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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?

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1 Answer 1

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The second term in OP's formula (1) for the Noether charge has a sign mistake. The second term should be $$\delta\psi^{\mu} \frac{\partial_L L}{ \partial \dot{\psi}^{\mu}} ~=~(-\epsilon \dot{X}^{\mu})(-i\psi_{\mu}) ~=~i \epsilon \dot{X}^{\mu} \psi_{\mu} ~=~(i\psi_{\mu})(-\epsilon \dot{X}^{\mu}) ~=~\frac{\partial_R L}{ \partial \dot{\psi}^{\mu}}\delta\psi^{\mu} ,$$ depending on where we use a left (right) derivative, i.e. the derivative acts from left (right), respectively. As a result the Noether charge becomes non-zero: $$Q~=~2i\psi_{\mu} \dot{X}^{\mu}.\tag{1.30'}$$ The overall factor $2i$ has to do with a strange normalization.

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