Consider the following action with a fermionic field $\psi$ and a scalar field $\sigma$,

$S = \int d^dx \{ -\bar{\psi}(\gamma^\mu \partial_\mu +\sigma )\psi + \Lambda^{d-4}[ \frac{(\partial_\mu \sigma )^2 + m^2\sigma^2 }{2g^2 } + \frac{\lambda \sigma^4 }{4!g^4 } ] \} - (N'-1)Trln(\gamma^\mu \partial_\mu + \sigma )$

Assuming that this has a large-N saddle with uniform $\sigma$ one gets the large-N free energy density as,

$E(\sigma) = \Lambda^{d-4}(\frac{m^2\sigma^2}{2g^2} + \frac{\lambda \sigma^4 }{4!g^4 } ) - \frac{N}{2}\int^\Lambda \frac{d^dq}{(2\pi)^d}ln [\frac{q^2 + \sigma^2 }{q^2 } ]$

And the large-N saddle value of $\sigma$ is determined by the large-N gap equation, $E'(\sigma)=0$

Now from here how do the following conclusions come?

- Firstly that a non-trivial solution to the gap equation exists only when, $\frac{m^2}{g^2} < N\Lambda^{4-d}(\frac{1}{(2\pi)^d} \int^\Lambda \frac{d^dk }{ k^2 } ) $

How does this one come?

- Secondly from this apparently follows that the inverse $\sigma$ propagator in the massive phase is,

$\Delta_\sigma^{-1}(p) = \Lambda^{d-4}(\frac{p^2}{g^2} + \frac{\lambda \sigma^2}{3g^4} ) + \frac{N(p^2+4\sigma^2) } {2(2\pi)^d}\int^\Lambda \frac{d^dq }{(q^2+\sigma^2)((p+q)^2 + \sigma^2)} $

How does this equation come?

Now from this one can show that $\Delta_\sigma \sim \frac{2}{N b(d) p^{d-2} }$

From the above it follows that the canonical dimension of $\sigma$ is 1. How does one understand that the mass dimension of the field $\sigma$ does not depend on the space-time dimension?

Now I don't understand this argument which says that now since $[\sigma] =1$, both the terms $(\partial_\mu \sigma)^2$ and $\sigma^4$ are of dimension $4$ and hence for $2\leq d \leq 4$ these terms vanish in the IR critical theory? For this argument to work was it necessary that the IR theory was critical?

This post imported from StackExchange Physics at 2014-08-11 14:51 (UCT), posted by SE-user user6818