Why Do I Get Two Different Answers for Na|n\rangle in a Fermionic Oscillator?

[pellman]

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.

Homework Statement

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

Homework Equations

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

\{a,a^\dag\}=aa^\dag+a^\dag a=1
\{a,a\}=0
\{a^\dag,a^\dag\}=0.

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

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

The Attempt at a Solution

The effect of a 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 N|n\rangle=n|n\rangle.

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

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

Na|n\rangle=(a^\dag a) a|n\rangle
=(\{a,a^\dag\}-aa^\dag)a|n\rangle
=(1\cdot a - a(a^\dag a))|n\rangle
=a(1-N)|n\rangle
=(1-n)a|n\rangle

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

Can anyone see where I went wrong?

 

[diazona]

Well, if n=0, a\left| n\rangle 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.

 

[pellman]

Thanks for the quick response.

Well, if n=0, a\left| n\rangle 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.

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

 

[diazona]

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

 

[pellman]

Well... OK, that makes sense. But wouldn't the zero state have all numbers as eigenvalues,

. 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.

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 Na\vert 0\rangle = 0.

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