14.3. Some quantum mechanical operators

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## 14.3. Some quantum mechanical operators

### General

In Quantum mechanics dynamics is described by $$-i\hbar \psi_t + \mathsf{H} \psi=0 \label{eq-14.3.1}$$ 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

$$\mathsf{H} := \frac{1}{2m}(-i\hbar\nabla)^2 +eV(\mathbf{x}) = -\frac{\hbar^2}{2m}\Delta +V(\mathbf{x}) \label{eq-14.3.2}$$ 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 $\mathbf{P}= -i\hbar\nabla- e\mathbf{A}(\mathbf{x})$: $$\mathsf{H} := \frac{1}{2m}(-i\hbar\nabla- e\mathbf{A}(\mathbf{x}))^2 +eV(\mathbf{x}) \label{eq-14.3.3}$$

### Schrödinger-Pauli operator

Particles with spin $\frac{1}{2}$ are described by $$\mathsf{H} := \frac{1}{2m}\bigl((-i\hbar\nabla- e\mathbf{A}(\mathbf{x})\cdot\boldsymbol{\sigma}\bigr)^2 +eV(\mathbf{x}) \label{eq-14.3.4}$$ where $\boldsymbol{\sigma})=(\sigma_1,\sigma_2,\sigma_3)$ are $2\times 2$ Pauli matrices: namely Hermitian matrices satisfying $$\sigma_j\sigma_k + \sigma_k\sigma_j =2\delta_{jk}I\qquad j,k=1,2,3, \label{eq-14.3.5}$$ $I$ is as usual a unit matrix, and $\psi$ is a complex $2$-vector, $$(-i\hbar\nabla- e\mathbf{A}(\mathbf{x})\cdot\boldsymbol{\sigma} := \sum_{1\le k\le 3} (-i\hbar\partial_{x_k}- eA_k(\mathbf{x})\sigma_k. \label{eq-14.3.6}$$

Remark 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. 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<k} [\sigma_j,\sigma_k] [P_j,P_k], \label{eq-14.3.10}\\ [P_j,P_k]= -i \hbar e(\partial_j A_k- \partial_k A_j) \label{eq-14.3.11} \end{gather} and $$[\sigma_1,\sigma_2]=i\sigma_3, \quad [\sigma_2,\sigma_3]=i\sigma_1,\quad [\sigma_3,\sigma_1]=i\sigma_2. \label{eq-14.3.12}$$ Then $$\bigl((-i\hbar\nabla- e\mathbf{A}(\mathbf{x})\cdot\boldsymbol{\sigma}\bigr)^2 = (-i\hbar\nabla- e\mathbf{A}(\mathbf{x}))^2 - e\hbar \mathbf{B}(\mathbf{x})\cdot\boldsymbol{\sigma}. \label{eq-14.3.13}$$ where $\mathbf{B}=\nabla \times \mathbf{A}$ is a vector intensity of magnetic field.

### Dirac operator

$$\mathsf{H} := (-i\hbar\nabla- e\mathbf{A}(\mathbf{x})\cdot\boldsymbol{\gamma}+ m\gamma_4 +eV(\mathbf{x}) \label{eq-14.3.14}$$ where $\boldsymbol{\gamma})=(\gamma_1,\gamma_2,\gamma_3)$ are $4\times 4$ Dirac matrices: namely Hermitian matrices satisfying (\ref{eq-14.3.5}) for $j,k=1,2,3,4$. Respectively $\psi$ is $4$-vector.

As $V=0$ $$\mathsf{H} ^2= \bigl((-i\hbar\nabla- e\mathbf{A}(\mathbf{x})\cdot\boldsymbol{\gamma}\bigr)^2+ m^2 I \label{eq-14.3.15}$$

Again, the precise choice of Dirac matrices is of no importance and one can assume that $\gamma_j = \begin{pmatrix} \sigma_j & 0\\ 0 &\sigma_j \end{pmatrix}$ for $j=1,2,3$ and $\gamma_4= \begin{pmatrix} 0 & I\\ -I &0 \end{pmatrix}$ (all blocks are $2\times 2$ matrices). Then $$\mathsf{H} ^2= (-i\hbar\nabla- e\mathbf{A}(\mathbf{x}))^2-\mathbf{B}(\mathbf{x})\cdot\boldsymbol{\gamma}+ m^2 I. \label{eq-14.3.16}$$

Exercise 1 Check all matrix calculations.