# (Anti)commutation of creation and annhilation operators for different fermion fields

The Fourier expansion of the fermion field operator is such that

$$\hat\psi=\int\!d^3p\,\left[ f_b(p)\hat b(p) +f_d(p)\hat d^\dagger\!(p) \right] ~~,$$

for some sufficiently complicated $$f_b$$ and $$f_d$$. The operators $$\hat b^\dagger$$ and $$\hat b$$ create and destroy electrons respectively, and $$\hat d^\dagger,\,\hat d$$ work the same for positrons. I haven't seen it stated explicitly, but I assume both $$b$$ operators commute with each $$d$$ operator. What is an efficient way to show, for instance,

$$\hat b^\dagger\hat d^\dagger\big|0\big\rangle= \hat d^\dagger\hat b^\dagger\big|0\big\rangle~~?$$

Any demonstration that $$b$$ and $$d$$ commute would suffice if another is more direct, $$[\hat b,\hat d]=0$$ for instance.

• Careful, if you're talking about a fermion field you will be dealing with anti-commutators rather than commutators. – Charlie Dec 3 '20 at 12:35
• My question is whether or not $\hat b$ commutes with $\hat d$ and $\hat d^\dagger$. – hodop smith Dec 3 '20 at 12:44

Actually you (and Tony Zee) are correct and I am being an idiot. You do need a different $$b$$ and $$d$$! Obviously (in retrospect) $$\psi\ne \psi^\dagger$$. The $$d$$ and $$b$$ anticommute $$\{b_k,d_{k'}\}=0\\ \{b^\dagger_k,d^\dagger_{k'}\}=0\\ \{b^\dagger_k, d_{k'}\}= 0\\ \{d^\dagger_k, b_{k'}\}= 0\\ \{b^\dagger_k, b_{k'}\}= (2\pi)^32 E_k \delta^3(k-k')\\ \{d^\dagger_k, d_{k'}\}= (2\pi)^32 E_k \delta^3(k-k')$$
What you are calculating $$\langle P_1|\psi\gamma^\mu \psi|P_2\rangle$$ should end up as something like $$v^\dagger(P_1)\gamma^0 \gamma^\mu v(P_2)$$.
• How about $\{b_k,b^\dagger_{k'}\}= 2E_k \delta^3(k-k')$, and $\{b_k,b_{k'}\}=0$? As I said, there are no $d$'s as for electron and positrons they are the same as the $b$'s, but with $b$ and $b^\dagger$ interchanged. – mike stone Dec 3 '20 at 13:00
• No. They anticommute, and moving a $b$ past a $b^\dagger$ gives you an additional c-number. – mike stone Dec 3 '20 at 13:38
• There is no obvious reason for it to be zero. Who says that it is? It is not a term that arises from $\bar\psi \gamma^\mu \psi$ as it includes four Fermi fields rather than two. – mike stone Dec 3 '20 at 13:48
• Note that the expression for the electron field $\psi$ in your original question is incorrect. The $d^\dagger$ should be a $b^\dagger$. – mike stone Dec 3 '20 at 13:50