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I'm wondering if $\phi^5$ is a descendant in $\phi^4$-theory in $d = 4 - \epsilon$ at the conformal Wilson-Fisher fixed point, where the coupling constant is $\lambda$. The e.o.m. tells us that $\phi^3$ is a total derivative (and therefore also a descendant) of $\phi$ (let us not care about the numerical factors, and suppress the spacetime indices on the derivatives that is summed over) \begin{equation} \partial^2\phi \sim \lambda\phi^3 \ . \end{equation} Now regarding $\phi^5$, we find from $\partial^2\phi^3$ and the e.o.m. that $\phi(\partial\phi)^2$ is related to $\phi^5$ \begin{equation} \partial^2\phi^3 \sim \phi(\partial\phi)^2 + \lambda\phi^5 \quad\text{or}\quad \phi(\partial\phi)^2 \sim \lambda\phi^5 + \partial^2\phi^3 \ . \end{equation} Since $\partial^2\phi^3$ is a descendant (which in turn is a descendant of $\phi$ due to the e.o.m.), we should only treat either $\phi^5$ or $\phi(\partial\phi)^2$ as a primary.

However, if we now study $\partial^4\phi$ using the equation above, we find \begin{equation} \partial^4\phi \sim \lambda^2\phi^5 + \lambda\partial^2\phi^3 \quad\text{or}\quad \lambda^2\phi^5 \sim \lambda\partial^2\phi^3 + \partial^4\phi \ . \end{equation} From this equation it looks like $\phi^5$ is a descendant of $\phi$. Is this correct?

If so, we can use the reasoning for $\phi^6$ (by studying $\partial^2\phi^4$ and $\partial^4\phi^2$) and find that $\phi^6$ is a descendant.

If my reasoning is incorrect, which ones are the primaries with scaling dimension $\Delta = 5 + \mathcal{O}(\epsilon)$?

Qmechanic
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A.Dunder
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1 Answers1

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Your last equation does not follow from the others.

The combinations of fields and derivatives which give primary operators in Wilson Fisher are almost the same as those for the free theory. Due to multiplet recombination, there is one exception at spin zero ($\phi^3$) and one exception at each odd spin $\geq 5$. If $\phi^5$ were to be eaten, the free theory would need to have a primary at the unitarity bound of dimension $5 - n$ which is killed by $n$ derivatives and not by any smaller number of derivatives. There is no such operator so $\phi^5$ stays primary.

Connor Behan
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  • What's wrong with my last equation? Using the e.o.m. $\partial^4\phi \sim \partial^2(\lambda\phi^3)$ and thus it follows. Or could it be that e.o.m. only applies to $\partial^2\phi$ and not higher derivatives?

    But assuming my last equation is wrong, then should $\phi^5$ be the only primary with scaling dimension $\Delta = 5 + \mathcal{O}(\epsilon)$?

     – A.Dunder Feb 17 '23 at 08:10
  • I doubt it. Three fields, two derivatives or two fields, three derivatives can probably get you another primary. – Connor Behan Feb 17 '23 at 08:49
  • Your last equation should say $\partial^4 \phi \sim \partial^2 (\lambda \phi^3) \sim \lambda^2 \phi^5 + \lambda \phi (\partial \phi)^2$ which is not what you wrote. – Connor Behan Feb 17 '23 at 08:53
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    I found the error now. In my last eq. I used my second eq. However, what I missed is that in my second eq. there's a minus sign in front of $\lambda\phi^5$. Due to this minus sign I find $\partial^4\phi \sim \lambda\partial^2\phi^3$. Thus $\phi^5$ is the only scalar primary with $\Delta = 5 + \mathcal{O}(\epsilon)$. Then as you say, there are other non-scalar primaries with $\Delta = 5 + \mathcal{O}(\epsilon)$ which contains single and cubic derivatives, but for my purposes I'm only interested in scalars (and thus even number of derivatives). Thank you for the help! – A.Dunder Feb 17 '23 at 09:12