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##[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