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$\begingroup$

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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    $\DeclareMathOperator{\curl}{curl}\DeclareMathOperator{\del}{del}$ This is terribly formatted. Did you mean $$(-i\del-eA) \times (-i \del - eA) = ie\del \times A = ie \curl (A)?$$2010-12-19
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    Yes that is right2010-12-19
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    He means $(-i\nabla-eA)\times(-i\nabla-eA) = (ie\nabla \times A)$. Hint: apply on a test function.2010-12-19
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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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    Is there a way to see the source (latex code) of answers ?2010-12-19
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    I don't know. But here's the code for the last line: (-i\nabla-eA)\times(-i\nabla-eA) = (ie\nabla \times A)2010-12-19
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    @kallemann: Right-click on any $\LaTeX$ entity you see; click on "Show Source", and you shall see the $\LaTeX$ code.2010-12-19
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    Thanks J.M.! Is this a feature of math.stackexchange only or is this generally implemented according to some standard?2010-12-19
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    @Ras: It's a [MathJax](http://www.mathjax.org/) feature, actually. :)2010-12-19