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Expectation value, harmonic oscillator

  1. May 28, 2005 #1

    I have to find the expectation values of xp and px for nth energy eigenstate in the 1-d harmonic oscillator. If I know <xp> I can immediately find <px>since [x,p]=ih. I use the ladder operators [tex]a_{\pm}=\tfrac1{\sqrt{2\hslash m\omega}}(\mp ip+m\omega x)[/tex] to find <xp>, but I get a complex value, <xp>=ih/2. It doesn't seem right in the context of the rest of the exercise...
    Last edited: May 28, 2005
  2. jcsd
  3. May 28, 2005 #2


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    Why not...?Why should it be real...?:wink:

  4. May 28, 2005 #3


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    What's your expression for xp in terms of the ladder operators?
  5. May 28, 2005 #4
    Well, xp is not hermitian, I see your point, dexter. My expression for xp is [tex]i\hslash/2({a_+}^2 + {a_-}^2 + 1)[/tex].
  6. May 28, 2005 #5


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  7. Nov 28, 2009 #6
    Could I be so bold as to ask how you got that expression for xp ?
  8. Nov 28, 2009 #7
    I got it. I convert p to ladder operators using this formula:

    [tex]\hat a_\pm=\frac{1}{\sqrt{2mh\omega}}(m\omega \hat x \mp i\hat p)[/tex]

    Just isolate p. Then we use this formula for x in ladder operators:

    [tex]\hat{x}=\sqrt{\frac{\hbar}{2m\omega}}(\hat a_++\hat a_-)[/tex]

    Then we simply multiply x and p.

    But what i get is a little bit different than above:

    [tex]\hat x \hat p=\frac{i\hbar}{2}(\hat a_+^2-\nf\hat a_-^2+1)[/tex]

    (note the - in the last formula)

    But the final result is the same
    [tex]<\hat x \hat p> = \frac{i\hbar}{2}[/tex]
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