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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 $$ h_{\alpha\beta}=\eta_{\alpha\beta} $$

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  • $\begingroup$ Toss the fermions for a moment, and consider the stress-energy tensor for the free X s. Is it correct? Is it conserved? $\endgroup$ – Cosmas Zachos Apr 22 at 0:44
  • $\begingroup$ @CosmasZachos oh wow, thank you for pointing that out. I am missing a term which makes the free X s zero. I'll edit it right away. Sorry for the mistake! $\endgroup$ – redhood Apr 22 at 1:28
  • $\begingroup$ Now ignore the bosons and consider the free Dirac action and the proper stress-energy. $\endgroup$ – Cosmas Zachos Apr 22 at 2:41
  • $\begingroup$ @CosmasZachos thank you for your comment. After lots of mistakes, I managed to figure out the energy-momentum tensor and it's conserved! I'll type my calculations in an answer, I think it will be useful for others(and probably future me) $\endgroup$ – redhood Apr 22 at 3:49
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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!

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