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.


1 Answer 1

  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.

  • $\begingroup$ 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. $\endgroup$ Nov 1, 2018 at 18:53

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