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 relatively 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: 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.
Note that the argument explicitly uses the fact that the supercurrent is conserved. Without the assumption, the "ugly" integral does not vanish, and the transformation law is ugly. This is not a bad thing; without the conservation law, charges are not Lorentz covariant, as they are defined by an "ugly" time+space splitting.
