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Fermionic oscillator contradiction

  1. Dec 21, 2009 #1
    This is a question with regard to a specific step in a problem. I don't think it is necessary to elaborate the whole problem.

    1. The problem statement, all variables and given/known data

    Resolve this apparent contradiction. I get two different answers for [tex]Na|n\rangle[/tex]

    2. Relevant equations

    For a fermionic oscillator, we have raising and lowering operators [tex]a^\dag[/tex] and [tex]a[/tex] that obey

    [tex]\{a,a^\dag\}=aa^\dag+a^\dag a=1[/tex]

    The last two amount to [tex]a^2=0[/tex] and [tex](a^\dag)^2=0[/tex].

    [tex]N=a^\dag a[/tex] is the number operator. It can be shown to have only two eigenvalues: 0 and +1.

    3. The attempt at a solution

    The effect of [tex]a[/tex] on an eigenstate of N is

    * if the eigenvalue is 1, a lowers it to 0, or
    * if the eigenvalue is 0, a annihilates it.

    Proof: Suppose [tex]N|n\rangle=n|n\rangle[/tex].

    [tex]Na|n\rangle = a^\dag a a|n\rangle = a^\dag(a^2|n\rangle)=0[/tex] since [tex]a^2=0[/tex]. So either the eigenvalue of [tex]a|n\rangle[/tex] is 0 or [tex]a|n\rangle=0[/tex].

    Ok. It isn't a proof. Just consistent. But what about this?

    [tex]Na|n\rangle=(a^\dag a) a|n\rangle[/tex]
    [tex]=(1\cdot a - a(a^\dag a))|n\rangle[/tex]

    So if [tex]n=0[/tex], the number eigenvalue of [tex]a|n\rangle[/tex] is 1? That can't be right. We get the same relation for the eigenvalue [tex]a^\dag|n\rangle[/tex] (which is what we actually expect).

    Can anyone see where I went wrong?
  2. jcsd
  3. Dec 21, 2009 #2


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    Homework Helper

    Well, if [itex]n=0[/itex], [itex]a\left| n\rangle[/itex] isn't even a state, so you can't talk about its eigenvalue with any sense. This is like an operator version of those "paradoxes" that try to show that 1=2 or something by hiding a division by zero under a bunch of algebra.
  4. Dec 21, 2009 #3
    Thanks for the quick response.

    Isn't the Hilbert space of states necessarily a vector space and have a zero (identity vector). When [tex]a[/tex] annihilates [tex]|0\rangle[/tex], the result is the zero (state) vector of the Hilbert space, isn't it? We write it as [tex]a|0\rangle=0[/tex] as if it were scalar, but it is a state vector, though a trivial one.
  5. Dec 21, 2009 #4


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    Homework Helper

    Well... OK, that makes sense. But wouldn't the zero state have all numbers as eigenvalues, since
    [tex]A\vert z\rangle = \vert z\rangle[/tex]
    for all operators [itex]A[/itex]? And (as a special case)
    [tex]n\vert z\rangle = \vert z\rangle[/tex]
    for all numbers [itex]n[/itex]? Then
    [tex]A\vert z\rangle = n\vert z\rangle[/tex]
    for any choice of [itex]A[/itex] and [itex]n[/itex]. Which would imply that yes, the number eigenvalue of [itex]a\vert 0\rangle[/itex] is 1, but it is also 0, and 7.5, and [itex]\pi[/itex]... so it's kind of meaningless.

    I guess the important thing to think about is this: in that last calculation in your original post, none of the individual expressions is inconsistent with the statement that [itex]Na\vert 0\rangle = 0[/itex].
  6. Dec 21, 2009 #5
    . Yep. Just like a N-vector with all zero components is a eigenvector of any NxN matrix, with eigenvalues consisting of all numbers. It's the trivial case.

    Bingo! I couldn't see the forest for the trees. Thanks!
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