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Checking that a coherent state is an eigenfunction of an operator

  1. Oct 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Hey guys, I'll type this thing up in Word.

    http://imageshack.com/a/img716/8219/wycz.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 30, 2013 #2

    hilbert2

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    What is the physical system in question, is it a harmonic oscillator? If it is, then the lowering operator acts on the energy eigenstates as ##\hat{a}\psi_{n}(x)=\sqrt{n}\psi_{n-1}(x)##. This is the only info you need in order to solve the problem.
     
  4. Oct 30, 2013 #3
    The question doesn't say that it is a harmonic oscillator, but it does say that "these states closely resemble classical particles" so I think you're right. If I do as you say, I end up with this:
    http://imageshack.com/a/img69/8361/loix.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  5. Oct 30, 2013 #4

    hilbert2

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    The lowering operator annihilates the ground state: ##\hat{a}\psi_{0}(x)=0##. Also, you can change the variable over which the summation is, e.g. ##k=n-1##. That way you should be able to show that acting on the coherent state with the lowering operator is equivalent to multiplying with a constant.
     
  6. Oct 30, 2013 #5
    Okay I'm lost...T_T Is this what you mean?
    http://imageshack.com/a/img703/889/xvhg.jpg [Broken]

    I feel kinda stupid now ¬_¬
     
    Last edited by a moderator: May 6, 2017
  7. Oct 30, 2013 #6

    hilbert2

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    After you change the index, the summation should be from ##k=0## to ##k=\infty## and the exponent of ##\lambda## becomes ##k+1##...
     
  8. Oct 30, 2013 #7
    So you mean this?
    http://imageshack.com/a/img842/8932/5a4q.jpg [Broken]

    which means that the eigenvalue is just...[itex]λ[/itex]?
     
    Last edited by a moderator: May 6, 2017
  9. Oct 30, 2013 #8

    hilbert2

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    Yes, that's the correct answer.
     
  10. Oct 30, 2013 #9
    Wow thank you :d !!!
     
  11. Oct 30, 2013 #10
    that was supposed to be all caps but i guess it got filtered :(
     
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