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  SUSY Kinetic and W potential terms: RG flow --- free or interacting

+ 1 like - 0 dislike
73 views

In this Seiberg's SUSY lecture

the professor said that the following theory with Kinetic and W potential terms:

$$
K=|\phi|^2
$$
$$
W=m\phi^2+g \phi^3
$$
1. "It is not a valid theory in $4d$, but it is infrared (IR) free, thus it is not an interacting QFT."

2. "In $2d$ and $3d$, they are valid interacting QFTs."


Could someone explain the logic? 

- what do free (quadratic lagrangian?) and interacting (higher order non-quadratic lagrangian?) mean respect to UV or respect to IR?

- I thought the $K=|\phi|^2$ requires derivative to be a kinetic term? Is he incorrect?

- I thought the $W=m\phi^2+g \phi^3$ are both relevant operators in the IR in 4d. Thus $g \phi^3$ changes the IR dynamics? Should this lead to an interacting QFT in 4d at IR?

- I thought the $m\phi^2$ is relevant and $g \phi^3$ is a marginal operator in the IR in 3d. Thus $g \phi^3$ again changes the IR dynamics? Should this lead to an interacting QFT in 3d?


- I thought the $m\phi^2$ is marginal and $g \phi^3$ is an irrelevant operator in the IR in 2d. Should this lead to a free QFT in 2d?

 ;


[![enter image description here][1]][1]


  [1]:

asked Apr 14 in Theoretical Physics by annie marie heart (1,205 points) [ revision history ]

here is the lecture link  ;

1 Answer

+ 1 like - 0 dislike

I'm struggling with my SUSY course right now so I'm not the most qualified person, but I think I can answer some of these at least. I'm sure the phi here are chiral superfields, i.e. 

\[\Phi = \phi + i\theta\sigma^\mu\bar\theta \partial_\mu\phi + \frac{1}{4}\theta^2\bar\theta^2 \partial^2\phi + \sqrt2\theta\psi - \frac{i}{\sqrt2}\theta^2\partial_\mu\psi\sigma^\mu\bar\theta + \theta^2 F.\]

So the kinetic term is something like

\[\int d^4\theta \ \bar\Phi \Phi = \bar F F - \partial_\mu\bar\phi\partial^\mu\phi+i\partial_\mu\bar\psi\bar\sigma^\mu\psi.\]

So they are indeed all derivatives. Similarly the potential term is

\[\int d^2\theta \ \frac{1}{2}m\Phi^2 + \frac{1}{3}\lambda\Phi^3 = mF\phi - \frac{m}{2}\psi\psi + \lambda\phi^2F - \lambda\phi\psi^2, \]

F has scaling dimension 2. So all terms are marginal in 4D. So yeah no I don't know why he said they're not interacting QFT, or that they're valid in 2/3D. I suspect it has something to do with the non renormalisation theory, which says the coefficients in the superpotential do not get quantum corrections at all. So if the theory is weakly interacting in the UV it is weakly interacting in the IR, in fact apparently it is weaker in the IR due to wavefunction renormalisation.

Hope this helps, but I'd like to know some proper explanation as well

answered Apr 20 by Wan [ no revision ]

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