# About unitarity and R-charge in 2+1 superconformal field theory

• How does unitarity require that every scalar operator in a $2+1$ SCFT will have to have a scaling dimension $\geq \frac{1}{2}$ ?

• Why is an operator with scaling dimension exactly equal to $\frac{1}{2}$ said to be "free (i.e decoupled from the rest of the theory)" ?

• Let us say that we know in some such theory the R-charge is monotonically decreasing with increasing coupling constant (say $\lambda$). Then define $\lambda _n ^f$ to be that value of the coupling constant at which the R-charge of the operator $Tr[\phi ^n]$ becomes $= \frac{1}{2n}$. Let $\lambda _n ^m$ be that that value of the coupling at which the R-charge of the same operator becomes $= \frac{2}{n}$ (..marginal ?..) Then clearly $\lambda _n ^m < \lambda_n ^f$.

How does this imply that there has to exist a $\lambda _c \leq \lambda_2^f$ where the theory might undergo a phase transition?

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## 1 Answer

### Why is the scale dimension bigger than 1/2?

The reason is the (Kallen-Lehman) spectral representation. Let

$$G(s-s') = \langle0|\phi^\dagger(s)\phi(s')|0\rangle$$

For any field $\phi$, and expand G's Fourier transform in energy states:

$$G(k) = \int ds {\rho(s)\over k^2-s + i\epsilon}$$

In other words, G is a superposition of propagators of mass-squared s with coefficient $\rho(s)$. then $\rho(s)>0$, because it is a positive definite norm of the state which is the Fourier transform of the field $\phi$ acting on the vacuum.

If $\rho(s)$ is a delta function at zero, the field has a massless free field propagator of (Euclidean) 1/r, and this falloff is as the square of the dimension of the field, so the field is dimension 1/2, the canonical free-field dimension. If you want another dimension for the field, you have to add together free field propagators with positive coefficients. In order to get a scale-free distribution, as required in a conformal theory, you must use a power-law superposition: $\rho(s) = s^{-\alpha}$, and for physical reasons $\alpha>0$, otherwise the $\rho$ is growing at large k. A growing density of states at large k means that the theory has an infinite number of different species at short distances.

So the falloff in k is (by dimensional analysis) ${1\over k^{2- 2\alpha}}$, which means the falloff in x is (again by dimensional analysis) ${1\over x^{1+2\alpha}}$, which means that the scale dimension is ${1\over 2}+\alpha$.

You can understand this intuitively by knowing that free propagators fall off as $1\over k^2$ and any faster falloff requires cancellations, which are forbidden in a quantum field theory by positivity.

### Why does scale dimension 1/2 imply free

the reason is that in a conformal theory, there is no scale for $\rho$, so it cannot have any shape other than a power law. So if the scale dimension is exactly 1/2, $\rho$ can only be a delta function at 0, and the operator has a free massless propagator. If it were interacting, it would have nonzero matrix elements with n-particle states which would give a positive $\rho$ somewhere away from zero.

### When are there allowed phase transitions?

I think that the criterion used here for allowing phase transitions is that you can deform the theory with an operator of dimensions less than or equal to 3 which makes the fields get a VEV. This also breaks the conformal invariance.

I believe that the operators that are being considered here are $tr(\bar{\phi}^2) + tr(\phi^2)$ and $|tr(\phi^2)|^2$. The dilatation generators are squares of the supersymmetry generators, so their scale transformation properties are determined from the R charge to go from 2 to 1/2 and from 4 to 1. So at some point, the $|tr(\phi^2)|$ term is going through dimension 3 and at bigger couplings than this you can stably deform the action to get a phase transition by adding the two terms and giving the trace of $\phi^2$ a VEV.

I am not sure about this, because I am not sure if the phase transition is off the conformal point. More context, like saying what theory you have in mind, would be helpful.

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