- #1

pellman

- 684

- 5

## Homework Statement

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

## Homework 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]

[tex]\{a,a\}=0[/tex]

[tex]\{a^\dag,a^\dag\}=0[/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.

## 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]=(\{a,a^\dag\}-aa^\dag)a|n\rangle[/tex]

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

[tex]=a(1-N)|n\rangle[/tex]

[tex]=(1-n)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?