# The Heisenberg Picture: Equations of Motion for Operators

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

$|\Psi(t)\rangle=\hat{U}(t,t_0)|\Psi(t_0)\rangle,$

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

$i\hbar\frac{\partial}{\partial t}\hat{U}(t,t_0)|\Psi(t_0)\rangle = \hat{H}(t)\hat{U}(t,t_0)|\Psi(t_0)\rangle.$

Since this equation will hold regardless of our choice of $|\Psi(t_0)\rangle,$ we conclude that

$i\hbar\frac{\partial}{\partial t}\hat{U}(t,t_0)=\hat{H}(t)\hat{U}(t,t_0).$

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:

$\hat{U}^\dagger(t,t_0)\hat{U}(t,t_0)=\hat{I}$

If we know the time evolution operator, $\hat{U},$ 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:

$\hat{U}(t,t_0)|\Psi(t_0)\rangle=|\Psi(t)\rangle.$

Therefore, if we know the initial state of a system, we can also obtain the expectation value of an operator $\hat{A}$ at some later time:

$\langle\hat{A}\rangle(t) = \langle\Psi(t)|\hat{A}|\Psi(t)\rangle= \langle \Psi(t_0)|\hat{U}^{\dagger}(t,t_0)\hat{A}\hat{U}(t,t_0)|\Psi(t_0)\rangle$

We may view the above equation in two ways. One way is to think of the operator $\hat{A}$ 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 $\hat{A}$ is given by

$\hat{A}_H(t)=\hat{U}^{\dagger}(t,t_0)\hat{A}\hat{U}(t,t_0)$

and our state vector is just the initial state vector $|\Psi(t_0)\rangle.$ 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

$\hat{U}(t,t_0)=e^{-i\hat{H}(t-t_0)/\hbar}.$

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

$|\Psi(t)\rangle=e^{-i\hat{H}(t-t_0)/\hbar}|\Psi(t_0)\rangle$

and that of an operator in the Heisenberg picture is

$\hat{A}_H(t)=e^{i\hat{H}(t-t_0)/\hbar}\hat{A}e^{-i\hat{H}(t-t_0)/\hbar}.$

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:

\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,

$\hat{H}=\frac{\hat{\mathbf{p}}^2}{2m}+V(\hat{\mathbf{r}}).$

then we can find the Heisenberg equations of motion for $\hat{\mathbf{p}}$ and $\hat{\mathbf{r}}.$

The equation for the position operator is

$\frac{d\hat{\mathbf{r}}_H(t)}{dt}=-\frac{i}{\hbar}\left (\left [\hat{\mathbf{r}},\hat{H}\right ]\right )_H.$

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

$\frac{1}{2m}\left [\hat{\mathbf{r}},\hat{\mathbf{p}}^2\right ]=\frac{i\hbar}{m}\hat{\mathbf{p}}$

and

$\left [\hat{\mathbf{r}},V(\hat{\mathbf{r}})\right ]=0.$

The equation of motion for $\hat{\mathbf{r}}$ is therefore

$\frac{d\hat{\mathbf{r}}(t)_{H}}{dt} = \frac{\hat{\mathbf{p}}(t)_{H}}{m}.$

We may see this as defining a velocity operator,

$\hat{\mathbf{v}}=\frac{\hat{\mathbf{p}}}{m}.$

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

$\left [\hat{\mathbf{p}},V(\hat{\mathbf{r}})\right ]=-i\hbar\nabla V(\mathbf{r}).$

The equation of motion for the momentum is thus

$\frac{d\hat{\mathbf{p}}_H}{dt}=-\nabla V(\hat{\mathbf{r}}_H).$

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 $\omega=\sqrt{\frac{k}{m}},\!$ we obtain

$\frac{d\hat{x}_H }{dt}=\frac{\hat{p}_H }{m}$

and

$\frac{d\hat{p}_H}{dt}=-k{\hat{x}_H}.$

Solving the above equations of motion, we obtain

$\hat{x}_H(t)=\hat{x}_H(0)\cos(\omega t)+\frac{\hat{p}_H(0)}{m\omega}\sin(\omega t)$

and

$\hat{p}_H(t)=\hat{p}_H(0)\cos(\omega t)-\hat{x}_H(0)m\omega\sin(\omega t).\!$

It is important to stress that the above solution is for the position and momentum operators. Also, note that $\hat x_H(0)\!$ and $\hat p_H(0)\!$ are just the time independent operators $\hat{x}$ and $\hat{p}.$

## An Example: Charged Particle in an Electromagnetic Field

Recall that the Hamiltonian of a particle of charge $e\!$ and mass $m\!$ in an external electromagnetic field, which may be time-dependent, is given by

$H=\frac{1}{2m}\left(\hat{\mathbf{p}}-\frac{e}{c}\mathbf{A}(\hat{\mathbf{r}},t)\right )^2+e\phi(\hat{\mathbf{r}},t),$

where $\mathbf{A}(\mathbf{r},t)\!$ is the vector potential and ${\phi(\mathbf{r},t)}\!$ 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, $\hat{\mathbf{r}}.$ We determine the equation of motion as follows:

\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 $\hat{\mathbf{v}},$

$\hat{\mathbf{v}}=\frac{1}{m}\left(\hat{\mathbf{p}}-\frac{e}{c}\mathbf{A}(\hat{\mathbf{r}},t)\right ).$

The Hamiltonian can be rewritten as

$H=\tfrac{1}{2}m\hat{\mathbf{v}}\cdot\hat{\mathbf{v}}+e\phi.$

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

\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{align} (\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^2c} \epsilon_{ijk}\left(p_j A_k(\hat{\mathbf{r}},t) + A_j(\hat{\mathbf{r}},t)p_k\right) \\ &= -\frac{e}{m^2c}\epsilon_{ijk}p_jA_k(\hat{\mathbf{r}},t) - \frac{e}{m^2c} \epsilon_{ijk} A_j(\hat{\mathbf{r}},t) p_k \\ &= -\frac{e}{m^2c}\epsilon_{ijk} p_j A_k(\hat{\mathbf{r}},t)-\frac{e}{m^2c} \epsilon_{ikj} A_k(\hat{\mathbf{r}},t) p_j \\ &= -\frac{e}{m^2c}\epsilon_{ijk} p_j A_k(\hat{\mathbf{r}},t) + \frac{e}{m^2c} \epsilon_{ijk} A_k(\hat{\mathbf{r}},t) p_j \\ &= -\frac{e}{m^2c}\epsilon_{ijk}\left[p_j,A_k(\hat{\mathbf{r}},t)\right] \\ &= -\frac{e}{m^2c}\epsilon_{ijk}\frac{\hbar}{i} \nabla_j A_k(\hat{\mathbf{r}},t) \\ &= i\hbar\frac{e}{m^2c}\left(\nabla \times \bold A\right)_i \end{align}

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

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

this becomes

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

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

\begin{align} \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{align}

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.