2
$\begingroup$

There is something I don't understand at page 36 of these lecture notes (Author: Fiorenzo Bastianelli from the university of Bologna, title: Path integrals for fermions and supersymmetric quantum mechanics.) I'll summarize it here but I linked them anyway in case someone want to check them.

So we're trying to build a supersymmetric action, we work in the super space $D=1$ and $N=2$ with one spacetime coordinate $t$ and two Grassman coordinates $\theta$ and $\bar{\theta}$.

The generator of time translation is $$H= i \frac{\partial}{\partial t}$$ The generators of supersymmetry transformation, that are translations in the anticommuting directions are $$Q= \frac{\partial}{\partial \theta} + i \bar{\theta} \frac{\partial}{\partial t} $$ and $$\bar{Q}= \frac{\partial}{\partial \bar{\theta}} + i \theta \frac{\partial}{\partial t} $$ We define a scalar, Grassman even superfield $X(t,\theta, \bar{\theta})$ which, under supersymmetry transformation transforms in this way $$\delta_S X(t,\theta, \bar{\theta}) = (\epsilon \bar{Q} + \bar{\epsilon} Q)\, X(t,\theta, \bar{\theta}) $$

With $\epsilon$ and $\bar{\epsilon}$ Grassmann parameters.

Now we define covariant derivatives $$ D= \frac{\partial}{\partial \theta} - i \bar{\theta} \frac{\partial}{\partial t}$$ $$ \bar{D}= \frac{\partial}{\partial \bar{\theta}} - i \theta \frac{\partial}{\partial t} $$

so that the covariant derivative of a superfield is still a superfield, which means

$$ \delta_S DX = (\epsilon \bar{Q} + \bar{\epsilon} Q)\, DX $$

All commutators and anticommutators are null beside these ones $$ \{ Q,\bar{Q} \} = 2H$$ $$\{D,\bar{D}\} = -2i \partial_t$$

Now we Say that a Lagrangian $L=L(X,DX,\bar{D}X)$ that depends only implicitly on the coordinates of the superspace through the superfield and its covariant derivatives can give you a supersymmetric action. And this is because it transforms under supersymmetry transformation as a total derivative. The exact form of the Lagrangian variation under supersymmetry transformation is this:

$$ \delta_S L(X,DX, \bar{D}X) = (\epsilon \bar{Q} + \bar{\epsilon} Q) \, L(X,DX, \bar{D}X) $$

Now the things I don't understand are these two:

  1. Why does the Lagrangian transforms like that under supersymmetry transformation? I am not able to prove it, I can provide a sketch of my attempt of working out its transformation if requested, but it doesn't really anything I think.

  2. Assuming that's the right transformation law of the Lagrangian, why is that a total derivative? It looks to me that it just transforms like a super field, but I don't see why that's a total derivative.

$\endgroup$
2
$\begingroup$
  1. Use that the covariant derivatives $D$ and $\bar{D}$ anticommute with the generators $Q$ and $\bar{Q}$ of SUSY$^1$ $$ \delta_S L ~=~\delta_SX~ \frac{\partial_L L}{\partial X} +D\delta_SX ~\frac{\partial_L L}{\partial DX} +\bar{D}\delta_SX~ \frac{\partial_L L}{\partial \bar{D}X} $$ $$~=~(\epsilon \bar{Q} + \bar{\epsilon} Q)X~ \frac{\partial_L L}{\partial X} +D(\epsilon \bar{Q} + \bar{\epsilon} Q)X ~ \frac{\partial_L L}{\partial DX} +\bar{D}(\epsilon \bar{Q} + \bar{\epsilon} Q)X~ \frac{\partial_L L}{\partial \bar{D}X} $$ $$~=~(\epsilon \bar{Q} + \bar{\epsilon} Q)X~ \frac{\partial_L L}{\partial X} +(\epsilon \bar{Q} + \bar{\epsilon} Q)DX ~ \frac{\partial_L L}{\partial DX} +(\epsilon \bar{Q} + \bar{\epsilon} Q)\bar{D}X~ \frac{\partial_L L}{\partial \bar{D}X} ~=~ (\epsilon \bar{Q} + \bar{\epsilon} Q)L.$$

  2. In the SUSY varied action $$\delta_SS~=~\int \!\mathrm{d}t~\mathrm{d}\theta~\mathrm{d}\bar{\theta}~(\epsilon \bar{Q} + \bar{\epsilon} Q)L $$ perform the Grassmann-odd Berezin integrations (which are the same as Grassmann-odd differentiations) to see that only a total time-derivative survives. (Recall that if we differentiate wrt. the same Grassmann-odd variable twice we get zero.)

--

$^1$ The subscript "$L$" on a partial derivative indicates left derivatives, i.e. a differentiation acting from left.

$\endgroup$
  • $\begingroup$ Thanks, I'll try it as soon as I can get my hands on a pen and a piece of paper, I'll ask here for clarification if I am not able to find the answer and editing the question with my attempt. $\endgroup$ – Run like hell Nov 1 '18 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.