Pauli Equation - Simple operator algebra question

[karkas]

Homework Statement

I am watching a course on Relativistic Quantum Mechanics to freshen up, and I have found to have some issues regarding simple operator algebra. This particular issue on the Pauli Equation (generalization of the Schrodinger equation that includes spin corrections) in an electromagnetic field, arises from the cross product of a linear combination of operators, as shown below (and in the video link - time 18:18)



I understand that when the operators don't commute, the cross product isn't zero, but how is it shown explicitly that the only non-trivial effect is the \vec{p}\times\vec{A} that survives? What about the other term diagonal in the two operators?

Homework Equations

\left(\vec{p}-\frac{e}{c}\vec{A}\right)\left(\vec{p}-\frac{e}{c}\vec{A}\right)=\frac{i\hbar e}{c}<br /> \vec{\nabla}\times\vec{A}

The Attempt at a Solution

I've been arguing why the second term \propto \vec{A}\times\vec{p} won't contribute but can't seem to reach it. Thinking about a crutch function didn't quite help as well. I'm pretty sure this is simple but can't think of it.

 

[karkas]

Seems I've pretty much forgot all of my vector calculus as well as operator priorities. I got the right thing though after some research on my textbooks

Rule 7 on Griffiths E-M: \vec{\nabla}\times\left(f\cdot\vec{A}\right)=f\left(\vec{\nabla}\times\vec{A}\right)-\vec{A}\times\left(\vec{\nabla}\cdot f\right)
Also the priority of operator action starts from the left and encompasses all through to right, so our action should be:
\left[\left(\vec{p}-\frac{e}{c}\vec{A}\right)\left(\vec{p}-\frac{e}{c}\vec{A}\right)\right]g=-\frac{e}{c}\vec{p}\times\left(\vec{A}g\right)-\frac{e}{c}\vec{A}\times\left(\vec{p}g\right)=i\hbar\frac{e}{c}\vec{\nabla}\times\left(\vec{A}g\right)+i\hbar\frac{e}{c}\vec{A}\times\left(\vec{\nabla}g\right)=i\hbar\frac{e}{c}g\left(\vec{\nabla}\times\vec{A}\right)-i\hbar\frac{e}{c}\vec{A}\times\left(\vec{\nabla}g\right)+i\hbar\frac{e}{c}\vec{A}\times\left(\vec{\nabla}g\right)=i\hbar\frac{e}{c}\left(\vec{\nabla}\times\vec{A}\right)g

Correct me if I'm wrong, otherwise I think it's done with.

 

[blue_leaf77]

If the vector multiplication in the left side of
$$
\left(\vec{p}-\frac{e}{c}\vec{A}\right)\left(\vec{p}-\frac{e}{c}\vec{A}\right)=\frac{i\hbar e}{c}
\vec{\nabla}\times\vec{A}
$$
is actually a cross product, then your derivation is correct.

 

[karkas]

If the vector multiplication in the left side of
$$
\left(\vec{p}-\frac{e}{c}\vec{A}\right)\left(\vec{p}-\frac{e}{c}\vec{A}\right)=\frac{i\hbar e}{c}
\vec{\nabla}\times\vec{A}
$$
is actually a cross product, then your derivation is correct.

Ah, a misprint, will fix, thank you for the confirmation!