The Heisenberg Picture: Equations of Motion for Operators

Quantum Mechanics A

Schrödinger Equation The most fundamental equation of quantum mechanics; given a Hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} , it describes how a state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle} evolves in time.

Physical Basis of Quantum Mechanics

Basic Concepts and Theory of Motion UV Catastrophe (Black-Body Radiation) Photoelectric Effect Stability of Matter Double Slit Experiment Stern-Gerlach Experiment The Principle of Complementarity The Correspondence Principle The Philosophy of Quantum Theory

Schrödinger Equation

Brief Derivation of Schrödinger Equation Relation Between the Wave Function and Probability Density Stationary States Heisenberg Uncertainty Principle Some Consequences of the Uncertainty Principle

Operators, Eigenfunctions, and Symmetry

Linear Vector Spaces and Operators Commutation Relations and Simultaneous Eigenvalues The Schrödinger Equation in Dirac Notation Transformations of Operators and Symmetry Time Evolution of Expectation Values and Ehrenfest's Theorem

Motion in One Dimension

One-Dimensional Bound States Oscillation Theorem The Dirac Delta Function Potential Scattering States, Transmission and Reflection Motion in a Periodic Potential Summary of One-Dimensional Systems

Discrete Eigenvalues and Bound States

Harmonic Oscillator Spectrum and Eigenstates Analytical Method for Solving the Simple Harmonic Oscillator Coherent States Charged Particles in an Electromagnetic Field WKB Approximation

Time Evolution and the Pictures of Quantum Mechanics

The Heisenberg Picture: Equations of Motion for Operators The Interaction Picture The Virial Theorem

Angular Momentum

Commutation Relations Angular Momentum as a Generator of Rotations in 3D Spherical Coordinates Eigenvalue Quantization Orbital Angular Momentum Eigenfunctions

Central Forces

General Formalism Free Particle in Spherical Coordinates Spherical Well Isotropic Harmonic Oscillator Hydrogen Atom WKB in Spherical Coordinates

The Path Integral Formulation of Quantum Mechanics

Feynman Path Integrals The Free-Particle Propagator Propagator for the Harmonic Oscillator

Continuous Eigenvalues and Collision Theory

Differential Cross Section and the Green's Function Formulation of Scattering Central Potential Scattering and Phase Shifts Coulomb Potential Scattering

There are several ways to mathematically approach the change of a quantum mechanical system with time. The Schrödinger picture considers wave functions which change with time while the Heisenberg picture places the time dependence on the operators and deals with wave functions that do not change in time. The interaction picture, also known as the Dirac picture, is somewhere between the other two, placing time dependence on both operators and wave functions. The two "pictures" correspond, theoretically, to the time evolution of two different things. Mathematically, all methods should produce the same result.

Definition of the Heisenberg Picture

The time evolution operator, defined as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi(t)\rangle=\hat{U}(t,t_0)|\Psi(t_0)\rangle,}

is at the heart of the "evolution" of a state, and is the same in each picture. The evolution operator is obtained from the time dependent Schrödinger equation as follows. We start by rewriting the equation as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar\frac{\partial}{\partial t}\hat{U}(t,t_0)|\Psi(t_0)\rangle = \hat{H}(t)\hat{U}(t,t_0)|\Psi(t_0)\rangle. }

Since this equation will hold regardless of our choice of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi(t_0)\rangle,} we conclude that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar\frac{\partial}{\partial t}\hat{U}(t,t_0)=\hat{H}(t)\hat{U}(t,t_0). }

It follows from the fact that the Schrödinger equation preserves the normalization of the state vector for all times that the time evolution operator is unitary:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{U}^\dagger(t,t_0)\hat{U}(t,t_0)=\hat{I}}

If we know the time evolution operator, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{U},} and the initial state of a particular system, all that is needed to determine the state at a later time is to apply the time evolution operator to the initial state vector:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{U}(t,t_0)|\Psi(t_0)\rangle=|\Psi(t)\rangle. }

Therefore, if we know the initial state of a system, we can also obtain the expectation value of an operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} at some later time:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle\hat{A}\rangle(t) = \langle\Psi(t)|\hat{A}|\Psi(t)\rangle= \langle \Psi(t_0)|\hat{U}^{\dagger}(t,t_0)\hat{A}\hat{U}(t,t_0)|\Psi(t_0)\rangle}

We may view the above equation in two ways. One way is to think of the operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} as time-independent and to consider all of the time dependence of its expectation value as coming from the state vector. This point of view is known as the Schrödinger picture. Alternatively, we may think of the operator as evolving in time, while the state vector stays constant. This is known as the Heisenberg picture. In this picture, the time evolution of the operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}_H(t)=\hat{U}^{\dagger}(t,t_0)\hat{A}\hat{U}(t,t_0)}

and our state vector is just the initial state vector Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi(t_0)\rangle.} Note that the only difference between the two pictures is in the way that we assign the time dependence of the system; i.e., whether we think of the state as evolving or of the operators as evolving.

In the special case that the Hamiltonian is independent of time, we may obtain an explicit expression for the time evolution operator. If we solve the differential equation that this operator satisfies, we find that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{U}(t,t_0)=e^{-i\hat{H}(t-t_0)/\hbar}.}

Therefore, the time evolution of a state vector is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi(t)\rangle=e^{-i\hat{H}(t-t_0)/\hbar}|\Psi(t_0)\rangle}

and that of an operator in the Heisenberg picture is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}_H(t)=e^{i\hat{H}(t-t_0)/\hbar}\hat{A}e^{-i\hat{H}(t-t_0)/\hbar}.}

The Heisenberg picture is useful because we can see a closer connection to classical physics than with the Schrödinger picture. In classical physics, we describe the evolution of a system in terms of the time evolution of the observables, such as position or angular momentum, as dictated by the classical equations of motion. Classical mechanics does not include the concept of state vectors, as quantum mechanics does.

The Heisenberg Equation of Motion

In the Heisenberg picture, the quantum mechanical observables change in time as dictated by the Heisenberg equations of motion. We can study the evolution of a Heisenberg operator by differentiating it with respect to time:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{d}{dt}\hat{A}_H(t) &= \frac{\partial\hat{U}^{\dagger}}{\partial t}\hat{A}\hat{U}+\hat{U}^{\dagger}\frac{\partial\hat{A}}{\partial t}\hat{U}+\hat{U}^{\dagger}\hat{A}\frac{\partial\hat{U}}{\partial t} \\ &=-\frac{1}{i\hbar}\hat{U}^{\dagger}\hat{H}\hat{U}\hat{U}^{\dagger}\hat{A}\hat{U}+\left(\frac{\partial\hat{A}}{\partial t}\right)_H+\frac{1}{i\hbar}\hat{U}^{\dagger}\hat{A}\hat{U}\hat{U}^{\dagger}\hat{H}\hat{U} \\ &=-\frac{i}{\hbar}\left(\left[\hat{A},\hat{H}\right]\right)_H+\left(\frac{\partial\hat{A}}{\partial t}\right)_H. \end{align} }

As an example, let us consider the Hamiltonian,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{H}=\frac{\hat{\mathbf{p}}^2}{2m}+V(\hat{\mathbf{r}}).}

then we can find the Heisenberg equations of motion for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{p}}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{r}}.}

The equation for the position operator is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\hat{\mathbf{r}}_H(t)}{dt}=-\frac{i}{\hbar}\left (\left [\hat{\mathbf{r}},\hat{H}\right ]\right )_H.}

The commuators of the position operator with respect to the kinetic and potential terms are

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2m}\left [\hat{\mathbf{r}},\hat{\mathbf{p}}^2\right ]=\frac{i\hbar}{m}\hat{\mathbf{p}}}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left [\hat{\mathbf{r}},V(\hat{\mathbf{r}})\right ]=0.}

The equation of motion for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{r}}} is therefore

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\hat{\mathbf{r}}(t)_{H}}{dt} = \frac{\hat{\mathbf{p}}(t)_{H}}{m}.}

We may see this as defining a velocity operator,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{v}}=\frac{\hat{\mathbf{p}}}{m}.}

We may find the equation for the momentum similarly. The commutator of the momentum with the kinetic term is obviously zero, while that with the potential term is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left [\hat{\mathbf{p}},V(\hat{\mathbf{r}})\right ]=-i\hbar\nabla V(\mathbf{r}).}

The equation of motion for the momentum is thus

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\hat{\mathbf{p}}_H}{dt}=-\nabla V(\hat{\mathbf{r}}_H).}

These are just the equations satisfied by the corresponding classical quantities, as expected.

In particular, if we apply these equations to a harmonic oscillator with natural frequency Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega=\sqrt{\frac{k}{m}},\!} we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\hat{x}_H }{dt}=\frac{\hat{p}_H }{m}}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\hat{p}_H}{dt}=-k{\hat{x}_H}.}

Solving the above equations of motion, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{x}_H(t)=\hat{x}_H(0)\cos(\omega t)+\frac{\hat{p}_H(0)}{m\omega}\sin(\omega t)}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{p}_H(t)=\hat{p}_H(0)\cos(\omega t)-\hat{x}_H(0)m\omega\sin(\omega t).\!}

It is important to stress that the above solution is for the position and momentum operators. Also, note that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat x_H(0)\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat p_H(0)\!} are just the time independent operators Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{x}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{p}.}

An Example: Charged Particle in an Electromagnetic Field

Recall that the Hamiltonian of a particle of charge Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e\!} and mass Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\!} in an external electromagnetic field, which may be time-dependent, is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H=\frac{1}{2m}\left(\hat{\mathbf{p}}-\frac{e}{c}\mathbf{A}(\hat{\mathbf{r}},t)\right )^2+e\phi(\hat{\mathbf{r}},t),}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{A}(\mathbf{r},t)\!} is the vector potential and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\phi(\mathbf{r},t)}\!} is the Coulomb potential of the electromagnetic field. We now wish to determine the Heisenberg equations of motion for the position and velocity operators.

We first turn our attention to the position operator, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{r}}.} We determine the equation of motion as follows:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{d\hat{\mathbf{r}}}{dt} &= \frac{1}{i\hbar}\left[\hat{\mathbf{r}},\hat{H}\right] \\ &= \frac{1}{i\hbar} \left[\hat{\mathbf{r}},\frac{1}{2m}\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)^2 + e\phi(\hat{\mathbf{r}},t)\right] \\ &= \frac{1}{2im\hbar} \left[\hat{\mathbf{r}},\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)^2\right] \\ &= \frac{1}{2im\hbar} \left[\hat{\mathbf{r}},\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)\right]\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) + \frac{1}{2im\hbar} \left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) \left[\hat{\mathbf{r}},\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)\right] \\ &= \frac{1}{2im\hbar} \left[\hat{\mathbf{r}}, \hat{\mathbf{p}}\right] \left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) + \frac{1}{2im\hbar} \left(\hat{\mathbf{p}} - \frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) \left[\hat{\mathbf{r}},\hat{\mathbf{p}}\right] \\ &= \frac{1}{2im\hbar}i\hbar \left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) + \frac{1}{2im\hbar} \left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)i\hbar \\ &= \frac{1}{m}\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right).\end{align} }

This equation defines the velocity operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{v}},}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf{v}}=\frac{1}{m}\left(\hat{\mathbf{p}}-\frac{e}{c}\mathbf{A}(\hat{\mathbf{r}},t)\right ).}

The Hamiltonian can be rewritten as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H=\tfrac{1}{2}m\hat{\mathbf{v}}\cdot\hat{\mathbf{v}}+e\phi.}

We now determine the equation of motion for the velocity operator:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{d\hat{\mathbf{v}}}{dt} &=\frac{1}{i\hbar}\left[\hat{\mathbf{v}},\hat{H}\right]+\frac{\partial\hat{\mathbf{v}}}{\partial t} \\ &= \frac{1}{i\hbar}\left[\hat{\mathbf{v}},\tfrac{1}{2}m\hat{\mathbf{v}}\cdot\hat{\mathbf{v}}\right]+\frac{1}{i\hbar}\left[\hat{\mathbf{v}},e\phi\right]-\frac{e}{mc}\frac{\partial\mathbf{A}}{\partial t} \end{align} }

Note that ${\hat {\mathbf {p} }}$ does not depend on $t\!$ expicitly.

Using the identity,

$\left[{\hat {\mathbf {v} }},{\hat {\mathbf {v} }}\cdot {\hat {\mathbf {v} }}\right]={\hat {\mathbf {v} }}\times \left({\hat {\mathbf {v} }}\times {\hat {\mathbf {v} }}\right)-\left({\hat {\mathbf {v} }}\times {\hat {\mathbf {v} }}\right)\times {\hat {\mathbf {v} }},$

we obtain

${\frac {d{\hat {\mathbf {v} }}}{dt}}={\frac {1}{i\hbar }}{\tfrac {1}{2}}m\left({\hat {\mathbf {v} }}\times ({\hat {\mathbf {v} }}\times {\hat {\mathbf {v} }})-({\hat {\mathbf {v} }}\times {\hat {\mathbf {v} }})\times {\hat {\mathbf {v} }}\right)+{\frac {1}{i\hbar }}e[{\hat {\mathbf {v} }},\phi ]-{\frac {e}{mc}}{\frac {\partial \mathbf {A} }{\partial t}}.$

We now evaluate ${\hat {\mathbf {v} }}\times {\hat {\mathbf {v} }}$ and $[{\hat {\mathbf {v} }},\phi ].$ The former is evaluated as follows:

${\begin{aligned}({\hat {\mathbf {v} }}\times {\hat {\mathbf {v} }})_{i}&=\epsilon _{ijk}v_{j}v_{k}\\&=\epsilon _{ijk}{\frac {1}{m}}\left(p_{j}-{\frac {e}{c}}A_{j}({\hat {\mathbf {r} }},t)\right){\frac {1}{m}}\left(p_{k}-{\frac {e}{c}}A_{k}({\hat {\mathbf {r} }},t)\right)\\&=-{\frac {e}{m^{2}c}}\epsilon _{ijk}\left(p_{j}A_{k}({\hat {\mathbf {r} }},t)+A_{j}({\hat {\mathbf {r} }},t)p_{k}\right)\\&=-{\frac {e}{m^{2}c}}\epsilon _{ijk}p_{j}A_{k}({\hat {\mathbf {r} }},t)-{\frac {e}{m^{2}c}}\epsilon _{ijk}A_{j}({\hat {\mathbf {r} }},t)p_{k}\\&=-{\frac {e}{m^{2}c}}\epsilon _{ijk}p_{j}A_{k}({\hat {\mathbf {r} }},t)-{\frac {e}{m^{2}c}}\epsilon _{ikj}A_{k}({\hat {\mathbf {r} }},t)p_{j}\\&=-{\frac {e}{m^{2}c}}\epsilon _{ijk}p_{j}A_{k}({\hat {\mathbf {r} }},t)+{\frac {e}{m^{2}c}}\epsilon _{ijk}A_{k}({\hat {\mathbf {r} }},t)p_{j}\\&=-{\frac {e}{m^{2}c}}\epsilon _{ijk}\left[p_{j},A_{k}({\hat {\mathbf {r} }},t)\right]\\&=-{\frac {e}{m^{2}c}}\epsilon _{ijk}{\frac {\hbar }{i}}\nabla _{j}A_{k}({\hat {\mathbf {r} }},t)\\&=i\hbar {\frac {e}{m^{2}c}}\left(\nabla \times {\mathbf {A}}\right)_{i}\end{aligned}}$

Noting that the curl of the vector potential is just the magnetic field,

$\mathbf {B} =\nabla \times \mathbf {A} ,$

this becomes

$\left({\hat {\mathbf {v} }}\times {\hat {\mathbf {v} }}\right)=i\hbar {\frac {e}{m^{2}c}}\mathbf {B} .$

The second commutator, $[{\hat {\mathbf {v} }},\phi ],$ is

${\begin{aligned}\left[{\hat {\mathbf {v} }},\phi \right]&={\frac {1}{m}}\left[{\hat {\mathbf {p} }}-{\frac {e}{c}}\mathbf {A} ({\hat {\mathbf {r} }},t),\phi ({\hat {\mathbf {r} }},t)\right]\\&={\frac {1}{m}}\left[{\hat {\mathbf {p} }},\phi ({\hat {\mathbf {r} }},t)\right]\\&={\frac {1}{m}}{\frac {\hbar }{i}}\nabla \phi \end{aligned}}$

Substituting and rearranging, we get

$m{\frac {d{\hat {\mathbf {v} }}}{dt}}={\frac {e}{2c}}\left({\hat {\mathbf {v} }}\times \mathbf {B} -\mathbf {B} \times {\hat {\mathbf {v} }}\right)+e\mathbf {E} ,$

where the electric field $\mathbf {E}$ is given by

$\mathbf {E} =-\nabla \phi -{\frac {1}{c}}{\frac {\partial \mathbf {A} }{\partial t}}.$

This is similar to the classical equation for the motion of a particle under the influence of electric and magnetic (Lorentz) forces. The reason why the Lorentz force term is slightly different from the classical expression, $\mathbf {F} ={\frac {e}{c}}\mathbf {v} \times \mathbf {B} ,$ is because the components of the velocity operator do not commute with those of the magnetic field.