# How to use Wick's theorem to compute this matrix element?

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 $\lambda$

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

where

$$\mathcal{L}_{\mathrm{int}}=-\dfrac{\lambda}{4}\phi^2\psi^2.$$

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?

• 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. May 13, 2017 at 22:07