We follow Ref. [4] [T]  [T] Keep in mind that $e$ is take to be negative in that reference, whereas we use $e>0$ ., since the approach above is somewhat unsatisfactory; if we take the view that these results should be a direct outcome of the minimal coupling prescription discussed in the previous chapter, any such term must be found from shifting $p_{\mu}\rightarrow p_{\mu}-qA_{\mu}$ . Now that looks impossible when working with a Hamiltonian of the form $\hat{H}=\hat{\vec{p}}^{2}/2m$ , but there is a crafty way out. First note that for particles with spin $1/2$ with a wavefunction of the form $$\braket{\vec{r}}{\psi}=\left(\begin{array}{c} \psi_{1}(\vec{r})\\ \psi_{2}(\vec{r})\end{array}\right)$$ we really have to use a matrix Hamiltonian operator, $$\hat{H}=\hat{\vec{p}}^{2}/2m\: I_{2}=\left(\begin{array}{cc} \hat{\vec{p}}^{2}/2m & 0\\ 0 & \hat{\vec{p}}^{2}/2m\end{array}\right),$$ and we have carelessly ignored this. This can usually be done safely, since spin decouples from the spatial motion, but that is not the case in a magnetic field! We can now use the identity discussed in appendix A.9↓, to show that we have two equivalent forms, $$\hat{H}_{\text{free}}=\frac{\hat{\vec{p}}^{2}}{2m}I_{2}=\frac{(\vec{\sigma}\cdot\vec{\hat{p}})^{2}}{2m}.$$ These two forms are of course identical, but they behave differently under the minimal coupling substitution--for the subtle reason that $$\left[\vec{\hat{p}},\vec{A}\right]\neq0.$$ Evaluating the spin-dependent form of the Hamiltonian with minimal coupling we find $$\begin{align*} \hat{H}_{\text{EM}} & =\frac{1}{2m}\left(\vec{\sigma}\cdot\left(\vec{\hat{p}}-q\vec{A}\right)\right)^{2}\\ & =\frac{1}{2m}\left(\vec{\hat{p}}-q\vec{A}\right)^{2}I_{2}+\frac{i}{2m}\vec{\sigma}\cdot\left(\left(\vec{\hat{p}}-q\vec{A}\right)\times\left(\vec{\hat{p}}-q\vec{A}\right)\right)\\ & =\frac{1}{2m}\left(\vec{\hat{p}}-q\vec{A}\right)^{2}I_{2}-\frac{\hbar q}{2m}\vec{\sigma}\cdot\vec{B}.\end{align*}$$ Here we used the fact that in a term of the form $\hat{\vec{p}}\times\vec{A}\phi$ , $\hat{p}$ can either act on $\vec{A}$ or the wave function $\phi$ . In full operatorial notation, $$\hat{\vec{p}}\times\vec{A}=-\vec{A}\times\hat{\vec{p}}-i\hbar(\vec{\nabla}\times A)=-\vec{A}\times\hat{\vec{p}}-i\hbar\vec{B}.$$ We thus would argue that the Schrödinger equation for a particle in a field should be replaced by the [U]  [U] This is also called the Pauli equation↓