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Expectation values calc.

  1. Jan 3, 2010 #1
    Hello,
    Can someone explain to me how the expectation values are calculated in the following picture:
    untitled.JPG
    I mean , What did they do after the brackets? What did they multiply with what?
    thanks
     
  2. jcsd
  3. Jan 3, 2010 #2
    what kind of states are those [tex]|z>[/tex] states?

    If those were enrgy-eigen-states of the harmonic oscillator, [tex]<z|a^\dagger a|z> \neq |z|^2[/tex].

    [tex]<z|a^\dagger a|z> = z[/tex] would be correct.
     
  4. Jan 3, 2010 #3
    Yes, I'm talking about harmonic oscillator. I don't understand how did they get the avg expression with the zs. (from the question mark to the right)

    For exmple, what's the value of <z|a|z> , <z|a+|z> etc. and why?
     
  5. Jan 3, 2010 #4
    <z|a|z> =<z|a+|z>=0

    since a and a+ are creation and annihilation operators:
    a|z> is proportional to |z-1>
    a+|z> is proportional to |z+1>

    for further explanation please see this pdf file: http://iftia9.univ.gda.pl/~sjk/skok/ou01.pdf [Broken]
     
    Last edited by a moderator: May 4, 2017
  6. Jan 3, 2010 #5
    Thank but I don't see there how to calculate mean value (expectation value) of x and p for exmple. Can you try to explain to me how did they got what they wrote in the picture I attached earlier?

    should I calculate them with integral accordering to the formula <a|x|a>=integral(a*xa)?
    But it seems like they did it in a much simpler and easy way.
     
  7. Jan 3, 2010 #6
    You can write x and p in terms of a and a+

    since by definition:

    [tex]a := \sqrt{\frac{m\omega}{2\hbar}}(x+\frac{ip}{m\omega})[/tex]
    and
    [tex]a^\dagger := \sqrt{\frac{m\omega}{2\hbar}}(x-\frac{ip}{m\omega})[/tex]

    You can invert these two formulas to get x an p:

    [tex]x = \sqrt{\frac{\hbar}{2m\omega}}(a+a^\dagger)[/tex]
    [tex]p = \sqrt{\frac{\hbar m \omega}{2}}i(a^\dagger-a)[/tex]

    Now you can use formulas 8.38a and 8.38b from the pdf-file to calculate x and p.
     
  8. Jan 3, 2010 #7
    o.k I'll try it, and is true :
    untitled.JPG
     
  9. Jan 3, 2010 #8
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