If we let field operators $$\psi(x)=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}e^{ip\cdot x}\sum_s(a^s_pu^s(p)+b^{s}_{-p}v^s(-p)).$$ Then the commutator of field operators will be $$[\psi,\psi^{\dagger}]=\int\frac{d^3pd^3q}{(2\pi)^6}\frac{1}{\sqrt{4E_pE_q}}e^{i(p\cdot x-q\cdot y)}\sum_{r,s}(a^r_pu^r(p)a^{s\dagger}_qu^{s\dagger}(q)-a^{s\dagger}_qu^{s\dagger}(q)a^r_pu^r(p)+b^r_{-p}v^r(-p)b^{s\dagger}_{-q}v^{s\dagger}(-q)-b^{s\dagger}_{-q}v^{s\dagger}(-q)b^r_{-p}v^r(-p))$$ How does it become $$\int\frac{d^3pd^3q}{(2\pi)^6}\frac{1}{\sqrt{4E_pE_q}}e^{i(p\cdot x-q\cdot y)}\sum_{r,s}([a^r_p,a^{s\dagger}_q]u^r(p)u^{s\dagger}(q)+[b^r_{-p},b^{s\dagger}_{-q}]v^r(-p)v^{s\dagger}(-q))?$$
5
-
1$\begingroup$ Since the Dirac field obeys fermi-dirac statistics, one would expect to see an anticommutator. Why are you considering a commutator as one would use with bose-einstein statistics? BTW, the $u$'s and $v$'s are not operators. So, you can factor them out and the rest would become commutators. $\endgroup$– flippiefanusCommented Jul 9, 2021 at 3:07
-
$\begingroup$ @flippiefanus Is the order of $u$ and $u^{\dagger}$not important? $\endgroup$– AsungCommented Jul 9, 2021 at 5:34
-
1$\begingroup$ @Asung - no. $u$ and $u^\dagger$ are just numbers you can move them around as you wish $\endgroup$– PraharCommented Jul 9, 2021 at 6:41
-
$\begingroup$ @Prahar Mitra I thought $u$ as column vector, and $u^{\dagger}$ as row vector. Then is $uu^{\dagger}=u^{\dagger}u$ valid? $\endgroup$– AsungCommented Jul 9, 2021 at 8:08
-
2$\begingroup$ @Asung - no! When I say you can move them around as you wish, I meant as operators. You are getting a bit confused because you are not putting the explicit matrix indices for the fields. The anti-commutator that you need to be evaluating is $\{ \psi_i , \psi_j^\dagger \}$. The same matrix index is there in $u_i$ and $v_i$ as well. $\endgroup$– PraharCommented Jul 9, 2021 at 11:21
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
While I'm not sure why you would evaluate the commutator of a fermionic field, the process is the same for the anticommutator so the next reasoning is general.
If you take the form of the field you gave, then the commutator (taken at two different spacetime points)
$$\left[\psi(x),\psi^\dagger(y)\right] = \left[\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}e^{ip\cdot x}\sum_s(a^s_pu^s(p)+b^{s}_{-p}v^s(-p))\,,\,\int\frac{d^3q}{(2\pi)^3}\frac{1}{\sqrt{2E_q}}e^{-iq\cdot y}\sum_{s^\prime}(a^{s^\prime\dagger}_qu^{s^\prime\dagger}(q)+b^{s^\prime\dagger}_{-q}v^{s^\prime\dagger}(-q)). \right]$$
the only relevant quantity for the (anti-)commutator are the creation and annihilation operators, the spinor fields you can think of as just "numbers". Then
$$ = \int\frac{d^3p}{(2\pi)^3}\frac{d^3q}{(2\pi)^3}\frac{1}{\sqrt{4E_pE_q}}\sum_{s,s^\prime}e^{i(px-qy)}\left[a^s_pu^s(p)+b^{s}_{-p}v^s(-p), a^{s^\prime\dagger}_qu^{s^\prime\dagger}(q)+b^{s^\prime\dagger}_{-q}v^{s^\prime\dagger}(-q)\right]$$
I think that now you should be able to go on by yourself given the properties of the (anti-)commutators and the canonical (anti-)commutator relations (some of them are zero).
answered Jul 9, 2021 at 10:56
-
$\begingroup$ How would you calculate $[a, a^\dagger]$, if you only know $\{ a, a^\dagger \}$ ? $\endgroup$ Commented Oct 17, 2021 at 10:14
-
$\begingroup$ @Quantumwhisp as per the rules of the site, you should ask a separate question. $\endgroup$ Commented Oct 17, 2021 at 16:49