Math in deriving Pauli Equation When deriving the Pauli Equation, it has the following step:
$$i\frac{e}{c}\hbar[\vec{A}\times\nabla+\nabla\times\vec{A}]\phi=i\frac{e}{c}\hbar\space curl\vec{A}\cdot \phi$$
$\phi$ is one of the spinors in the bispinor of the electron. $\vec{A}$ is the vector potential. How does it go from the LHS to the RHS? I thought $\nabla\times\vec{A}$ is just $\text{curl}\ \vec{A}$?
 A: You need to show $\overbrace{\vec{A} \times \nabla \phi + \nabla \times (\vec{A}\phi)}^{LHS}=\overbrace{\phi \nabla\times \vec A}^{RHS} $
or equivalently $ \nabla \times (\vec{A}\phi)=\phi \nabla\times \vec A -\vec{A} \times \nabla \phi$
To show this we write
\begin{eqnarray}
\nabla \times (\vec{A}\phi) &=& \varepsilon_{ijk }\partial_i (A_j\,\phi)\;,
\\&=&\Big[\varepsilon_{ijk }\partial_i (A_j)\,\phi+\varepsilon_{ijk }A_j\,\partial_i (\phi)\Big]\;,\\
&=&\Big[\varepsilon_{ijk }\partial_i (A_j)\,\phi-\varepsilon_{jik }A_j\,\partial_i (\phi)\Big]\;,\\
&=&\Big[\varepsilon_{ijk }\partial_i (A_j)\,\phi-\varepsilon_{ijk }A_i\,\partial_j (\phi)\Big]\;,\\
&\equiv& (\nabla \times\vec A)\,\phi -\vec A \times \nabla \phi
\end{eqnarray}
Your $\cdot$ in the right hand side may also cause confusion since it look likes the dot product but indeed it is surely a scalar multiplication.
A: I'm pretty sure they are trying to use the second identity here. The way you have your equation written it is difficult to see where the differential operators are acting. So I will state it more clearly. Your left hand side is
$$\vec{A} \times \nabla \phi + \nabla \times \left( \phi \vec{A}\right)$$
and your right hand side is
$$\left(\nabla \times \vec{A}\right) \phi.$$
When you use the identity to expand the second term on the LHS, you $\nabla \phi \times \vec{A}$ term cancel with the first term in the LHS, and you are left with the RHS.
A: I will omit the physical constants here, since they are not so relevant.
LHS $=\vec{A} \times \nabla \phi + \nabla \times (\vec{A}\phi)$ 
(use vector calculus)
$= \vec{A}\times \nabla \phi + (\nabla \phi) \times \vec{A} + \phi (\nabla \times \vec{A})$
$=\vec{A}\times \nabla \phi - \vec{A}\times \nabla \phi + \phi (\nabla \times \vec{A})$ 
($\vec{A} \times \vec{B} = - \vec{B} \times \vec{A}$)
$= \phi (\nabla \times \vec{A}) = \phi \cdot curl \, \vec{A}$ = RHS
QED
