Does the momentum operator commute with the Pauli matrix? I tried to calculate the effect of spin orbit coupling $H_s=\alpha(\sigma_xp_y-\sigma_yp_x)$ as in the Rashba effect.
But I just found out that it is not hermitian. Some paper propose some way by adding the hermitian conjugate to ensure the hermicity.
However, when I tried to add a h.c. term, the Hamiltonian turned out to be 0.
$H_s=\alpha(\sigma_xp_y-\sigma_yp_x)=\alpha\left(
\begin{array}{cccc}
0&\frac{\hbar}{i}\frac{\partial}{\partial y}+\hbar\frac{\partial}{\partial x}\\\frac{\hbar}{i}\frac{\partial}{\partial y}-\hbar\frac{\partial}{\partial x}&0
\end{array}\right)$
This matrix is just anti-hermitian and will end up to be 0 after adding up its hermit conjugate. Am I doing this correctly?
 A: The momentum and spin operators do commute. Since $H_s$ is a sum of products of commuting Hermitian operators, it is Hermitian (assuming $\alpha$ is real).
The matrix you have written to represent $H_s$ is correct and Hermitian. But you're right that it appears to be anti-Hermitian. To see what's wrong, consider the simpler $1\times 1$ matrix
$$
\frac{\hbar}i \frac{\partial}{\partial x}
$$
This looks anti-Hermitian too, since taking the conjugate transpose seems to give the same thing back with a minus sign. But of course we know this thing is really Hermitian because it's the operator representing $x$-momentum, which is an observable.
Here's what we've been missing. $\frac{\partial}{\partial x}$ is not just a number. It's an operator that acts on states and returns different states. When we do the Hermitian conjugate we have to remember to take the transpose of this operator. And the transpose of $\frac{\partial}{\partial x}$ is in fact $-\frac{\partial}{\partial x}$. To see this, let $|\psi\rangle$ and $|\psi'\rangle$ be any two states, with wavefunctions $\psi$, $\psi'$. The transpose of $\frac{\partial}{\partial x}$ is defined by
$$
\langle{\psi'}|\left(\frac{\partial}{\partial x}\right)^T|\psi\rangle
=\langle{\psi}|\left(\frac{\partial}{\partial x}\right)|\psi'\rangle^*
$$
We can write the right hand side in terms of the wavefunctions like so:
$$
\langle{\psi'}|\left(\frac{\partial}{\partial x}\right)^T|\psi\rangle
=\int dx \psi(x)\frac{\partial}{\partial x}\psi'^*(x)
$$
Integrating this by parts gives
$$
\langle{\psi'}|\left(\frac{\partial}{\partial x}\right)^T|\psi\rangle
=-\int dx \psi'^*(x)\frac{\partial}{\partial x}\psi(x)
$$
which, translated back into bras and kets, reads
$$
\langle{\psi'}|\left(\frac{\partial}{\partial x}\right)^T|\psi\rangle
=-\langle{\psi'}|\left(\frac{\partial}{\partial x}\right)|\psi\rangle.
$$
So since the states were arbitrary
$$
\left(\frac{\partial}{\partial x}\right)^T = -\left(\frac{\partial}{\partial x}\right).
$$
