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Showing there are no eigenvectors of the annhilation operator

  1. Nov 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Show there are no eigenvectors of [itex] a^{\dagger}[/itex] assuming the ground state |0> is the lowest energy state of the system.


    2. Relevant equations
    Coherent states of the SHO satisfy:
    a|z> = z|z>


    3. The attempt at a solution
    Based on the hint that was given (assume there is such an eigenvector like the coherent state above and expand the state in the basis |n>) I tried this, but it seems too simple.

    [itex]a^{\dagger}|0> = k|0> [/itex] (is this expansion in the |n> basis? or |0> basis?)

    then conjugate transpose both sides

    [itex] a<0|=\bar{k}<0|[/itex]

    the lhs vanishes as a is acting on the ground state ket

    Is this correct? Any help is greatly appreciated
     
  2. jcsd
  3. Nov 26, 2013 #2
    Hello Hakkinen!

    You are not taking the conjugate transpose correctly - the left hand side should read [itex]\langle 0 \lvert a[/itex]. That is, it should look like a row vector multiplying a matrix. The operator [itex]a[/itex] "acts to the left" as a creation operator, not as an annihilation operator.
     
  4. Nov 26, 2013 #3
    Thanks for the reply! sorry for the simple mistakes

    So is this not the correct approach to show this? If it is, should I start with [itex] a|0> = k|0>[/itex] and then conjugate, leaving the annihilation operator acting on the ground state bra?

    so [itex]0 = <0|\bar{k}[/itex]

    but this doesn't necessarily show there are no eigenvectors of the ann. operator does it?
     
  5. Nov 26, 2013 #4
    Well [itex]a\lvert 0 \rangle = 0[/itex], since [itex]\lvert 0 \rangle[/itex] is the ground state. We need a new name for the hypothetical eigenstate of [itex]a^{\dagger}[/itex], let's just call it [itex]\lvert \psi \rangle[/itex]. Then

    [itex]a^{\dagger}\lvert \psi \rangle = \lambda \lvert \psi \rangle,[/itex]

    where [itex]\lambda[/itex] is some number (the eigenvalue). Try expanding [itex]\lvert \psi \rangle[/itex] in terms of the [itex]\lvert n \rangle[/itex] and see what you get.
     
  6. Nov 26, 2013 #5
    Note that people usually call [itex]a[/itex] the annihilation operator, and [itex]a^\dagger[/itex] the creation operator. The reason it is worth to avoid confusion here, is that the annihilation operator [itex]a[/itex] (which satisfies [itex]a|0\rangle=0[/itex], where [itex]|0[/itex] is the ground state) actually has an eigenstate, while the creation operator does not.
     
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