Tree-level of two-point function

Question

I am currently learning AQFT (advanced QFT) with the lecture notes of Osborn (I've had a course on QFT but this was not with the path integral formalism) and at some point he says the following

$$\mathcal{L} = - \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2 \phi^2 - \frac{1}{6} g \phi^3 - \frac{\lambda}{24}\phi^4.\tag{4.39}$$

Now we look at the tree-level of the two point function and we get $$\tau^{(0)}_2(p, -p) = -p^2-m^2\tag{4.40a}$$ but I do not see how they got there. In my head its just a "line" so I would take the propagator $\frac{1}{k^2+m^2}$ but this is not right. Now I have looked around here and elsewhere and I found something saying you just have to look at the quadratic part of $\mathcal{L}$ but I don't see why and thus wont take it as a proper response.

Any help would be greatly apreciated.

Edit: for 4-point it says $$\tau_4^{(0)}(p_1,p_2,p_3,p_4) = -\lambda\tag{4.40b}$$ and for 3-point $$\tau_3^{(0)}(p_1,p_2,p_3) = -g.\tag{4.40c}$$

The rules given are

1. consider all 1PI graphs with $n$ external lines

2. internal line add factor $\frac{1}{k^2+m^2}$

3. each $n$ vertex add $-V_n(0)$

4. do integral with loop momenta

5. symmetry factor

Edit 2: I saw the post How are propagator and two-point function related? but it doesn't really help me. I guess I can kind of follow but when I do i get confused as to why all the other rules stay the same and how it is possible that they are physically different (give a different cross section).

Answer 1

Suppressing possible overall $i$-factors the logic is essentially as follows:

• $\tau^{(L)}_n$ presumably denotes the $n$-point $L$-loop 1PI correlation function in momentum space (without the delta function imposing momentum conservation).

• Note that the tree $\tau^{(0)}_n$ cannot contain any free propagators, since cutting any of them would disconnect the diagram, contradicting the 1PI property.

• Diagrammatically, $\tau^{(0)}_n$ must be just an amputated $n$-vertex.

• If we attach $n$ free propagators $G_0=\frac{1}{k^2+m^2}$ to the amputated $n$-point function $\tau^{(0)}_n$, we apparently get a connected tree-level $n$-point function $$G^{(0)}_{c,n}~=~\underbrace{G_0\ldots G_0}_{n\text{ factors}} \tau^{(0)}_n.$$

• Since the interaction part of the Lagrangian (4.39) does not$^1$ have an interaction 2-vertex, the connected tree-level $2$-point function $$G^{(0)}_{c,2}~=~G_0\tau^{(0)}_nG_0~=~G_0$$ is apparently just a free propagator.

• Altogether it follows$^1$ that $\tau^{(0)}_2=G_0^{-1}=k^2+m^2$ is an inverse free propagator, cf. OP's question.

Related Phys.SE post: How are propagator and two-point function related?

References:

1. Hugh Osborn, Advanced QFT notes, 2019.

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$^1$ This conclusion depends on how the Lagrangian (4.39) is split into a free and an interaction part in perturbation theory. Example: One could e.g. assign the free propagator $G_0=\frac{1}{k^2}$ to be massless. Then the mass term in eq. (4.39) constitutes a 2-vertex interaction term. $\tau^{(0)}_2=k^2+m^2$ still has the same value, but would now be a sum of an inverse free propagator $G_0^{-1}=k^2$ and a mass term $m^2$, because now there is 2 relevant perturbative Feynman diagrams.

Answer 2

Basically it all comes down to Gaussian integrals. Generally we don't know how to evaluate integrals of the form (written somewhat loosely)

$$\int \prod_i d\phi_i e^{-\frac{1}{2} \phi^T\Sigma^{-1}\phi - g\phi^3 + ...} $$

but we do know how to calculate Gaussian integrals. So we expand everything beyond the quadratic term as a power series (assuming that $g$ is small enough and the series converges)

$$ = \int \prod_i d\phi_i e^{-\frac{1}{2} \phi^T\Sigma^{-1}\phi}(1 - g\phi^3 + \frac{g^2}{2!}\phi^6 + ...)$$

such that we can calculate any expectation in terms of the Gaussian higher-order moments. For example

$$\langle \phi^2 \rangle = \langle \phi^2 \rangle_F - g\langle \phi^5 \rangle_F + \frac{g^2}{2!}\langle \phi^8 \rangle_F + ...$$

where $\langle \cdot \rangle_F$ denotes expectation with respect to the Gaussian (aka free theory) distribution. Note that the Gaussian two-point correlation function is just

$$ \langle \phi_i \phi_j \rangle_F = \int \prod_k d\phi_k (\phi_i \phi_j)e^{-\frac{1}{2} \phi^T\Sigma^{-1}\phi } = \Sigma_{ij}$$

which is why the free propagator is the inverse of the quadratic coefficient in the Lagrangian. The higher order Gaussian moments can all be expressed in terms of the covariance matrix $\Sigma$ (the free propagator) using Isserlis theorem, which can also be represented diagrammatically with Feynman diagrams. This is where the rest of the Feynman rules comes from.

(Note: I've written the path integral in a discretized form for simplicity, and also ignored normalization factors, but the moral of the story is the same.)