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Determine the mean of the position <x> in a state

  1. Aug 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Hello everyone...

    I'm kinda stuck with a problem I'm trying to do.
    The problem states:

    Express the operator [itex]\hat{x}[/itex] by the ladder operators [itex]a_{+}[/itex] and [itex]a_{-}[/itex], and determine the mean of the position [itex]\left\langle x \right\rangle[/itex] in the state [itex]\left| \psi \right\rangle[/itex].


    2. Relevant equations
    [tex]\left| \psi \right\rangle = \frac{1}{\sqrt{2}}(\left| 3 \right\rangle + \left| 2 \right\rangle)[/tex]

    Hamiltonian for a one dimensional harmonic oscillator:

    [tex]\hat{H} = \frac{\hat{p}^{2}}{2m}+\frac{1}{2}mw^{2}\hat{x}^{2},[/tex]
    where [itex]w[/itex] is the oscillators frequency, and [itex]x[/itex] and [itex]p[/itex] are the operators for position and momentum. The normalized energy eigenfunctions for [itex]H[/itex] is denoted [itex]\left| n \right\rangle[/itex], where [itex]n = 0,1,2,....[/itex] so that:

    [tex]\hat{H}\left| n \right\rangle = (n + \frac{1}{2})\hbarw\left| \psi \right\rangle[/tex]


    3. The attempt at a solution
    The first is easy, since:

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

    My problem is finding the mean of the position.

    I tried to do it like this:

    [tex]\left\langle x \right\rangle = \sqrt{\frac{\hbar}{2mw}}\int \psi^{*}_{n}(a_{+} + a_{-})\psi_{n} dx[/tex]

    And that didn't go well. It got very confusing, so I was not sure if I was on the right track or not. So here I am.

    I know the answer should be:

    [tex]\left\langle x \right\rangle = \sqrt{\frac{3}{2}}\sqrt{\frac{\hbar}{mw}},[/tex]
    but again, I'm kinda lost atm.

    So I was hoping any of you could give me a clue. Something in my heads tells me it's pretty simple, but I really can't figure it out right now, so :)


    Regards
     
  2. jcsd
  3. Aug 11, 2010 #2

    diazona

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    Hint: what are
    [tex]a_+\lvert n\rangle[/tex]
    and
    [tex]a_-\lvert n\rangle[/tex]
    equal to?
     
  4. Aug 11, 2010 #3
    [tex]a_+ \lvert n\rangle = \sqrt{n+1} \lvert n+1\rangle[/tex]
    and
    [tex]a_- \lvert n\rangle = \sqrt{n} \lvert n-1\rangle[/tex]

    I think I did something like that, but as I said. I may have messed it up. But I was right with the thought I had ? I just need to redo the math perhaps ?
     
  5. Aug 12, 2010 #4

    vela

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    The idea is to do the problem without resorting to doing any integrals. Just use the orthogonality of the eigenstates. For example,

    [tex]\langle 1|a_+|0 \rangle = \langle 1|\sqrt{1}|1\rangle = 1[/tex]
     
  6. Aug 12, 2010 #5
    Hmmm... I think I may have it now. But what if the ladder operators are used on state |3> fx. ? At least the a+ operator. Then it would raise it to state 4, but that state isn't represented here, so will that just equal 0, or... ?
     
  7. Aug 12, 2010 #6

    vela

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    Yes, that term will drop out.
     
  8. Aug 12, 2010 #7
    Thank you very much. I thought it was what I did to start with, but I just messed it up.
    But I got it now :)

    Thank you.
     
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