I have a doubt about the following formula (label (1))

$\langle \Omega | T \{ \phi(x) \phi(y) \} | \Omega \rangle = \lim_{t\rightarrow \infty (1-i \epsilon)} \frac{ \langle 0 | T [ \phi_i(x) \phi_i(y) \exp \{ -i \int_{-t}^{t} dt V(t) \} ] | 0 \rangle}{\langle 0 | U(t,-t) | 0 \rangle}$

in perturbation theory, where $H=H_0 + V$, $|\Omega \rangle$ is $H$ ground state, $| \ 0 \rangle $ is $H_0$ ground state, $V$ a weak perturbation so that $\langle \Omega | 0 \rangle \neq 0$, $T[]$ is the time-ordered product and $\phi_i(x)$ denotes the scalar quantum field in the interaction picture. In the weak interaction and $t \rightarrow \infty$ approximations, it is found that $| \Omega \rangle$ can be obtained from the ground state of the free hamiltonian through a time evolution, i.e. $| \Omega \rangle = \lim_{t\rightarrow \infty (1-i \epsilon)} (e^{-iE_0 t} \langle \Omega | 0 \rangle )^{-1} e^{-iHt} | 0 \rangle$, where $ H_0 | \Omega \rangle =E_0 | \Omega \rangle $.

In Peskin-Schroeder it is explained how the denominator leads to the exclusion of the disconnected parts of Feynman diagram. Now, my doubt is: if the (n-points) Green function is given by

$G^{(n)}(x_1, ..., x_n) = \langle T [ \phi (x_1) ... \phi(x_n)] \rangle _0$

and $|\Omega \rangle$ is given by the evolution of $|0 \rangle$, how come equation (1) gives the connected parts of the diagrams?

Explicitly calculating it, it can be found that the denominator is a consequence of rewriting the above brackets with whole hamiltonian ground state in terms of the free hamiltonian ground state. So I was wondering if obtaining the connected Green function is a mere consequence of the two approximations and therefore of the theory being studied.


I assume that we are working with a fixed order in perturbation theory, so that the set of all possible Feynman graphs is finite. I also assume that the theory has been regularized such that all graphs evaluate to finite amplitudes.

We call a bubble graph a Feynman diagram with no external legs.

We call a connected graph a Feynman diagram such that among its connected components there are no bubbles. Important: note that this definition, despite being used heavily in many QFT textbooks, is somewhat counterintuitive. Usually a connected graph is a graph with a single connected component. Not so here!

Consider an arbitrary Feynman graph $G$. We can naturally split it into the connected subgraph $G_0$ (in the sense defined above) and the bubble graph $G_b$, which in the general case is a union of bubbles. It can be obtained directly from the Feynman rules that

$$ \text{Amp}(G) = \text{Amp}(G_0) \cdot \text{Amp}(G_b). $$

Now consider a sum of all possible Feynman graphs with some fixed set of external legs. This sum is over the Cartesian product of connected graphs (in the sense defined above) and bubble graphs. Thus the sum factors out as

$$ \sum \text{graphs} = \sum \text{connected graphs} \cdot \sum \text{bubble graphs}. $$

This result is referred to as "bubble factorization formula".

Now note that the denominator in your formula is just a special case of the numerator where there's no field operators before $U(t,-t) = T \exp \{-i \int_{-t}^t dt V(t) \}$, or in our diagrammatic terminology, where there's no external legs. There's only one connected graph with no external legs which is the empty graph $\emptyset$ with no vertices and no edges, for which (again, directly from Feynman rules)

$$ \text{Amp} (\emptyset) = 1. $$

Thus we conclude that the denominator is just a sum of bubble graphs, and from the bubble factorization formula we conclude that the fraction is equal to the sum of connected graphs.

All this reasoning can be made precise if we introduce an upper order of perturbation theory and a regularization scheme, as it was assumed in my answer.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.