Which of these two different forms of spin-orbit interaction is correct? I am seeing the spin-orbit interaction in two different ways:


*

*$\lambda [\mathbf{p} \times \nabla V]\cdot \sigma$

*$\lambda [\nabla V \times \mathbf{p}]\cdot \sigma$


I don't see how these two expressions can be equivalent though; in (1) there will be derivative operators from momentum acting on the gradient whereas in (2) this will not be the case.
Which one of these is correct?
 A: Recall that (times $\lambda$)
$$\vec{\nabla} V \times \vec{p} =-\vec{p} \times \vec{\nabla} V$$
Call $\vec{a} = \vec{\nabla} V \times \vec{p}$. You can check that:
$$\vec{a} \cdot \vec{\sigma}=(-\vec{a})\cdot\vec{\sigma}$$
A: Good question. In fact, you can check that $\triangledown V\times \mathbf{P}=-\mathbf{P}\times\triangledown V$. Here is the brief derivation:
$\mathbf{P}\times\triangledown V=(p_y\partial _zV-p_z\partial _yV,p_z\partial _xV-p_x\partial _zV,p_x\partial _yV-p_y\partial _xV)$. Let's take the first component $p_y\partial _zV-p_z\partial _yV$ for example, it's obvious that
$$p_y\partial _zV-p_z\partial _yV=-i\frac{\partial^2}{\partial _y\partial _z}V+\partial _zVp_y-(-i\frac{\partial^2}{\partial _z\partial _y}V+\partial _yVp_z)$$, 
but $\frac{\partial^2}{\partial _y\partial _z}V=\frac{\partial^2}{\partial _z\partial _y}V$, thus 
$$p_y\partial _zV-p_z\partial _yV=\partial _zVp_y-\partial _yVp_z$$. While the first component of $\triangledown V\times \mathbf{P}$ reads $\partial _yVp_z-\partial _zVp_y$, therefore we arrive at $\triangledown V\times \mathbf{P}=-\mathbf{P}\times\triangledown V$.
