0
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ 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
3
$\begingroup$

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 $v^\dagger(P_1)\gamma^0 \gamma^\mu v(P_2)$.

$\endgroup$
11
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.