Fermion propagator as Green's function of Dirac operator

Question

On book "A modern introduction to quantum field theory" by Maggiore, page 133, he mentions that $S(x-y)$ defined as: $$S_{ab}(x-y)\equiv\langle 0|T\{\Psi_a(x)\bar\Psi_b(y)\}|0\rangle$$ it's the Green's function of the Dirac operator: $$(i\not\partial_x -m)_{ab}S_{bc}(x-y) = i\delta^4(x-y)\delta{ac}$$

To me it's not obvious from that definition of $S(x-y)$ that it's the Green's function of the Dirac operator.

So am I missing some concept that makes such deduction immediate?

Below the sketch of the (long) calculations I had to do to get it, which makes me suspect that I have taken an unnecessarily long or even wrong path:

$$ S_{ab}(x-y) = \langle 0|\theta(x^0-y^0) \Psi_a(x)\bar\Psi_b(y) - \theta(y^0-x^0) \bar\Psi_b(y)\Psi_a(x) | 0 \rangle ~~~~~(1)$$

use the explicit formula for $\Psi$, $\bar\Psi$ and the identity $\sum_{s=1,2} u_a^s(p)\bar u_b^s(p)= (\not p + m )_{ab}$ to get

$$ \langle 0|\Psi_a(x)\bar\Psi_b(y)|0\rangle = \int {{d^3p}\over{(2\pi)^3}}{{1}\over{2E_p}}e^{-ip(x-y)}(\not p +m)_{ab}~~~~~(2) $$

$$ \langle 0|\bar\Psi_b(y)\Psi_a(x)|0\rangle = \int {{d^3p}\over{(2\pi)^3}}{{1}\over{2E_p}}e^{ip(x-y)}(\not p -m)_{ab}~~~~~(3) $$

substitute $(2)$ and $(3)$ in $(1)$ $$ S_{ab}(x-y) = \int {{d^3p}\over{(2\pi)^3}} {{1}\over{2E_p}} \left(\theta(x^0-y^0)e^{-ip(x-y)}(\not p +m)_{ab} + \theta(y^0-x^0)e^{ip(x-y)}(-\not p +m)_{ab} \right) ~~~~~(4)$$

I have been stuck here for long till I realized to use the following identity valid for $f(p)$ polynomial (it comes from complex residual calcolous, see Srednicki pag.269)

$$ \int {{d^4p}\over{(2\pi)^4}}{{i}\over{p^2-m^2+i\epsilon}}e^{-ip(x-y)}f(p) = \int {{d^3p}\over{(2\pi)^3 2E_p}}\left(\theta(x^0-y^0)e^{-ip(x-y)}f(p)+\theta(y^0-x^0)e^{ip(x-y)}f(-p)\right) ~~~~~(5)$$

to get

$$ S_{ab}(x-y) = \int {{d^4p}\over{(2\pi)^4}}{{i}\over{p^2-m^2+i\epsilon}}e^{-ip(x-y)}(\not p +m)_{ab} ~~~~~(6)$$

which can be rewritten

$$ S_{ab}(x-y)=(i\not\partial_x + m)_{ab} \int {{d^4p}\over{(2\pi)^4}}{{i}\over{p^2-m^2+i\epsilon}}e^{-ip(x-y)} ~~~~(7)$$

now using $(i\not\partial_x -m)_{ab} (i\not\partial_x + m)_{bc} = -(\square_x + m^2)_{ac}$ I have

$$(i\not\partial_x -m)_{ab}S_{bc}(x-y) = -(\square_x + m^2)_{ac} \int {{d^4p}\over{(2\pi)^4}}{{i}\over{p^2-m^2+i\epsilon}}e^{-ip(x-y)}\\ = \left(\int {{d^4p}\over{(2\pi)^4}}ie^{-ip(x-y)}\right) \delta_{ac} = i\delta^4(x-y)\delta_{ac} $$

Answer 1

I) Let us denote $K_{ab}(x,y)$ as: $$ K_{ab}(x,y)=\theta(x^0-y^0)\langle0|\hat{\Psi}_a(x)\hat{\Psi}_b^+(y)|0\rangle-\theta(y^0-x^0)\langle0|\hat{\Psi}_b^+(y)\hat{\Psi}_a(x)|0\rangle~~~~~~(1) $$ $K_{ab}(x,y)$ differs from $S_F(x,y)$ in that we use $\hat{\Psi}_b^+(y)$ instead of $\hat{\bar{\Psi}}(y)$; so we miss one $\gamma^0$ matrix.

We will calculate: $$ (i\not\partial_x-m)_{ca}K_{ab}(x,y)~~~~~~~~~(2) $$

II) Let us consider $x^0>y^0$.

At $x^0>y^0$, $K_{ab}(x,y)$ reduces to: $$ K_{ab}(x,y)|_{x^0>y^0}=\langle0|\hat{\Psi}_a(x)\hat{\Psi}_b^+(y)|0\rangle~~~~~~(3) $$ Now, let us recall that $\hat{\Psi}_a(x)$ operator satisfies the Dirac equation: $$ (i\not\partial_x-m)_{ca}\hat{\Psi}_a(x)=0~~~~~~~(4) $$ Because of (4), for $x^0>y^0$ we will obtain: $$ x^0>y^0:(i\not\partial_x-m)_{ca}\langle0|\hat{\Psi}_a(x)\hat{\Psi}_b^+(y)|0\rangle=0~~~~~~(5) $$ and, equivalently: $$ x^0>y^0:(i\not\partial_x-m)_{ca}K_{ab}(x,y)=0~~~~~~~(6) $$ III) In the similar way, we find for $x^0<y^0$ that: $$ x^0<y^0:(i\not\partial_x-m)_{ca}K_{ab}(x,y)=0~~~~~~~(7) $$ IV) Now we should consider $x^0=y^0$.

$\not\partial_x$ includes: $\partial_{x^0}$, $\partial_{x^1}$, $\partial_{x^2}$, $\partial_{x^3}$. The problematic thing at $x^0=y^0$ is $\partial_{x^0}$. It is differentiating (1) by $x^0$ that produces singular contributions at $x^0=y^0$.

The singular contribution to $\partial_{x^0}K_{ab}(x,y)$ at $x^0=y^0$ is: $$ \delta(x^0-y^0)\langle0|\hat{\Psi}_a(x^0,\mathbf{x})\hat{\Psi}_b^+(x^0,\mathbf{y})|0\rangle+\delta(x^0-y^0)\langle0|\hat{\Psi}_b^+(x^0,\mathbf{y})\hat{\Psi}_a(x^0,\mathbf{x})|0\rangle~~~~~~(8) $$ In (7), we have equal time anticommutator, that is (see Peskin, Schroeder, (3.102)): $$ \{\hat{\Psi}_a(\mathbf{x}),\hat{\Psi}_b^+(\mathbf{y})\}=\delta_{ab}\delta(\mathbf{x}-\mathbf{y})~~~~~(9) $$ The form of (9) applicable to (8) is: $$ \langle0|\{\hat{\Psi}_a(x^0,\mathbf{x}),\hat{\Psi}_b^+(x^0,\mathbf{y})\}|0\rangle=\delta_{ab}\delta(\mathbf{x}-\mathbf{y})~~~~~~~~~(10) $$ The singular contribution to $\partial_{x^0}K_{ab}(x,y)$ at $x^0=y^0$ is: $$ (8)=\delta(x^0-y^0)\delta_{ab}\delta(\mathbf{x}-\mathbf{y})~~~~~~~~(11) $$ In $(i\not\partial_x-m)$, $\partial_{x^0}$ is also multiplied by $i$ and $\gamma^0$. So, the singular contribution to $(i\not\partial_x-m)_{ca}K_{ab}(x,y)$ at $x^0=y^0$ is: $$ i\gamma^0_{ca}\delta(x^0-y^0)\delta_{ab}\delta(\mathbf{x}-\mathbf{y})=i\gamma^0_{cb}\delta(x^0-y^0)\delta(\mathbf{x}-\mathbf{y})~~~~~~(12) $$ V) So, for $(i\not\partial_x-m)_{ca}K_{ab}(x,y)$ we obtain: $$ (i\not\partial_x-m)_{ca}K_{ab}(x,y)= \begin{cases} 0, & \text{if}~~x^0 > y^0\\ 0, & \text{if}~~x^0 < y^0\\ i\gamma^0_{cb}\delta(x^0-y^0)\delta(\mathbf{x}-\mathbf{y}), & \text{if}~~x^0 = y^0~~\text{(singular contribution)} \end{cases}~~~(13) $$ Overall, (13) can be rewritten as: $$ (i\not\partial_x-m)_{ca}K_{ab}(x,y)=i\gamma^0_{cb}\delta^{(4)}(x-y)~~~~~~(14) $$ VI) In $S_F(x,y)$, we have extra $\gamma^0$ matrix in comparison with $K_{ab}(x,y)$. This extra $\gamma^0$ matrix will compensate the $\gamma^0$ matrix in (14). So, for $S_F(x,y)$ we will obtain the desired relation: $$ (i\not\partial_x-m)S_F(x,y)=i\delta^{(4)}(x-y)~~~~~~(15) $$ VII) The full layout of the above consideration is also not that short. However, one can: look at equation (1); apply relations (4) and (9) (which are quite basic); the sketch of how (14) can be derived takes shape. For me, this still seems shorter than direct calculation, that involves the definition of $\hat{\Psi}_a(x)$.