Quantization of Dirac Field
Canonical Quantization of Dirac Spinor Field
The Hamiltonian of Dirac Spinor Field is:
$$H = \int \td^3 \bm{x} \Big\{-\sum_{i=1}^3 \ti \bar{\psi}\gamma^i \partial_i \psi + m \bar{\psi}\psi \Big\}$$
Where the $\psi$ should be a 4-component-Dirac-spinor. And the canonical momentum is $\pi=\ti \bar{\psi}\gamma^0=\ti\psi^\dagger$ . Then with the anti-commutation relationship we need:
$$\{\pi_\alpha (\bm{x}),\pi_\beta(\bm{x}')\}=\{\psi_\alpha(\bm{x}),\psi_\beta(\bm{x}')\}=0 \ ; \ \{\psi_\alpha(\bm{x}),\pi_\beta(\bm{x}')\}=\ti \delta_{\alpha\beta}\delta^3(\bm{x}-\bm{x}')$$
Which means that:
$$\begin{aligned}[] \{\psi_\alpha(\bm{x}),\psi_\beta(\bm{x}')\}&=0 \\ \{\psi_\alpha(\bm{x}),\psi_\beta(\bm{x}')^\dagger\}&=\delta_{\alpha\beta}\delta^3(\bm{x}-\bm{x}') \end{aligned}$$
One can rewrite them into the plane wave expansion. In Schrodinger Picture:
$$\psi(\bm{x})=\int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \sqrt{\frac 1 {2\omega_{|\bm{k}|}}} \sum_{s=1,2}\Big(e^{\ti \bm{k}\cdot \bm{x}} u^{(s)}(\bm{k}) a(\bm{k},s)+e^{-\ti \bm{k}\cdot \bm{x}} v^{(s)}(\bm{k}) b(\bm{k},s)^\dagger \Big)$$
Where $a,b$ , are operators, and $\omega_{|\bm{k}|}=\sqrt{m^2+|\bm{k}|^2}$ . Spinors:
$$u^{(s)}(\bm{k})=\frac {m+k_{\mu}\gamma^\mu} {\sqrt{2m(m+\omega_{|\bm{k}|})}} \begin{bmatrix} \chi^{(s)} \\ \chi^{(s)}\end{bmatrix} \ ; \ v^{(s)}(\bm{k})=\frac {m-k_{\mu}\gamma^\mu} {\sqrt{2m(m+\omega_{|\bm{k}|})}} \begin{bmatrix} \xi^{(s)} \\ -\xi^{(s)}\end{bmatrix}$$
With the 2-component quantities $\chi,\xi$ satisfy $\chi^{(r)\dagger}\chi^{(s)}=\xi^{(r)\dagger}\xi^{(s)}=\delta_r^s m$ .
We can write the inverse transform, which will represent $a,b$ , by $\psi,\psi^\dagger$ :
$$\begin{aligned} a(\bm{k},s)&= \frac 1 {\sqrt{2\omega_{|\bm{k}|}}} \int \frac {\td^3 \bm{x}} {(2\pi)^{3/2}} e^{-\ti \bm{k}\cdot \bm{x}} u^{(s)\dagger}(\bm{k}) \psi(\bm{x}) \\ b(\bm{k},s)&= \frac 1 {\sqrt{2\omega_{|\bm{k}|}}} \int \frac {\td^3 \bm{x}} {(2\pi)^{3/2}} e^{-\ti\bm{k}\cdot \bm{x}}\psi (\bm{x})^\dagger v^{(s)}(\bm{k}) \end{aligned}$$
And their commutation relation:
$$\begin{aligned} \{a(\bm{k},s),b(\bm{k}',s')\}&=\{a(\bm{k},s)^\dagger,b(\bm{k}',s')^\dagger\}=0 \\ \{a(\bm{k},s),a(\bm{k}',s')\}&=\{b(\bm{k},s),b(\bm{k}',s')\}=0 \\ \{a(\bm{k},s),b(\bm{k}',s')^\dagger\}&=\{a(\bm{k},s)^\dagger,b(\bm{k}',s')\}=0 \\ \{a(\bm{k},s),a(\bm{k}',s')^\dagger\}&=\{b(\bm{k},s),b(\bm{k}',s')\}=\delta^{s,s'}\delta^3(\bm{k}-\bm{k}') \end{aligned}$$
Proof
Using the 3-d delta function: $\delta^3(\bm{x}-\bm{y})=\int \frac {\td^3 \bm{k}} {(2\pi)^3} e^{\ti \bm{k}\cdot (\bm{x}-\bm{y})}$ ,, we have: $$\begin{aligned} \frac 1 {\sqrt{2\omega_{|\bm{k}|}}} \sum_{s=1,2} \Big(u^{(s)}(\bm{k})a(\bm{k},s)+v^{(s)}(-\bm{k})b(-\bm{k},s)^\dagger\Big)&= \int \frac {\td^3 \bm{x}} {(2\pi)^{3/2}} e^{-\ti \bm{k}\cdot \bm{x}} \psi(\bm{x}) \\ \frac 1 {\sqrt{2\omega_{|\bm{k}|}}} \sum_{s=1,2} \Big(u^{(s)}(-\bm{k})^\dagger a(-\bm{k},s)^\dagger +v^{(s)}(\bm{k})^\dagger b(\bm{k},s)\Big)&= \int \frac {\td^3 \bm{x}} {(2\pi)^{3/2}} e^{-\ti \bm{k}\cdot \bm{x}} \psi(\bm{x})^\dagger \end{aligned}$$ With the orthogonal property of $u,v$ :: $$\begin{aligned} u^{(s')\dagger}(\bm{k})u^{(s)}(\bm{k})&=\bar{u}^{(s')}(\bm{k})\gamma^0 u^{(s)}(\bm{k}) =2 k^0 \delta^{s,s'} \\ v^{(s')\dagger}(\bm{k})v^{(s)}(\bm{k})&=\bar{v}^{(s')}(\bm{k})\gamma^0 v^{(s)}(\bm{k}) =2 k^0 \delta^{s,s'} \end{aligned}$$ And using that $(m+k_\mu \gamma^\mu)^{\dagger}=(m+(k^0,-\bm{k})\cdot \gamma)$ and $(m+k_\mu \gamma^\mu)(m-k_\mu \gamma^\mu)=0$ , one can prove that: $$u^{(s')\dagger}(\bm{k})v^{(s)}(-\bm{k})=0$$ Then noting that $\gamma$ matrices and spinors are commutable to operators $a,b$ , , we can write down: $$\begin{aligned} a(\bm{k},s)&= \frac 1 {\sqrt{2\omega_{|\bm{k}|}}} \int \frac {\td^3 \bm{x}} {(2\pi)^{3/2}} e^{-\ti \bm{k}\cdot \bm{x}} u^{(s)\dagger}(\bm{k}) \psi(\bm{x}) \\ b(\bm{k},s)&= \frac 1 {\sqrt{2\omega_{|\bm{k}|}}} \int \frac {\td^3 \bm{x}} {(2\pi)^{3/2}} e^{-\ti\bm{k}\cdot \bm{x}}\psi (\bm{x})^\dagger v^{(s)}(\bm{k}) \end{aligned}$$ Then we can check the commutation relationship. Obviously: $$\begin{aligned}[] \{a(\bm{k},s),a(\bm{k}',s')\}&=\{b(\bm{k},s),b(\bm{k}',s')\}=0 \\ \{a(\bm{k},s),b(\bm{k}',s')^\dagger\}&=\{a(\bm{k},s)^\dagger,b(\bm{k}',s')\}=0 \\ \end{aligned}$$ Then: $$\begin{aligned}[] \{a(\bm{k},s),a(\bm{k}',s')^\dagger\}&= \frac 1 {\sqrt{4\omega \omega'}} \int \frac {\td^3 \bm{x}\td^3 \bm{x}'} {(2\pi)^3} e^{-\ti \bm{k}\cdot \bm{x}+\ti \bm{k}'\cdot \bm{x}'} \{u^{(s)\dagger}(\bm{k})\psi(\bm{x}),\psi(\bm{x}')^\dagger u^{(s')}(\bm{k}')\} \\ &= \frac 1 {\sqrt{4\omega \omega'}} \int \frac {\td^3 \bm{x}\td^3 \bm{x}'} {(2\pi)^3} e^{-\ti \bm{k}\cdot \bm{x}+\ti \bm{k}'\cdot \bm{x}'} \delta^3(\bm{x}-\bm{x}') u^{(s)\dagger}(\bm{k})u^{(s')}(\bm{k}') \\ &= \frac 1 {\sqrt{4\omega \omega'}} u^{(s)\dagger}(\bm{k})u^{(s')}(\bm{k}') \delta^3(\bm{k}-\bm{k}') \\ &=\delta^{s,s'}\delta^3(\bm{k}-\bm{k}') \end{aligned}$$ Similarly, we can prove that: $\{b(\bm{k},s),b(\bm{k}',s')\}=\delta^{s,s'}\delta^3(\bm{k}-\bm{k}')$ . And: $$\begin{aligned}[] \{a(\bm{k},s),b(\bm{k}',s')\}&= \frac 1 {\sqrt{4\omega \omega'}} \int \frac {\td^3 \bm{x}\td^3 \bm{x}'} {(2\pi)^3} e^{-\ti \bm{k}\cdot \bm{x}-\ti \bm{k}'\cdot \bm{x}'} \{u^{(s)\dagger}(\bm{k})\psi(\bm{x}),\psi(\bm{x}')^\dagger v^{(s')}(\bm{k}')\} \\ &=\frac 1 {\sqrt{4\omega \omega'}} \int \frac {\td^3 \bm{x}\td^3 \bm{x}'} {(2\pi)^3} e^{-\ti \bm{k}\cdot \bm{x}-\ti \bm{k}'\cdot \bm{x}'} \delta^3(\bm{x}-\bm{x}') u^{(s)\dagger}(\bm{k})v^{(s')}(\bm{k}') \\ &=\frac 1 {\sqrt{4\omega \omega'}}u^{(s)\dagger}(\bm{k})v^{(s')}(\bm{k}') \delta^3(\bm{k}+\bm{k}') \\ &=0 \end{aligned}$$ q.e.d.With this result and the hold-order discussion in Classical theory, we can write the Hamiltonian as:
$$\hat H = \sum_s \int \td^3 \bm{k} \ \omega_{|\bm{k}|} \Big[a(\bm{k},s)^\dagger a(\bm{k},s) + b(\bm{k},s)^\dagger b(\bm{k},s) \Big]$$
There is also a $\infty$ quantity at the right-hand-side. But like what in Klein-Gordon Field, we ignore it.
Result with usual commutation relationship
With usual commutation relationship, we have the Hamiltonian as: $$\begin{aligned} \hat H &= \sum_s \int \td^3 \bm{k} \omega_{|\bm{k}|} \Big[a(\bm{k},s)^\dagger a(\bm{k},s) - b(\bm{k},s) b(\bm{k},s)^\dagger \Big] \\ & \sim \sum_s \int \td^3 \bm{k} \omega_{|\bm{k}|} \Big[a(\bm{k},s)^\dagger a(\bm{k},s) - b(\bm{k},s)^\dagger b(\bm{k},s) \Big]+ \int \td^3 \bm{k} \ \omega_{|\bm{k}|}\delta^3(\bm{0}) \end{aligned}$$ Even we ignore the $\infty$ quantity, the Hamiltonian still has no lower bound, which is caused by the `negative-energy` electrons. The anti-commutation relationship will let Hamiltonian has a lower bound. In original Dirac theory, the vacuum is defined as the state of all negative energy levels are occupied by negative-energy electrons, the anti-particle is treated as the hole in negative-energy electrons sea. Here $b^\dagger$ will generate an anti-electron automatically.And one can also write the conserved charge of Dirac Field:
$$\hat Q =\sum_{s=1,2}\int \td^3 \bm{k} \Big(a(\bm{k},s)^\dagger a(\bm{k},s)-b(\bm{k},s)^\dagger b(\bm{k},s)\Big) $$
Which shows that the particles created by $a,b$ , have different charge.
One can also write these operators in Heisenberg Picture :
$$\begin{aligned} \psi(x)&=\int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \sqrt{\frac 1 {2\omega_{|\bm{k}|}}} \sum_{s=1,2}\Big(e^{-\ti k\cdot x} u^{(s)}(\bm{k}) a(\bm{k},s)+e^{\ti k\cdot x} v^{(s)}(\bm{k}) b(\bm{k},s)^\dagger \Big)\\ \bar \psi(x)&=\int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \sqrt{\frac 1 {2\omega_{|\bm{k}|}}} \sum_{s=1,2}\Big(e^{\ti k\cdot x} \bar u^{(s)}(\bm{k}) a(\bm{k},s)^\dagger+e^{-\ti k\cdot x} \bar v^{(s)}(\bm{k}) b(\bm{k},s) \Big) \end{aligned}$$
Proof
We can also check that: $$[a(\bm{k},s),\hat H]=\omega_{|\bm{k}|}a(\bm{k}) \ ; \ [b(\bm{k},s),\hat H]=\omega_{|\bm{k}|}b(\bm{k})$$ And: $$a_H(\bm{k})=e^{\ti t \hat H} a(\bm{k})e^{-\ti t \hat H} = e^{-\ti \omega_{|\bm{k}|}t} a(\bm{k})$$ same for $b$ . . Then: $$\psi(x)=e^{\ti t \hat H} \psi(\bm{x})e^{-\ti t \hat H}=\int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \sqrt{\frac 1 {2\omega_{|\bm{k}|}}} \sum_{s=1,2}\Big(e^{-\ti k\cdot x} u^{(s)}(\bm{k}) a(\bm{k},s)+e^{\ti k\cdot x} v^{(s)}(\bm{k}) b(\bm{k},s)^\dagger \Big)$$Hilbert Space of Dirac Spinor Field
[Definition] : vacuum of the Dirac Field is the ground state of it. Denoting it as $\ket{0}$ , we will find:
$$a(\bm{k},s)\ket{0}=b(\bm{k},s)\ket{0}=0 \ ; \ \forall \bm{k},s$$
[Definition] : 1-particle state is defined as:
$$\ket{\bm{k},s,\text{Fermion}} = \sqrt{2\omega_{|\bm{k}|}} a(\bm{k},s)^\dagger \ket{0}$$
And 1-anti-particle state is defined as:
$$\ket{\bm{k},s,\text{anti-Fermion}} = \sqrt{2\omega_{|\bm{k}|}} b(\bm{k},s)^\dagger \ket{0}$$
With this definition, we can prove(like what we did for Scalar field):
$$\bra{\Lambda_{\bm{k}\rightarrow \bm{0}}\bm{k},s}\Lambda_{\bm{k}\rightarrow \bm{0}} \bm{k}',s'\rangle = \bra{\bm{k},s}\bm{k}',s'\rangle=2\omega_{|\bm{k}|}\delta^{s,s'}\delta^3(\bm{k}-\bm{k}')$$
Where $\Lambda_{\bm{k}\rightarrow 0}k=(m,0,0,0)$ only contains the boost. The behavior of spin under spatial-rotation will be discussed in following.
And the complete relation in 1-(anti-)particle subspace is also:
$$1_{\text{one-(anti-)particle}} = \sum_{s=1,2} \int \frac {\td^3 \bm{k}} {2\omega_{|\bm{k}|}} \ket{\bm{k},s,(\text{anti})}\bra{\bm{k},s,(\text{anti})}$$
Also has the covariant integral measure:
$$\int \frac {\td^3 \bm{p}} {2\omega_{|\bm{p}|}} = \int \td^4 p \delta(p_\mu p^\mu -m^2) \Theta(p^0)$$
[Definition] : Consider the coordinate transformation(special Lorentz Transformation) $x\rightarrow x'=\Lambda(\omega) x$ , the corresponding operator on Hilbert space is $U(\Lambda(\omega))$ , then we have:
$$U(\Lambda(\omega))\ket{\bm{k},s}=\ket{\Lambda(\omega) \bm{k},s}$$
The vacuum is unique $U(\Lambda)\ket{0}=\ket{0}$ . Then:
$$U(\Lambda(\omega)) a(\bm{k},s)^\dagger U(\Lambda(\omega))^{-1} = \sqrt {\frac {\omega_{|\bm{\Lambda(\omega)\bm{k}}|}} {\omega_{|\bm{k}|}} } a(\Lambda(\omega)\bm{k},s)^\dagger$$
same for anti-Fermion operator $b$ . . Then one can check that the field operator will be:
$$\psi'(x'=\Lambda(\omega)x) \equiv U(\Lambda(\omega))\psi(x)U(\Lambda(\omega))^{-1}=\Lambda_{1/2}(\omega)^{-1} \psi(x'=\Lambda(\omega)x)$$
Where $\Lambda_{1/2}(\omega)=\exp(-\frac \ti 4 \sum_{\mu\nu}\sigma_{\mu\nu}\omega_{\mu\nu})=\exp(-\ti\omega_{\mu\nu}S^{\mu\nu}/2)$ .
Proof
The proof is similar to what we did for Klein-Gordon Field. Just noting that the property of $u,v$ : : $$u^{(s)}(\bm{k})=\Lambda_{\bm{0}\rightarrow \bm{k}}\Lambda_{\bm{k}'\rightarrow \bm{0}} u^{(s)}(\bm{k}')=\Lambda_{\bm{k}'\rightarrow \bm{k}}u^{(s)}(\bm{k}')$$ Same for $v$ . Where $\Lambda_{\bm{k}'\rightarrow \bm{k}}=\Lambda_{1/2}(\omega)$ with the $\omega$ that satisfies $\Lambda(\omega)\bm{k}'=\bm{k}$ . Then: $$\begin{aligned} U(\Lambda)\psi(x)U(\Lambda)^{-1} &= \int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \frac {\sqrt{2\omega_{|\Lambda \bm{k}|}}} {2\omega_{|\bm{k}|}} \sum_{s=1,2}\Big(e^{-\ti k\cdot x} u^{(s)}(\bm{k}) a(\Lambda\bm{k},s)+e^{\ti k\cdot x} v^{(s)}(\bm{k}) b(\Lambda\bm{k},s)^\dagger \Big) \\ &=\Lambda_{1/2}(\omega)^{-1} \int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \frac {\sqrt{2\omega_{|\Lambda \bm{k}|}}} {2\omega_{|\bm{k}|}} \sum_{s=1,2}\Big(e^{-\ti k\cdot x} u^{(s)}(\Lambda\bm{k}) a(\Lambda\bm{k},s)+e^{\ti k\cdot x} v^{(s)}(\Lambda\bm{k}) b(\Lambda\bm{k},s)^\dagger \Big) \end{aligned}$$ Then similarly using that $k_\mu x^\mu=(\Lambda k)_\mu (\Lambda x)^\mu$ and the covariant integral measure, we have: $$U(\Lambda)\psi(x)U(\Lambda)^{-1}=\Lambda_{1/2}(\omega)^{-1}\psi(\Lambda x)$$Causality of Dirac Field
Our quantization will not violent to the causality demanded by special Relativistic Theory. That is to say:
[Theorem] : In Heisenberg Picture, for any two points $x,y$ with spacelike interval: $\|x-y\|^2\lt 0$ , .. Then the field operators are commutable:
$$\{\psi_\alpha(x),\psi_\beta(y)^\dagger\}=0 \ ; \ \|x-y\|^2\lt 0$$
Proof
With the plane wave expansion: $$\psi(x)=\int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}}\sqrt {\frac 1 {2\omega_{|\bm{k}|}}} \sum_{s=1,2}\Big(e^{-\ti k\cdot x} u^{(s)}(\bm{k}) a(\bm{k},s) + e^{\ti k\cdot x} v^{(s)}(\bm{k}) b(\bm{k},s)^\dagger \Big)$$ And the anti-commutation relation: $$\{a(\bm{k},s),a(\bm{k}',s')^\dagger\}=\delta^{s,s'}\delta^3 (\bm{k}-\bm{k}')$$ Same for $b$ . , and vanish for all of other types. Then we have: $$\{\psi_\alpha(x),\psi_\beta(y)^\dagger\} = \int \frac {\td^3 \bm{k}} {(2\pi)^3} \frac 1 {2\omega_{|\bm{k}|}} \sum_{s=1,2} \Big(e^{-\ti k\cdot (x-y)} u^{(s)}_\alpha(\bm{k})u^{(s)}_\beta(\bm{k})^* + e^{\ti k\cdot (x-y)} v^{(s)}_\alpha(\bm{k})v^{(s)}_\beta(\bm{k})^* \Big)$$ With $$\begin{aligned} \sum_{s=1,2} u^{(s)}_\alpha(\bm{k})\bar u^{(s)}_\beta(\bm{k})&= \sum_{s=1,2} (u^{(s)}(\bm{k})\bar u^{(s)}(\bm{k}))_{\alpha\beta}=(m+k_\mu\gamma^\mu)_{\alpha\beta} \\ \sum_{s=1,2} v^{(s)}_\alpha(\bm{k})\bar v^{(s)}_\beta(\bm{k})&= \sum_{s=1,2} (v^{(s)}(\bm{k})\bar v^{(s)}(\bm{k}))_{\alpha\beta}=-(m-k_\mu\gamma^\mu)_{\alpha\beta} \\ \end{aligned}$$ Then with that $\bar x = x^\dagger \gamma^0$ , we have: $$\begin{aligned} \sum_{s=1,2} u^{(s)}_\alpha(\bm{k}) u^{(s)}_\beta(\bm{k})^*&= \sum_{s=1,2} (u^{(s)}(\bm{k})\bar u^{(s)}(\bm{k})\gamma^0)_{\alpha\beta}=(m\gamma^0+k_\mu\gamma^\mu\gamma^0)_{\alpha\beta} \\ \sum_{s=1,2} v^{(s)}_\alpha(\bm{k}) v^{(s)}_\beta(\bm{k})^*&= \sum_{s=1,2} (v^{(s)}(\bm{k})\bar v^{(s)}(\bm{k})\gamma^0)_{\alpha\beta}=(-m\gamma^0+k_\mu\gamma^\mu\gamma^0)_{\alpha\beta} \\ \end{aligned}$$ Then we have: $$\begin{aligned} \{\psi_\alpha(x),\psi_\beta(y)^\dagger\} &= \int \frac {\td^3 \bm{k}} {(2\pi)^3} \frac 1 {2\omega_{|\bm{k}|}} \Big(e^{-\ti k\cdot (x-y)} (m\gamma^0+k_\mu\gamma^\mu \gamma^0) - e^{\ti k\cdot (x-y)} (m\gamma^0-k_\mu\gamma^\mu \gamma^0) \Big)_{\alpha\beta} \\ &=(m\gamma^0+\ti\gamma^\mu\gamma^0 \partial_{x,\mu})\int \td^4 k \ \delta(k_\mu k^\mu-m^2)\Theta(k^0)\Big(e^{-\ti k\cdot(x-y)}-e^{\ti k\cdot (x-y)}\Big)\\ &=(m\gamma^0+\ti\gamma^\mu\gamma^0 \partial_{x,\mu}) (D(x-y)-D(y-x)) \end{aligned}$$ Where $\partial_{x,\mu}=\frac {\partial} {\partial x^\mu}$ , and $$D(x)=\int \td^4 k \ \delta(k_\mu k^\mu-m^2)\Theta(k^0) e^{-\ti k_\mu x^\mu}$$ Is invariant under Lorentz transform. And we have proved that when $\|x-y\|^2\lt 0$ , , we have $D(x-y)=D(y-x)$ in the discussion of causality of Scalar field. Then we proved when $\|x-y\|^2\lt 0$ , , $\{\psi_\alpha(x),\psi_\beta(y)^\dagger\}=0$ q.e.d.We need explain why this equality means the causality of Dirac Field. The core reason is that local measurable operators are products of even numbers of fermionic fields and their derivatives . Then the commutator of them can be written as a summation of some anti-commutator of fermionic fields’ operator, like:
$$[\hat A \hat B,\hat C\hat D]=\hat A\{\hat B,\hat C\}\hat D-\hat A\hat C\{\hat B,\hat D\}-\{\hat A,\hat C\}\hat D\hat B+\hat C\{\hat A,\hat D\}\hat B$$
which will be zero for spacelike interval. That is to say the two local observables with spacelike interval can be measured exactly on the same system(Heisenberg’s inequality) . And this is what the causality means.
Propagator of Dirac Field
For fermionic operators, we need adjust the definition of time-ordered product:
$$\mathcal{T}\{A(x)B(y)\}=\Theta(x^0-y^0)A(x)B(y)-\Theta(y^0-x^0)B(y)A(x)$$
[Theorem] the Feynman propagator of Dirac Field is:
$$\ti \Delta_{F}(x-y)_{\alpha\beta}:=\bra{0}\mathcal{T}\{\psi_\alpha(x)\bar{\psi}_\beta(y)\}\ket{0}$$
One can also write it as the matrix form:
$$\ti \Delta_F(x-y)= \int \frac{\td^4 p} {(2\pi)^4} \frac {\ti (\gamma^\mu p_\mu +m)} {p_\mu p^\mu -m^2 +\ti \epsilon} e^{-\ti p\cdot (x-y)}$$
Proof
We compute these two terms in time-ordered product: $$\begin{aligned} \bra{0}\psi_\alpha (x)\bar{\psi}_\beta(y)\ket{0} &= \int \frac {\td^3 \bm{k}_1\td^3 \bm{k}_2} {(2\pi)^3 \sqrt{4\omega_1\omega_2}} \sum_{s_1,s_2} e^{-\ti (k_1\cdot x-k_2\cdot y)} u^{(s_1)}_\alpha(\bm{k}_1) \bar {u}^{(s_2)}_\beta(\bm{k}_2) \bra{0} a(\bm{k}_1,s_1) a(\bm{k}_2,s_2)^\dagger\ket{0} \\ &= \int \frac {\td^3 \bm{k}} {(2\pi)^3 2\omega_{|\bm{k}|}} \sum_{s=1,2} e^{-\ti k\cdot (x-y)} u_\alpha^{(s)}(\bm{k}) \bar u_\beta^{(s)}(\bm{k}) \\ &=\int \frac {\td^3 \bm{k}} {(2\pi)^3 2\omega_{|\bm{k}|}} e^{-\ti k\cdot (x-y)} (m+k_\mu \gamma^\mu)_{\alpha\beta} \\ \bra{0}\bar{\psi}_\beta(y)\psi_\alpha (x)\ket{0} &= \int \frac {\td^3 \bm{k}_1\td^3 \bm{k}_2} {(2\pi)^3 \sqrt{4\omega_1\omega_2}} \sum_{s_1,s_2} e^{-\ti (k_1\cdot y-k_2\cdot x)} \bar v^{(s_1)}_\beta(\bm{k}_1) v^{(s_2)}_\alpha(\bm{k}_2) \bra{0} b(\bm{k}_1,s_1) b(\bm{k}_2,s_2)^\dagger\ket{0} \\ &= \int \frac {\td^3 \bm{k}} {(2\pi)^3 2\omega_{|\bm{k}|}} \sum_{s=1,2} e^{\ti k\cdot (x-y)} v_\alpha^{(s)}(\bm{k}) \bar v_\beta^{(s)}(\bm{k}) \\ &=-\int \frac {\td^3 \bm{k}} {(2\pi)^3 2\omega_{|\bm{k}|}} e^{\ti k\cdot (x-y)} (m-k_\mu \gamma^\mu)_{\alpha\beta} \\ \end{aligned}$$ Then we have: $$\ti \Delta_F(x-y) = \int \frac {\td^3 \bm{k}} {(2\pi)^3 2\omega_{|\bm{k}|}}\Big(e^{-\ti k\cdot (x-y)} \Theta(x^0-y^0) (m+k_\mu \gamma^\mu) +e^{\ti k\cdot (x-y)}\Theta(y^0-x^0)(m-k_\mu\gamma^\mu)\Big)$$ For the integral: $$\int \frac {\td^4 p} {(2\pi)^4} e^{-\ti p\cdot x} \frac {\ti f(p)} {p_\mu p^\mu-m^2+\ti \epsilon}=\int_{p^0=\omega_{|\bm{p}|}} \frac {\td^3 \bm{p}} {(2\pi)^3 2\omega_{|\bm{p}|}} \Big(\Theta(x^0)e^{-\ti p\cdot x}f(p^0=\omega_{|\bm{p}|},\bm{p})+\Theta(-x^0)e^{\ti p\cdot x} f(p^0=-\omega_{|\bm{p}|},-\bm{p}) \Big)$$ We can write that: $$\ti \Delta_F(x-y) = \int \frac {\td^4 p} {(2\pi)^4} \frac {\ti (m+p_\mu \gamma^\mu)} {p_\mu p^\mu -m^2 +\ti \epsilon} e^{-\ti p\cdot (x-y)} $$ Some times we also write it as : $$\ti \Delta_F(x-y) = \int \frac {\td^4 p} {(2\pi)^4} \frac {\ti} {\gamma^\mu p_\mu -m +\ti \epsilon} e^{-\ti p\cdot (x-y)} $$Discrete Symmetry of Dirac Field
Spatial Inversion
We hope the spatial inversion induces a unitary operator so as to:
$$U_P \psi(x^0,\bm{x}) U_P^{-1} = A_P \psi(x^0,\bm{x})$$
Where $A_P$ is a unitary matrix in Dirac-Spinor space. This is a little different to what for Scalar field. And $U_P\ket{0}=\ket{0}$ .
For creation operators, we demand the unitary operator of spatial inversion will cause the direction-inverse of momentum:
$$U_P a(\bm{k},s) U_P^{-1} = \eta_a a(-\bm{k},s)\ ; \ U_P b(\bm{k},s) U_P^{-1} = \eta_b b(-\bm{k},s)$$
Where $\eta_a,\eta_b$ are c-numbers. If we demand all observables(even number of fermionic operators’ product) invariant under two spatial inversion:
$$\eta_a^2=\pm 1\ ; \ \eta_b^2 =\pm1$$
In early discussion, we know that under parity-transformation, Weyl spinor will change the chirality. That means $A_P\propto \gamma^0$ . That means we should let $\eta_b^*=-\eta_a$ . And:
$$\begin{aligned} U_P \psi(x^0,\bm{x}) U_P^{-1} &= \eta_a \gamma^0 \psi(x^0,-\bm{x})\\ U_P \bar \psi(x^0,\bm{x}) U_P^{-1} &= \eta_a^* \psi(x^0,-\bm{x})\gamma^0 \end{aligned}$$
Charge Conjugation
We hope the charge conjugation induces a unitary operator which can exchange the fermion and anti-fermion:
$$U_C a(\bm{k},s)U_C^{-1} = b(\bm{k},s) \ ; \ U_C b(\bm{k},s)U_C^{-1} = a(\bm{k},s)$$
With the convention of $\chi,\xi$ : $\xi^{(s)}=(-\ti\sigma^2) (\chi^{(s)})^*$ , we can prove that:
$$\begin{aligned} U_C \psi U_C^{-1} &= -\ti (\bar{\psi}\gamma^0\gamma^2)^T \\ U_C \bar{\psi} U_C^{-1} &= (-\ti \gamma^0 \gamma^2 \psi)^T \end{aligned}$$
Proof
With the property of $u,v$ : : $$u^{(s)}(\bm{k})=-\ti \gamma^2\Big(v^{(s)}(\bm{k}) \Big)^* \ ; \ v^{(s)}(\bm{k})=-\ti \gamma^2 \Big(u^{(s)}(\bm{k}) \Big)^*$$ we have: $$\begin{aligned} U_C \psi U_C^{-1} &=\sum_s \int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \frac 1 {\sqrt{2\omega_{|\bm{k}|}}} \Big(e^{-\ti k\cdot x} u^{(s)}(\bm{k}) U_C a(\bm{k},s)U_C^{-1}+e^{\ti k\cdot x} v^{(s)}(\bm{k}) U_C b(\bm{k},s)^\dagger U_C^{-1} \Big) \\ &= \sum_s\int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \frac 1 {\sqrt{2\omega_{|\bm{k}|}}} \Big(e^{-\ti k\cdot x} u^{(s)}(\bm{k}) b(\bm{k},s)+e^{\ti k\cdot x} v^{(s)}(\bm{k}) a(\bm{k},s)^\dagger \Big) \\ &=-\ti \gamma^2\sum_s \int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \frac 1 {\sqrt{2\omega_{|\bm{k}|}}} \Big(e^{-\ti k\cdot x} (v^{(s)})^* (\bm{k}) b(\bm{k},s)+e^{\ti k\cdot x} (u^{(s)}(\bm{k}))^* a(\bm{k},s)^\dagger \Big) \\ &=-\ti \gamma^2 \gamma^0 \bar {\psi}^T \\ &= -\ti (\bar{\psi}\gamma^0\gamma^2)^T \end{aligned}$$ Another equality's proof is similar.Time Reversal
ime reversal will induce an anti-unitary operator. With conjugation operator $K$ we can write it as $U_T K$ , where $U_T$ is a unitary operator. We hope the time reversal will inverse the spin and momentum. Where the spin-inversion should be $x^{(\bar s)}=-\ti \sigma^2 (x^{(s)})^*$ . Then we demand:
$$U_T K a(\bm{k},s) (U_T K)^{-1} = a(-\bm{k},\bar{s}) \ ; \ U_T K b(\bm{k},s) (U_T K)^{-1} = b(-\bm{k},\bar{s})$$
Where $a(\bm{k},\bar 1)=a(\bm{k},2)\ ; \ a(\bm{k},\bar 2)=-a(\bm{k},1)$ , same for $b$ . .
Then we can prove that:
$$\begin{aligned} U_T K \psi(x^0,\bm{x}) (U_T K)^{-1} &= \gamma^1\gamma^3 \psi(-x^0,\bm{x}) \\ U_T K \bar \psi(x^0,\bm{x}) (U_T K)^{-1} &= \bar \psi(-x^0,\bm{x}) \gamma^1\gamma^3 \\ \end{aligned}$$