# 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:

## Solving equations of motions in some gravity background

Posted in AdS/CFT, String Theory by Index Guy on August 31, 2008

I would like to consider the gravity background:

$ds^2 = A dx^2 + B dr^2 ,$

with the case that $A = B^{-1}.$ We saw previously that the equations of motion were given by:

$\partial_{i}\left(A\partial_{i}x^a\right) = 0$     and     $-2\partial_{i}^{2}r = \displaystyle\frac{d \left(\ln{B}\right)}{d r}\left(-\eta_{ab}\partial_{i}y^{a}\partial_{i}y^{b} +\partial_{i}r\partial_{i}r\right).$

Now we take the warp factor to have the form $B = r^{\beta}$. Then we have

$-2 r \partial_{i}^{2}r = \beta\displaystyle\left(-\eta_{ab}\partial_{i}y^{a}\partial_{i}y^{b} +\partial_{i}r\partial_{i}r\right).$

We will introduce a new symbol,

$\eta_{ab}\partial_{i}y^{a}\partial_{i}y^{b} = -W^{2}$     with $\displaystyle W(u_{1}, u_{2})$ a function of the worldsheet coordinates.

The differential equation now looks like:

$2 r \partial_{i}^{2}r + \beta \partial_{i}r\partial_{i}r = \beta W^{2}.$

We now assume that $r$ can be factored into functions of each coordinate, $r = T(u_{1}) S(u_{2})$. Then we can solve this PDE when

$k^2 = \displaystyle\frac{\beta W^2}{2 T^{2} S^{2}}$     with $k$ a constant.

Separation of variables gives the following differential equation:

$\displaystyle\frac{d^2 Y}{du_{i}^2} + \alpha \left(\displaystyle\frac{d Y}{d u_{i}}\right)^{2} = k_{i}^{2},$

with

$\alpha = 1 + \displaystyle\frac{\beta}{2}$     and     $T = \exp{Y}.$

Inserting this into Mathematica gives:

$Y(u) = C_{1} + \displaystyle\frac{1}{\alpha}\ln{\left(\cosh{\left(k_{i}\sqrt{\alpha}\left[u_{i} + C_{2}\right]\right)}\right)}.$     which means     $T(u) = C_{1} \left(\cosh{\left(k_{i}\sqrt{\alpha}\left[u_{i} + C_{2}\right]\right)} \right)^{1/\alpha}.$

Notice that the case $\beta = -2$ is interesting: for the constraints we have used the solution is an exponential function of a quadratic polynomial.

## A curved lagrangian in terms of a flat one

Posted in Relativity, String Theory by Index Guy on August 22, 2008

Let us consider the following gravitational background:

$ds^2 = A(r)\eta_{ab}dx^a dx^b + B(r)dr^2,$

and the Polyakov lagrangian in conformal gauge with Euclidean Lorentzian signature:

$L_{p} = A \eta_{ab}\partial_{i}x^{a}\partial_{i}x^{b} + B\partial_{i}r\partial_{i}r.$

The Euler-Lagrange equations can be found to be:

$\partial_{i}\left(A\partial_{i}x^a\right) = 0$     and     $2B\partial_{i}^{2}r = \displaystyle\frac{d A}{d r} \eta_{ab}\partial_{i}x^{a}\partial_{i}x^{b} - B\displaystyle\frac{d \left(\ln{B}\right)}{d r}\partial_{i}r\partial_{i}r.$

We look at this and think how can we make these equations simpler. The first equation can be solved with:

$\partial_{i} y^{a} = A\epsilon_{ij}\partial_{j}x^{a}.$

This is a “T-duality” transformation. For the second equation we see that if $A = B^{-1}$ then we get

$\displaystyle\frac{d\left(\ln{B}\right)}{d r} = \displaystyle\frac{\partial_{i}^{2}r}{L_{f}(y^{a}, r)},$

with $L_{f}$ the flat-space Polyakov lagrangian in terms of the new coordinates $y^{a}$:

$L_{f}\left(y^{a},r\right) = \eta_{ab}\partial_{i}y^{a}\partial_{i}y^{b} + \partial_{i}r\partial_{i}r.$

We can integrate this equation to find an expression for $B$ in terms of the solutions of the equations of motion and the flat-space lagrangian:

$B(r) = B(r_{0}) \exp{\left(-2\displaystyle\int_{r_{0}}^{r}d\tilde{r}\displaystyle\frac{\partial^{2}\tilde{r}}{L_{f}(\tilde{y}^{a} ,\tilde{r})}\right)}.$

Then, pluging this back into the classical action we get

$S = \displaystyle\int d^2 u B(r_{0}) L_{f}(y^{a}, r) \exp{\left(-2\displaystyle\int_{r_{0}}^{r}d\tilde{r}\displaystyle\frac{\partial^{2}\tilde{r}}{L_{f}(\tilde{y}^{a}, \tilde{r})}\right)}.$

Finally we can write

$d\tilde{r} = \partial_{j}\tilde{r} du_{j},$

so we can write the expression in the exponential as a sum of integrals over the worldsheet coordinates.

On the other hand, if instead we have $A = B$ then we can write:

$A(r) = A(r_{0}) = \exp{\left(-2\displaystyle\int_{r_{0}}^{r}d\tilde{r}\displaystyle\frac{\partial^{2}\tilde{r}}{K_{f}(\tilde{x}^{a} ,\tilde{r})}\right)},$

with

$K_{f}(x^{a} ,r) = 2\partial_{i}r\partial_{i}r - L_{f}(x^{a} , r).$

The lagrangian can be writen as

$L_{p} = A(r_{0}) L_{f}(x^{a}, r) \exp{\left(-2\displaystyle\int_{r_{0}}^{r}d\tilde{r}\displaystyle\frac{\partial^{2}\tilde{r}}{K_{f}(\tilde{x}^{a} ,\tilde{r})}\right)}.$

We have found an expression for the classical lagragian in some (not-so) arbitrary gravitational background in terms of the coordinates and the flat-space lagrangian. The problem is we do not get any information about the solution for the equations of motion.

## Ego-breaking

Posted in Organizational by Index Guy on August 12, 2008

The hardest part of research is to overcome one’s ego and know when one’s ideas are leading nowhere. Also, to accept advice from others.

## Minimal area in arbitrary background (I)

Posted in Gravity, String Theory by Index Guy on August 7, 2008

$ds^2 = A\eta_{mn}dx^m dx^n + Bdr^2 ,$

where the functions $A$ and $B$ are functions of the extra coordinate $r$. The case of anti-de Sitter space corresponds to

$A = B = \displaystyle\frac{R^2}{r^2}.$

Under the “T-duality” transformation

$\partial_{a}y^m = i A \epsilon_{ab}\partial_{b}x^m$   and   $\rho = \displaystyle\frac{R^2}{r}$

the metric in the dual space takes the form

$ds^2 = \tilde{A}\eta_{mn}dy^m dy^n + \tilde{B}d\rho^2 ,$

but now

$\tilde{A} = \left[A(\rho)\right]^{-1}$   and   $\tilde{B} = \displaystyle\frac{R^4}{\rho^4}B(\rho).$

Since we can always bring the metric in this form, we will just consider the initial case and see what can we do with it. Note that it could be the case that the change of variables between $(r,\rho)$ could be of the general form

$\rho = Ar$   which means   $dr^2 = \left(r\displaystyle\frac{\partial A}{\partial r} + A\right)^{-2}d\rho^2 .$

This case could be more complicated… For now we will just stick with the first change of variables introduced.

## Some progress towards something

Posted in Relativity, String Theory by Index Guy on August 4, 2008

On Thursday night I was reading Polyakov’s contribution to the book 50 years of Yang-Mills theory. This is my version of a bed time story. ;-)

Polyakov mentions some of his earlier work and how he solved different problems. I actually got tired after a while, and decided to work and solve my own problem. I had been sort of running away from it with feelings of overwhelming difficulty. There is also a chance that reading this post on the (now dead) string coffee table provided some motivation. In any case, here we go…