Hint for showing the identity $(\nabla\psi\cdot\nabla)\nabla\psi^*+(\nabla\psi^*\cdot\nabla)\nabla\psi=\nabla|\nabla \psi|^2$

Question

I need some help to show this following identity. $$(\nabla\psi\cdot\nabla)\nabla\psi^*+(\nabla\psi^*\cdot\nabla)\nabla\psi=\nabla|\nabla \psi|^2$$ My attempt: $$ \partial_i\psi\partial_i\partial_k\psi^*+\partial_i\psi^*\partial_i\partial_k\psi\\ \partial_i(\psi\partial_i\partial_k\psi^*+\psi^*\partial_i\partial_k\psi) $$ I term inside looks like a product rule but except there are two derivative. I am not sure what should I do next. Any hint will help. Thank you.
Update: I have made some progress $$ \partial_i\partial_k(\psi \psi^*)=0\\\partial_i(\psi^*\partial_k\psi+\psi\partial_k\psi^*)=0\\\partial_i\psi^*\partial_k\psi+\psi^*\partial_i\partial_k\psi+\partial_i\psi^*\partial_k\psi+\psi\partial_i\partial_k \psi^*=0 $$ So, $$ \partial_i\psi^*\partial_k\psi+\partial_i\psi^*\partial_k\psi=-(\psi^*\partial_i\partial_k\psi+\psi\partial_i\partial_k \psi^*) $$ So $$ \partial_i(\psi\partial_i\partial_k\psi^*+\psi^*\partial_i\partial_k\psi)\\\implies\partial_i(\partial_i\psi^*\partial_k\psi+\partial_i\psi^*\partial_k\psi)\\\implies -\partial_i\partial_i(\psi^*\partial_k\psi+\psi\partial_k\psi^*) $$ Is that the correct way?

Answer 1

Using the summation convention,

$$\begin{aligned} R H S_i&=\partial_i \frac{\partial \psi}{\partial x_j} \frac{\partial \psi^*}{\partial x_j} \\ & =\frac{\partial \psi}{\partial x_j} \frac{\partial^2 \psi^*}{\partial x_i \partial x_j} +\frac{\partial\psi^*}{\partial x_j} \frac{\partial^2 \psi}{\partial x_i \partial x_j} \\ & =\frac{\partial \psi}{\partial x_j} \partial_j \frac{\partial \psi^*}{\partial x_i} +\frac{\partial \psi^*}{\partial x_j} \partial_j \frac{\delta \psi}{\partial x_i} \\ & = L H S_i \\ \end{aligned}$$

Answer 2

Think about the form of the right hand side. It is ultimately the gradient of a scalar field, so you will want to manipulate your expression for the left hand side using index notation to start with $\partial_k$. Commutation relationships will be useful for this means since $\partial_i \partial_k = \partial_k \partial_i +[\partial_i,\partial_k]$, for example. Especially be mindful of the product rule since it will be understood that $\partial_k$ acts on every scalar field to the right of it. It should feel ideologically similar to integration by parts.

Alternatively, starting from the right hand side may be more natural here.