Conserved quantities in supersymmetric theory Given the following supersymmetric action
$$
\mathcal{S}=-\frac{T}{2}\int{d^2x\;\eta^{\alpha\beta}\eta_{\mu\nu}\left(\partial_\alpha X_\mu\partial_\beta X^\nu-i\bar{\psi}^\mu\rho_\alpha\partial_\beta\psi^\nu\right)}
$$
I want to show that the following quantity is conserved
$$
T_{\alpha\beta}=\partial_\alpha X^\mu\partial_\beta X_\mu
-\frac{1}{2}\eta_{\alpha\beta}\eta^{\gamma\delta}\partial_\gamma X^\mu \partial_\delta X_\mu
+\frac{i}{4}\bar{\psi}^\mu\rho_\alpha\partial_\beta\psi_\mu
+\frac{i}{4}\bar{\psi}^\mu\rho_\beta\partial_\alpha\psi_\mu
$$
My attempt was to compute $\partial_\alpha T^{\alpha\beta}$ and show that it's zero using the equations of motion which are:
$$
\rho^\alpha\partial_\alpha\psi^\mu=0\\\partial_\alpha\partial^\alpha X^\mu = 0
$$
My calculation:
$$
\partial_\alpha T^{\alpha\beta}=
\partial_\alpha\left(\partial^\alpha X^\mu\partial^\beta X_\mu\right)
-\frac{1}{2}\partial_\alpha\left(\eta^{\alpha\beta}\eta^{\gamma\delta}\partial^\gamma X^\mu\partial^\delta X_{\mu}\right)
+\frac{i}{4}(\partial_\alpha\bar{\psi}^\mu)\rho^\alpha\partial^\beta\psi_\mu+\frac{i}{4}\bar{\psi}^\mu\rho^\alpha\partial_\alpha\partial^\beta\psi_\mu+\frac{i}{4}(\partial_\alpha\bar{\psi}^\mu)\rho^\beta\partial^\alpha\psi_\mu+\frac{i}{4}\bar{\psi}^\mu\rho^\beta\partial^\alpha\partial_\alpha\psi_\mu
$$
Now the first 2 terms cancel each other out when we combine them with the eom $\partial_\alpha\partial^\alpha X^\mu=0$
Then we are left with the fermionic part
$$
\partial_\alpha T^{\alpha\beta}=
\frac{i}{4}(\partial_\alpha\bar{\psi}^\mu)\rho^\alpha\partial^\beta\psi_\mu+
\frac{i}{4}\bar{\psi}^\mu\rho^\alpha\partial_\alpha\partial^\beta\psi_\mu+
\frac{i}{4}(\partial_\alpha\bar{\psi}^\mu)\rho^\beta\partial^\alpha\psi_\mu+
\frac{i}{4}\bar{\psi}^\mu\rho^\beta\partial^\alpha\partial_\alpha\psi_\mu
$$
The first term is zero from the Dirac equation hence we are left with
$$
\partial_\alpha T^{\alpha\beta}=
\frac{i}{4}\bar{\psi}^\mu\rho^\alpha\partial_\alpha\partial^\beta\psi_\mu+
\frac{i}{4}(\partial_\alpha\bar{\psi}^\mu)\rho^\beta\partial^\alpha\psi_\mu+
\frac{i}{4}\bar{\psi}^\mu\rho^\beta\partial^\alpha\partial_\alpha\psi_\mu
$$
Edit:
Definition of $\rho$:
$$
\{\rho^\alpha,\rho^\beta\}=-2\eta^{\alpha\beta}
$$
The usual Clifford algebra.
Another note: we set the induced metric to the minkowski metric due to residual gauge transformations 
$$
h_{\alpha\beta}=\eta_{\alpha\beta}
$$
 A: Okay this is a long one, but I believe it will come handy in the future.
Definitions:
First we need to note some definitions that will make everything clearer but very tedious!
$$
\eta^{\mu\nu}=\left(\matrix{-1&0\\0&1}\right)\;\;\psi^\mu=\left(\matrix{\psi_-\\\psi_+}\right)\;\;X^\mu=\left(\matrix{X^0\\X^1}\right)
$$
We now rewrite the action(we omit the $\frac{T}{2}$ since it's just a prefactor it follows through the computations trivially)
$$
\mathcal{S}=-\int{d^2x\;\eta^{\alpha\beta}\eta_{\mu\nu}\left(\partial_\alpha X^\mu\partial_\beta X^\nu-i\bar{\psi}^\mu\rho_\alpha\partial_\beta\psi^\nu\right)}
$$
by substituting the form of the metric and expanding $X^\mu$ and $\psi^\mu$
$$
\mathcal{S}=\int{d^2x\;\eta^{\alpha\beta}\left(-\partial_\alpha X^0\partial_\beta X^0+\partial_\alpha X^1 \partial_\beta X^1+i\bar{\psi}_-\rho_\alpha\partial_\beta\psi_- -i\bar{\psi}_+\rho_\alpha\partial_\beta\psi_+\right)}
$$
Now we use the standard form of the energy-momentum tensor$$
T^{\gamma\delta}=\frac{\partial\mathcal{L}}{\partial(\partial_\gamma\phi_i)}\;\partial^\delta\phi^i-\eta^{\gamma\delta}\mathcal{L}
$$
where the $\phi^i$ fields are summed over and in our case those are $X^0,X^1,\psi_-,\psi_+,\bar{\psi}_-,\bar{\psi}_+$.
Now calculating the terms we get
$$
T^{\gamma\delta}=-\eta^{\gamma\beta}\partial_\beta X^0\partial^\delta X^0-\eta^{\alpha\gamma}\partial_\alpha X^0 \partial^\delta X^0+\eta^{\gamma\beta}\partial_\beta X^1\partial^\delta X^1+\eta^{\alpha\gamma}\partial_\alpha X^1\partial^{\delta} X^1+\eta^{\gamma\delta}\eta^{\alpha\beta}\partial_\alpha X^0\partial_\beta X^0-\eta^{\gamma\delta}\eta^{\alpha\beta}\partial_\alpha X^1\partial_\beta X^1+i\eta^{\alpha\gamma}\bar{\psi}_-\rho_\alpha\partial^\delta\psi_--i\eta^{\alpha\gamma}\bar{\psi}_+\rho_\alpha\partial^\delta\psi_+-i\eta^{\alpha\beta}\eta^{\gamma\delta}\bar{\psi}_-\rho_\alpha\partial_\beta\psi_-+i\eta^{\alpha\beta}\eta^{\gamma\delta}\bar{\psi}_+\rho_\alpha\partial_\beta\psi_+
$$
Now to check that $T^{\gamma\delta}$ is conserved we calculate $\partial_\gamma T^{\gamma\delta}$ which must be zero if the quantity is to be conserved. Calculating that we have 
$$
\partial_\gamma T^{\gamma\delta} = -\eta^{\gamma\beta}\partial_\gamma\partial_\beta X^0\partial^\delta X^0-\eta^{\gamma\beta}\partial_\beta X^0\partial_\gamma\partial^\delta X^0-\eta^{\alpha\gamma}\partial_\gamma\partial_\alpha X^0\partial^\delta X^0-\eta^{\alpha\gamma}\partial_\alpha X^0\partial_\gamma\partial^\delta X^0+\eta^{\gamma\beta}\partial_\gamma\partial_\beta X^1\partial^\delta X^1+\eta^{\gamma\beta}\partial_\beta X^1 \partial_\gamma\partial^\delta x^1+\eta^{\alpha\gamma}\partial_\gamma\partial_\alpha X^1 \partial^\delta X^1+\eta^{\alpha\gamma}\partial_\alpha X^1\partial_\gamma\partial^\delta X^1+\eta^{\gamma\delta}\eta^{\alpha\beta}\partial_\gamma\partial_\alpha X^0 \partial_\beta X^0+\eta^{\gamma\delta}\eta^{\alpha\beta}\partial_\alpha X^0\partial_\gamma\partial_\beta X^0-\eta^{\gamma\delta}\eta^{\alpha\beta}\partial_\gamma\partial_\alpha X^1\partial_\beta X^1-\eta^{\gamma\delta}\eta^{\alpha\beta}\partial_\alpha X^1\partial_\gamma\partial_\beta X^1+i\eta^{\alpha\gamma}\partial_\gamma\bar{\psi}_-\rho_\alpha\partial^\delta\psi_-+i\eta^{\alpha\gamma}\bar{\psi}_-\rho_\alpha\partial_\gamma\partial^\delta\psi_--i\eta^{\alpha\gamma}\partial_\gamma\bar{\psi}_+\rho_\alpha\partial^\delta\psi_+-i\eta^{\alpha\gamma}\bar{\psi}_+\rho_\alpha\partial_\gamma\partial^\delta\psi_+-i\eta^{\alpha\beta}\eta^{\gamma\delta}\partial_\gamma\bar{\psi}_-\rho_\alpha\partial_\beta\psi_--i\eta^{\alpha\beta}\eta^{\gamma\delta}\bar{\psi}_-\rho_\alpha\partial_\gamma\partial_\beta\psi_-+i\eta^{\alpha\beta}\eta^{\gamma\delta}\partial_\gamma\bar{\psi}_+\rho_\alpha\partial_\beta\psi_++i\eta^{\alpha\beta}\eta^{\gamma\delta}\bar{\psi}_+\rho_\alpha\partial_\gamma\partial_\beta\psi_+
$$
Then we use the equations of motion 
$$
\rho^{\alpha}\partial_\alpha\psi^\mu=0\\\partial_\alpha\partial^\alpha X^\mu = 0
$$
Which simplifies the above to 
$$
=\partial^\gamma X^0\partial_\gamma\partial^\delta X^0
-\partial^\gamma X^0\partial_\gamma\partial^\delta X^0
+\partial^\gamma X^1\partial_\gamma\partial^\delta X^1
+\partial^\gamma X^1\partial_\gamma\partial^\delta X^1
+\partial^\delta\partial^\gamma X^0\partial_\gamma X^0
+\partial^\gamma X^0\partial^\delta\partial_\gamma X^0
-\partial^\delta\partial^\gamma X^1\partial_\gamma X^1
-\partial^\gamma X^1 \partial^\delta\partial_\gamma X^1
+i\bar{\psi}_-\rho^\gamma\partial_\gamma\partial^\delta\psi_-
-i\bar{\psi}_+\rho^\gamma\partial_\gamma\partial^\delta\psi_+
-i\partial^\delta\bar{\psi}_-\rho^\alpha\partial_\alpha\psi_-
-i\bar{\psi}_-\rho^\alpha\partial^\delta\partial_\alpha\psi_-
+i\partial^\delta\bar{\psi}_+\rho^\beta\partial_\beta\psi_+
+i\bar{\psi}_+\rho^\alpha\partial^\delta\partial_\alpha\psi_+=0
$$
Hence the quantity $T^{\gamma\delta}$ is conserved!
