I am still struggling to get my head around QFT and whilst I think I understand the method of generating functionals to compute correlation functions (as in my question here), my course notes often use Wick's theorem for calculations, and I'd really like to understand how that works.
So if I return to the standard example of scalar Yukawa theory with interaction Lagrangian:
$$\mathcal{L}_{\text{int}}=-g\psi^{\dagger}\psi\phi$$
I would like to know how to compute quantities like:
$$\langle \Omega|T[\psi^{\dagger}(x)\psi(y)\phi(z)]|\Omega\rangle$$
Where $T[…]$ represents the time-ordering operator. My understanding is that this can be expressed as:
$$\langle \Omega|T[\psi^{\dagger}(x)\psi(y)\phi(z)]|\Omega\rangle = \lim_{t\to\infty}\frac{\langle0|T[\psi_{I}^{\dagger}(x)\psi_{I}(y)\phi_{I}(z)\exp\left(-i\int_{-t}^{t}\mathcal{H}_{\text{int}}\:\mathrm{d}^{4}s\right)]|0\rangle}{\langle0|T[\exp\left(-i\int_{-t}^{t}\mathcal{H}_{\text{int}}\:\mathrm{d}^{4}s\right)]|0\rangle}$$
We can expand the exponential to (arbitrarily) first-order in $g$:
$$\langle0|T[\psi_{I}^{\dagger}(x)\psi_{I}(y)\phi_{I}(z)\exp\left(-i\int_{-t}^{t}\mathcal{H}_{\text{int}}\:\mathrm{d}^{4}s\right)]|0\rangle = \langle 0|T\left[\psi_{I}^{\dagger}(x)\psi_{I}(y)\phi_{I}(z)(1-ig\int_{-t}^{t}\psi_{I}^{\dagger}(s)\psi_{I}(s)\phi_{I}(s)\:\mathrm{d}^{4}s)\right]|0\rangle$$
By Wick's theorem, the zeroth order in $g$ has only 3 operators and so will inevitably yield a contribution of 0. The second gives:
$$\langle0|\int_{-t}^{t}T[\psi_{I}^{\dagger}(x)\psi_{I}(y)\phi_{I}(z)\psi_{I}^{\dagger}(s)\psi_{I}(s)\phi_{}(s)]\:\mathrm{d}^{4}s|0\rangle$$
However, I'm not sure what the best way to proceed is. I can use Wick's theorem to expand the time-ordered string of operators in terms of products of operators, but then where do I go?
Best Answer
In your example there's two contractions giving two terms. Its amplitude is
$$ \left< \psi^{\dagger}_a(x) \psi^b(y) \phi(z) \right> = -i \lambda \int d^4 s \, \Delta_M(z - s) \Delta_m(y-s)^b_{\;c} \Delta_m(x-s)^c_{\;a} $$ $$ - i \lambda \int d^4 s \Delta_M(x - y) ^b_{\;a} \Delta_M(0)^c_{\;c} \Delta_m(s-x) + \mathcal{O}\left(\lambda^2\right). $$
How can I see that this is true? The algorithm is pretty simple, actually:
You have a term corresponding to each possible contraction. A contraction is a diagram where pairs of $(\psi^{\dagger}, \psi)$ and $(\phi, \phi)$ are replaced by the corresponding propagators.
You have to exclude the diagrams containing bubble subgraphs. A bubble graph is a graph with no external legs. An external leg is a contraction which has one of the fields appearing in the correlation bracket ($\psi^{\dagger}(x), \psi(y), \phi(z)$). This is because we want to account for the normalization $\mathcal{N}/\mathcal{N}_0$ as I mentioned in the answer to your previous question. Proofs of this can be found in any QFT textbook, e.g. Peskin-Schreder.
For each internal (interaction) vertex we have a factor of $$-i \lambda \int d^4 s. $$
Each term is a product of integrals over spacetime positions of internal (interaction) vertices and propagators.
In your example there's two $\phi$ fields, two $\psi$ fields and two $\psi^{\dagger}$ fields. Therefore we must contract $\phi$ with $\phi$, but we have a choice of which $\psi^{\dagger}$ gets contracted with specific $\psi$.
These correspond to the two diagrams below:
Both don't contain bubble subgraphs, however the second contains the tadpole divergence $\Delta_m(0)$. These divergences arise in QFT often. They have to be renormalized by requiring that $\left< \phi \right> = 0$, which is eqiuvalent to throwing away the tadpole contribution. (Actually, we don't just throw away mathematical expressions; we absorb them in the redefinition of fields).
The relevant part of your vertex amplitude is thus $$ \left< \psi^{\dagger}_a(x) \psi^b(y) \phi(z) \right> = -i \lambda \int d^4 s \, \Delta_M(z - s) \Delta_m(y-s)^b_{\;c} \Delta_m(x-s)^c_{\;a}. $$
Let me know if you have further questions.
P.S. oh and I supposed that $\psi$ and $\psi^{\dagger}$ are field multiplets, thus the internal indices $a, b, c$ labeling the components of the multiplet. If it's just a complex number then just throw these indices away :)