I am considering a scalar field theory with $\sim\phi^3$ interaction term, with Lagrangian \begin{equation} \mathcal{L} = \frac{1}{2}\left( \partial_\mu\phi\right)^2 - \frac{m^2}{2}\phi^2 - \frac{\eta}{3!}\phi^3.\tag{1} \end{equation} In the interaction picture, this gives the `interaction Hamiltonian density' as \begin{equation} \mathcal{H}_I = \frac{\eta}{3!}\phi^3,\tag{2} \end{equation} which I use to expand the $S$-matrix, \begin{equation} \hat{S} = T\left[ 1-\frac{i\eta}{3!}\int\mathrm{d}^4z\;\phi(z)^3 +\frac{(-i)^2}{2!}\left(\frac{\eta}{3!}\right)^2\int\mathrm{d}^4z\mathrm{d}^4w\;\phi(z)^3\phi(w)^3 +\ldots\right].\tag{3} \end{equation} I then consider the various contributions to the amplitude $$\mathcal{A}=\langle q|\hat{S}|p\rangle\tag{4}$$ using Wick's theorem. There are no first order (in $\eta$) terms.


Omitting prefactors, one of the second-order terms goes as \begin{equation} \mathcal{A}^{(2)}_1 \sim \langle 0|:\hat{a}_q\phi(z):|0\rangle \langle0|:\phi(z)\phi(z):|0\rangle\langle0|:\phi(w)\phi(w):|0\rangle\langle0|:\phi(w)\hat{a}^\dagger_p:|0\rangle,\tag{5} \end{equation} where $:\ldots:$ denotes a Wick contraction and $a_q/a^\dagger_p$ are annihilation/creation operators. This gives an amplitude \begin{equation} \mathcal{A}_1^{(2)} = \frac{(-i\eta)^2}{8}\int\mathrm{d}^4z\mathrm{d}^4w\;e^{iqz}\Delta(z-z)\Delta(w-w)e^{-ipw},\tag{6} \end{equation} where $\Delta(x-y)$ is the Feynman propagator.

What is the meaning of this term? It appears that I have an incoming particle with momentum $p$, which then turns into a bubble and vanishes from existence... likewise a bubble appears from nowhere and turns into a particle with momentum $q$. However, were I to carry out the $z$ integration, for example, would I not have a factor of $\delta(q)$? What is the interpretation of these delta functions?

  • $\begingroup$ These are disconnected diagrams and they do not contribute to the connected component of the $S$ matrix. Each of these diagrams on its own contributes to the one-point function of the scalar field so what you are really calculating here has the form $\langle \phi(p) \phi(q) \rangle = \langle \phi(p) \rangle \langle \phi(q) \rangle$. $\endgroup$
    – Prahar
    Commented Jul 1, 2021 at 6:43
  • $\begingroup$ Sure, I know that these types of diagrams don't contribute to the connected component, but I still struggle to visalise exactly what they do mean. $\endgroup$
    – dsfkgjn
    Commented Jul 1, 2021 at 7:30
  • $\begingroup$ OP's application (5) of Wick's theorem seems incorrect (v2). The RHS of eq. (5) is manifestly zero because $\langle 0|:\{f(a,a^{\dagger})-f(0,0)\}:|0\rangle=0$, where $f:\mathbb{C}^2\to \mathbb{C}$ is an arbitrary function. See also my Phys.SE answer here. $\endgroup$
    – Qmechanic
    Commented Jul 1, 2021 at 7:34
  • $\begingroup$ Could you define f, or explain this result in words? $\endgroup$
    – dsfkgjn
    Commented Jul 1, 2021 at 7:48

1 Answer 1


The offending amplitude is zero unless $p=q=0$. To see this we can carry out the integrations over $w$ and $z$, which gives us 4D delta functions: $$ \mathcal{A}_1^{(2)} = \frac{(-i\eta)^2}{8}\delta^{(4)}(p)\Delta(0)\delta^{(4)}(q)\Delta(0) $$ In this form it is clear that the term vanishes if $p\neq0$ or $q\neq0$.

The corresponding diagram (in ASCII-art) is --O O--. It describes a particle disappearing into the vacuum and another particle popping out of the vacuum again. By momentum conservation this is only possible if all components of the four-momenta of the two particles vanish. (Hence the two delta functions in the amplitude.) Note that if your particles are massive this amplitude always vanishes on-shell, so it does not contribute to physical processes.

  • $\begingroup$ I'd obtained these delta functions, but didn't know what to make of them. Regarding your last sentence, is this because there is a `zero point energy' which prevents massive particles from having zero momentum? $\endgroup$
    – dsfkgjn
    Commented Jul 5, 2021 at 7:48
  • 1
    $\begingroup$ On-shell particles have $p^2=m^2$, so their energy is at least $m$. It's not zero point energy, it's just their mass. But yes, they cannot have zero four-momentum. $\endgroup$ Commented Jul 5, 2021 at 20:32

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