Fermionic oscillator contradiction

[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!