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Inner Product in this step of the working

  1. Jan 21, 2014 #1
    Hi guys, I'm not sure how to evaluate this inner product at step (3.8)

    I know that:

    ##\hat {H} |\phi> = E |\phi>##

    2cna71v.png

    [tex] <E_n|\frac{\hat H}{\hbar \omega} + \frac{1}{2}|E_n> [/tex]

    [tex] <E_n| \frac{\hat H}{\hbar \omega}|E_n> + <E_n|\frac{1}{2}|E_n> [/tex]

    I also know that ##<\psi|\hat Q | \psi>## gives the average value of observable to ##\hat Q##. In this case, it's not ##\psi## but ##E_n##, does the same principle hold?
     
    Last edited: Jan 21, 2014
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  3. Jan 21, 2014 #2

    George Jones

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    Use equation (3.4).
     
  4. Jan 21, 2014 #3

    dextercioby

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    Hi George, I can only conclude that you figured out where he picked his 2 questions from. :biggrin:.
     
  5. Jan 22, 2014 #4
    Okay, I got step 3.8. But For Step 3.9, I'm having a slight issue here:

    [tex]E_n = <E_n|\hat H|E_n> = \frac{1}{2m}<E_n|(m\omega \hat x)^2 + \hat {p}^2|E_n>[/tex]
    [tex] = \frac{1}{2m}(m^2\omega^2)<E_n|\hat x \hat x|E_n> + \frac{1}{2m}<E_n|\hat p \hat p|E_n> [/tex]

    Using ##\hat x|\phi> = x|\phi>## and ##\hat p |\phi> = p|\phi>##,

    [tex] = \frac{1}{2m}(m^2\omega^2)<E_n|\hat x x|E_n> + \frac{1}{2m}<E_n|\hat p p|E_n>[/tex]

    Removing the last "hats" from ##\hat p## and ##\hat x## and Using orthogonality ##<E_n|E_n> = 1 ##:

    [tex] = \frac{1}{2m}\left ( (m\omega x)^2 + p^2 \right ) [/tex]

    Is it wrong to assume that the energy eigenkets have norm = 1? That would be strange because later we show that ##E_n = (n + \frac{1}{2})\hbar \omega##

    And, why is there an additional factor of ##\frac{1}{\omega}## in the expression?
     
  6. Jan 22, 2014 #5

    dextercioby

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    Energy eigenkets for the harmonic oscillator do have norm =1.
     
  7. Jan 22, 2014 #6
    Ok, then I have no idea what's gone wrong with my working..
     
  8. Jan 22, 2014 #7

    dextercioby

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    The energy eigenket is not an eigenket of either x nor p. You need the ladder operators to evaluate the matrix elements.
     
  9. Jan 22, 2014 #8
    How did they get 3.9 then?
     
  10. Jan 22, 2014 #9

    dextercioby

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    <p E_n, p E_n> = ||p |E_n> ||^2 = <E_n, p^2 E_n>, because the eigenkets of energy are in the domain of both p and p^2, on which the 2 operators are essentially self-adjoint. p^2 is a positive operator, hence the inequality at the end.

    The same goes for x.
     
  11. Jan 22, 2014 #10
    I just don't get how ##<E_n|\hat x \hat x|E_n> = |x|E_n>|^2## and ##<E_n|\hat p \hat p|E_n> = |p|E_n>|^2##
     
  12. Jan 22, 2014 #11

    ChrisVer

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    hmm because in the same way you have in the complex numbers that:
    [itex] z^{*} z = |z|^{2}[/itex]
    In fact the ket can be interpreted as a vector on Hilber space, while the bra as its dual.
    So in the case of this, you can write:

    [itex] < E_{n}| \hat{x} \hat{x} |E_{n}>= (\hat{x} |E_{n}>)^{t} \hat{x} |E_{n}>[/itex]
    using that x operator is self adjoint. The same goes for p... with the "t" I denoted the adjoint conjugate operation
     
  13. Jan 22, 2014 #12
    That is true, but ##\hat x## and ##\hat p## are operators.. so the x and p in the norm should have hats?

    ##<E_n|\hat x \hat x|E_n> = |\hat x|E_n>|^2## and ##<E_n|\hat p \hat p|E_n> = |\hat p|E_n>|^2##
     
  14. Jan 22, 2014 #13

    ChrisVer

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    they do have hats... the person who wrote the things in the image you posted, doesn't use hats so much...

    You can get a feeling there must be hats because otherwise there would be no reason to use the eigenvectors in kets... (he'd get 1)
     
  15. Jan 22, 2014 #14
    This clears things up a little, thanks!
     
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