Curl of an inner product with a curl: $\nabla \times \left<A|\nabla A\right>$

Question

This seems extremely trivial, but I'd like some clarification on why the following statement is true: \begin{equation} \nabla \times \left<A|\nabla A\right> = \left<\nabla A\right|\times\left|\nabla A\right>. \end{equation} I tried interpreting the braket as an inner product, but the outer product of an inner product (i.e. an outer product of vector and a scalar) yields no meaningful answer. Thanks!

Answer 1

Just guessing, but perhaps this is the idea:

$\nabla \times \left<A|\nabla A\right>$

$ = \nabla \times \int A^* \nabla A$

$ = \int \nabla \times (A^* \nabla A)$

$ = \int (\nabla A^*) \times (\nabla A)$

$ = \left<\nabla A\right|\times\left|\nabla A\right>$