Fermionic scattering in Yukawa Theory The scattering process I'd like to describe is the following fermion-fermion scattering in Yukawa Theory
$\psi_{p1} + \psi_{p2} \rightarrow \psi_{p1'} + \psi_{p2'}$
The interaction Hamiltonian for this system is 
$H_{INT} = \int g\overline\psi(x) \psi(x) \phi(x)d^3x$
In order to calculate the invariant element of the $S$ matrix, which is defined by 
$⟨p_1',p_2'|iT|p_1,p_2⟩ = (2\pi)^4 \delta^4(p_1'+p_2'-p_1-p_2)iM$, 
my textbook, Peskin & Schroesder, 
uses Wick's theorem. The theorem is applied to the following expression 
$⟨p_1',p_2'|iT|p_1,p_2⟩ = ⟨p_1',p_2'|\frac{(-ig)^2}{2}T\{\int \overline\psi(x) \psi(x) \phi(x) d^4x\int \overline\psi(y) \psi(y) \phi(y)d^4y\}|p_1,p_2⟩$
where T{...} is the time ordered product. I don't really know how to continue with the explicit calculation, how should I apply the contractions and how should I evaluate them?
Could you please show me the explicit calculation needed to get a final answer for the S matrix?
A great thank to whoever will take the time to help me!
 A: The first thing to do is contract the two instances of the scalar field. The scalar operator field has the expansion,
$$
\phi(t,x^{r})=\frac{1}{(2\pi)^{3/2}}\int\frac{d^{3}p}{2p^{0}}\left(\eta_{p}e^{-ip^{r}x^{r}}+\eta^{\dagger}_{p}e^{ip^{r}x^{r}}\right)
$$
Here $\eta_{p},\eta^{\dagger}_{p}$ are emission and absorption operators respectively (notation from Dirac's 1966 book "Lectures on Quantum Field Theory"). Also, I'm using Lorentz invariant measure in the integral. You have to expand the field operators into emission and absorption parts and the use the commutators to move the absorption operators to the right (normal ordering). If you do this this by hand you get an instance of Wick's theorem,
$$
\mathcal{T}(\phi(x)\phi(y))=\mathcal{N}(\phi(x)\phi(y))+D_{F}(x^{\mu}-y^{\mu})
$$
where $\mathcal{T}$ is time ordering and $\mathcal{N}$ is normal ordering and $D_{F}$ is the Feynman prescription propagator.  The in-going state is,
$$
|p_{1}\lambda_{1},p_{2}\lambda_{2}\rangle=\eta_{e^{-}p_{1}\lambda_{1}}\eta_{e^{-}p_{2}\lambda_{2}}|S\rangle
$$
where $\eta_{e^{-}p\lambda}$ is an emission operator for an electron with 3-momentum $p$ and helicity (spin along the momentum vector) $\lambda$ and $|S\rangle$ is the vacuum state of the free theory. The emission and absorption operators for the scalar particle commute with the fermionic operators so that the absorption operators for the scalar particle are moved to the right where they eventually kill the vacuum state. So the normally ordered bit vanishes and the invariant matrix element just gets a  factor $D_{F}(x-y)$.
Now the same thing has to be done for the fermion operator field. The expansion is,
$$
\psi(x)=\int\frac{d^{3}p}{(2\pi)^{3/2}}\frac{\sqrt{m}}{2p^{0}
}u_{e^{-}p\lambda}\eta^{\dagger}_{e^{-}p\lambda}e^{-ip_{\mu}x^{\mu}}
+\int\frac{d^{3}p}{(2\pi)^{3/2}}\frac{\sqrt{m}}{2p^{0}}u_{e^{+}p\lambda}\eta^{\dagger}_{e^{+}p\lambda}e^{ip_{\mu}x^{\mu}}
$$
The coefficients $u_{e^{-}p\lambda},u_{e^{+}p\lambda}$ are four component spinors (spinor index suppressed) and come from the plane wave solutions of the classical Dirac equation so that these four component spinors are completey known. $\eta^{\dagger}_{e^{-}p\lambda}$ is the absorption operator for an electron and $\eta^{\dagger}_{e^{+}p\lambda}$ is the emission operator for a positron. In the expansion, there is a sum over the repeated helicity labels $\lambda=\pm1/2$.In the calculations, I found it helpful to write this expansion as,
$$
\psi(x)=\psi_{e^{-}}(x)+\psi_{e^{+}}(x)
$$
so that $\psi_{e^{-}}(x)$ absorbs electrons and $\psi_{e^{+}}(x)$ emits positrons. Similarly,
$$
\bar{\psi}(x)=\psi^{\dagger}(x)\gamma^{0}=\bar{\psi}_{e^{-}}(x)+\bar{\psi}_{e^{+}}(x)
$$
and $\bar{\psi}_{e^{-}}(x)$ emits electrons and $\bar{\psi}_{e^{+}}(x)$ absorbs positrons.
We're trying to evaluate,
$$
\langle p'_{1}\lambda'_{1}p'_{2}\lambda '_{2}|\mathcal{T}\frac{(-ig)^{2}}{2!}\int d^{4}xd^{4}y\bar{\psi}(x)\psi(x)\phi(x)\bar{\psi}(y)\psi(y)\phi(y)|p_{1}\lambda_{1}p_{2}\lambda_{2}\rangle
$$
and the scalar field has already been contracted to $\mathcal{T}\phi(x)\phi(y)=D_{F}(x-y)$. Now we have to multiply out,
$$
\bar{\psi}(x)\psi(x)\bar{\psi}(y)\psi(y)=(\bar{\psi}_{e^{-}}(x)+\bar{\psi}_{e^{+}}(x))(\psi_{e^{-}}(x)+\psi_{e^{+}}(x))(\bar{\psi}_{e^{-}}(y)+\bar{\psi}_{e^{+}}(y))(\psi_{e^{-}}(y)+\psi_{e^{+}}(y))
$$
and consider the two time orderings $x^{0}>y^{0}$ and $y^{0}>x^{0}$. Consider just one term made randomly picking a term from each factor. It might be,
$$
\bar{\psi}_{e^{+}}(x)\psi_{e^{-}}(x)\bar{\psi}_{e^{-}}(y)\psi_{e^{+}}(y)
$$
Starting from the left these are absorb,absorb,emit,emit. This term is not normally ordered. We start to bring it into normally ordered form by moving the absorption term $\psi_{e^{-}}(x)$ one place to the right. This becomes,
$$
-\bar{\psi}_{e^{+}}(x)\bar{\psi}_{e^{-}}(y)\psi_{e^{-}}(x)\psi_{e^{+}}(y)+\bar{\psi}_{e^{+}}(x)[\psi_{e^{-}}(x),\bar{\psi}_{e^{-}}(y)]_{+}\psi_{e^{+}}(y)
$$
For $x^{0}>y^{0}$ the anticommutator is the propagator $S_{F}(x-y)$. The presence of the propagator means that we've done an internal contraction. However, in this problem we can forget about the internal contractions of fermion operators. In the example term, the contraction leaves the term $\bar{\psi}_{e^{+}}(x)\psi_{e^{+}}(y)$. The absorption operator $\bar{\psi}_{e^{+}}(x)$ can be moved to the right, anticommuting as it goes, until it kills the vacuum $|S\rangle$. By examining the different cases, it turns out that all cases of an internal fermionic contraction leave behind operators which either kill the vacuum state or give rise to forward scattering which is of no interest. 
So, when multiplying out the fermionic terms, we just move the operators by anticommuting and ignore any anticommutators that appear because these terms will be zero or give forward scattering. Furthermore, we can ignore any positron operators because the absorption operator $\bar{\psi}_{e^{+}}$ can be moved to the right until it hills the vacuum whilst emission operator $\psi_{e^{+}}$ can be moved to the left by anticommutation until it kills the vacuum bra $\langle S|$. 
The problem now looks tractable because we only have to deal with electron operators and we don't have to bother with internal contractions,
$$
\langle p'_{1}\lambda '_{1}p'_{2}\lambda '_{2}|\mathcal{T}\bar{\psi}_{e^{-}}(x)\psi_{e^{-}}(x)\bar{\psi}_{e^{-}}(y)\psi_{e^{-}}(y)|p_{1}\lambda _{1}p_{2}\lambda_{2}\rangle
$$
For $x^{0}>y^{0}$, normal ordering by moving the absorption operator $\psi_{e^{-}}(x)$ to the left,
$$
(-1)\langle p'_{1}\lambda '_{1}p'_{2}\lambda '_{2}|\bar{\psi}_{e^{-}}(x)\bar{\psi}_{e^{-}}(y)\psi_{e^{-}}(x)\psi_{e^{-}}(y)|p_{1}\lambda_{1}p_{2}\lambda_{2}\rangle
$$
We now have to contract the absorption operators with the in-going state.
$$
\psi_{e^{-}}(x)\psi_{e^{-}}(y)|p_{1}\lambda_{1}p_{2}\lambda_{2}\rangle=\psi_{e^{-}}(x)\psi_{e^{-}}(y)\eta_{e^{-}p_{1}\lambda_{1}}\eta_{e^{-}p_{2}\lambda_{2}}|S\rangle=\\\psi_{e^{-}}(x)\int\frac{d^{3}p}{(2\pi)^{3/2}}\frac{\sqrt{m}}{2p^{0}}u_{e^{-}p\lambda}\eta^{\dagger}_{e^{-}p\lambda}e^{-ip_{\mu}x^{\mu}}\eta_{e^{-}p_{1}\lambda_{1}}\eta_{e^{-}p_{2}\lambda_{2}}|S\rangle
$$
Moving the absorption operator $\eta^{\dagger}_{e^{-}p\lambda}$ to the left using the anticommutator,
$$
[\eta^{\dagger}_{e^{-}p\lambda},\eta_{e^{-}p'\lambda '}]_{+}=\langle p\lambda|p'\lambda'\rangle=2p^{0}\delta^{3}(p-p')\delta_{\lambda,\lambda'}
$$
we eventually get,
$$
\psi_{e_{-}}(x)\psi_{e^{-}}(y)|p_{1}\lambda_{1}p_{2}\lambda_{2}\rangle=\frac{m}{(2\pi)^{3}}\left(u_{e^{-}p_{1}\lambda_{1}}e^{-i(p_{1})_{\mu}y^{\mu}}u_{e^{-}p_{2}\lambda_{2}}e^{-i(p_{2})_{\mu}x^{\mu}}-u_{e^{-}p_{2}\lambda_{2}}e^{-i(p_{2})_{\mu}y^{\mu}}u_{e^{-}p_{1}\lambda_{1}}e^{-i(p_{1})_{\mu}x^{\mu}}\right)|S\rangle
$$
We get a similar expression for $\langle p'_{1}\lambda '_{1}p'_{2}\lambda '_{2}|\bar{\psi}_{e^{-}}(x)\bar{\psi}_{e^{-}}(y)$. Then we sandwich these last two expressions together and integrate over $d^{4}xd^{4}y$. The exponentials integrate to 4-d delta functions with momentum arguments. This is for $x^{0}>y^{0}$. The result for $y^{0}>x^{0}$ turns out to be the same expression obtained by swopping $x$ and $y$. So, the integrals go,
$$
\int_{x^{0}>y^{0}} d^{4}xd^{4}y<\text{expression}>+\int_{y^{0}>x^{0}}d^{4}xd^{4}y<\text{same expression}>\\
=\int_{\text{all spactime}}d^{4}xd^{4}y<\text{expression}>
$$
The final result is,
$$
\langle p'_{1}\lambda '_{1}p'_{2}\lambda '_{2}|\mathcal{T}\frac{(-ig)^{2}}{2!}\int d^{4}xd^{4}y\bar{\psi}(x)\psi(x)\phi(x)\bar{\psi}(y)\psi(y)\phi(y)|p_{1}\lambda_{1}p_{2}\lambda_{2}\rangle\\=\frac{(-ig)^{2}m^{2}}{(2\pi)^{2}}\left(\bar{u}_{e^{-}p'_{2}\lambda'_{2}}u_{e^{-}p_{2}\lambda_{2}}\bar{u}_{e^{-}p'_{1}\lambda '_{1}}u_{e^{-}p_{1}\lambda_{1}}\frac{i}{(p_{\mu}p^{\mu}-m_{\phi}^{2})}|_{p=p'_{1}-p_{1}=p_{2}-p'_{2}}\\-\bar{u}_{e^{-}p'_{1}\lambda '_{1}}u_{e^{-}p_{2}\lambda_{2}}\bar{u}_{e^{-}p'_{2}\lambda'_{2}}u_{e^{-}p_{1}\lambda_{1}}\frac{i}{(p_{\mu}p^{\mu}-m_{\phi}^{2})}|_{p=p'_{1}-p_{2}=p_{1}-p'_{2}}\right)
$$
