# Noether current associated with transformation $\delta \psi=i\alpha \psi$

I'm doing problem 3 from sheet 2 of David Tong's lecture notes. We have given the complex field $$\psi(x)$$ which is governed by the Lagrangian $$\mathcal{L}=\partial_\mu \psi^*\partial^\mu \psi -m^2\psi^*\psi -\frac{\lambda}{2}(\psi^*\psi)^2.$$

Using the Euler-Lagrange equation, I find (see calculations) the equations for the system (for $$\psi$$) $$\partial_\mu \partial_\mu\psi^*\cdot \eta^{\mu\mu}+\lambda (\psi^*)^2\psi+m^2\psi^*=0$$

Next, I find the change in Lagrangian under transformation $$\delta\psi=i\alpha \psi$$ which is given by $$\delta \mathcal{L}=\alpha^2(\cdots )\sim 0.$$ Next, I determined the Noether current associated with this transformation $$\mathcal{J}^\mu=\frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi)}\delta \psi=\partial_\mu \psi^*\cdot \eta^{\mu\mu}i\alpha \psi.$$ In the next part, we need to show that this is indeed conserved. To do that $$\partial_\mu \mathcal{J}^\mu = \partial_\mu \partial_\mu \psi^* n^{\mu\mu}i\alpha \psi+\partial_\mu \psi^*\cdot \eta^{\mu\mu}i\alpha \partial_\mu \psi.$$ I don't know how to show that this is zero.

Calculation for Equation of motion

The EL equation given by $$0=\frac{\partial \mathcal{L}}{\partial \psi}-\partial_\mu\left(\frac{\partial \mathcal{L}}{\partial (\partial_\mu \psi)}\right).$$ The first term is trivial to determined, the later can be found as $$\frac{\partial }{\partial (\partial_\mu \psi)}\partial_\mu \psi^*\partial^\mu \psi=\partial_\mu \psi^*\eta^{\mu\nu}\delta_{\mu\nu}=\partial_\mu \psi^*\eta^{\mu\mu}.$$ Here, In the last step, I wrote $$\partial^\mu \psi$$ as $$\eta^{\mu\nu}\partial_\nu\psi$$. Here $$\eta$$ is minkowski metric.

• As an aside: under the summation convention, an index should never appear more than twice – either it's free (in which case it appears once), or it's summed over (in which case it appears twice). May 9, 2022 at 15:44

Mindful of @Robbie's comment, let's first rewrite your calculations as \begin{align}0&=\partial_\nu\frac{\partial\mathcal{L}}{\partial\partial_\nu\psi}-\frac{\partial\mathcal{L}}{\partial\psi}\\&=\partial_\nu\partial^\nu\psi^\ast+\lambda\psi^{\ast2}\psi+m^2\psi^\ast,\\\mathcal{J}^\nu&=\frac{\partial\mathcal{L}}{\partial\partial_\nu\psi}i\alpha\psi\\&=i\alpha\partial^\nu\psi^\ast\psi,\\\partial_\nu\mathcal{J}^\nu&=i\alpha(\partial_\nu\partial^\nu\psi^\ast\psi+\partial^\nu\psi^\ast\partial_\nu\psi)\\&=i\alpha(-\lambda\psi^{\ast2}\psi^2-m^2\psi^\ast\psi+\partial^\nu\psi^\ast\partial_\nu\psi).\end{align}But to answer your question, we have to recognize this definition of $$\mathcal{J}^\nu$$ is wrong. What you actually want is$$\mathcal{J}^\nu=\frac{\partial\mathcal{L}}{\partial\partial_\nu\psi}\delta\psi+\frac{\partial\mathcal{L}}{\partial\partial_\nu\psi^\ast}\delta\psi^\ast=i\alpha(\partial^\nu\psi^\ast\psi-\partial^\nu\psi\psi^\ast).$$(If you wonder why $$\psi,\,\psi^\ast$$ should be treated as independent fields in the above formula, revisit why $$z,\,z^\ast$$ are treated as independent variables for the derivatives in complex analysis.) Hence$$\partial_\nu \mathcal{J}^\nu=i\alpha(\partial_\nu\partial^\nu\psi^\ast\psi-\partial_\nu\partial^\nu\psi\psi^\ast)=i\alpha((-\lambda\psi^{\ast2}\psi-m^2\psi^\ast)\psi+(\lambda\psi^\ast\psi^2+m^2\psi)\psi^\ast)=0.$$
• How have you differentiated $\partial^\mu \psi$ with respect to $\partial_\nu\psi$? Don't we need to change $\partial^\nu$ to $\eta^{\nu \lambda }\partial_\lambda$. May 9, 2022 at 16:28
• @YoungKindaichi Example:$$\frac{\partial}{\partial\partial_\nu\psi}(\partial_\mu\psi^\ast\partial^\mu\psi)=\partial_\mu\psi^\ast\eta^{\mu\nu}=\partial^\nu\psi^\ast.$$Or equivalently,$$\frac{\partial}{\partial\partial_\nu\psi}(\partial^\mu\psi^\ast\partial_\mu\psi)=\partial^\mu\psi^\ast\delta_\mu^\nu=\partial^\nu\psi^\ast.$$