# Commutator of fermionic operators

The fermionic creation/annihilation operators are defined by the anti-commutation relations:

$$\{a_k^{\dagger},a_q^{\dagger}\} = 0 = \{a_k,a_q \}$$ $$\{a_k^{\dagger},a_q\} = \delta_{kq} \, .$$

I want to know what their commutation relations are:

$$[a_k^{\dagger},a_q^{\dagger}] = ?$$ $$[a_k,a_q] = ?$$ $$[a_k^{\dagger},a_q] = ?$$

My first thought was:

$$\{a_k,a_q \} = a_ka_q + a_qa_k = 0 \;\Longrightarrow\; a_ka_q-a_qa_k = -2a_qa_k = [a_k,a_q]$$

But this is obviously not the whole story, since $[a_k,a_k]$ should equal $0$, rather than $-2a_ka_k$, so there should be a $\delta_{kq}$ that I'm missing [unless it's okay to assume that $a_ka_k = 0$ (because of Pauli exclusion principle?). Where did the $\delta$ go? What are the correct commutation relations?

## 1 Answer

your result is correct $$[a_k, a_q] = -2 a_k a_q$$ which is consistent with $$[a_k, a_k ]= - 2 a_k a_k = 0$$ because $$a_k a_k = \frac{1}{2}\{a_k, a_k \} = 0$$

And in general you can use $$[A,B] = 2AB - \{A,B\}$$ which would also give $$[a_k^\dagger, a_q] = 2a_k^\dagger a_q - \delta_{kq}$$

• I don't agree with your last two results. Shouldn't it be: $[A,B]=AB-BA=\{A,B\}-2BA$, so $[a_k^{\dagger},a_q]=\delta_{kq}-2a_qa_k^{\dagger}$? – alexvas Apr 16 '15 at 19:04
• you can easily show that the two results are equivalent – Ali Moh Apr 16 '15 at 19:09