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What would be the Feynman Graphs be like for scalar fields like $\phi$ $\phi$ interactions? Also, how can it be summed up to orders of Mass^6.

I know it is the 2 pt function of the scalar fields. I don't understand how can I use only this interaction to draw and evaluate the feynmann graphs.

Can someone clearly explain this concept?

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  • $\begingroup$ Do you know what Feynman diagrams are? For free theories, they are not required, since the 2-point function can be evaluated exactly. $\endgroup$ – Prof. Legolasov Mar 31 '18 at 14:15
  • $\begingroup$ What if I have to draw the degenerate looking feynman diagrams for the 2 point function of the $\phi$ fields but with the mass term $\dfrac{1}{2} m^{2} \phi^{2}$ in the Lagrangian acting as an interaction term instead of simply a kinetic term? $\endgroup$ – SSS Mar 31 '18 at 14:27
  • $\begingroup$ Then you should’ve emphasized that in your question. The only allowed connected diagrams in this case are the “bamboo” graphs which sum giving the geometric series. $\endgroup$ – Prof. Legolasov Mar 31 '18 at 14:32
  • $\begingroup$ Do you mean the "bamboo" graphs will give me the sum up to orders of $m^{6}$? $\endgroup$ – SSS Mar 31 '18 at 14:36
  • $\begingroup$ No, they will give you an infinite geometric series which is equal to the full massive 2-point function. $\endgroup$ – Prof. Legolasov Mar 31 '18 at 14:43
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I assume you start with a massive Klein-Gordon QFT:

$$ S[\phi] = \int d^4 x \left( \frac{1}{2} (\partial \phi)^2 - \frac{1}{2}m^2 \phi^2 \right), $$

and you want to treat the massive term as a perturbation around the massless Klein-Gordon QFT just for the heck of it (even though an exact nonperturbative solution exists in this case).

Deriving Feynman rules shouldn't be a big problem. The easiest way is to use path integrals. The derivation then goes in three steps:

Step 1: you write down a generic path integral for some expectation functional $\Omega[\phi]$: $$ \left< \Omega \right> = \int D\phi e^{i S[\phi]} \Omega[\phi]. $$

Now (even though this integral can be computed rigorously) you wish to split $S[\phi]$ into the free (massless) and interaction parts, ending up with $$ \left< \Omega \right> = \int D\phi e^{i S_0[\phi]}\left( 1 + i\int d^4 x \left( - \frac{1}{2} m^2 \phi^2 \right) + \frac{1}{2!}i^2 \left[ \int d^4 x \left( - \frac{1}{2} m^2 \phi^2 \right)\right]^2 + \dots \right) \Omega[\phi]. $$

You then observe that the evaluation of the expectation $\Omega$ in the full theory is equivalent to the evaluation of the infinite series of the complex expectations containing powers of the space-time integrals in the "free" theory with action $S_0[\phi] = \int d^4 x \frac{1}{2} (\partial \phi)^2$.

You come up with a graphical way to represent the terms in your expansion. Since each of the functionals in the series is evaluated in a free theory, the Wick theorem holds, meaning that it is a sum of contractions. In the contractions, two types of vertices contribute: the external vertices coming from $\Omega$, and internal vertices coming from the integrals containing $m$. Moreover, the factor of $\frac{1}{2}$ cancels out $2!=2$ ways to contract a 2-valent vertex, so the expression for the internal (interaction) vertex is:

$$ A(v) = \int d^4 x_v i m^2. $$

Different terms in the series then correspond to diagrams with different number of interaction vertices.

Step 2: one remembers that the expectation can be normalized to give $\left<1\right> = 1$, if one does not consider vacuum bubble diagrams. This is standard QFT diagrammatics, textbook authors usually refer to it as "bubble graph exponentiation".

Step 3: one writes propagators of the free theory in terms of the momentum integrals:

$$ \Delta(x, y) = \int \frac{d^4 p}{(2\pi)^4} \frac{i e^{i p x}}{p^2 + i \varepsilon}.$$

Notice the absense of $m$.

Now we can perform the vertex integrals, ending up with the momentum-space Feynman rules:

  • For every propagator with momentum $p$ we associate $$\frac{i}{p^2 + i \varepsilon},$$
  • For every vertex we associate $$i m^2,$$
  • We integrate over undetermined loop momenta, and impose momemtum conservation.

Now one can calculate the actual 2-point function using the rules which she just derived. Consider $\Omega[\phi] = \phi(x) \phi(y)$ for $x, y$ spacetime points. The only contributing diagram is a bamboo graph with $n$ interacting vertices. Its contribution is given by

$$ \left< \phi(x) \phi(y) \right> = \sum_n \int \frac{d^4 p}{(2\pi)^4} e^{i p x} \left( \frac{i}{p^2 + i \varepsilon} \right)^{n+1} \cdot \left( i m^2 \right)^n = \int \frac{d^4 p}{(2\pi)^4} e^{i p x} \frac{i}{p^2 - m^2 + i \varepsilon},$$

where the last equation is the sum of the geometric series. Thus, we obtain the usual massive propagator. Perturbative and nonperturbative methods agree in this case, which is kinda nice.

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  • $\begingroup$ Thank you @Solenodon for this response I appreciate it. If I have to do the same thing, but using LSZ formalism and not Path integrals formalism, then how it should be dealt with? $\endgroup$ – SSS Apr 1 '18 at 12:03

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