Fermion creation and annihilation operators

[daudaudaudau]

Hi.

If $c$ and $c^\dagger$ are fermion annihilation and creation operators, respectively, we know that $cc^\dagger+c^\dagger c=1$ and $cc=0$ and $c^\dagger c^\dagger=0$. I can use this to show the following
$$[c^\dagger c,c]=c^\dagger cc-c c^\dagger c=-cc^\dagger c=-c(1-cc^\dagger)=-c$$

But on the other hand I have
$$[c^\dagger c,c]=c^\dagger[c,c]+[c^\dagger,c]c=[c^\dagger,c]c$$

Does this not imply that $[c^\dagger,c]=-1$ and consequently that BOTH the commutator and anti-commutator of $c$ and $c^\dagger$ is equal to unity?

 

[xepma]

No, it does not imply that. Namely, the commutator is:

$$[c^\dag , c] = c^\dag c - cc^\dag = c^\dag c - 1 + c^\dag c = 2c^\dag c - 1$$

As you can see, the first term gives zero when you act with this commutator on the operator $$c$$. In other words,

$$[c^\dag , c]c = (2c^\dag c - 1)c = -c$$

Which is your result. But for operators you cannot use that if AB = CB, then C = A. The reason is that the operator B is not always invertible (as in this case), but is nilpotent instead. So you have to be careful with these types of manipulations.

 

[daudaudaudau]

I see. Thanks.