2
$\begingroup$

In the one-loop renormalization of $\phi^4$-theory, only 1PI vertex functions $\Gamma^{(2)}$ and $\Gamma^{(4)}$ are regularized and renormalized. But they do not exhaust all the irreducible connected diagrams at one loop. One can have a diagram, for example, with one-loop, 3 vertices and 6 external lines, or with one-loop, 4 vertices and 8 external lines and so on. What about these diagrams? They respectively correspond to $\Gamma^{(6)}$ and $\Gamma^{(8)}$. What about these 1PI diagrams with one-loop? Shouldn't they require renormalization as well? In fact these diagrams contribute to the effective potential.

EDIT : arxiv.org/abs/hep-ph/9901312 This might be an useful reference. Please look at the one-loop diagrams in the calculation of the effective potential in $ϕ^4$-theory.

$\endgroup$
  • 3
    $\begingroup$ Alright, then this is a round-about way of asking how we know that $\phi^4$-theory is renormalizable (i.e., once when you take care of $\phi^2$ and $\phi^4$ operators, everything else is alright). $\endgroup$ – innisfree Sep 23 '15 at 6:58
3
$\begingroup$

The naive power counting approach for a $d$-dimensional theory with coupling constant $\lambda$ tells us that the amplitude of diagrams with $E$ external lines and $V$ vertices behaves with the cutoff $\Lambda$ as $\propto\Lambda^D$ with $$ D = d - [\lambda]V - \frac{d-2}{2}E$$ where $[\dot{}]$ is the mass dimension. Since $\phi^4$ in four dimension has a dimensionless coupling, $$ D = 4-E $$ and since only diagrams with $D \geq 0$ need renormalization, the only diagrams needing it in 4D $\phi^4$ are those with $E \leq 4$. All diagrams with an odd number of external lines vanish due to the $\phi\mapsto-\phi$ symmetry, so what's left to renormalize is $E=0,2,4$, which are the vacuum energy, the propagator, and the 4-vertex, respectively.

The diagrams you ask about exist, but have $D < 0$, and do not need to be renormalized, since they are not diverging when we take the cutoff to infinity.

$\endgroup$
  • $\begingroup$ @ACM Can you please make a contrast with $\phi^6$ theory? In particular, how can I understand that infinitely many diagrams have to be renormalized here? $\endgroup$ – SRS Mar 1 '17 at 8:58
  • $\begingroup$ @SRS All you need to do is plug in the different value of $[\lambda]$ and then think about which diagrams have $D\geq 0$ with this new formula. $\endgroup$ – ACuriousMind Mar 1 '17 at 12:16
  • $\begingroup$ @ACuriousMind I think the formula $D = d - [\lambda]V - [\phi]E$ holds true only for $\lambda \phi^N$ theories. How does one get the corresponding formula for more complicated theories? Or, is this a general formula for all theories? $\endgroup$ – Nanashi No Gombe Jul 23 '17 at 13:02
  • $\begingroup$ @NanashiNoGombe Of course it's not a general formula for all theories (how could it be, what would "$\lambda$" be in a theory with more than one coupling constant?), but the generalization is pretty straightforward (if awkward to write down) - you need to introduce different counts $V_i$ for different types of vertices with coupling constants $\lambda_i$. $\endgroup$ – ACuriousMind Jul 24 '17 at 8:01

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.