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Calculation of S-matrix elements in quantum field theory
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[QUOTE="samalkhaiat, post: 5465659, member: 35381"] Energy-momentum conserving delta function results from the space-time integration in the S-matrix expansion [tex]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 \} .[/tex] For QED [tex]\mathcal{H}_{I}(x) = -e N \big \{ \bar{\psi}(x) \gamma^{\mu}\psi (x) A_{\mu}(x) \big \} .[/tex] Expanding the field operators in terms of positive and negative frequency modes, and doing the normal ordering, we obtain [tex] \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*} [/tex] The first line contains the four elementary processes of photon-absorption, whereas the terms in the second line represent [itex]\gamma-[/itex]emission. These are not real physical processes because they don’t conserve both energy and momentum for physical photons [itex]k^{2} = 0[/itex], and fermions [itex]p^{2} = m^{2}[/itex]. Nevertheless, we can use them to see where the delta function comes from. Let us consider the pair creation process [tex]\gamma (\mathbf{k} , l ) \to e^{-}(\mathbf{p_{1}} , s) e^{+}(\mathbf{p_{2}}, r) .[/tex] So, we have [tex]| i \rangle = | \gamma ; \mathbf{k} , l \rangle = a^{\dagger}_{l}(\mathbf{k}) | 0 \rangle ,[/tex] [tex]| 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 .[/tex] Thus, the matrix element for this first-order process is given by [tex]\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 .[/tex] Now, I leave you to do the rest. Just substitute the following expansions, and use the anti-commutation relations for the fermionic operators, [itex]c , c^{\dagger} , d[/itex] and [itex]d^{\dagger}[/itex], and the commutation relations for the bosonic operators, [itex]a^{\dagger}[/itex] and [itex]a[/itex], to get rid of all operators and the vacuum state [tex]\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 } ,[/tex] [tex]\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 } .[/tex] [tex]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} .[/tex] Then, do the x-integration to obtain the delta function. The final result should look like [tex]\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} ,[/tex] where [tex]\mathcal{M} = i e \ \bar{u} (\mathbf{p}_{1}; s) \gamma^{\mu} \epsilon_{\mu} (\mathbf{k}; l ) v (\mathbf{p}_{2}; r ) .[/tex] [/QUOTE]
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