How Does Position Interact with Spin Angular Momentum in Quantum Mechanics?

[dnl914]

TL;DR Summary

How do the position operator x and the spin angular momentum operator S commute?

I know how position and momentum commute, but now I have the spin angular momentum operator involved as well as a dot product. Specifically, what would the commutation [x,S·p] be?

 

[PeterDonis]

I know how position and momentum commute, but now I have the spin angular momentum operator involved as well as a dot product. Specifically, what would the commutation [x,S·p] be?

The spin angular momentum operator commutes with both position and momentum (i.e., its commutator with those operators vanishes), since it operates on a different part of the Hilbert space from those operators (the spin operator operates on the spin degrees of freedom, not the configuration space degrees of freedom).

Note that the total angular momentum operator, which includes orbital angular momentum as well as spin, does not commute with position or momentum (i.e., its commutator with those operators does not vanish).

 

[vanhees71]

TL;DR Summary: How do the position operator x and the spin angular momentum operator S commute?

I know how position and momentum commute, but now I have the spin angular momentum operator involved as well as a dot product. Specifically, what would the commutation [x,S·p] be?

The answer depends on whether you work in non-relativistic quantum mechanics of relativistic quantum-field theory.

In non-relativistic quantum mechanics the spin is just an additional "intrinsic angular momentum" degree of freedom, which is implement by a set of self-adjoint operators, obeying the angular-momentum commutation relations,
$$[\hat{s}_j,\hat{s}_k]=\mathrm{i} \hbar \epsilon_{jkl} \hat{s}_l.$$
Since it's providing entirely independent degrees of freedom, the spin operators commute with both position and momentum operators,
$$[\hat{s}_j,\hat{x}_k]=0,\quad [hat{s}_j,\hat{p}_k]=0.$$
You get the wave-mechanics description by choosing as a complete set of compatible observables the position, $\hat{\vec{s}}^2$, and $\hat{s}_3$.

A particle, in addition of mass, has the spin-quantum number $s \in \{0,1/2,1 \ldots \}$ as and additional intrinsic property. I.e., for a certain kind of partice you have only states with one $s$. So a complete basis is $|\vec{x},m_s \rangle$ with $m_s \in \{-s,-s+1,\ldots,s-1,s \}$ fulfilling the eigenvalue equations
$$\hat{\vec{x}} |\vec{x},m_s \rangle=\vec{x} |\vec{x},m_s \rangle, \quad \hat{s}_3 |\vec{x},m_s \rangle=\hbar m_s |\vec{x},m_s \rangle, \quad \hat{\vec{s}}^2 |\vec{s},m_s \rangle=\hbar^2 s(s+1) |\vec{x},m_s \rangle.$$
The wave function is now a $(2s+1)$ component "spinor":
$$\psi(\vec{x})=\begin{pmatrix} \langle \vec{x},s|\psi \rangle \\ \langle \vec{x},s-1|\psi \rangle \\ \vdots \\ \langle \vec{x},-s|\psi \rangle \end{pmatrix} = \begin{pmatrix} \psi_s(\vec{x}) \\ \psi_{s-1}(\vec{x}) \\ \vdots \\ \psi_{-s}(\vec{x}) \end{pmatrix}.$$
The spin operators are represented as $(2s+1) \times (2s+1)$-dimensional self-adjoint matrices
$$\vec{s}_{m_s m_s'}=\langle m_s|\hat{\vec{s}}|m_s' \rangle.$$
The spin entirely acts on the components of the spinor-valued wave function and does nothing related $\vec{x}$
$$[\hat{\vec{s}} \psi(\vec{x})]_{m_s}=\sum_{m_s'=-s}^s \vec{s}_{m_s m_s'} \psi_{m_s'}(\vec{x}).$$
The position operator for wave functions is simply the multiplication with $\vec{x}$, which commutes with the matrix multiplication of the wave funtion for spin. Also momentum is given as an operator acting on position-wave functions as $\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}$, which also commutes with the matrix multiplication of the wave function with the spin matrices.

In relativistic QFT it's an entirely different business. There you need the more complicated theory of the unitary representation of the Poincare group.