesn't. Let's find out!
    What is the eigenvalue of [itex]H[/itex] associated with the eigenvector [itex]\phi_0[/itex]?
     
  14. Jun 7, 2012 #13
    0=[itex]\frac{1}{2}[/itex]ℏωϕ0

    So, E0=[itex]\frac{1}{2}[/itex]ℏω

    => Hϕn+1 = ℏωϕn+1+Enϕn+1
    ?
     
    Last edited: Jun 7, 2012
  15. Jun 7, 2012 #14

    George Jones

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  16. Jun 7, 2012 #15
    can i simply say after that:
    the energies are:

    En=hw(n+[itex]\frac{1}{2}[/itex])

    ?
     
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