# Index Concordia

## Advanced Quantum Field Theory II – Week 15

Posted in Quantum Field Theory, Supersymmetry by Index Guy on May 4, 2009

The last week of the semester was dedicated to Feynman rules in superspace. Before we go into that, it is good to review and set our conventions in superspace.

A remark: as in any quantum field theory analysis, here we will see the need for regularization. Now we need to to preserve the supersymmetry of the theory and the task of finding a regulator that obeys this criteria is non-trivial. Nevertheless, there exist such regularization schemes, namely dimensional reduction.

In any case, as action we will take the Wess-Zumino model coupled to super Yang-Mills. Of course this theory only has simple supersymmetry, but actually this is one of the only cases where superspace methods are useful. I do not want to rule out superspace methods for extended supersymmetry since one never knows what one might end up doing for his/her Ph.D. thesis. 😉 The action has the form:

$S = \displaystyle\int d^{4} x d^{4} \theta \bar{\phi}e^{V}\phi + \int d^{4}x d^{2}\theta W\left(\theta\right) + h.c. + S_{SYM}$.

The prepotential is

$W\left(\theta\right) = \displaystyle\frac{1}{2}m \phi^{2} + \frac{1}{3!}\lambda \phi^{3}$.

In four-dimensional Minkowski spacetime the Lorentz group $SO(1,3)$ is doubly covered by $SL(2, \mathbb{C})$. We will label spinors with a Weyl index $\alpha, \dot{\alpha}$. The supercovariant derivatives, (giving objects that are covariant under supersymmetric transformations) are defined as follows:

$D_{\alpha} = \displaystyle \frac{\partial}{\partial \theta^{\alpha}} - i \sigma^{a} _{\alpha \dot{\beta}}\theta^{\dot{\beta}}\partial_{a} \qquad \bar{D}_{\dot{\alpha}} = \displaystyle\frac{\partial}{\partial \bar{\theta}^{\dot{\alpha}}} - i \sigma^{a} _{\beta \dot{\alpha}}\theta^{\beta}\partial_{a}$

Now we set the convention for raising and lowering Weyl indices (which actually we do not follow in the definition of the supercovariant derivatives): Weyl indices are raised with the two-dimensional antisymmetric symbol in the ¨from north-west to south-east¨ fashion. Namely,

$\psi^{\alpha} = \epsilon^{\alpha \beta} \psi_{\beta} \qquad \bar{\psi}^{\dot{\alpha}} = \epsilon^{\dot{\alpha} \dot{\beta}} \bar{\psi}_{\dot{\beta}}$

The bars on spinors with dotted indices will be omited in what follows (they are redundant anyway). For example, we have the anticommutators for the supercovariant derivatives:

$\displaystyle\left\{ D_{\alpha}, D_{\dot{\beta}} \right\} = -2i \sigma^{a}_{\alpha \dot{\beta}}\partial_{a} \qquad \left\{D_{\alpha}, D_{\beta}\right\} = 0 \qquad \left\{ D_{\dot{\alpha}}, D_{\dot{\beta}}\right\} = 0$

$\displaystyle \int d^{2} \theta \theta^{2} = \int \frac{1}{2}d\theta^{1}d\theta^{2} \theta^{\alpha}\theta_{\alpha} = 1$

Similarly for the dotted coordinates:

$\displaystyle \int d^{2} \bar{\theta} \bar{\theta}^{2} = \int \frac{1}{2}d\bar{\theta}^{2}d\bar{\theta}^{1} \theta^{\dot{\alpha}}\theta_{\dot{\alpha}} = 1$

This expression follows from the single integration:

$\displaystyle \int d\theta^{\alpha} \theta^{\beta} = \delta^{\alpha \beta}$

The integral of anticommuting variables can be written in terms of the derivative.

$\displaystyle \int d^{2} \theta \left(\cdot\right) = -\frac{1}{4}\partial^{\alpha}\partial_{\alpha}\left(\cdot\right) \qquad \int d^{2} \bar{\theta} \left(\cdot\right) = -\frac{1}{4}\partial^{\dot{\alpha}}\partial_{\dot{\alpha}}\left(\cdot\right)$

This will be useful later. Total supercovariant derivatives integrate to zero:

$\displaystyle \int d^{4} x d^{4}\theta D_{\alpha}\left(\cdot\right) = 0$

Here we have ignore boundary terms. Integration over all superspace can be expressed as

$\displaystyle \int d^{4}x d^{2}\theta d^{2}\bar{\theta}\left(\cdot\right) = \int d^{4} x \left(-\frac{1}{4}D^{2}\right)\left(-\frac{1}{4}\bar{D}^{2}\right)\left(\cdot\right)$

Now we turn to chiral superfields. A superfield is a function that is defined on superspace (i.e. it has dependence on the spacetime coordinates and the supercoordinates also). A chiral superfield satisfies the constraint:

Either   $\bar{D}_{\dot{\alpha}} \phi = 0$   or   $D_{\alpha} \phi = 0$   but not both.

With chiral superfields one can do wonders. For example, sticking to the first choice for constraint defining a chiral superfield,  we have the property

$\bar{D}^{2}D^{2}\phi = 16 \Box \phi$

This is useful when re-writing chiral integrals as integrals over the whole supercoordinates:

$\displaystyle \int d^{4} x d^{2} \phi = \int d^{4} x \left(-\frac{D^{2}}{4}\right) \frac{\bar{D}^{2} D^{2}}{16 \Box}\phi = \int d^{4} x d^{4} \theta \left(-\frac{D^{2}}{4 \Box}\right) \phi$

We have forgotten about Dirac delta functions for supercoordinates! For a single supercoordinate we define:

$\displaystyle \int d \theta \delta\left(\theta - \theta' \right) f\left(\theta\right) = f\left(\theta'\right)$

In particular, for the case of the identity function we get

$\displaystyle\frac{\partial}{\partial \theta}\delta\left(\theta - \theta' \right) = 1 \Rightarrow\delta\left(\theta - \theta' \right) = \theta - \theta'$

This result can be generalized to the full superspace integral:

$\displaystyle\frac{1}{16} \partial^{\alpha}\partial_{\alpha}\partial^{\dot{\beta}}\partial_{\dot{\beta}}\delta^{4}\left(\theta_{1} - \theta_{2} \right) = 1$

(Writing the integral as derivatives as discussed above.) The solution to this equation is

$\delta^{4}\left(\theta_{1} - \theta_{2} \right) = \left(\theta_{1} - \theta_{2} \right)^{2} \left(\bar{\theta}_{1} - \bar{\theta}_{2} \right)^{2} = \delta_{12}$

We can mention some properties related to this delta function: