# Lorentz Transformation of Supercharges in $d=1+1$

I am currently learning $$\mathcal{N}=(2,2)$$ SUSY and have come upon another probably silly issue.

I am following chapter $$12$$ of Mirror Symmetry by Hori et. al. and am currently trying to derive the transformation properties of the supercharges in the $$\mathcal{N}=(2,2)$$ chiral scalar theory under the Lorentz group.

The Lagrangian density is as follows: $$\mathcal{L} = |\partial_{0}\phi|^{2}-|\partial_{1}\phi|^{2}-|W'(\phi)|^{2}+i\bar{\psi}_{-}(\partial_{0}+\partial_{1})\psi_{-}+i\bar{\psi}_{+}(\partial_{0}-\partial_{1})\psi_{+}-W''(\phi)\psi_{+}\psi_{-}-\bar{W}''(\bar{\phi})\bar{\psi}_{-}\bar{\psi}_{+}+|F+\bar{W}'(\bar{\phi})|^{2}$$ for $$\phi$$ a complex scalar field, $$\psi_{\pm}$$ a Dirac spinor, $$W$$ a holomorphic function and $$F$$ an auxilliary field.

From here I have used Noether's theorem to find the conserved currents corresponding to the $$\mathcal{Q}_{+}$$ supersymmetry: \begin{align*} & G^{0}_{+} = 2\partial_{+}(\bar{\phi})\psi_{+}-i\bar{\psi}_{-}\bar{W}'(\bar{\phi}) \\ & G^{1}_{+} = -2\partial_{+}(\bar{\phi})\psi_{+}-i\bar{\psi}_{-}\bar{W}'(\bar{\phi}) \end{align*} where $$\partial_{\pm}$$ is the derivative with respect to $$x^{\pm}=x^{0}\pm x^{1}$$.

We then define the $$Q_{+}$$ supercharge as the corresponding conserved charge: $$Q_{+} = \int dx^{1}G^{0}_{+}$$

I want to show that this supercharge transforms as a spinor, namely that under a Lorentz transformation of parameter $$\gamma$$: $$Q_{+}\mapsto e^{-\gamma/2}Q_{+}$$ This should be realtively easy, but I am having some issues. I will now detail my attempt so far.

Firstly, note that the Grassmann numbers transform as: $$\theta^{\pm}\mapsto e^{\pm\gamma/2}\theta^{\pm} \text{ , and } \bar{\theta}^{\pm}\mapsto e^{\pm\gamma/2}\bar{\theta}^{\pm}$$ Then from the fact that the chiral superfield is a Lorentz scalar, I was able to deduce from terms such as $$\theta^{+}\psi_{+}$$ in the superfield expansion that we must have: $$\psi_{\pm}\mapsto e^{\mp\gamma/2}\psi_{\pm} \text{ , and } \bar{\psi}_{\pm}\mapsto e^{\mp\gamma/2}\bar{\psi}_{\pm}$$ this then tells me that: $$G^{0}_{+} \mapsto e^{-\gamma/2} \left( e^{-\gamma}2\partial_{+}(\bar{\phi})\psi_{+}-e^{\gamma}i\bar{\psi}_{-}\bar{W}'(\bar{\phi}) \right) = e^{-\gamma/2}(\cosh(\gamma)G^{0}_{+}+\sinh(\gamma)G^{1}_{+})$$ and in fact: $$\begin{pmatrix} G_{+}^{0}\\ G_{-}^{1} \end{pmatrix} \mapsto e^{-\gamma/2} \begin{pmatrix} \cosh(\gamma)&\sinh(\gamma)\\ \sinh(\gamma)&\cosh(\gamma) \end{pmatrix} \begin{pmatrix} G_{+}^{0}\\ G_{-}^{1} \end{pmatrix}$$

I now want to know how the integration measure $$dx^{1}$$ transforms. From demanding that $$\int dx^{0,1}x^{0,1}$$ is Lorentz invariant, I was able to deduce the opposite transformation property, loosely speaking: $$\begin{pmatrix} \int dx^{1}& \int dx^{0} \end{pmatrix} \mapsto \begin{pmatrix} \int dx^{1}& \int dx^{0} \end{pmatrix} \begin{pmatrix} \cosh(-\gamma)&\sinh(-\gamma)\\ \sinh(-\gamma)&\cosh(-\gamma) \end{pmatrix}$$ This gives the adjacent result: $$Q_{+}+\int dx^{0}G_{+}^{1} \mapsto e^{-\gamma/2} \left( Q_{+}+\int dx^{0}G_{+}^{1} \right)$$

This is almost the desired result, but when I try to transform $$Q_{+}$$ alone I just get a mess. From here I have been unable to conclude that $$Q_{+}$$ alone transforms in the desired way. Any help would be much appreciated.

A generic integral of the form $$Q=\int J^0\mathrm d\boldsymbol r$$ does not have "nice" transformation properties under Lorentz transformations. Why -- because we chose an unnatural splitting into time and space. The key point is, $$Q$$ is a Noether charge, so $$J$$ is conserved and this simplifies matters quite a bit. In particular, assuming reasonable boundary conditions, $$Q$$ is time-independent.
So, the simplification that OP expects is just not true for generic fields; here one must use the non-trivial fact that $$J$$ is conserved. Otherwise, the transformation law will be messy.
Consider the supercurrent $$G$$. It is closed, meaning $$\frac{\partial}{\partial t}G^0_\alpha=\frac{\partial}{\partial x}G^1_\alpha$$ where $$\alpha$$ is a spinor index. In particular, the "ugly" integral in the OP $$\int dx^{0}G_{+}^{1}$$ is easily shown to be independent of $$x$$: \begin{align} \frac{\partial}{\partial x}\int G_{+}^{1}dt&=\int \frac{\partial}{\partial x}G_{+}^{1}dt\\ &=\int \frac{\partial}{\partial t}G_{+}^{0}dt\\ &=0 \end{align} being a total derivative.
As this integral is independent of $$x$$, we can take $$x\to\infty$$. Again, assuming reasonable boundary conditions, the fields vanish in this limit, and therefore the integral itself vanishes. Therefore, the "ugly" integral is zero.