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In the paper by y Nathan Seiberg, T. Senthil, Chong Wang and Edward Witten,

A Duality Web in 2+1 Dimensions and Condensed Matter Physics

it is claimed on page 1 that the two theories

$$|D_{B}\phi|^{2}-g|\phi|^{4}\longleftrightarrow\frac{-1}{4e^{2}}\hat{f}_{\mu\nu}\hat{f}^{\mu\nu}+|D_{\hat{b}}\hat{\phi}|^{2}-\hat{g}|\hat{\phi}|^{4}+\frac{1}{2\pi}\epsilon^{\alpha\beta\gamma}\hat{b}_{\alpha}\partial_{\beta}B_{\gamma}$$

are dual and flow to Wilson-Fisher fixed point in the IR, where $\hat{f}_{\mu\nu}=\partial_{\mu}\hat{b}_{\nu}-\partial_{\nu}\hat{b}_{\mu}$. The classical mass dimensions are $ [ \phi ]=[ \hat{\phi} ]=1/2 $, $ [B]=[ \hat{b} ]=1 $, $ [g]=[ \hat{g} ]=1 $, and $ [e]=1/2 $. In the IR limit, I expect that $e\rightarrow\infty$ and $g,\hat{g}\rightarrow\infty$, so I can drop the kinetic term of $\hat{b}$ field $d\hat{b}\wedge\ast d\hat{b}$, and the theories become strongly coupled.

However, from this bachelor thesis Functional Renormalization Group for scalar field theories by Arthur Vereijken, the exact $\beta$-function of real $\phi^{4}$ scalar is computed via using the Wetterich's exact RG flow equation, which is widely used in the quantum gravity community.

It shows that the theory

$$S=\int d^{D}x \left\{\frac{1}{2}\phi(x)\left(-\partial^{2}+m^{2}\right)\phi(x)+\frac{\lambda}{4!}\phi(x)^{4}\right\}$$

$$\equiv\int d^{D}x\left\{\frac{1}{2}\phi(x)\left(-Z_{\Lambda}\partial^{2}+\Lambda^{2}\tilde{m}_{\Lambda}^{2}\right)\phi(x)+\frac{\Lambda^{4-D}}{4!}\lambda_{\Lambda}\phi(x)^{4}\right\}$$

has a Wilson-Fisher fixed point at

$$Z_{\ast}=1$$

$$\tilde{m}_{\ast}=\frac{D-4}{16-D}$$

$$\tilde{\lambda}_{\ast}=\frac{9\cdot 2^{D+5}\pi^{D/2}\Gamma(D/2+1)(4-D)}{(16-D)^{3}}$$

In D=2+1, the Wilson-Fisher fixed point has a finite coupling, with negative mass-squared $-1/13$, and so has a spontaneous symmetry breaking.

However, in the paper by Nathan Seiberg, T. Senthil, Chong Wang and Edward Witten, it clearly says that the Wilson-Fisher fixed point in 2+1 dimensions is massless.

Am I misunderstanding anything here?

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This is a difference in language, not in physics. QFTs at a critical point are by definition massless: two-point correlation functions of appropriately renormalized fields decay like a power law in position space, which in momentum space corresponds to propagators of the form $$ \Pi(p) = \frac{c}{(p^2)^{1+\gamma}} $$ for some number $\gamma$. The above function has no poles at finite $p^2$, so the theory is massless.

The second formula you're discussing computes something entirely different, namely which counterterms you need to add to the $\phi^4$ Lagrangian to flow to the critical point. The mass term $$ \Lambda^2 m_{\Lambda}^2 \phi^2$$ is just a bare coupling, a UV parameter of the theory. If you actually compute correlation functions in the theory, you will discover that the mass gap is zero, not $O(m_\Lambda)$.

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  • $\begingroup$ But the critical point computed in the second paper shows that it has negative mass squared. I am confused. $\endgroup$ – The Last Knight of Silk Road Jul 5 '18 at 15:57
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    $\begingroup$ @NewStudent: What you call the negative mass squared, i.e., the coefficient of $\phi^2$ in the Lagrangian is not the true mass. That's what TempAccount2020 was trying to explain. $\endgroup$ – Abdelmalek Abdesselam Jul 5 '18 at 18:52
  • $\begingroup$ I fully agree with @AbdelmalekAbdesselam. In a Gaussian QFT the $m^2$ coupling in the Lagrangian happens to be the physical mass, but in an interacting theory that's just not right - the mass gets renormalized due to interactions. That's the whole point of computing loop diagrams. I'm sure the Wetterich-style paper you quote has an extensive derivation of the value of $\tilde{m}_\Lambda$, and this derivation will show that it's the correct choice to get massless physics. $\endgroup$ – TempAccount2020 Jul 6 '18 at 3:31
  • $\begingroup$ As far as I understand from RG flow, integrating high frequency modes leads to quantum corrections to the bare couplings. At a certain energy scale (critical point), the couplings fix. This is the point where the beta function is zero. From the second paper, the Wilson-Fisher fixed point has a spontaneous symmetry breaking. Am I misunderstanding anything? Would you please elaborate? $\endgroup$ – The Last Knight of Silk Road Jul 6 '18 at 9:22

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