Stern Gerlach term in the Pauli Equation

[Mantella]

Where does the Stern Gerlach term in the Pauli equation come from? Taken from http://en.wikipedia.org/wiki/Pauli_equation. Following wikipedia's steps the Stern Gerlach term pops out when you apply the Pauli vector identity. I don't understand this step. It seems as if there should be no Stern Gerlach term.

Here are my steps starting with the Pauli vector identity,
$(\boldsymbol{\sigma} \cdot \boldsymbol{a})(\boldsymbol{\sigma} \cdot \boldsymbol{b}) = \boldsymbol{a} \cdot \boldsymbol{b} + i\boldsymbol{\sigma} \cdot (\boldsymbol{a} \times \boldsymbol{b})$

$(\boldsymbol{\sigma} \cdot (\boldsymbol{p} - e\boldsymbol{A}))^2 = (\boldsymbol{\sigma} \cdot (\boldsymbol{p} - e\boldsymbol{A}))(\boldsymbol{\sigma} \cdot (\boldsymbol{p} - e\boldsymbol{A})) = (\boldsymbol{p} - e\boldsymbol{A})^2 + i\boldsymbol{\sigma} \cdot ((\boldsymbol{p} - e\boldsymbol{A}) \times (\boldsymbol{p} - e\boldsymbol{A}))$

Shouldn't

$\boldsymbol{v} \times \boldsymbol{v} = \boldsymbol{0}$

and

$(\boldsymbol{\sigma} \cdot (\boldsymbol{p} - e\boldsymbol{A}))^2 = (\boldsymbol{p} - e\boldsymbol{A})^2$

I did it out in individual components as well, and came to the same conclusion. What am I missing?

 

[king vitamin]

You almost get zero, but you end up with one term because momentum does not commute with the vector potential in general.

$$i\boldsymbol{\sigma} \cdot ((\boldsymbol{p} - e\boldsymbol{A}) \times (\boldsymbol{p} - e\boldsymbol{A})) = -\hbar e \epsilon_{jkl}\sigma_j \partial_k A_l = - e \hbar \mathbf{\sigma} \cdot \mathbf{B}$$

I used p = -i hbar ∇. If you look back over your work when you did the calculation by components, you'll probably find the step where you moved the momentum operator past the vector potential.

 

[MisterX]

Note that unlike in conventional vector algebra, in this algebra of vector operators
$\mathbf{p}\times \mathbf{A} + \mathbf{A} \times \mathbf{p} \neq 0$

$$[\mathbf{A} \times \mathbf{p}]_i = \sum_{jk} \epsilon_{ijk}A_jp_k = \sum_{jk} \epsilon_{ijk}p_kA_j - [p_k, A_j] = \sum_{jk} \epsilon_{ijk}p_kA_j +i\hbar \frac{\partial A_j}{\partial x_k} $$ $$\mathbf{p}\times \mathbf{A} + \mathbf{A} \times \mathbf{p}= i\hbar \nabla \times \mathbf{A} $$

An important takeaway from this is that identities from vector calculus will not always be true for vector operators.

Edit: Have I made a sign error?