Editing LSZ reduction formula (section)

==Reduction formula for fermions==
Recall that solutions to the quantized free-field [[Dirac equation]] may be written as

:<math>\Psi(x)=\sum_{s=\pm}\int\!\mathrm{d}\tilde{p}\big(b^s_\textbf{p}u^s_\textbf{p}\mathrm{e}^{ip\cdot x}+d^{\dagger s}_\textbf{p}v^s_\textbf{p}\mathrm{e}^{-ip\cdot x}\big),</math>

where the metric signature is mostly plus, <math> b^s_\textbf{p}</math> is an annihilation operator for b-type particles of momentum <math>\textbf{p}</math> and spin <math>s=\pm</math>, <math>d^{\dagger s}_\textbf{p}</math> is a creation operator for d-type particles of spin <math>s</math>, and the spinors <math>u^s_\textbf{p}</math> and <math>v^s_\textbf{p}</math> satisfy <math>(p\!\!\!/+m)u^s_\textbf{p}=0</math> and <math>(p\!\!\!/-m)v^s_\textbf{p} = 0</math>. The Lorentz-invariant measure is written as <math>\mathrm{d}\tilde{p}:=\mathrm{d}^3 p/(2\pi)^3 2\omega_\textbf{p}</math>, with <math>\omega_\textbf{p} = \sqrt{\textbf{p}^2+m^2}</math>. Consider now a scattering event consisting of an ''in'' state <math>|\alpha\ \mathrm{in}\rangle</math> of non-interacting particles approaching an interaction region at the origin, where scattering occurs, followed by an ''out'' state <math>|\beta\ \mathrm{out}\rangle</math> of outgoing non-interacting particles. The probability amplitude for this process is given by

:<math>\mathcal{M} = \langle \beta\ \mathrm{out}|\alpha\ \mathrm{in}\rangle,</math>

where no extra time-ordered product of field operators has been inserted, for simplicity. The situation considered will be the scattering of <math>n</math> b-type particles to <math>n'</math> b-type particles. Suppose that the ''in'' state consists of <math>n</math> particles with momenta <math>\{\textbf{p}_1,...,\textbf{p}_n\}</math> and spins <math>\{s_1,...,s_{n}\}</math>, while the ''out'' state contains particles of momenta <math>\{\textbf{k}_1,...,\textbf{k}_{n'}\}</math> and spins <math>\{\sigma_1,...,\sigma_{n'}\}</math>. The ''in'' and ''out'' states are then given by

:<math>|\alpha\ \mathrm{in}\rangle = |\textbf{p}_1^{s_1},...,\textbf{p}_n^{s_n}\rangle\quad\text{and}\quad|\beta\ \mathrm{out}\rangle = |\textbf{k}_1^{\sigma_1},...,\textbf{k}_{n'}^{\sigma_{n'}}\rangle.</math>

Extracting an ''in'' particle from <math>|\alpha\ \mathrm{in}\rangle</math> yields a free-field creation operator <math>b^{\dagger s_1}_{\textbf{p}_1,\mathrm{in}}</math> acting on the state with one less particle. Assuming that no outgoing particle has that same momentum, we then can write

:<math>\mathcal{M} = \langle\beta\ \mathrm{out}|b^{\dagger s_1}_{\textbf{p}_1,\mathrm{in}}-b^{\dagger s_1}_{\textbf{p}_1,\mathrm{out}}|\alpha'\ \mathrm{in}\rangle,</math>

where the prime on <math>\alpha</math> denotes that one particle has been taken out. Now recall that in the free theory, the b-type particle operators can be written in terms of the field using the inverse relation

:<math>b^{\dagger s}_\textbf{p} = \int\!\mathrm{d}^3 x\;\mathrm{e}^{ip\cdot x}\bar{\Psi}(x)\gamma^0 u^s_\textbf{p},</math>

where <math>\bar{\Psi}(x)=\Psi^\dagger(x)\gamma^0</math>. Denoting the asymptotic free fields by <math>\Psi_\text{in}</math> and <math>\Psi_\text{out}</math>, we find

:<math>\mathcal{M} = \int\!\mathrm{d}^3x_1\;\mathrm{e}^{ip_1\cdot x_1}\langle\beta\ \mathrm{out}|\bar{\Psi}_\text{in}(x_1)\gamma^0 u^{s_1}_{\textbf{p}_1}-\bar{\Psi}_\text{out}(x_1)\gamma^0 u^{s_1}_{\textbf{p}_1}|\alpha'\ \mathrm{in}\rangle.</math>

The weak asymptotic condition needed for a Dirac field, analogous to that for scalar fields, reads

:<math>\lim_{x^0\rightarrow-\infty}\int\!\mathrm{d}^3 x\langle \beta|\mathrm{e}^{ip\cdot x}\bar{\Psi}(x)\gamma^0 u^{s}_{\textbf{p}}|\alpha\rangle=\sqrt{Z}\int\!\mathrm{d}^3 x\langle \beta|\mathrm{e}^{ip\cdot x}\bar{\Psi}_\text{in}(x)\gamma^0 u^{s}_{\textbf{p}}|\alpha\rangle,</math>

and likewise for the ''out'' field. The scattering amplitude is then

:<math>\mathcal{M} = \frac{1}{\sqrt{Z}}\Big(\lim_{x_1^0\rightarrow-\infty}-\lim_{x^0_1\rightarrow+\infty}\Big)\int\!\mathrm{d}^3 x_1\;\mathrm{e}^{ip_1\cdot x_1}\langle\beta\ \mathrm{out}|\bar{\Psi}(x_1)\gamma^0 u^{s_1}_{\textbf{p}_1}|\alpha'\ \mathrm{in}\rangle,</math>

where now the interacting field appears in the inner product. Rewriting the limits in terms of the integral of a time derivative, we have

:<math>\mathcal{M} = -\frac{1}{\sqrt{Z}}\int\!\mathrm{d}^4 x_1\partial_0\big(\mathrm{e}^{ip_1\cdot x_1}\langle\beta\ \mathrm{out}|\bar{\Psi}(x_1)\gamma^0 u^{s_1}_{\textbf{p}_1}|\alpha'\ \mathrm{in}\rangle\big)</math>

::<math> =-\frac{1}{\sqrt{Z}}\int\!\mathrm{d}^4 x_1(\partial_0\mathrm{e}^{ip_1\cdot x_1}\eta(x_1)+\mathrm{e}^{ip_1\cdot x_1}\partial_0\eta(x_1)\big)\gamma^0 u^{s_1}_{\textbf{p}_1},</math>

where the row vector of matrix elements of the barred Dirac field is written as <math>\eta(x_1):=\langle\beta\ \mathrm{out}|\bar{\Psi}(x_1)|\alpha'\ \mathrm{in}\rangle</math>. Now, recall that <math>\mathrm{e}^{ip\cdot x}u^s_\textbf{p}</math> is a solution to the Dirac equation:

:<math>(-i\partial\!\!\!/+m)\mathrm{e}^{ip\cdot x}u^s_\textbf{p}=0.</math>

Solving for <math>\gamma^0\partial_0 \mathrm{e}^{ip\cdot x}u^s_\textbf{p}</math>,  substituting it into the first term in the integral, and performing an integration by parts, yields

:<math>\mathcal{M} = \frac{i}{\sqrt{Z}}\int\!\mathrm{d}^4x_1\mathrm{e}^{ip_1\cdot x_1}\big(i\partial_\mu\eta(x_1)\gamma^\mu + \eta(x_1)m\big)u^{s_1}_{\textbf{p}_1}.</math>
Switching to Dirac index notation (with sums over repeated indices) allows for a neater expression, in which the quantity in square brackets is to be regarded as a differential operator:

:<math>\mathcal{M} = \frac{i}{\sqrt{Z}}\int\!\mathrm{d}^4x_1\mathrm{e}^{ip_1\cdot x_1}[(i{\partial\!\!\!/}_{x_1} + m)u^{s_1}_{\textbf{p}_1}]_{\alpha_1}\langle\beta\ \mathrm{out}|\bar{\Psi}_{\alpha_1}(x_1)|\alpha'\ \mathrm{in}\rangle.</math>

Consider next the matrix element appearing in the integral. Extracting an ''out'' state creation operator and subtracting the corresponding ''in'' state operator, with the assumption that no incoming particle has the same momentum, we have

:<math>\langle\beta\ \mathrm{out}|\bar{\Psi}_{\alpha_1}(x_1)|\alpha'\ \mathrm{in}\rangle = \langle\beta'\ \mathrm{out}|b^{\sigma_1}_{\textbf{k}_1,\mathrm{out}}\bar{\Psi}_{\alpha_1}(x_1) - \bar{\Psi}_{\alpha_1}(x_1)b^{\sigma_1}_{\textbf{k}_1,\mathrm{in}}|\alpha'\ \mathrm{in}\rangle.</math>

Remembering that <math>(\bar{\Psi}\gamma^0u^s_\textbf{p})^\dagger = \bar{u}^s_\textbf{p}\gamma^0\Psi</math>, where <math>\bar{u}^s_\textbf{p}:=u^{\dagger s}_\textbf{p}\beta</math>, we can replace the annihilation operators with ''in'' fields using the adjoint of the inverse relation. Applying the asymptotic relation, we find

:<math>\langle\beta\ \mathrm{out}|\bar{\Psi}_{\alpha_1}(x_1)|\alpha'\ \mathrm{in}\rangle =\frac{1}{\sqrt{Z}}\Big(\lim_{y^0_1\rightarrow\infty}-\lim_{y^0_1\rightarrow-\infty}\Big)\int\!\mathrm{d}^3 y_1\mathrm{e}^{-ik_1\cdot y_1}[\bar{u}^{\sigma_1}_{\textbf{k}_1}\gamma^0]_{\beta_1}\langle\beta'\ \mathrm{out}|\mathrm{T}[\Psi_{\beta_1}(y_1)\bar{\Psi}_{\alpha_1}(x_1)]|\alpha'\ \mathrm{in}\rangle.</math>

Note that a time-ordering symbol has appeared, since the first term requires <math>\Psi_{\beta_1}(y_1)</math> on the left, while the second term requires it on the right. Following the same steps as before, this expression reduces to

:<math>\langle\beta\ \mathrm{out}|\bar{\Psi}_{\alpha_1}(x_1)|\alpha'\ \mathrm{in}\rangle =\frac{i}{\sqrt{Z}}\int\!\mathrm{d}^4y_1\mathrm{e}^{-ik_1\cdot y_1}[\bar{u}^{\sigma_1}_{\textbf{k}_1}(-i\partial\!\!\!/_{y_1}+m)]_{\beta_1}\langle\beta'\ \mathrm{out}|\mathrm{T}[\Psi_{\beta_1}(y_1)\bar{\Psi}_{\alpha_1}(x_1)]|\alpha'\ \mathrm{in}\rangle.</math>

The rest of the ''in'' and ''out'' states can then be extracted and reduced in the same way, ultimately resulting in

:<math>\langle \beta\ \mathrm{out}|\alpha\ \mathrm{in}\rangle=\int\!\prod_{j=1}^n \mathrm{d}^4 x_j \frac{i\mathrm{e}^{-ip_j x_j}}{\sqrt{Z}} [(i{\partial\!\!\!/}_{x_j}+m)u^{s_j}_{\textbf{p}_j}]_{\alpha_j}\prod_{l=1}^{n'}\mathrm{d}^4 y_l\frac{i\mathrm{e}^{ik_l y_l}}{\sqrt{Z}}[\bar{u}^{\sigma_l}_{\textbf{k}_l}(-i{\partial\!\!\!/}_{y_l}+m)]_{\beta_l} \langle 0| \mathrm{T}[\Psi_{\beta_1}(y_1)...\Psi_{\beta_{n'}}(y_{n'})\bar{\Psi}_{\alpha_1}(x_1)...\bar{\Psi}_{\alpha_n}(x_n)]|0\rangle.</math>

The same procedure can be done for the scattering of d-type particles, for which <math>u^s_\textbf{p}</math>'s are replaced by <math>v^s_\textbf{p}</math>'s, and <math>\Psi</math>'s and <math>\bar{\Psi}</math>'s are swapped.