# Wick contraction in proton-pion production

Proton-pion production $\gamma + p \rightarrow \pi^0 + p$ occurs through the interaction hamiltonian $$\mathcal H_{int} = ig \bar \psi^{(p)} \gamma_5 \psi^{(p)} \phi + e \bar \psi^{(p)} \gamma_{\mu} \psi^{(p)} A^{\mu}$$ with $\phi$ the hermitean field of the $\pi^0$. I just want to ask a few things about Wick contractions that arise in this example.

The interaction hamiltonian tells us that there is no local $\gamma pp \pi^0$ vertex but that the process in question arises through $s$ and $t$ channel like diagrams mediated by a proton. At the lowest order, the first term appearing in the S matrix expansion is $$S^{(2)} = -\frac{i}{2} \int \text{d}^4 x_1 \text{d}^4 x_2\, T(\mathcal H(x_1) \mathcal H(x_2))$$ where the integrand may be evaluted using Wick's theorem. Explicitly, $$T(\mathcal H(x_1) \mathcal H(x_2)) = :(ig \bar \psi^{(p)} \gamma_5 \psi^{(p)} \phi + e \bar \psi^{(p)} \gamma_{\mu} \psi^{(p)} A^{\mu})_{x_1} (ig \bar \psi^{(p)} \gamma_5 \psi^{(p)} \phi + e \bar \psi^{(p)} \gamma_{\mu} \psi^{(p)} A^{\mu})_{x_2} : + \text{contractions}$$

The relevant terms are ones giving rise to a fermionic propagator namely terms involving a contraction of $\bar \psi$ and $\psi$. Terms that are contracted I will denote capatalised (If someone knows how to denote a contraction in the usual way please feel free to edit, I couldn't get the latex to work). So the relevant contractions are $$: (ig \bar{\Psi}^{(p)} \gamma_5 \psi^{(p)} \phi)_{x_1}( e \bar \psi^{(p)} \gamma_{\mu} \Psi^{(p)} A^{\mu})_{x_2} + (e \bar \Psi^{(p)} \gamma_{\mu} \psi^{(p)} A^{\mu})_{x_1} (ig \bar \psi^{(p)} \gamma_5 \Psi^{(p)} \phi)_{x_2} +$$ $$(ig \bar{\psi}^{(p)} \gamma_5 \Psi^{(p)} \phi)_{x_1} (e \bar \Psi^{(p)} \gamma_{\mu} \psi^{(p)} A^{\mu})_{x_2} + (e \bar \psi^{(p)} \gamma_{\mu} \Psi^{(p)} A^{\mu})_{x_1}(ig \bar{\Psi}^{(p)} \gamma_5 \psi^{(p)} \phi)_{x_2}:$$

Let's label the terms $(a), (b), (c)$ and $(d)$ respectively. Apparently, terms $(a)$ and $(c)$ are equal under integration over $x_1$ and $x_2$ and similarly terms $(b)$ and $(d)$ are equal. Can someone explain why? It is precisely this factor of 2 which cancels the factor of 2 in the pre factor to the first term above in the Dyson expansion.

• Because under the integral sign you can use the change of variable $x_1\leftrightarrow x_2$ in one of the terms, and it'll look exactly like the other? – AccidentalFourierTransform Apr 26 '16 at 20:09
• So many $p$'s... – Your Majesty Apr 26 '16 at 20:33
• @AccidentalFourierTransform: I was just wondering how we expect them to be equal in the first place - it seems to imply that $$:(ig \bar \Psi \gamma_5 \psi \phi)_{x_1} (e \bar \psi \gamma_{\mu} \Psi A^{\mu})_{x_2}: = :(ig \bar \psi \gamma_5 \Psi \phi)_{x_2} (e \bar \Psi \gamma_{\mu} \psi A^{\mu})_{x_1}:$$ but why is this true? Thanks! – CAF Apr 27 '16 at 6:14
• @caf no, what you wrote is not true in general. Let $a_i$ be a series of real numbers. Then in general $$a_i\neq a_j$$ but $$\sum_i a_i=\sum_j a_j$$ Its only when you sum over all $i,j$ that the series agree. The same is going on here: you have two functions of $x_1,x_2$ that are not equal, but if you integrate over $x_1,x_2$ you do get an equality. – AccidentalFourierTransform Apr 27 '16 at 12:32
• I see. But why is it that they contribute the same under integral? The contraction in term (a) involves an outgoing proton at $x_1$ and an incoming proton at $x_2$. The contraction in term (c) has an outgoing proton at $x_2$ and an incoming proton at $x_1$. Is it because only under the integral we can swap $x_1 \leftrightarrow x_2$ and thus have the proton going from $x_1$ to $x_2$ in both cases say? Thanks! – CAF Apr 27 '16 at 16:11