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How to show that:

$ (-i\nabla-eA)\times(-i\nabla-eA) = (ie\nabla \times A) $

i and e are constants
A is a vector field
$\nabla$ = vector differential operator

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    Raskolnikov already gave you the answer. But I just want to add the comment that the identity you wrote down is also the definition for the [curvature form of a connection on a vector bundle](http://en.wikipedia.org/wiki/Connection_%28vector_bundle%29#Curvature)2010-12-19

1 Answers 1

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As I suggested in my hint, apply the operator on a test function $f$. Then

$(-i\nabla-eA)\times(-i\nabla-eA)f = (-i\nabla-eA)\times(-i\nabla f-e A f)$

Working this out we get

$(-i\nabla-eA)\times(-i\nabla f-e A f)= -\nabla \times \nabla f + ie A\times \nabla f +i e \nabla \times (A f) + e^2 A \times A f$

The first and last term are zero. The third term can be expanded using the last identity (or simply working it out in terms of partial derivatives). This will give the result.

$(-i\nabla-eA)\times(-i\nabla-eA) = (ie\nabla \times A)$

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    @Ras: It's a [MathJax](http://www.mathjax.org/) feature, actually. :)2010-12-19