Green's function in path integral approach (QFT)

Question

After having studied canonical quantization and feeling (relatively) comfortable with it, I have now been studying the path integral approach. But I don't feel entirely comfortable with.

I have the feeling that the main objective of the path-integral approach is to calculate the Green's function: \begin{equation} G^{(n)}(x_1,\ldots,x_n) = \langle 0 | \mathcal{T} \{\phi(x_1) \cdots \phi(x_n) \} |0\rangle = \left(\frac{1}{i}\right)^n \frac{\delta^n W[J]}{\delta J(x_1) \ldots \delta J(x_n)}\biggr|_{J=0} \end{equation} where, for simplicity, I have considered the neutral scalar field $\phi$ and $\mathcal{T}$ denotes the time-ordering operator. I have troubles understanding the physical meaning of the Green's function.

I understand that for the canonical quantization procedure, i.e. when $\phi$ is a field operator, $G^{(n)}(x_1,\ldots,x_n)$ is the vacuum expectation value. However, if I understand it correctly, in the path-integral approach we consider $\phi$ to be a classical field. I don't understand how rhyme these two different pictures.

Furthermore, for the canonical quantization formalism, we can represent the S-matrix: \begin{equation} S_{fi} = \langle f | S | i \rangle \end{equation} by Feynman diagrams. On the other hand, for the path-integral approach we seem to represent $G^{(n)}(x_1,\ldots,x_n)$ by Feynman diagrams. Do these Feynman diagram for the two different approaches somehow represent the same scattering amplitude?

Basically, I feel like I can't see the forest for the trees, and I am hoping someone can clarify the above described problems.

P.S. we have derived the LSZ reduction formula, and thus I understand that in the canonical quantization formalism we can express the S-matrix elements in terms of $G^{(n)}(x_1,\ldots,x_n)$. However, our lecturer told us that no one really uses the LSZ formula for practical purposes, and thus I don't think this answers my questions.

Answer 1

Good question; I remember spending hours trying to understand this when I first learned QFT. Let's address your two main points in turn. First, you say

I don't understand how rhyme these two different pictures.

Let's outline how to connect the two pictures in steps. It's a good exercise to try and work through all of the gory details yourself, so I encourage you to try!

1. For each admissible classical field configuration $\varphi:\mathbb R^3\to \mathbb R$, let $|\varphi,t\rangle$ denote a field configuration eigenstate at time $t$. Namely, \begin{align} \hat \phi(t,\mathbf x)|\varphi,t\rangle = \varphi( \mathbf x)|\varphi,t\rangle. \end{align} Take special note of the fact that $\hat\phi$ and $\varphi$ are different. The former is a operator valued distribution defined on spacetime, while the latter is a classical field configuration defined on space only.

2. Show that given admissible classical field configurations $\varphi_a,\varphi_b:\mathbb R^3\to \mathbb R$, there is a simple functional integral expression for the time-ordered expectation value from $|\varphi_a, t_a\rangle$ to $|\varphi_b, t_b\rangle$ of the product of a finite sequence of field operators: \begin{align} \langle \varphi_b, t_b| T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]&|\varphi_a, t_a\rangle \\ &= \int\limits_{\phi(t_a,\mathbf x) = \varphi_a(\mathbf x)}^{\phi(t_b,\mathbf x) = \varphi_b(\mathbf x)} \mathscr D\phi \,\phi(x_1)\cdots \phi(x_n)\,e^{iS_{t_a,t_b}[\phi]} \tag{$\star$} \end{align} where we have defined \begin{align} S_{t_a,t_b}[\phi] = \int_{t_a}^{t_b}dt\int d^3\mathbf x \,\mathscr L_\phi(t) \end{align} and $\mathscr L_\phi$ is the Lagrangian density of the theory.

3. Show that the expectation value on the left hand side of $(\star)$ can be used to compute a corresponding vacuum expectation value (vev); \begin{align} \lim_{t\to(1-i\epsilon)\infty} \frac{\langle \varphi_b, t| T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]|\varphi_a, -t\rangle}{\langle \varphi_b, t |\varphi_a, -t\rangle} &= \langle 0|T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]|0\rangle \end{align} where $\epsilon$ is a "positive infinitesimal" (namely you take the $\epsilon\to 0$ limit at the end). This is often called the $i\epsilon$ prescription; notice it's basically a clever trick for projecting out the ground state from a general expectation value.

4. Notice that the functional integration on the right hand side of $(\star)$ can be written as \begin{align} \int\limits_{\phi(t_a,\mathbf x) = \varphi_a(\mathbf x)}^{\phi(t_b,\mathbf x) = \varphi_b(\mathbf x)} \mathscr D\phi \,\phi(x_1)\cdots \phi(x_n)\,e^{iS_{t_a,t_b}[\phi] }=\left(\frac{1}{i}\right)^n \frac{\delta^n Z_{t_a, \varphi_a, t_b, \varphi_b}[J]}{\delta J(x_1)\cdots \delta J(x_n)}\bigg|_{J=0} \end{align} where we have defined \begin{align} Z_{t_a, \varphi_a, t_b, \varphi_b}[J] = \int\limits_{\phi(t_a,\mathbf x) = \varphi_a(\mathbf x)}^{\phi(t_b,\mathbf x) = \varphi_b(\mathbf x)} \mathscr D\phi \,e^{iS_{t_a,t_b}[\phi]+ i\int_{t_a}^{t_b}\int d^3\mathbf x\,J(x)\phi(x)} \end{align}

5. Combine steps 2-4 to show that if we define \begin{align} W[J] = \lim_{t\to(1-i\epsilon)\infty}\frac{Z_{t_a, \varphi_a, t_b, \varphi_b}[J]}{Z_{t_a, \varphi_a, t_b, \varphi_b}[0]}, \end{align} then we obtain our desired expression which gives vacuum expectation values in terms of path integrals: \begin{align} \langle 0|T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]|0\rangle &= \left(\frac{1}{i}\right)^n \frac{\delta^n W[J]}{\delta J(x_1)\cdots \delta J(x_n)}\bigg|_{J=0} \end{align}

Notice that \begin{align} G^{(n)}(x_1, \dots, x_n) = \langle 0|T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]|0\rangle \end{align} is just a suggestive definition which makes us think of Green's functions. It's suggestive because, for example, $G^{(2)}(x_1, x_2)$, the so called "two-point function," is the Green's function for the corresponding classical field theory.

Second, you ask

Do these Feynman diagram for the two different approaches somehow represent the same scattering amplitude?

The LSZ reduction formula is the answer to the question of how vevs, or equivalently Green's functions, are related to the $S$-matrix and scattering amplitudes, and above we have argued how the canonical formalism (which is formulated in terms of vevs) is related to the functional integral formalism, so we have found how the functional integral formalism allows us to compute the $S$-matrix. In practice, it's true, that you don't see people explicitly using the LSZ reduction formula, but that's because although it conceptually underlies the connection between Green's functions and the $S$-matrix, in practice people have already used LSZ to justify codified rules, namely Feynman rules, that allow one to go directly from Feynman diagrams (which simply represent terms in the perturbative expansions of Feynman integrals) to scattering amplitudes.