# Calculation of S-matrix elements in quantum field theory

Consider the following extract taken from page 60 of Matthew Schwartz's 'Introduction to Quantum Field Theory':

We usually calculate ##S##-matrix elements perturbatively. In a free theory, where there are no interactions, the ##S##-matrix is simply the identity matrix ##\mathbb{1}##. We can therefore write ##S=\mathbb{1}+i\mathcal{T}##, where ##\mathcal{T}## is called the transfer matrix and describes deviations from the free theory. Since the ##S##-matrix should vanish unless the initial and final states have the same total ##4##-momentum, it is helpful to factor an overall momentum-conserving ##\delta##-function: ##\mathcal{T}=(2\pi)^{4}\delta^{4}(\sum p)\mathcal{M}##. Here, ##\delta^{4}(\sum p)## is shorthand for ##\delta^{4}(\sum p^{\mu}_{i} - \sum p^{\mu}_{f})##, where ##p^{\mu}_{i}## are the initial particles' momenta and ##p^{\mu}_{f}## are the final particles' momenta. In this way, we can focus on computing the non-trivial part of the ##S##-matrix, ##\mathcal{M}##. In quantum field theory, "matrix elements" usually means ##\langle f|\mathcal{M}|i\rangle##. Thus we have

$$\langle f|S-\mathbb{1}|i\rangle = i(2\pi)^{4}\delta^{4}(\sum p)\langle f|\mathcal{M}|i\rangle.$$

Now, it might seem worrisome at first that we need to take the square of a quantity with a ##\delta##-function. However, this is actually simple to deal with. When integrated over, one of the delta functions in the square is sufficient to enforce the desired condition; the remaining ##\delta##-function will always be non-zero and formally infinite, but with our finite time and volume will give ##\delta^{4}(0)=\frac{TV}{(2\pi)^{4}}##. For ##|f\rangle \neq |i\rangle## (the case ##|f\rangle = |i\rangle##, for which nothing happens, is special),

$$|\langle f|S|i\rangle |^{2}=\delta^{4}(0)\delta^{4}(\sum p)(2\pi)^{8}|\langle f|\mathcal{M}|i\rangle|^{2} = \delta^{4}(\sum p)TV(2\pi)^{4}|\mathcal{M}|^{2},$$

where ##| \mathcal{M} |^{2} \equiv | \langle f | \mathcal{M} | i \rangle |^{2}.##

Why can we factor out the momentum-conserving delta function? Are we integrating the transfer matrix ##\mathcal{T}## over the some integration variable so that the delta function helps to conserve the ##4##-momentum?

samalkhaiat
Energy-momentum conserving delta function results from the space-time integration in the S-matrix expansion $$S = \sum_{n=0}^{\infty} S^{(n)} = \sum_{n=0}^{\infty} \frac{(-i)^{n}}{n!} \int d^{4}x_{1} \cdots d^{4}x_{n} \ T \big \{ \mathcal{H}_{I}(x_{1}) \cdots \mathcal{H}_{I}(x_{n}) \big \} .$$ For QED
$$\mathcal{H}_{I}(x) = -e N \big \{ \bar{\psi}(x) \gamma^{\mu}\psi (x) A_{\mu}(x) \big \} .$$
Expanding the field operators in terms of positive and negative frequency modes, and doing the normal ordering, we obtain
$$\begin{equation*} \begin{split} \mathcal{H}(x) =& -e \left[ \bar{\psi}^{+}\gamma^{\mu}\psi^{+} + \bar{\psi}^{-}\gamma^{\mu}\psi^{-} + \bar{\psi}^{-}\gamma^{\mu}\psi^{+} - \psi^{-}_{b}\gamma^{\mu}_{ab}\bar{\psi}^{+}_{a} \right] A^{+}_{\mu} \\ & - e \left[ \bar{\psi}^{+}\gamma^{\mu}\psi^{+} + \bar{\psi}^{-}\gamma^{\mu}\psi^{-} + \bar{\psi}^{-}\gamma^{\mu}\psi^{+} - \psi^{-}_{b}\gamma^{\mu}_{ab}\bar{\psi}^{+}_{a} \right] A^{-}_{\mu} . \end{split} \end{equation*}$$
The first line contains the four elementary processes of photon-absorption, whereas the terms in the second line represent $\gamma-$emission. These are not real physical processes because they don’t conserve both energy and momentum for physical photons $k^{2} = 0$, and fermions $p^{2} = m^{2}$. Nevertheless, we can use them to see where the delta function comes from. Let us consider the pair creation process $$\gamma (\mathbf{k} , l ) \to e^{-}(\mathbf{p_{1}} , s) e^{+}(\mathbf{p_{2}}, r) .$$ So, we have $$| i \rangle = | \gamma ; \mathbf{k} , l \rangle = a^{\dagger}_{l}(\mathbf{k}) | 0 \rangle ,$$ $$| f \rangle = | e^{-}; \mathbf{p_{1}}, s \rangle | e^{+}; \mathbf{p_{2}} , r \rangle = c^{\dagger}_{s}(\mathbf{p_{1}}) d^{\dagger}_{r}(\mathbf{p_{2}}) | 0 \rangle .$$
Thus, the matrix element for this first-order process is given by
$$\langle f | S^{(1)} | i \rangle = i e \int d^{4}x \ \langle 0 | c_{s}(\mathbf{p}_{1}) d_{r}(\mathbf{p}_{2}) \bar{\psi}^{-}(x) \gamma^{\mu} \psi^{-}(x) A^{+}_{\mu}(x) a^{\dagger}_{l}(\mathbf{k}) | 0 \rangle .$$
Now, I leave you to do the rest. Just substitute the following expansions, and use the anti-commutation relations for the fermionic operators, $c , c^{\dagger} , d$ and $d^{\dagger}$, and the commutation relations for the bosonic operators, $a^{\dagger}$ and $a$, to get rid of all operators and the vacuum state
$$\bar{ \psi }^{-}(x) = \sum_{\bar{\mathbf{p}}_{1} , \bar{s}} \left( \frac{m}{V E( \bar{\mathbf{p}}_{1} )} \right)^{1/2} c^{\dagger}_{\bar{s}}(\bar{\mathbf{p}}_{1}) \ \bar{u}_{\bar{s}}(\bar{\mathbf{p}}_{1}) \ e^{i \bar{p}_{1}x } ,$$
$$\psi^{-}(x) = \sum_{\bar{\mathbf{p}}_{2} , \bar{r}} \left( \frac{m}{V E(\bar{\mathbf{p}}_{2})} \right)^{1/2} d^{\dagger}_{\bar{r}}(\bar{\mathbf{p}}_{2}) \ v_{\bar{r}}(\bar{\mathbf{p}}_{2}) \ e^{i \bar{p}_{2}x } .$$
$$A^{+}_{\mu}(x) = \sum_{\bar{\mathbf{k}} , \bar{l}} \left( \frac{1}{2V \omega (\bar{\mathbf{k}})} \right)^{1/2} \epsilon_{\mu}(\bar{\mathbf{k}};\bar{l}) \ a_{\bar{l}} (\bar{\mathbf{k}}) \ e^{- i \bar{k}x} .$$
Then, do the x-integration to obtain the delta function. The final result should look like
$$\langle f | S^{(1)} | i \rangle = \left( \frac{m}{V E(\mathbf{p}_{1})} \right)^{1/2} \left( \frac{m}{V E(\mathbf{p}_{2})} \right)^{1/2} \left( \frac{1}{2V \omega (\mathbf{k})} \right)^{1/2} \ (4\pi)^{4} \delta^{4}( p_{1} + p_{2} - k ) \ \mathcal{M} ,$$ where
$$\mathcal{M} = i e \ \bar{u} (\mathbf{p}_{1}; s) \gamma^{\mu} \epsilon_{\mu} (\mathbf{k}; l ) v (\mathbf{p}_{2}; r ) .$$

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