Second Quantization: Explaining c^\dagger_ic_j = \delta_{i,j}c_jc^\dagger_i

[Niles]

Hi

Say I have the following two fermionic creation/annihilation operators

$$c^\dagger_ic_j$$

1) Yesterday, my lecturer said that the following is valid

$$c^\dagger_ic_j = \delta_{i,j}c_jc^\dagger_i$$

Can you guys explain to me, where this formula comes from? I originally thought that it was one of the anti-commutator relations, but it cannot come from there.
2) Say I have en expression of the form

$$c_{k+q}^\dagger c_{k-q'} c^\dagger_{k'-q'}c_{k'}$$

If the operators are fermionic, then if I want to have all dagger-operators on the LHS and non-dagger operators on the RHS, then do I have to use anti-commutator relatations in order to rewrite the expression?

Likewise, if they were bosonic operators, then I would have to use commutator relations in order to rewrite the expression?Niles.

 

[Avodyne]

1) Your lecturer is wrong.

2) Yes.

 

[Niles]

Thank you, it is kind of you to answer so quickly.

 

[A. Neumaier]

$$c^\dagger_ic_j = \delta_{i,j}c_jc^\dagger_i$$

probably he meant c^*_ic_j = \delta_{i,j}+c_jc^*_i
which is the CCR for bosons.

 

[Niles]

Thanks. I have another question related to fermionic operators, so I'll just ask it here. It is regarding the relation

$$\delta (t - t')\left\langle {\left\{ {c_i (t),c_i^\dag (t')} \right\}} \right\rangle = \delta (t - t')$$

The curly brackets denote an anti-commutator. Is there an easy way of showing this? The way I would show this is to look at the case t = t' and the case t != t'.

 

[A. Neumaier]

probably he meant c^*_ic_j = \delta_{i,j}+c_jc^*_i
which is the CCR for bosons.

Sorry, this is not quite true; $$c^*_ic_j = -\delta_{i,j}+c_jc^*_i$$

 

[A. Neumaier]

Thanks. I have another question related to fermionic operators, so I'll just ask it here. It is regarding the relation

$$\delta (t - t')\left\langle {\left\{ {c_i (t),c_i^\dag (t')} \right\}} \right\rangle = \delta (t - t')$$

The curly brackets denote an anti-commutator. Is there an easy way of showing this? The way I would show this is to look at the case t = t' and the case t != t'.

You cannot distinguish cases in this way since this is meant in the sense of distributions. Thus you need to multiply both sides by a function f(t,t'), integrate over t and t', and simplify before you can interpret the statement.