# Deducing charges from a Lagrangian, in supersymmetric quantum mechanics

I'm trying to read through this paper called "Supersymmetric Ground State Wavefunctions". Half way down page 4, is says

"We begin by reviewing Witten's original model which is defined by the Lagrangian \begin{align*} L = \frac{1}{2} \dot{x}^2 - \frac{1}{2}(W')^2 + i \hbar \bar{\psi} \dot{\psi} - \hbar W'' \bar{\psi}\psi.\tag{2.4} \end{align*} Here $$x$$ is the real coordinate and $$W(x)$$ the superpotential. Also, we use overdots to denot the time differntiation while primes denote differentiation with respect to $$x$$. From Noether's theorem, we obtain the charges \begin{align*} Q &= \psi(p + iW') \\ \bar{Q} &= \bar{\psi}(p - i W'),\tag{2.5} \end{align*} where $$p = \dot{x}$$ is the momentum conjugate to $$x$$."

Here is the Witten paper they reference when they mention his original model.

1. I would like to verify these charges do follow from Noether's theorem. To deduce those charges, must we not first note the symmetry of $$L$$? If so, what is it in this case?

2. I'm also a bit confused about how to think about this Lagrangian. Is it a single boson and two fermions? The boson is $$x(t)$$? Are $$\psi, \bar{\psi}$$ functions of $$t$$ (or $$x$$)?

(3.) Incidentally, I don't see this exact Lagrangian in the referenced Witten paper. The closest thing I see starts on page 24, but looks to me to be quite different to the above. If I'm just not seeing the link, I would appreciate someone pointing the connection out to me please.

If it's not already clear, I'm only starting to learn about SUSY. So there's no doubt plenty of basic stuff I am not aware of.

• Linked. To find the susy symmetries, (anti-)commute Q with your variables.... Commented Oct 28, 2023 at 16:23
• (2.11) of C&H is (29) of W, but you must redefine W' to W ... Commented Oct 28, 2023 at 20:50
• Thanks, but I don't understand your first comment. And for the second, the two equations also differ in that one has matrices, and one has grassmann variables. Commented Oct 29, 2023 at 11:06
• The ψs in W are wavefunctions acted upon, normalized to 1, but omitted: $\sigma_1, \sigma_2$ are perfectly fine Grassman variables, so the supercharges Q are fermionic. In C&H, the Grassman variables include this normalization in their different ψs, as the extant answer indicates: a representation. The first comment asks you to compute $[Q,\psi]\propto \delta \psi$, etc, to verify Noether's theorem, that you appear conflicted about... Commented Oct 29, 2023 at 13:49

1. For conservation laws, it is easier (and more precise!) to consider the quantum mechanical Hamiltonian operator formulation than the Lagrangian classical formulation. Due to $$2\hat{H}~=~\{\hat{Q},\hat{\overline{Q}}\}_+~\stackrel{(2.5)+(2.6)}{=}~\hat{p}^2 + W^{\prime}(\hat{x})^2 +\hbar W^{\prime\prime}(\hat{x})[\hat{\overline{\psi}},\hat{\psi}],\tag{2.7}$$ and $$\hat{Q}^2~\stackrel{(2.5)+(2.6)}{=}~0~\stackrel{(2.5)+(2.6)}{=}~\hat{\overline{Q}}^2,\tag{2.1}$$ it follows from the super Jacobi identity for the super-commutator that the supercharges $$\hat{Q}$$ and $$\hat{\overline{Q}}$$ commute with the Hamiltonian operator \begin{align} 4[\hat{Q},\hat{H}]~\stackrel{(2.7)}{=}~&2[\hat{Q},\{\hat{Q},\hat{\overline{Q}}\}_+]~\stackrel{\text{Jac.Id.}}{=}~[\{\hat{Q},\hat{Q}\}_+,\hat{\overline{Q}}]~\stackrel{(2.1)}{=}~0,\cr 4[\hat{\overline{Q}},\hat{H}]~\stackrel{(2.7)}{=}~&2[\hat{\overline{Q}},\{\hat{\overline{Q}},\hat{Q}\}_+]~\stackrel{\text{Jac.Id.}}{=}~[\{\hat{\overline{Q}},\hat{\overline{Q}}\}_+,\hat{Q}]~\stackrel{(2.1)}{=}~0, \end{align} and are hence conserved quantities.
2. $$x(t)$$ is a boson, while $$\psi(t)$$ and $$\overline{\psi}(t)$$ are fermions in 0+1D.