Consider the $0+1$ dimensional Lagrangian
$$L=\frac{1}{2}\dot{X}^2(t)+i \psi(t) \dot{\psi}(t).\tag{1.24}$$
Essentially this the Lagrangian of a particle moving in one dimension, $X$, with an additional degree of freedom $\psi$. This can be thought of as a Lagrangian for a spinning particle moving in one dimension.
Define the supersymmetry transformations (and think of $\delta$ as a fermionic operator on the fields) as
$$\delta X=2i \epsilon \psi\tag{1.28a}$$ and $$\delta \psi=- \epsilon \dot{X}.\tag{1.28b}$$
Noting that $\psi$ and $\delta$ anticommute, $X$ and $\delta$ commute, and also that $\delta$ is a linear operator, we can easily see that
$$\delta L = i \epsilon \frac{d}{dt}(\psi \dot{X}).\tag{1.29}$$
Thus, the action is invariant since the Lagrangian changes only by a total derivative, under the above transformation. The conserved 'current' (in fact in one dimension it is the conserved charge) gives, by Noether's theorem,
$$\epsilon Q=\frac{\partial L}{ \partial \dot{X}} \delta X+\frac{\partial L}{ \partial \dot{\psi}} \delta \psi-i \epsilon \psi \dot{X}=2i\epsilon \dot{X} \psi-i \epsilon \dot{X} \psi-i \epsilon \psi \dot{X}=0!\tag{1}$$
So the charge turns out to be trivial. However, in these notes, in equation (1.30) it is claimed that the supercharge is, in fact,
$$Q=\psi \dot{X}.\tag{1.30}$$
What am I missing?
