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.

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    $\begingroup$ Careful, if you're talking about a fermion field you will be dealing with anti-commutators rather than commutators. $\endgroup$ – Charlie Dec 3 '20 at 12:35
  • $\begingroup$ My question is whether or not $\hat b$ commutes with $\hat d$ and $\hat d^\dagger$. $\endgroup$ – 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 $v^\dagger(P_1)\gamma^0 \gamma^\mu v(P_2)$.

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    $\begingroup$ 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. $\endgroup$ – mike stone Dec 3 '20 at 13:00
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    $\begingroup$ Yes, You will need to include a charge conjugation matrix of course. You will need to read about that in a field-theory textbook. $\endgroup$ – mike stone Dec 3 '20 at 13:21
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    $\begingroup$ No. They anticommute, and moving a $b$ past a $b^\dagger$ gives you an additional c-number. $\endgroup$ – mike stone Dec 3 '20 at 13:38
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    $\begingroup$ 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. $\endgroup$ – mike stone Dec 3 '20 at 13:48
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    $\begingroup$ Note that the expression for the electron field $\psi$ in your original question is incorrect. The $d^\dagger$ should be a $b^\dagger$. $\endgroup$ – mike stone Dec 3 '20 at 13:50

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