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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?

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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 $$[a_k^\dagger, a_q] = 2a_k^\dagger a_q - \delta_{kq}$$

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

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