If you have a self-interacting Lagrangian for a scalar field theory:

$$L= L_0 + L_I = \frac{1}{2} (\partial_\mu\phi)^2 - \frac{1}{2} m^2\phi^2- \frac{g}{4!}\phi^4$$

where $g$ is the coupling constant, the the lowest order contribution is given by:

$$S^{(1)} = -i \int{d^4 x\;T(H_I(x))}$$

where $H_I$ is the interaction Hamiltonian and $T$ the time ordering operator.

Suppose you want to compute this for two body scattering: $$s(p_1) + s(p_2) \rightarrow s(p_3) + s(p_4) $$

then one needs to compute $\langle p_4, p_3 \vert S \vert p_1,p_2\rangle$. My understanding was that this can be done by computing the following:

$$T \langle 0 \vert a_{p_4} a_{p_3} \phi(x)^4 a^\dagger_{p_2}a^\dagger_{p_1} \lvert0\rangle$$

To do this then I would compute all the contraction between the fields $\phi(x)$ and the creation and annhilation operators.

However, my lecture notes mention the following (the field is expanded in terms of creation and annhilation):

The Wick's theorem gives $T(\phi(x)^4) = :\phi^4 (x) :$, i.e. no contractions are necessary and we have $$: \phi \phi\phi\phi : = \phi^{+4} + 4\phi^-\phi^{+3} + 6\phi^{-2}\phi^{+2} + 4\phi^{-3}\phi^+ + \phi^{-4}$$. The only relevant term is $6\phi^{-2}\phi^{+2}$.

And then they proceed to calculate:

$$\langle p_3 p_4 \lvert 6\phi^{-2}\phi^{+2} \lvert p_1p_2 \rangle$$

However, I don't understand why does this follow and how this "shortcut" way of doing it works. Is my idea wrong?


1 Answer 1


$$ \mathcal M\sim\langle 0 |\hat{a}_{p_3}\hat{a}_{p_4}\hat S\hat{a}_{p_2}^\dagger\hat{a}_{p_1}^\dagger|0\rangle \\\mathcal M\sim \mathcal T\left[\langle 0 |\hat{a}_{p_3}\hat{a}_{p_4}\hat{a}_{p_2}^\dagger\hat{a}_{p_1}^\dagger|0\rangle+\frac{-ig}4\int\mathrm d^4z\langle 0 |\hat{a}_{p_3}\hat{a}_{p_4}\hat{\phi}(z)^4\hat{a}_{p_2}^\dagger\hat{a}_{p_1}^\dagger|0\rangle+\dots\right] $$ $\langle 0 |\hat{a}_{p_3}\hat{a}_{p_4}\hat{\phi}(z)^4\hat{a}_{p_2}^\dagger\hat{a}_{p_1}^\dagger|0\rangle$ can indeed be calculated by considering all possible contractions between the field, creation and annihilation operators, although this method is rather tedious. Since the annihilation and creation operators act at $t=\infty$ and $t=-\infty$ respectively, we only need to consider the time ordering of the field operators $\phi$.

Recall that in the quantum harmonic oscillator, an amplitude that looks like $$ \langle n|P(a^\dagger, a)|n\rangle $$ where $|n\rangle$ is the $n$'th energy eigenstate and $P$ is a polynomial, is only non-zero for the terms that have an equal number of $\hat a^\dagger$ and $\hat a$ (see this by commuting all the $\hat a$ past the $\hat a^\dagger$ until all the $\hat a$ act on $|n\rangle$ and the $\hat a^\dagger$ act on $\langle n|$, or vice versa). By an analogous argument, the only amplitude in QFT computations like these that is non-zero has an equal number of creation and annihilation components wedged between two asymptotic states that have an equal number of particles (if they don't, you need to compensate correspondingly with c/a operator(s) so that the number of particles matches).

Hence $\langle p_3p_4|\phi^{-}\phi^{-}\phi^{+}\phi^{+}|p_1p_2\rangle\sim \langle p_3p_4|\hat{a}^\dagger\hat{a}^\dagger\hat{a}\hat{a}|p_1p_2\rangle$ is the only non-zero term in the first-order scattering amplitude. In principle, this works even for more complicated interactions involving different fields (imposing the above argument separately for each particle type), although in practice this is again tedious to do because e.g. photon and spinor fields have multiple components.

  • $\begingroup$ Ok, I see that makes sense. Is this true for all interaction theories or only for scalar interactions? $\endgroup$
    – mathripper
    Apr 30, 2021 at 15:29
  • $\begingroup$ @mathripper in principle it works for all fields that split as $\hat\psi\sim(\hat a+\hat{a}^\dagger)$, although it becomes more complicated with e.g. spinor and photon fields that have multiple components $\endgroup$ Apr 30, 2021 at 16:24

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