I wanted to see how to use Wick's theorem in practice (I know with Feynman diagrams it is better, but here I want to do this with Wick's theorem only), so I considered computing the matrix element for the process $\phi(p_1)+\psi(p_2)\mapsto \phi(p_3)+\psi(p_4)$ considering following lagrangian at first order in $\lambda$

$$\mathcal{L}=\partial_\mu\phi \partial ^\mu\phi -\dfrac{m^2}{2}\phi^2+\partial_\mu\psi \partial^\mu \psi-\dfrac{\mu^2}{2}\psi^2+\mathcal{L}_{\mathrm{int}}$$



so we have two scalar fields with masses $m$ and $\mu$ which interact through this lagrangian.

We have an initial state with two particles with momenta $p_1$ and $p_2$ each of each field and similarly for the final state. The LSZ reduction formula tells then that

$$\langle f | S|i\rangle=\left[i\int d^{4}x_1 e^{-i p_1x_1}(\Box_1+m^2)\right] \left[i\int d^{4}x_2 e^{-i p_2x_2}(\Box_2+\mu^2)\right] \\ \left[i\int d^{4}x_3 e^{i p_3x_3}(\Box_3+m^2)\right]\left[i\int d^{4}x_4 e^{-i p_4x_4}(\Box_4+\mu^2)\right]\langle \Omega | T\{\phi_1\psi_2\phi_3\psi_4\}|\Omega\rangle$$

where I rename $\phi(x_i)=\phi_i$ and $\psi(x_i)=\psi_i$ for simplicity. Now I know that I can write

$$\langle \Omega | T\{\phi_1\psi_2\phi_3\psi_4\} |\Omega\rangle=\dfrac{\langle 0 |T\{\phi_1\psi_2\phi_3\psi_4 e^{i\int \mathcal{L}_{\mathrm{int}}}\} |0\rangle}{\langle 0|T\{e^{i \int \mathcal{L}_{\mathrm{int}}}\}|0\rangle}$$

Now I need to use Wick's theorem. Since I want the result for $O(\lambda)$ I expand the exponentials as

$$e^{i\int\mathcal{L}_{\mathrm{int}}}\approx1+i \int \mathcal{L}_{\mathrm{int}} d^4x=1-i\dfrac{\lambda}{4} \int \phi_{x}\phi_x \psi_x\psi_x d^4x.$$

Thus the numerator becomes

$$\mathcal{N}=\langle 0 | T\{\phi_1\psi_2\phi_3\psi_4\}|0\rangle-i\dfrac{\lambda}{4}\int d^4x \langle 0 |T\{\phi_1\phi_3\phi_x\phi_x\psi_2\psi_4\psi_x\psi_x\} |0\rangle$$

while the denominator becomes

$$\mathcal{D}=1-i\dfrac{\lambda}{4}\int d^4x\langle 0 |T\{\phi_x\phi_x\psi_x\psi_x\}|0\rangle=1-i\dfrac{\lambda}{4}\int d^4 x D_{xx}^\phi D_{xx}^\psi$$

with the notation $D_{ab}^{\circ}$ being the propagator for $\circ$ between $a$ and $b$.

For the numerator I've listed all possible full contractions and end up finding

$$\mathcal{N}=D_{13}^\phi D_{24}^\psi - \dfrac{i\lambda}{4}\int d^4 x\left[D_{13}^\phi D_{xx}^\phi D_{24}^\psi D_{xx}^\psi+2D_{13}^\phi D_{xx}^\phi D_{2x}^\psi D_{4x}^\psi+D_{1x}^\phi D_{3x}^\phi D_{24}^\psi D_{xx}^\psi+4D_{1x}^\phi D_{3x}^\phi D_{2x}^\psi D_{4x}^\psi\right]$$

and know we would need to let all those KG operators hit the object $\mathcal{N}/\mathcal{D}$ and I'm stuck because this got horrible.

It turns out that we can factor the denominator out of the first term of the numerator so that we get

$$\dfrac{\mathcal{N}}{\mathcal{D}}=\dfrac{D_{13}^\phi D_{24}^\psi \mathcal{D}-\dfrac{i\lambda}{2} \int d^4x D_{13}^\phi D_{xx}^\phi D_{2x}^\psi D_{4x}^\psi-\dfrac{i\lambda}{2}\int d^4x D_{1x}^\phi D_{3x}^\phi D_{24}^\psi D_{xx}^\psi-i\lambda \int d^4x D_{1x}^\phi D_{3x}^\phi D_{2x}^\psi D_{4x}^\psi}{\mathcal{D}}$$

I believe this is not the right way to do so. The denominator won't cancel.

What is the correct way to use Wick's theorem directly to compute this matrix element?

  • $\begingroup$ The correct way is to use Feynman diagrams and see that the denominator factorises out of the numerator when you consider terms of all order. $\endgroup$
    – gautampk
    May 13, 2017 at 22:07

1 Answer 1


That is not so horrible actually. Division by denominator just removes disconnected diagrams with vacuum-vacuum factors. So this is taken into account dropping from scratch such contributions. Once this is done, LSZ in momentum space is trivial, at least at tree level. Just amputate external propagators and put all on shell with suitable polarization vectors ( if spin > 0). All this is discussed in great details, e.g. , in Peskin's book.


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