# Confirming an action is invariant under a supersymmetric transformation

I am studying chapter 9 of the book Mirror Symmetry, available here. My question is relating to page 156/157 where Supersymmetry is being introduced for the first time in QFT in 0-dimensions.

We are given the following action with one boson and two fermions ($$h$$ is some function of $$X$$): \begin{align*} S(X, \Psi_1, \Psi_2) = \frac{1}{2} (\partial h)^2 - \partial^2 h \Psi_1 \Psi_2. \tag{9.29} \end{align*}

We are told to consider the following transformations:

\begin{align*} \delta_{\epsilon} X &= \epsilon^1 \Psi_1 + \epsilon^2 \Psi_2 \\ \delta\Psi_1 &= \epsilon^2 \partial h \\ \delta\Psi_2 &= - \epsilon^1 \partial h. \end{align*}\tag{9.30}

$$\epsilon^i$$ and $$\Psi_i$$ are Grassmann-odd variables, and so anticommute.

We are told it is easy to check the action is invariant under these transformations. I am having difficulty verifying this.

My attempt was:

\begin{align*} \delta S &= \frac{\partial S}{\partial X} \delta_{\epsilon}X + \frac{\partial S}{\partial \Psi_1} \delta \Psi_1 + \frac{\partial S}{\partial \Psi_2} \delta \Psi_2 \\ &= (\partial h \partial^2 h -\partial^3 h \Psi_1 \Psi_2)(\epsilon^1 \Psi_1 + \epsilon^2 \Psi_2) - \partial^2 h \Psi_2(\epsilon^2 \partial h) - \partial^2 h \Psi_1 \epsilon^1 \partial h. \end{align*}\tag{1}

The two terms with $$\partial^3 h$$ will give zero since $$(\Psi_i)^2 = 0$$. My issue is that I don't see why the remaining terms cancel, because although they seem to match up an cancel in pairs nicely, in order to make them match properly we would need to move epsilons past $$\Psi$$s, and then the terms add rather than cancel.

One further confusion I have is that this issue would have been avoided if I have chosen to write

\begin{align*} \delta S &= \frac{\partial S}{\partial X} \delta_{\epsilon}X + \delta \Psi_1 \frac{\partial S}{\partial \Psi_1} + \delta \Psi_2 \frac{\partial S}{\partial \Psi_2}. \end{align*}\tag{2}

Is this an implicit convention being used?

Could someone please let me know my mistake above, and also clarify about implicit conventions that one might need to choose when doing a linear Taylor-like expansion like above.

EDIT: One further issue I have found is that it is apparently a common convention in this context to consider derivatives as acting from the right. This, I believe, would again give relative minus signs compared with them acting from the left, as I had assumed.

Therefore, I believe my questions now boils down to, whether I am correct there are two conventional choices here: 1. Derivatives acting from left or right, and 2. Whether to put the $$\delta \Psi_i$$ on the right or left of the derivative in the formula for $$\delta S$$. And if I am correct that in both cases, the different choices correspond to different signs in the affected terms, and thus in this particular problem, we can only confirm $$\delta S = 0$$ if we are inspired to know the choices the author had in mind.

More generally, given a function $$f(\theta)~=~\theta a +b~=~(-1)^{|a|}a\theta +b,$$ of a Grassmann-odd variable $$\theta$$, where $$|a|$$ denotes the Grassmann-parity of the supernumber $$a$$, it follows from the definition of the left and right derivatives$$^1$$ $$\frac{\partial^Lf(\theta)}{\partial\theta}~=~ a, \qquad \frac{\partial^Rf(\theta)}{\partial\theta}~=~(-1)^{|a|}a,$$ that the term ordering of the differential is $$\mathrm{d}\theta\frac{\partial^Lf(\theta)}{\partial\theta} ~=~ \mathrm{d}f(\theta) ~=~ \frac{\partial^Rf(\theta)}{\partial\theta}\mathrm{d}\theta,\tag{1}$$ and not vice-versa. See also this related Phys.SE post.
$$^1$$ A left derivative means a derivative that acts from the left. A right derivative means a derivative that acts from the right.