Calculating Green's function of interacting field theory

Question

I am currently wrestling with computation of Green's functions in an interacting field theory.

If I use the Yukawa field theory with a scalar $\phi$ field and a complex $\psi$ field as a simple example, I have the following Lagrangian:

$$\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}-\frac{1}{2}M^{2}\phi^{2}+\frac{1}{2}(\partial_{\mu}\psi^{\dagger})(\partial^{\mu}\psi)-\frac{1}{2}m^{2}\psi^{\dagger}\psi-g\psi^{\dagger}\psi\phi$$

If I wanted to find a Green's function like:

$$G_{1}^{(3)}(x_{1},x_{2},x_{3})=\langle\Omega|T\phi(x_{1})\psi(x_{2})\psi^{\dagger}(x_{3})|\Omega\rangle\quad\text{or}\quad G_{2}^{(3)}(x_{1},x_{2},x_{3})=\langle\Omega|T\phi(x_{1})\psi^{\dagger}(x_{2})\psi^{\dagger}(x_{3})|\Omega\rangle$$

I can do this using a generating functional, $Z[J,\eta,\eta^{\dagger}]$:

$$G^{(3)}_{1}(x_{1},x_{2},x_{3})=\frac{1}{Z[0,0,0]}\left.\frac{\delta^{3}Z}{\delta J\delta \eta\delta \eta^{\dagger}}\right|_{J=\eta=\eta^{\dagger}=0},\quad G_{2}^{(3)}(x_{1},x_{2},x_{3})=\frac{1}{Z[0,0,0]}\left.\frac{\delta^{3}Z}{\delta J\delta \eta^{\dagger}\delta \eta^{\dagger}}\right|_{J=\eta=\eta^{\dagger}=0}$$

However, I'm not sure how to derive the generating functional in this case. All of the notes that I can find talk about deriving a generating functional for a free scalar field theory.

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I know that we can write a Green's function in terms of the sum of all connected graphs with $n$ external lines. However, in this case I don't have the Feynman rules and so (I think) this doesn't help me

Answer 1

The generating functional, as always, is given by

$$ Z[J, \eta, \eta^{\dagger}] = \int \mathcal{D}\phi \mathcal{D} \psi \mathcal{D} \psi^{\dagger} \exp \left[ \frac{i}{\hbar} \int d^4 x \, \left( \mathcal{L}(\phi, \psi, \psi^{\dagger}) +J\phi + \eta^{\dagger}\psi + \psi^{\dagger}\eta\right)\right] = $$ $$ \exp \left[ -\frac{i g}{\hbar} \int d^4 x \, \frac{\delta}{\delta J} \frac{\delta}{\delta \eta} \frac{\delta}{\delta \eta^{\dagger}} \right] \; Z_0[J, \eta, \eta^{\dagger}], $$

where $Z_0$ is the generating functional of the free theory:

$$ Z_0[J, \eta, \eta^{\dagger}] = \hbar \int d^4 x \, \int d^4 y \, \left( \frac{1}{2} J(x) \Delta_M(x, y) J(y) + \eta^{\dagger}(x) \Delta_m(x, y) \eta(y) \right) $$

with $\Delta_m$ the Klein-Gordon propagator with mass $m$.

Your formula for correlations (aka Green's functions of the interacting theory) is correct. You have to Tailor-expand the exponential of the interacting term to the appropriate order in perturbation theory.

Just do the math and you will arrive at the correct expressions for correlation functions. Spoiler alert: the second one vanishes.

Btw you don't need to explicitly write the generating functional to calculate correlations. There's a simpler way: first, note that you can calculate expressions like

$$ \left< F[\phi, \psi, \psi^{\dagger}] \right>_0 = \mathcal{N}_0 \int\mathcal{D}\phi \mathcal{D}\psi \mathcal{D}\psi^{\dagger} \exp\left[ \frac{i}{\hbar} \int d^4 x \, \mathcal{L}_0(\phi, \psi, \psi^{\dagger}) \right] \cdot F[\phi, \psi, \psi^{\dagger}] $$

where $F$ is polynomial in fields with help of Wick's theorem. Also note the diagrammatic interpretation of the terms in the Wick expansion.

Then define

$$ \left< F[\phi, \psi, \psi^{\dagger}] \right> = \mathcal{N} \int\mathcal{D}\phi \mathcal{D}\psi \mathcal{D}\psi^{\dagger} \exp\left[ \frac{i}{\hbar} \int d^4 x \, \mathcal{L}(\phi, \psi, \psi^{\dagger}) \right] \cdot F[\phi, \psi, \psi^{\dagger}] = $$ $$ \frac{\mathcal{N}}{\mathcal{N}_0} \left< F \cdot \exp \left[ - \frac{i g}{\hbar} \int d^4 x \, \phi \psi^{\dagger} \psi \right] \right>_0. $$

and Tailor-expand the exponential of the interacting term to any order (given in advance). You will arrive at the polynomial expression (because both $F$ and the truncated Tailor series are polynomials in $\phi$, $\psi$ and $\psi^{\dagger}$). We already know how to compute these:

The expectation bracket, to each order in the perturbative series, is given by a sum of terms. Each term can be represented graphically as a Feynman diagram.

The expectation brackets are by definition normalized such that $$ \left<1\right>_0 = \left<1\right> = 1, $$

which means that $\mathcal{N}_0$ and $\mathcal{N}$ are related to each other. There's a very generic result, which is: the ratio $\mathcal{N} / \mathcal{N_0}$ corresponds to the product of all bubble graphs (the ones without external legs). These nicely factor out, giving a convenient way to calculate correlations:

The appropriate normalization can be accounted for by simply disregarding graphs with disconnected bubbles.