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

This seems extremely trivial, but I'd like some clarification on why the following statement is true: $$\nabla \times \left<A|\nabla A\right> = \left<\nabla A\right|\times\left|\nabla A\right>.$$ 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!

• Where did this show up as I don't know if I've ever seen this in any QM text Commented Jan 31, 2017 at 13:16
• @Triatticus It's in Berry's original paper on the Berry Phase: jstor.org/stable/2397741?seq=1#page_scan_tab_contents on page 3 (47 in the journal) Commented Jan 31, 2017 at 14:10

Just guessing, but perhaps this is the idea:

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

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

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

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

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

• Why does line 3 --> 4 hold? That's the step I didn't get.. Commented Jan 31, 2017 at 14:10
• I'm aware of those rules but I still don't see it... $A^*$ isn't a scalar, neither is $\nabla A$ afaik? I feel kinda dumb ... :( Commented Jan 31, 2017 at 15:01
• $A$ (and hence $A^*$) is a scalar and $\nabla A$ is a vector. Where is the issue? Commented Jan 31, 2017 at 16:00
• Ah okay, I get it now. I would've suggested $\nabla \times (a \vec b) = -(\vec b \times \nabla a) + a(\nabla \times \vec b)$, taking $a = A^*$ and $\vec b = \nabla A$ instead, but this works for me now. The confusion arose from the tensor-rank of $A$. Thanks! Commented Jan 31, 2017 at 16:16
• Ah I meant curl only, my bad :( Will remove the faulty comment now Commented Jan 31, 2017 at 16:28