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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.