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Finding this eigenstate

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Given the a|n> = α|n-1>, show that α = √n :

    wa1pjt.png

    2. Relevant equations



    3. The attempt at a solution

    [tex]<n|a^{+}\hat {a}|n> = \alpha <n|a^{+}|n-1> = | \hat a|n>|^2 [/tex]

    [tex] \alpha = \frac{<n|a^{+}\hat {a}|n>}{<n|a^{+}|n-1>}[/tex]

    Taking the complex conjugate of both sides:

    [tex] \alpha* = \frac{<n|a^{+}\hat {a}|n>}{\alpha<n-1|n-1>} [/tex]
    [tex] |\alpha^2| = \frac{<n|a^{+}\hat {a}|n>}{<n-1|n-1>} [/tex]

    Where ##\hat {a} a^{+} = \frac{1}{2m\omega} \left( (m\omega x)^2 + p^2 + im\omega [\hat{x},\hat{p}] \right) ##
     
    Last edited: Jan 21, 2014
  2. jcsd
  3. Jan 21, 2014 #2

    BruceW

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    nice work. I think you're getting there. although remember that ##\alpha## isn't necessarily real, so when you took the complex conjugate of both sides, you should also take complex conjugate of ##\alpha##. For the final equation you wrote down, I think that is the right idea. But it's definitely missing some ##\hbar## and the square root should not still be there. it's just a minor mistake.
     
  4. Jan 21, 2014 #3
    The problem is, I have no idea how to proceed now..
     
  5. Jan 21, 2014 #4

    BruceW

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    once you've fixed the minor mistake, you have the operator ##aa^\dagger##. what other operator does this operator look like? hint: the physical model of this system is simple harmonic motion.
     
  6. Jan 21, 2014 #5
    Okay, The square root has gone away, due to it appearing twice from a and a+ and i've fixed the square term in the bracket.
     
  7. Jan 21, 2014 #6
    USing ##[\hat x,\hat p] = i\hbar##

    [tex]\hat a a^{+} = \frac{\hat H}{\hbar \omega} - \frac{1}{2} [/tex]

    [tex] |\alpha^2| = \frac{<n| \frac{\hat H}{\hbar \omega} - \frac{1}{2}|n>}{<n-1|n-1>}[/tex]

    How do I proceed from here?
     
  8. Jan 21, 2014 #7

    BruceW

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    hmm. wait hold on, since they ask you to consider ##|\hat{a}|n\rangle |^2## I think you should use the operator that you were originally working with, ##\hat{a}^\dagger \hat{a}##

    edit: sorry about that, the rest of your working looks good. you've got the right idea.
     
  9. Jan 21, 2014 #8
    How do I evaluate the inner product:

    [tex] |\alpha^2| = \frac{<n| \frac{\hat H}{\hbar \omega} - \frac{1}{2}|n>}{<n-1|n-1>} = \frac{<n|a^{+}\hat {a}|n>}{<n-1|n-1>}[/tex]


    It should give ##n## on the RHS.
     
  10. Jan 21, 2014 #9

    BruceW

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    you're almost there now. In the question, they tell you ##E_n = (n+1/2)\hbar \omega## they mean this is the energy of the ##|n\rangle## eigenstate. so this should be enough to evaluate that inner product.
     
  11. Jan 21, 2014 #10

    How do I evaluate the numerator?

    [tex] <n| \frac{\hat H}{\hbar \omega} - \frac{1}{2}|n> [/tex]
    [tex]= \frac{1}{\hbar \omega} <n|\hat H|n> - \frac{1}{2}<n|n> [/tex]
    [tex]= \frac{1}{\hbar \omega} <n|\hat H|n> - \frac{1}{2}<n|n> [/tex]
    [tex] = \frac{E_n}{\hbar \omega}<n|n> - \frac{1}{2}<n|n>[/tex]
    [tex] = \left (\frac{E_n}{\hbar \omega} - \frac{1}{2} \right) <n|n>[/tex]

    Thus,
    [tex]|\alpha|^2 = \left (\frac{E_n}{\hbar \omega} - \frac{1}{2} \right)\frac{ <n|n>}{<n-1|n-1>} [/tex]
    [tex] |\alpha|^2 = n \frac{ <n|n>}{<n-1|n-1>} [/tex]
     
    Last edited: Jan 21, 2014
  12. Jan 21, 2014 #11

    BruceW

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    yep. that is good so far for the numerator. But the denominator is not (n-1)2. each of the ##|n\rangle## are states, right? so what (conventionally), is the inner product of the state with itself?

    edit: and now, there is a similar question for what is left over in the numerator.
     
  13. Jan 21, 2014 #12
    Can I assume they are linearly independent, and have norm = 1?
     
  14. Jan 21, 2014 #13

    BruceW

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    yeah! I guess it is not obvious really. But I think in these kinds of problems, you can assume that if they give you a set of states, those states will be orthonormal. (unless they say otherwise).
     
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