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Harmonic Oscillator

  1. Dec 18, 2013 #1
    1. The problem statement, all variables and given/known data

    I need to show that for an eigen state of 1D harmonic oscillator the expectation values of the position X is Zero.

    2. Relevant equations

    Using

    a+=[itex]\frac{1}{\sqrt{2mhw}}[/itex]([itex]\hat{Px}[/itex]+iwm[itex]\hat{x}[/itex])
    a-=[itex]\frac{1}{\sqrt{2mhw}}[/itex]([itex]\hat{Px}[/itex]-iwm[itex]\hat{x}[/itex])

    3. The attempt at a solution

    <x>=<n|x|n>=[itex]\sqrt{\frac{h}{2mw}}[/itex]<n|(a-+a+)|n=??
     
  2. jcsd
  3. Dec 18, 2013 #2

    DrClaude

    User Avatar

    Staff: Mentor

    That doens't look like the standard form, see http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator#Ladder_operator_method

    You won't get ##\hat{x}## from ##a^+ + a^-## with those.

    What is your question exactly? Can you figure out what
    $$
    \left( a^- + a^+ \right) \left|n\right\rangle
    $$
    results in?
     
  4. Dec 19, 2013 #3
    I used the book "Fundamentals of Quantum Mechanics for Solid State Electronics Optics" - C.Tang
    Here is the link to the book :http://en.bookfi.org/book/1308543 [Broken] (page 79 (pdf)).


    Using the a+ and a- in the link you gave I get:
    <n|x|n>=[itex]\sqrt{\frac{h}{2mw}}[/itex]<n|a-+a+|n>

    Now using:
    a+|n>=[itex]\sqrt{n+1}[/itex]|n+1>
    <n|a-=a+|n>=[itex]\sqrt{n+1}[/itex]|n+1>

    I get:
    <n|x|n>=2*[itex]\sqrt{\frac{h}{2mw}}[/itex]{<n|[itex]\sqrt{n+1}[/itex]|n+1>}

    Now can I say that since all {n} vectors are orthogonal the expression<n|[itex]\sqrt{n+1}[/itex]|n+1>=[itex]\sqrt{n+1}[/itex]<n|n+1>=0?

    Thanks
     
    Last edited by a moderator: May 6, 2017
  5. Dec 19, 2013 #4

    vela

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yes, you can.
     
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