In solving the $\phi^4$ theory in Peskin, we define the contraction (on pg. 110 eq. 4.94)

$$C\left(\phi(x)|p \rangle \right)=\phi(x)|p \rangle . $$

That is, the contraction is just the action of $\phi(x)$ on the state $|p\rangle$. Peskin asserts that

$$\phi(x)|p\rangle =e^{-ipx}|0\rangle.$$

However, when I compute

\begin{align*} \phi(x)|p\rangle&=\int\frac{d^3p'}{(2\pi)^3}\left(a_{p'}e^{-ip'\cdot x}+ a_{p'}^\dagger e^{ip'\cdot x}\right)a^\dagger_{p}|0\rangle \\ &=\int\frac{d^3p'}{(2\pi)^3}\left([a_{p'},a^\dagger_p]e^{-ip'\cdot x}|0\rangle+ e^{ip'\cdot x} a_{p'}^\dagger a_p^\dagger|0\rangle\right)\\ &=\int\frac{d^3p'}{(2\pi)^3}\left((2\pi)^3\delta^3(p-p')e^{-ip'\cdot x}|0\rangle+ e^{ip'\cdot x} a_{p'}^\dagger a_p^\dagger|0\rangle\right)\\ &=e^{-ipx}|0\rangle + \int\frac{d^3p'}{(2\pi)^3}e^{ip'\cdot x} a_{p'}^\dagger a_p^\dagger|0\rangle \end{align*}

So, I am left with an extra term, but see no reason why it should vanish.

  • $\begingroup$ What edition of P&S are you using? In the one I have (1st), the equation you reference is different from the one you quote. Did you make a typo when transcribing it? $\endgroup$ May 30, 2018 at 3:25
  • $\begingroup$ Sorry, I made the proper edits. $\endgroup$ May 30, 2018 at 3:36
  • $\begingroup$ It's still different... $\endgroup$ May 30, 2018 at 3:40
  • $\begingroup$ I don’t have my textbook at the moment.. but I have the latest version.. I’ll update it soon $\endgroup$ May 30, 2018 at 3:57
  • $\begingroup$ The equation I wrote is how it’s written in the most recent version of Peskin. $\endgroup$ May 30, 2018 at 4:25

1 Answer 1


I ended up getting somewhere by myself on this one, so I figure I'd share in case somebody else had the same question. As Accidental Fourier Transform pointed out, the stated definition in my original question is not the proper definition of the contraction of a field with a momentum eigenstate. The correct definition is

$$ C(\phi | p \rangle) = e^{-ipx}\tag{a}$$

The correct way to think about this is that this contraction definition is just notation for the normal ordering of the uncontracted $\phi$ operators’s action on the momentum eigenstates.

Having said that, I still didn’t see the motivation for this definition. But it makes perfect sense. Here's the motivation. Using Wick's theorem in the $S$-matrix we have that

$$\langle p_1, p_2|N\left( -i\frac{\lambda}{4!}\int d^4x\ \phi^4 + \text{contractions}\right)|p_a,p_b\rangle.$$

Now, the "contractions" term just refers to the contractions of the $\phi$ operators amongst themselves, as usual. But as Peskin states on the previous page, these contractions only contribute to the trivial part of the $S$ -matrix, so we ignore these terms. This just leaves the uncontracted $\phi$s under the normal ordering operator.

To deal with this, we make use of the $\phi_+, \phi_-$ operators (as the way Peskin defines them). Plugging these in and expanding out, we get

$$-i\frac{\lambda}{4!}\int d^4x \langle p_1, p_2|N\left(\phi_-^4 \ldots + \phi_-\phi_-\phi_+\phi_+\ldots+\phi_+^4 \right)|p_a,p_b\rangle.\tag{1}$$

Now, using the fact that $[\phi_+,a^\dagger_p]=e^{-ipx}$ it is straight forward to show that

$$\phi_+\phi_+a^\dagger_{p_a}a^\dagger_{p_b}|0\rangle = 2e^{-ip_ax}e^{-ip_bx}|0\rangle \tag{2}$$.

Therefore, only terms with exactly two $\phi_+$ and $\phi_-$ survive. There are 6 such terms in equation (1). Recall that the $N$ operator makes them all $\phi_-\phi_-\phi_+\phi_+$. Using a similar relation as (2) for the two $\phi_-$, we get that

\begin{align*} \langle p_1, p_2|N\left(\phi_-^4 \ldots + \phi_-\phi_-\phi_+\phi_+\ldots+\phi_+^4 \right)|p_a,p_b\rangle &= (6)(2)(2)e^{-ip_ax}e^{-ip_bx}e^{ip_1x}e^{ip_2x}\\ &=4!e^{-ip_ax}e^{-ip_bx}e^{ip_1x}e^{ip_2x}, \end{align*}

which combinatorially justifies the contraction definition in (a).


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