# Fermionic creation and annihilation

1. Jan 3, 2014

### Abigale

Hi guys,
I consider fermionic holes and I know the creation and annihilation operators of them.

I have shown that $\vec{S}= \sum\limits_{k \nu \mu} \frac{1}{2} \bf{c}_{k \mu}^{\dagger} \vec{\sigma}_{\mu \nu} \bf{c}_{k \nu} =...= \sum\limits_{k \nu \mu} \frac{1}{2} \bf{h}_{-k \mu}^{\dagger} \vec{\sigma}_{\mu \nu} \bf{h}_{ -k \nu}^{ }$.
But I have to show that $\vec{S}=\sum\limits_{k \nu \mu} \frac{1}{2} \bf{h}_{k \mu}^{\dagger} \vec{\sigma}_{\mu \nu} \bf{h}_{ k \nu}^{ }$.

It would be nice if somebody has an idea how I can get rid of the minus sign.

THX
Abby

Is the sum maybe from minus infinity to plus infinity? Then I think I got it.

Last edited: Jan 3, 2014
2. Jan 3, 2014

### George Jones

Staff Emeritus
Yes, if there is no hard cut-off on the momenta.

3. Jan 3, 2014

### Abigale

So because k is from infinity to infinity I can write:
$\vec{S}= \sum\limits_{k \nu \mu} \frac{1}{2} \bf{c}_{k \mu}^{\dagger} \vec{\sigma}_{\mu \nu} \bf{c}_{k \nu} =...= \sum\limits_{k \nu \mu} \frac{1}{2} \bf{h}_{-k \mu}^{\dagger} \vec{\sigma}_{\mu \nu} \bf{h}_{ -k \nu}^{ } = \sum\limits_{-k \nu \mu} \frac{1}{2} \bf{h}_{k \mu}^{\dagger} \vec{\sigma}_{\mu \nu} \bf{h}_{ k \nu}^{ } \stackrel{(1)}{=} \sum\limits_{k \nu \mu} \frac{1}{2} \bf{h}_{k \mu}^{\dagger} \vec{\sigma}_{\mu \nu} \bf{h}_{ k \nu}^{ }$.

I would say (1) is true because if I regard the k from "-"infinity to "+"infinity:

$\sum\limits_{-k}^{}f(k)=\sum\limits_{k}^{}f(-k)= \sum\limits_{k}^{}f(k)$

Could somebody check my thoughts? Would be nice!!!
THX Abby