I am following David Tong's notes on QFT. On page 58, he applies Wick's theorem to $\psi\psi\rightarrow\psi\psi$ scattering for a scalar field with a Yukwara interaction term. The claim is, that by Wick's theorem, the only contribution to $$\int d^4x_1d^4x_2 T[\psi^\dagger(x_1)\psi(x_1)\phi(x_1)\psi^\dagger(x_2)\psi(x_2)\phi(x_2)]\tag{3.46}$$ is $$:\psi^\dagger(x_1)\psi(x_1)\psi^\dagger(x_2)\psi(x_2):\Delta_F(x_1,x_2).\tag{3.47}$$ That is, contracting on the two $\phi$ terms. However, I do not see why we don't get a contribution by contracting on a $\psi^\dagger$ and $\psi$, i.e, a contribution from

$$:\psi^\dagger(x_1)\psi(x_2):\Delta_F(x_1, x_2)\Delta_F(x_1,x_2).$$

when the one of incoming and outgoing momenta is the same. That is to say, if the initial state is $|pq>$ and the final state is $<pq|$, why do we not get a further second order contribution.

Indeed, if I try it, I get

$$<pq|:\psi^\dagger(x_1)\psi(x_2):|pq>$$ $$<0|2\sqrt{E_pE_q}b_{p}b_{q}^\dagger e^{i(-q.x_1+p.x_2)}|0>,$$ so

$$<pq|:\psi^\dagger(x_1)\psi(x_2)\Delta_F(x_1, x_2)\Delta_F(x_1,x_2):|pq>$$ $$=<p|\int \frac{d^4k}{2\pi^2} \frac{\delta(p-q)}{(k^2-m^2)((p-k)^2-m^2)}|q>,$$

which certainly gives a non-zero contribution when $p=q$.


1 Answer 1


Remmeber that if you contract two scalar fields you get a scalar propagator $\Delta_S(k)$. If you contract two fermionic fields, you get the fermion propagator $\Delta_F(k)=(\gamma^\mu k_\mu-m)\Delta_S(k)$. Upload your notation! On the ohter hand, the interaction process $\psi\psi\rightarrow \psi\psi$ requires the anihilation of two fermions and the creation of two fermions, so you need four fermionic fields uncontracted in terms that contribute to this process.

The term you mention is different since it has a loop on it, it is similar to the Fermion-Self energy diagram of QED, but instead of a photon line you have a scalar one. So the term you mention does not contribute to the process $\psi\psi\rightarrow \psi\psi$, instead they are corrections to the fermion propagator.

  • $\begingroup$ Thanks, this goes some way to helping me understand the problem. I have one further confusion: in the notes it is claimed that there is a 1-to-1 correspondence between Feynman diagrams and contributions from Wick's theorem. However, we seem here to be discounting particular diagrams and contributions from Wick (like in the question). Indeed, I can draw more Feynman diagrams with two vertices than contractions on $(3.46)$. So, is there a precise statement of this 1-to-1 relationship? How do we know which Feynman diagrams and which contractions to count? $\endgroup$
    – awsomeguy
    Jun 17, 2020 at 8:18
  • $\begingroup$ You are right, at second order in the Wick Expansion (WE), there are lots of terms that are non-zero, not only the ones that contribute to the process $\psi\psi\rightarrow\psi\psi$. All terms in the WE are valid as far as they respect conservation laws (momentum/charge), they just contribute to different processes. For example at second order, if you look at terms with one contraction, you have the process $\psi \phi \rightarrow \psi \phi$ too. Then, terms with two contractions yield to the processes $\psi\rightarrow\psi$ and $\phi\rightarrow\phi$ which are corrections to the propagator. $\endgroup$
    – vin92
    Jun 17, 2020 at 8:57
  • $\begingroup$ > "On the other hand, the interaction process $\psi \psi \to \psi \psi$ requires the anihilation of two fermions and the creation of two fermions." Why is that the case? Surely "annihilate just one fermion, then create one" and "do nothing at all" are consistent with $\psi \psi \to \psi \psi$. $\endgroup$
    – mpc
    Jan 14 at 18:45
  • $\begingroup$ You have your point but this is not how this process work as far as I know. C: $\endgroup$
    – vin92
    Jan 16 at 20:03

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