1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted