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Creation operator and eigenvalues

  1. Jun 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove that the creation operator [itex]a_+ [/itex] has no eigenvalues, for instance in the [itex]\vert n \rangle [/itex].

    2. Relevant equations
    Action of [itex]a_+ [/itex] in a harmonic oscillator eigenket [itex]\vert n \rangle [/itex]:
    [itex] a_+\vert n \rangle =\vert n +1\rangle [/itex]

    3. The attempt at a solution
    Calling a the eigenvalues of [itex]a_+ [/itex]
    [itex]a_+ \vert \Psi \rangle = a \vert \Psi \rangle = a \sum c_n \vert n \rangle = \sum a c_n \vert n \rangle[/itex]
    [itex]a_+ \vert \Psi \rangle = a_+ \sum c_n \vert n \rangle = \sum c_n a_+ \vert n \rangle = \sum c_n\vert n+1\rangle = \sum c_{n-1}\vert n\rangle[/itex]

    Equating both
    [itex]a_+ \vert \Psi \rangle = \sum a c_n \vert n \rangle= \sum c_{n-1}\vert n\rangle[/itex]

    We have
    [itex]a c_n = c_{n-1}[/itex].

    I think I can take the a factor out and then claim that eigenkets have to be linearly dependent, so their coefficients cannot be proportional to each other.

    However, I am not sure that this does prove that the creation operatro has no eigenvalues.
     
    Last edited: Jun 28, 2014
  2. jcsd
  3. Jun 28, 2014 #2
    Hi,
    ##a=1=c_n=c_{n-1}=...## is a solution of the equations for the coefficients. What happens when ##a_+## is applied on ##|\Psi>=|n>+|n-1>+...+|n-m>##?
     
  4. Jun 28, 2014 #3
    We get ##a_+|\Psi\rangle=|n+1>+|n\rangle+...+|n-m+1\rangle##. If the set of eigenvalues for the oscilator Hamiltonian is infinite (and it is, right?) the sum remains the same except for ##\vert 0 \rangle##, which becomes ##\vert 1 \rangle ##. Since this is the new ground state we have transformed the oscillator eigenkets into eigenkets for an oscillator with ##\hbar \omega ## more energy. Is that so?

    What now? I don't know what do you mean.
     
  5. Jun 28, 2014 #4
    Yes but you are more interested in eigenvector to express ##|\Psi>##.

    Yes and what is ##a|\Psi>## with a=1 in the basis ##\{|i>\}##?

    No, it's just some algebra to show that there is no vector ##|\Psi>## such that ##a_+|\Psi>=a|\Psi>##. The example with a=1 is just a simple example to see what happens.

    If you add the bounds to the sums in your attempt at a solution you will see that there is a problem.
     
  6. Jun 28, 2014 #5
    What do you mean with the bounds?
     
  7. Jun 28, 2014 #6
    Sorry for the delay. I will try another way: you expended ##|\Psi>## on the basis ##\{|n>:n=0,1,...\}##. On one hand you multiplied it by a and you obtained ##a|\Psi>## on that basis. On the other hand you applied ##a_+## on it and you obtained ##a_+|\Psi>## on the same basis ##\{|n>:n=0,1,...\}##. For the two vectors obtained to be equal the coefficients must be equal for all basis vectors. Your result ##a c_n=c_{n-1}## cannot hold for all n, do you see why?
     
  8. Jun 28, 2014 #7

    vela

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    Try writing out the first few terms of each summation.
     
  9. Jun 28, 2014 #8
    Wow, why do you apologyze? You're helping me! I really appreciate it, no matter how long does it take.
    I've just realized that ##\hat{a}^{\dagger}\vert n \rangle = \sqrt{n} \vert n \rangle ##, so my recursion formula should've been ## c_n a = c_{n-1} \sqrt{n} ##.

    Anyway, I don't see what's the deal with it. Is it with the ground state? It is 0 in one expansion and ##c_0 a ## in the other one.
     
  10. Jun 28, 2014 #9
    Exactly. And they have to be the same, so...?
     
  11. Jun 28, 2014 #10
    So ##c_0## is zero and so are all others. Thank you!
     
  12. Jun 28, 2014 #11
    Some clean up is needed(what if a=0? what if ##c_0##,...,##c_i##=0?) but you get it.
     
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