- #1

- 105

- 4

The energy functional is given by

\begin{equation}

E[\phi] = \int dx \frac{1}{2}\left( (\partial_{x}\phi(x) )^{2} + m^{2}\phi(x)^{2}\right) + V(\phi)

\end{equation}

Where ## x ## might as well be a n-dimensional vector and ## V(\phi) = \frac{\lambda}{4!}\phi^{4} ##. The partition function is then given by

\begin{equation}

Z[J] = \int D\phi \exp(-E[\phi] + J\phi) =\int D\phi \exp\left( -\int dx V(\frac{\delta}{\delta J(x)})\right) \exp(\frac{1}{2}JG_{0}(x,x')J)

\end{equation}

With ## J ## some current added for computational purposes. For the second equality sign I have rewritten the free part of the energy, completed the square and set the vacuum expectation energy to 1, and used(what in my notes is called) Wicks theorem on the interaction part. ## G_{0}(x,x') ## is the free propagator given by ## G_{0}(x,x') = \int \frac{dP}{(2\pi)^{D}} \frac{\exp(ip(x-x'))}{p^{2} + m^{2}}##. We now have the n-point correlation functions

\begin{equation}

G^{n}(x_{1},...,x_{n}) = \frac{\delta}{\delta J(x_{1})}...\frac{\delta}{\delta J(x_{n})} Z[J]\big|_{J=0}

\end{equation}

To calculate the 2-point function we expand equation 2 and performs the functional derivatives. To zeroth order in $ \lambda $ we (of cause) just gets

\begin{equation}

G^{2}_{0}=G_{0}

\end{equation}

To first order we get some 15 diagrams, some of which are tadpoles and some of which are disconnected. The first problem shows up now. I believe there is some easy argument as to why I can discard the disconnected diagrams, but I can not seem to remember what it is. Also is there any easy way to write the diagrams on this forum?\\

Lets use my intuition and throw away the disconnected diagrams. To first order we now have the diagramatic expansion.\\

Lets do the calculations in Fourier space. Using the usual Feynman rules the zeroth diagram becomes

\begin{equation}

I_{0} = \frac{1}{p^{2} + m^{2}}

\end{equation}

The first diagram hopefully becomes

\begin{equation}

I_{1} = \frac{\delta(p-p')}{p^{2} + m^{2}}\frac{\lambda}{(p')^{2}+m^{2}} \int \frac{dq}{(2\pi)^{D}} \frac{1}{q^{2} + m^{2}} = \frac{\delta(p-p')}{p^{2} + m^{2}}\frac{\lambda}{(p')^{2}+m^{2}} \frac{\Gamma (1-D/2)(m^{2})^{D/2-1}}{(4\pi)^{D/2}}

\end{equation}

So in three dimension I would be done by now?(besides having to Fourier transforming back again, and is that even possible?). Suppose 4 dimensions then. ## I_{1} ## diverges and we need to regularize. Using dimensional regularization we define ## \epsilon = 4-d ## and ## \tilde{\lambda} = \mu^{\epsilon}\lambda ##. The gamma-function part can now be written as

\begin{equation}

\frac{\lambda \Gamma(\epsilon/2 - 1)}{2(4\pi)^{2}} \left(\frac{4\pi \mu^{2}}{m^{2}}\right)^{\epsilon/2} = \frac{\lambda}{(4\pi)^{2}}m^{2}\left( \frac{1}{\epsilon} - \frac{1}{2}\log\left( \frac{m^{2}}{4\pi \mu^{2}}\right) + \frac{1}{2}\psi(2) + O(\epsilon) \right)

\end{equation}

With ## \psi(x) ## being the Digamma function.\\

This is as far as I got. Now what? I suppose I have to use some renormalization to throw away the divergent terms? And how about the renormalization group flow?