Fermion creation and annihilation operators

302
0
Hi.

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

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

Does this not imply that [itex][c^\dagger,c]=-1[/itex] and consequently that BOTH the commutator and anti-commutator of [itex]c[/itex] and [itex]c^\dagger[/itex] is equal to unity?
 
525
6
No, it does not imply that. Namely, the commutator is:

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

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

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

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.
 
302
0
I see. Thanks.
 

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