0
$\begingroup$

In deriving the expression for the exact propagator

$$G_c^{(2)}(x_1,x_2)=[p^2-m^2+\Pi(p)]^{-1}$$

for $\phi^4$ theory all books that i know use the following argument:

$$G_c^{(2)}(x_1,x_2)=G_0^{(2)}+G_0^{(2)}\Pi G_0^{(2)}+G_0^{(2)}\Pi G_0^{(2)}\Pi G_0^{(2)}+...$$

wich is a geometric series so the formula for the exact propagator.Here

$$G_0^{(2)}$$

is the free propagator and

$$\Pi=X+Y+Z+...+W$$

is the sum of all irreducible diagrams.Here the irreducible diagrams is represented by $X,Y,Z,...,W$.

Using the path integral i can see, that connected diagrams $D$ can be written in the form

$$D=G_0^{(2)}XG_0^{(2)}Z...G_0^{(2)}W$$

Question: But how to prove that there is no constant $C$ so that instead we have

$$D=G_0^{(2)}XG_0^{(2)}Z...G_0^{(2)}W$$

we would have

$$D=CG_0^{(2)}XG_0^{(2)}Z...G_0^{(2)}W~?$$

In the last case we would not have a geometrical serie. Can someone explain me i t please or give me another way to derive it.

$\endgroup$
1
$\begingroup$

It is probably not useful to think of $\Pi$ in terms of the separate diagrams. If you keep the $\Pi$ as it is you can see that the full propagator is like the summation of a geometric series, but where one treats $G_0$ and $\Pi$ like operators.

Take your second expression and multiply it by $G_0\Pi$. Then subtract this result from the second expression. The resulting equation reads

$ G_c - G_0\Pi G_c = G_0 .$

Now one can operate on both sides with $(1-G_0\Pi)^{-1}$ to get

$ G_c = (1-G_0\Pi)^{-1} G_0 = (G_0^{-1}-\Pi)^{-1} . $

Since $G_0^{-1}=p^2-m^2$, you get back the full propagator (accept for a minus sign in front of $\Pi$, which I think is a typo in your question).

$\endgroup$
  • $\begingroup$ My main problem is how to problem is how to prove that reducible diagrams can be obtained from irreducible diagrams $\endgroup$ – Paul Dirac Aug 4 '16 at 5:28
  • $\begingroup$ What i want is to prove that the second equation is correct $\endgroup$ – Paul Dirac Aug 4 '16 at 5:37

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.