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Average of Momentum for 1D Quantum Harmonic Oscillator

  1. Feb 15, 2016 #1
    For a 1D QHO we are given have function for ##t=0## and we are asked for expectation and variance of P at some time t.

    ##|\psi>=(1/\sqrt 2)(|n>+|n+1>)## Where n is an integer

    So my idea was to use Dirac operators ##\hat a## and ##\hat a^\dagger## and so I get the following solution

    ##<\hat P>=<\psi|\hat P|\psi>=i \sqrt {m\omega\hbar/2} <\psi|(\hat a^\dagger - \hat a^)|\psi>=##
    ##=i \sqrt {m\omega\hbar/2}(1/\sqrt 2) (1/\sqrt 2)(<n|+<n+1|(\hat a^\dagger-\hat a)|n>+|n+1>)##

    Knowing that
    ##\hat a^\dagger |\psi_n>= \sqrt {n+1} |\psi_{n+1}>##
    ##\hat a |\psi_n>= \sqrt {n} |\psi_{n-1}>##


    I eventually arrived at

    ##<\hat P>=i\sqrt{m\omega\hbar/8}(-\sqrt n <n|n> + \sqrt{n+1} <n+1|n+1>)##

    However the answer is supposed to be 0, did I do something wrong or could I perhaps make use of the fact that ##\sqrt{n+1}-\sqrt n ## for ##n\to\infty## equal to 0?
     
  2. jcsd
  3. Feb 15, 2016 #2

    blue_leaf77

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    It seems that you made a mistake when calculating ##\hat{a}|n+1\rangle##, what value did you get for this?
     
  4. Feb 15, 2016 #3
    That would be ##\sqrt n |n>## but I think I did forget the terms for ##\hat a^\dagger|n+1>## and <n+1| which would be ##\sqrt{n+2}|n+2>## and the corresponding bra.
     
  5. Feb 15, 2016 #4

    blue_leaf77

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    No, that's wrong. So, it's clear now that this is at least one of the mistakes.
    The above expression will give rise to a term proportional to ##|n+2\rangle## in the right part of the braket. Meanwhile in the left part of the braket there is no ##\langle n+2|##. This means, you can safely abandon the term ##\hat a^\dagger|n+1>##.
     
  6. Feb 15, 2016 #5
    I knew there was a reason I didn't include the |n+2> term in the first place. However what exactly is wrong about that other result?
    It follows from applying ##\hat a |\psi_n>= \sqrt {n} |\psi_{n-1}>## to |n+1>
     
  7. Feb 15, 2016 #6

    blue_leaf77

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    No, it doesn't. For instance, note that the index of the state in the left hand side is the same as the thing under the square root in the right hand side. Now what happen to this thing if the index of state in the left hand side is changed to ##n+1##?
     
  8. Feb 15, 2016 #7
    Oh I see, I found it weird because I have an example from class that makes uses it the way I said, but in the case we only had two possible n-states 0 and 1 instead of a generic integer n, I suposse that's what makes the difference.

    Okay, I redid the calculations and this pretty much breaks all parity (except for one that cancelled itself) and the result is 0. Thank you for the help.
     
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