In Peskin and Schroeder (PS), the Feynman rules for calculating correlation functions are first presented. Only terms involving all field contractions need to be considered.

In Section 4.6, this is then extended towards calculating S-matrix elements. Now terms that aren't fully contracted may contribute.

For example, consider $$\langle p_{1}p_{2}|T\left(-i\frac{\lambda}{4!}\int d^4{x}\phi^{4}(x)\right)|p_{A}p_{B}\rangle=\langle p_{1}p_{2}|N\left(-i\frac{\lambda}{4!}\int d^4{x}\phi^{4}(x)+\textrm{contractions}\right)|p_{A}p_{B}\rangle$$ as in Equation 4.92 of PS. If we contract only two field operators, clearly terms including $a^{\dagger}a$ will still contribute. We need a way to deal with this.

Given that $\phi(x)=\phi^{+}(x)+\phi^{-}(x)$ PS shows that $\phi^{+}(x)|p\rangle=e^{-ipx}|0\rangle$, before stating

'An uncontracted $\phi$ operator inside the N-product of (4.92) has two terms: $\phi^{+}$ on the far right and $\phi^{-}$ on the far left.'

This is just the definition of normal ordering, I believe.

'We get one contribution to the S-matrix element for each way of commuting the $a$ of $\phi^{+}$ past an initial-state $a^{\dagger}$, and one contribution for each way of commuting the $a^{\dagger}$ of $\phi^{-}$ past a final state $a$.'

I don't think I get this. What does it mean by 'way of commuting'? Does it mean that, because $|p_{A}p_{B}\rangle=a^{\dagger}(p_{A})a^{\dagger}(p_{B})|0\rangle=a^{\dagger}(p_{B})a^{\dagger}(p_{A})|0\rangle$ there are two 'ways' - either past $a^{\dagger}(p_{A})$ or $a^{\dagger}(p_{B})$?

It is natural, then, to define the contractions of field operators with external states as follows: $$C(\phi(x)|p\rangle)=e^{-ipx}|0\rangle$$ $$C(\langle p|\phi(x))=\langle 0|e^{ipx}$$ I have denoted a contraction using $C()$. I don't really see how this follows from the above, assuming my interpretation so far is correct.

Could anybody possibly give a bit more detail as to what is being explained here? Thank you!


This business of "contractions" is just an overly formal way of saying the following: "commute all ladder operators until they annihilate the vacuum; your matrix element is given by whatever delta functions you pick up along the way."

Commuting the $a$ in $\phi(x)$ with $a^\dagger$ in $| p \rangle$ produces a delta function, but $\phi$ integrates over the momentum, so you pick up the coefficient $e^{-ip\cdot x}$ of $a$. (The factor $1/\sqrt{2 \omega_p} $ is cancelled by that in $| p \rangle$).


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