- #1

JD_PM

- 1,131

- 158

- TL;DR Summary
- I want to understand how Feynman rules look in a weakly coupled theory that not only has an interaction term ##\phi^4## term but also a ##\phi^3## term (please see the Lagrangian density below). To do so I am comparing it to QED's Feynman rules (which I studied from Mandl & Shaw's second edition, section 7.3).

The weakly coupled theory is given by the Lagrange density$$\mathcal{L}=\mathcal{L}_0 + \mathcal{L}_I=\frac 1 2 \partial_{\mu} \phi \partial^{\mu} \phi - \frac{m^2}{2} \phi^2 - \frac{\lambda_3}{3!} \phi^3 - \frac{\lambda_4}{4!} \phi^4 \tag{1}$$

Where

\begin{equation*}

\mathcal{L}_0 = \frac 1 2 \partial_{\mu} \phi \partial^{\mu} \phi - \frac{m^2}{2} \phi^2

\end{equation*}

\begin{equation*}

\mathcal{L}_I = - \frac{\lambda_3}{3!} \phi^3 - \frac{\lambda_4}{4!} \phi^4

\end{equation*}

My aim is to derive what are the Feynman rules for such a theory.

I could go for the brute approach; evaluating the S-matrix expansion explicitly (where I use :: for normal ordering)

\begin{align*}

S_{fi}&=\langle f | T \left\{ \exp\left( i\frac{\lambda_3}{3!}\int d^4 x :\phi (x)\phi (x)\phi (x) : + i\frac{\lambda_4}{4!}\int d^4 x :\phi (x)\phi (x)\phi (x)\phi (x) : \right) \right\}| i \rangle \\

&= \langle f | i \rangle + i \frac{\lambda_3}{3!} \int d^4 x \langle f| :\phi (x)\phi (x)\phi (x): |i \rangle + i\frac{\lambda_4}{4!} \int d^4 x \langle f | :\phi (x)\phi (x)\phi (x)\phi (x) : | i \rangle \\

&+ \left(\frac{i\lambda_3}{3!}\right)^2\frac{1}{2!} \int d^4 x d^4 y \langle f| T\left\{ : \phi (x)\phi (x)\phi (x) :: \phi (y)\phi (y)\phi (y) : \right\}|i\rangle \\

&+ \left(\frac{i\lambda_4}{4!}\right)^2\frac{1}{2!} \int d^4 x d^4 y \langle f| T\left\{ : \phi (x)\phi (x)\phi (x)\phi (x) :: \phi (y)\phi (y)\phi (y)\phi (y) \right\}: |i\rangle + \mathcal{O}(\lambda_3^4, \lambda_4^4)

\end{align*}

Before evaluating the 4 terms above, let me first make a guess

As building blocks for Feynman diagrams, we will have 2 types of interaction vertices

## \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ##

Where I followed the convention of writing mesons with dotted lines.

The left diagram comes from ##i \frac{\lambda_3}{3!} \int d^4 x \langle f| :\phi (x)\phi (x)\phi (x): |i \rangle##. The right diagram comes from ##i\frac{\lambda_4}{4!} \int d^4 x \langle f | :\phi (x)\phi (x)\phi (x)\phi (x) : | i \rangle##

If we were to evaluate ##\left(\frac{i\lambda_3}{3!}\right)^2\frac{1}{2!} \int d^4 x d^4 y \langle f| T\left\{ : \phi (x)\phi (x)\phi (x) :: \phi (y)\phi (y)\phi (y) : \right\}|i\rangle## we would end up getting meson scattering i.e.

I have no guess for what we would get out of ##\left(\frac{i\lambda_4}{4!}\right)^2\frac{1}{2!} \int d^4 x d^4 y \langle f| T\left\{ : \phi (x)\phi (x)\phi (x)\phi (x) :: \phi (y)\phi (y)\phi (y)\phi (y) \right\}: |i\rangle##

So I would say that the Feynman rules basically stay the same except for the interaction vertices.

So my questions are:

Is there an alternative approach to derive the Feynman Rules for such a theory (which avoids the evaluation of the S-matrix)?

Do you agree with my guess?

Thank you

Where

\begin{equation*}

\mathcal{L}_0 = \frac 1 2 \partial_{\mu} \phi \partial^{\mu} \phi - \frac{m^2}{2} \phi^2

\end{equation*}

\begin{equation*}

\mathcal{L}_I = - \frac{\lambda_3}{3!} \phi^3 - \frac{\lambda_4}{4!} \phi^4

\end{equation*}

My aim is to derive what are the Feynman rules for such a theory.

I could go for the brute approach; evaluating the S-matrix expansion explicitly (where I use :: for normal ordering)

\begin{align*}

S_{fi}&=\langle f | T \left\{ \exp\left( i\frac{\lambda_3}{3!}\int d^4 x :\phi (x)\phi (x)\phi (x) : + i\frac{\lambda_4}{4!}\int d^4 x :\phi (x)\phi (x)\phi (x)\phi (x) : \right) \right\}| i \rangle \\

&= \langle f | i \rangle + i \frac{\lambda_3}{3!} \int d^4 x \langle f| :\phi (x)\phi (x)\phi (x): |i \rangle + i\frac{\lambda_4}{4!} \int d^4 x \langle f | :\phi (x)\phi (x)\phi (x)\phi (x) : | i \rangle \\

&+ \left(\frac{i\lambda_3}{3!}\right)^2\frac{1}{2!} \int d^4 x d^4 y \langle f| T\left\{ : \phi (x)\phi (x)\phi (x) :: \phi (y)\phi (y)\phi (y) : \right\}|i\rangle \\

&+ \left(\frac{i\lambda_4}{4!}\right)^2\frac{1}{2!} \int d^4 x d^4 y \langle f| T\left\{ : \phi (x)\phi (x)\phi (x)\phi (x) :: \phi (y)\phi (y)\phi (y)\phi (y) \right\}: |i\rangle + \mathcal{O}(\lambda_3^4, \lambda_4^4)

\end{align*}

Before evaluating the 4 terms above, let me first make a guess

As building blocks for Feynman diagrams, we will have 2 types of interaction vertices

Where I followed the convention of writing mesons with dotted lines.

The left diagram comes from ##i \frac{\lambda_3}{3!} \int d^4 x \langle f| :\phi (x)\phi (x)\phi (x): |i \rangle##. The right diagram comes from ##i\frac{\lambda_4}{4!} \int d^4 x \langle f | :\phi (x)\phi (x)\phi (x)\phi (x) : | i \rangle##

If we were to evaluate ##\left(\frac{i\lambda_3}{3!}\right)^2\frac{1}{2!} \int d^4 x d^4 y \langle f| T\left\{ : \phi (x)\phi (x)\phi (x) :: \phi (y)\phi (y)\phi (y) : \right\}|i\rangle## we would end up getting meson scattering i.e.

So I would say that the Feynman rules basically stay the same except for the interaction vertices.

So my questions are:

Is there an alternative approach to derive the Feynman Rules for such a theory (which avoids the evaluation of the S-matrix)?

Do you agree with my guess?

Thank you