$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ ##[14.3. Some quantum mechanical operators](id:sect-14.3) ----- > 1. [General](#:sect-14.3.1) > 2. [Classical Schrödinger operator](#sect-14.3.2) > 3. [Schrödinger-Pauli operator](#sect-14.3.3) > 4. [Dirac operator](#sect-14.3.4) ###[General](id:sect-14.3.1) In Quantum mechanics dynamics is described by \begin{equation} -i\hbar \psi\_t + \mathsf{H} \psi=0 \label{eq-14.3.1} \end{equation} where $\psi=\psi (\mathbf{x},t)$ is a *wave function* and $\mathsf{H}$ is a *quantum Hamiltonian*; $\hbar$ is *Planck's constant*. ###[Classical Schrödinger operator](id:sect-14.3.2) \begin{equation} \mathsf{H} := \frac{1}{2m}(-i\hbar\nabla)^2 +eV(\mathbf{x}) = -\frac{\hbar^2}{2m}\Delta +V(\mathbf{x}) \label{eq-14.3.2} \end{equation} Here $V(x)$ is electric potential, $e$ is particle's, $m$ is particle's mass, $-i\hbar\nabla$ is a (generalized) momentum operator. If we add *magnetic field* with vector potential $\mathbf{A}(\mathbf{x})$ we need to replace $-i\hbar\nabla$ by $\mathbf{P}= -i\hbar\nabla- e\mathbf{A}(\mathbf{x})$: \begin{equation} \mathsf{H} := \frac{1}{2m}(-i\hbar\nabla- e\mathbf{A}(\mathbf{x}))^2 +eV(\mathbf{x}) \label{eq-14.3.3} \end{equation} ###[Schrödinger-Pauli operator](id:sect-14.3.3) Particles with spin $\frac{1}{2}$ are described by \begin{equation} \mathsf{H} := \frac{1}{2m}\bigl((-i\hbar\nabla- e\mathbf{A}(\mathbf{x})\cdot\boldsymbol{\sigma}\bigr)^2 +eV(\mathbf{x}) \label{eq-14.3.4} \end{equation} where $\boldsymbol{\sigma})=(\sigma\_1,\sigma\_2,\sigma\_3)$ are $2\times 2$ Pauli matrices: namely Hermitian matrices satisfying \begin{equation} \sigma\_j\sigma\_k + \sigma\_k\sigma\_j =2\delta\_{jk}I\qquad j,k=1,2,3, \label{eq-14.3.5} \end{equation} $I$ is as usual a unit matrix, and $\psi$ is a complex $2$-vector, \begin{equation} (-i\hbar\nabla- e\mathbf{A}(\mathbf{x})\cdot\boldsymbol{\sigma} := \sum\_{1\le k\le 3} (-i\hbar\partial\_{x\_k}- eA\_k(\mathbf{x})\sigma\_k. \label{eq-14.3.6} \end{equation} **[Remark 1.](id:remark-14.3.1)** It does not really matter which matrices satisfying (\ref{eq-14.3.5}). More precisely, we can replace $\sigma\_k$ by $Q\sigma\_k$ and $\psi$ by $Q\psi$ where $Q$ is a unitary matrix and we can reduce our matrices to \begin{align} \sigma\_1 = \sigma\_x &= \begin{pmatrix} 0&1\\\\ 1&0 \end{pmatrix} ,\label{eq-14.3.7}\\\\ \sigma\_2 = \sigma\_y &= \begin{pmatrix} 0&-i\\\\ i&0 \end{pmatrix}, \label{eq-14.3.8}\\\\ \sigma\_3 = \sigma\_z &= \begin{pmatrix} 1&0\\\\ 0&-1 \end{pmatrix} .\label{eq-14.3.9} \end{align} Actually, it is not completely correct, as it may happen that instead of $\sigma\_3$ we get $-\sigma\_3$ but the remedy is simple: replace $x\_3$ by $-x\_3$ (changing orientation). We assume that our system is as written here. **[Remark 2.](id:remark-14.3.2)** Actually here and in Dirac operator $\psi$ are not vectors but (spinors)[https://en.wikipedia.org/wiki/Spinor] which means that under rotation of coordinate system components of $\psi$ are transformed in a special way. Observe that \begin{gather} \bigl((-i\hbar\nabla- e\mathbf{A}(\mathbf{x})\cdot\boldsymbol{\sigma}\bigr)^2 = (-i\hbar\nabla- e\mathbf{A}(\mathbf{x})^2 + \sum\_{j