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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!

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  • $\begingroup$ Where did this show up as I don't know if I've ever seen this in any QM text $\endgroup$
    – Triatticus
    Commented Jan 31, 2017 at 13:16
  • $\begingroup$ @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) $\endgroup$
    – user55789
    Commented Jan 31, 2017 at 14:10

1 Answer 1

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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>$

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  • $\begingroup$ Why does line 3 --> 4 hold? That's the step I didn't get.. $\endgroup$
    – user55789
    Commented Jan 31, 2017 at 14:10
  • $\begingroup$ 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 ... :( $\endgroup$
    – user55789
    Commented Jan 31, 2017 at 15:01
  • $\begingroup$ $A$ (and hence $A^*$) is a scalar and $\nabla A$ is a vector. Where is the issue? $\endgroup$ Commented Jan 31, 2017 at 16:00
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    $\begingroup$ 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! $\endgroup$
    – user55789
    Commented Jan 31, 2017 at 16:16
  • $\begingroup$ Ah I meant curl only, my bad :( Will remove the faulty comment now $\endgroup$ Commented Jan 31, 2017 at 16:28

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