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