Heisenberg Equation

Purpose

The Heisenberg equation of motion assigns closed-system time dependence to operators:

$$\frac{dA_H}{dt} = \frac{i}{\hbar}[H_H,A_H] +\left( \frac{\partial A_S}{\partial t} \right)_H.$$

Equivalently, with $[A,B]=AB-BA$,

$$\frac{dA_H}{dt} = \frac1{i\hbar}[A_H,H_H] +\left( \frac{\partial A_S}{\partial t} \right)_H.$$

The two commutator forms have the same sign because both the commutator order and prefactor have been reversed. This equation is the operator counterpart of Hamiltonian evolution.

The canonical derivation is Heisenberg Equations of Motion. This card collects sign conventions, special cases, solutions, and checks.

At a glance

Task	Formula
Transform an operator	$A_H(t)=U^\dagger(t,t_0)A_S(t)U(t,t_0)$
Transform a state	$\lvert\psi_H\rangle=U^\dagger(t,t_0)\lvert\psi_S(t)\rangle=\lvert\psi_S(t_0)\rangle$
General equation	$\dot A_H=(i/\hbar)[H_H,A_H]+(\partial_tA_S)_H$
No explicit time dependence	$\dot A_H=(i/\hbar)[H_H,A_H]$
Constant of motion	$\partial_tA_S=0$ and $[H_H,A_H]=0\ \Longrightarrow\ \dot A_H=0$
Time-independent finite evolution	$A_H(t)=e^{iH\tau/\hbar}A_Se^{-iH\tau/\hbar}$
Expectation-value motion	$d\langle A\rangle/dt=\langle\dot A_H\rangle_H$
Density-operator dual	$\dot\rho_S=-(i/\hbar)[H_S,\rho_S]$

Here $\tau=t-t_0$, and the subscript $H$ on an explicitly transformed Schrödinger operator means

$$(B_S)_H \equiv U^\dagger(t,t_0)B_S(t)U(t,t_0).$$

Picture convention

This card uses

$$A_H(t) = U^\dagger(t,t_0)A_S(t)U(t,t_0).$$

The corresponding Heisenberg state is

$$\begin{aligned} \lvert\psi_H\rangle &= U^\dagger(t,t_0)\lvert\psi_S(t)\rangle \\ &= \lvert\psi_S(t_0)\rangle, \end{aligned}$$

so it is fixed. Expectation values agree:

$$\langle\psi_S(t)\rvert A_S(t) \lvert\psi_S(t)\rangle = \langle\psi_H\rvert A_H(t) \lvert\psi_H\rangle.$$

This is a redistribution of time dependence, not a different physical theory. A picture is also distinct from a representation: position, momentum, and energy bases can be used in any picture.

At the reference time,

$$U(t_0,t_0)=I \quad\Longrightarrow\quad A_H(t_0)=A_S(t_0).$$

Changing $t_0$ changes representatives and intermediate formulas but not consistently computed predictions.

Derivation and sign

Differentiate

$$A_H=U^\dagger A_SU.$$

Using

$$i\hbar\dot U=H_SU, \qquad \dot U^\dagger = \frac{i}{\hbar}U^\dagger H_S,$$

gives

$$\begin{aligned} \dot A_H &= \dot U^\dagger A_SU +U^\dagger\dot A_SU +U^\dagger A_S\dot U \\ &= \frac{i}{\hbar} U^\dagger(H_SA_S-A_SH_S)U \\ &\quad+ U^\dagger \frac{\partial A_S}{\partial t} U. \end{aligned}$$

Since unitary conjugation preserves products and commutators,

$$\dot A_H = \frac{i}{\hbar}[H_H,A_H] +\left( \frac{\partial A_S}{\partial t} \right)_H.$$

With the convention $[A,B]=AB-BA$,

$$\frac{i}{\hbar}[H,A] = \frac1{i\hbar}[A,H].$$

The often-seen expression $-(i/\hbar)[H,A]$ has the wrong sign under this commutator convention.

Explicit time dependence

The partial derivative term records time dependence already present in the definition of $A_S(t)$, such as a moving detector axis or controlled observable. It is distinct from picture-induced motion.

For example, if

$$A_S(t)=f(t)B_S$$

with fixed $B_S$, then

$$\dot A_H = \frac{i}{\hbar}[H_H,A_H] +\dot f(t)(B_S)_H.$$

Even if $[H_S,B_S]=0$, the operator is not conserved when $\dot f\ne0$.

For the Hamiltonian itself,

$$\frac{dH_H}{dt} = \left( \frac{\partial H_S}{\partial t} \right)_H,$$

because $[H_H,H_H]=0$. Energy is conserved for a closed system with no explicitly time-dependent Hamiltonian. A driven Hamiltonian can change the system’s energy despite commuting with itself at the same instant.

Time-independent finite solution

For time-independent $H$ and an operator $A_S$ with no explicit time dependence,

$$A_H(t) = e^{iH\tau/\hbar} A_S e^{-iH\tau/\hbar}.$$

Hadamard’s lemma gives the nested-commutator expansion

$$\begin{aligned} A_H(t) &= A_S +\frac{i\tau}{\hbar}[H,A_S] \\ &\quad+ \frac1{2!} \left( \frac{i\tau}{\hbar} \right)^2 [H,[H,A_S]] \\ &\quad+ \frac1{3!} \left( \frac{i\tau}{\hbar} \right)^3 [H,[H,[H,A_S]]] +\cdots . \end{aligned}$$

If the nested commutators terminate, the series is finite. If they close on a small operator basis, the Heisenberg equation reduces to a finite system of linear ordinary differential equations.

The series and its convergence require the same care as other exponential conjugation formulas for unbounded operators. See Baker–Campbell–Hausdorff.

Hamiltonian derivation

Define the superoperator

$$\mathcal D_H(A) \equiv \frac{i}{\hbar}[H,A].$$

For operators without explicit time dependence,

$$\dot A_H=\mathcal D_H(A_H).$$

The map is a derivation:

$$\mathcal D_H(AB) = \mathcal D_H(A)B +A\mathcal D_H(B).$$

It also respects adjoints when $H=H^\dagger$:

$$\mathcal D_H(A^\dagger) = \mathcal D_H(A)^\dagger.$$

Therefore unitary Heisenberg evolution preserves operator products, commutation relations, adjoints, spectra, and algebraic identities. In particular,

$$[A_H(t),B_H(t)] = U^\dagger[A_S,B_S]U.$$

Canonical commutation relations are preserved in time under the same unitary evolution.

Constants of motion

If $A_S$ has no explicit time dependence and

$$[H_H,A_H]=0,$$

then

$$\dot A_H=0.$$

For a time-independent Hamiltonian, this is equivalently checked at the reference time:

$$[H,A_S]=0.$$

The conclusion is operator-level conservation. Every spectral projector and every moment of $A$ is then conserved for every initial state.

The converse needs care. One expectation value can be constant in one special state even when $A_H$ is not constant as an operator. Also, an explicitly time-dependent observable can be conserved through cancellation:

$$\frac{i}{\hbar}[H_H,A_H] +\left( \frac{\partial A_S}{\partial t} \right)_H = 0$$

without either term vanishing separately.

Canonical variables

For

$$H = \frac{P^2}{2m}+V(X)$$

and $[X,P]=i\hbar I$, the position equation is

$$\begin{aligned} \dot X_H &= \frac{i}{\hbar} \left[ \frac{P_H^2}{2m},X_H \right] \\ &= \frac{P_H}{m}. \end{aligned}$$

The momentum equation is

$$\begin{aligned} \dot P_H &= \frac{i}{\hbar} [V(X_H),P_H] \\ &= -V'(X_H). \end{aligned}$$

Thus

$$m\ddot X_H=-V'(X_H).$$

These resemble Hamilton’s equations but remain operator equations. Operator ordering and domains still matter.

For a free particle, $V=0$, so

$$P_H(t)=P_H(t_0)$$

and

$$X_H(t) = X_H(t_0) +\frac{\tau}{m}P_H(t_0).$$

The canonical bracket is preserved:

$$[X_H(t),P_H(t)]=i\hbar I.$$

Harmonic oscillator

For

$$H = \hbar\omega \left( a^\dagger a+\frac12 \right),$$

the commutators are

$$[H,a]=-\hbar\omega a, \qquad [H,a^\dagger]=\hbar\omega a^\dagger.$$

Therefore

$$\dot a_H=-i\omega a_H, \qquad \dot a_H^\dagger=i\omega a_H^\dagger,$$

with solutions

$$a_H(t)=e^{-i\omega\tau}a_H(t_0),$$ $$a_H^\dagger(t) = e^{i\omega\tau}a_H^\dagger(t_0).$$

The number operator is conserved:

$$N_H(t)=a_H^\dagger(t)a_H(t)=N_H(t_0).$$

The phase of $a_H$ is the compact algebraic form of oscillator rotation in phase space.

Spin precession

Take

$$H=-\gamma\,\mathbf B\cdot\mathbf S$$

with constant $\mathbf B$ and

$$[S_i,S_j] = i\hbar\epsilon_{ijk}S_k.$$

The Heisenberg equation gives

$$\frac{d\mathbf S_H}{dt} = \gamma\,\mathbf S_H\times\mathbf B.$$

The component parallel to $\mathbf B$ is conserved, while transverse components precess. The sign and interpretation of $\gamma$ depend on the particle and magnetic-moment convention; see Larmor Precession.

Expectation values and Ehrenfest bridge

Because the Heisenberg state is fixed,

$$\frac{d}{dt} \langle A\rangle = \langle\psi_H\rvert \dot A_H \lvert\psi_H\rangle.$$

Hence

$$\frac{d}{dt}\langle A\rangle = \frac{i}{\hbar} \langle[H,A]\rangle +\left\langle \frac{\partial A}{\partial t} \right\rangle,$$

where all objects may be evaluated consistently in either picture.

For the canonical particle,

$$\frac{d}{dt}\langle X\rangle = \frac{\langle P\rangle}{m},$$ $$\frac{d}{dt}\langle P\rangle = -\langle V'(X)\rangle.$$

In general,

$$\langle V'(X)\rangle \ne V'(\langle X\rangle),$$

so expectation values do not obey a closed classical trajectory equation for an arbitrary state and potential. The precise classical-limit discussion belongs to Ehrenfest Theorem.

State–observable duality

In the Schrödinger picture, a density operator obeys

$$\dot\rho_S = -\frac{i}{\hbar}[H_S,\rho_S].$$

An observable with no explicit time dependence obeys

$$\dot A_H = \frac{i}{\hbar}[H_H,A_H].$$

The signs differ because states and observables transform oppositely:

$$\rho_S(t)=U\rho_0U^\dagger, \qquad A_H(t)=U^\dagger A_SU.$$

This duality ensures

$$\operatorname{Tr}[\rho_S(t)A_S] = \operatorname{Tr}[\rho_0A_H(t)].$$

For a reduced open system, the Heisenberg description uses the adjoint of a quantum channel or master-equation generator. It is not generally unitary conjugation on the reduced Hilbert space.

Classical correspondence

Classical Hamiltonian motion reads

$$\frac{df}{dt} = \{f,H\}_{\mathrm{PB}} +\frac{\partial f}{\partial t}.$$

The formal correspondence is

$$\{f,g\}_{\mathrm{PB}} \longleftrightarrow \frac1{i\hbar}[\hat f,\hat g].$$

This analogy explains the shape of the Heisenberg equation, but it is not a complete quantization algorithm. Operator ordering, domains, boundary conditions, and higher quantum corrections remain.

Assumptions and domains

The standard equation assumes:

• closed-system unitary dynamics;
• a self-adjoint Hamiltonian generating the stated propagator;
• one consistent picture transformation;
• existence of the products $H_HA_H$ and $A_HH_H$ on the relevant vectors;
• differentiability of the transformed operator in the intended sense.

For bounded finite matrices, these conditions are usually automatic. For unbounded $X$, $P$, and $H$, a formal commutator can be correct on a dense core without defining an everywhere-valid operator identity. Boundary conditions can also change domains and invalidate naive integration by parts.

When only expectation values are required, weak or quadratic-form versions may suffice, but that relaxation must be stated rather than silently assumed.

Numerical checks

For a numerical Heisenberg evolution $\widetilde A_H(t)$, check:

• Hermiticity when $A_S$ is Hermitian;
• preservation of the spectrum under unitary conjugation;
• preservation of known commutators;
• constants of motion;
• agreement of $\operatorname{Tr}[\rho_0\widetilde A_H(t)]$ with the corresponding Schrödinger-picture calculation;
• convergence under time-step or basis refinement.

For a time-independent finite Hamiltonian, compare direct conjugation

$$\widetilde A_H(t) = \widetilde U^\dagger A_S\widetilde U$$

with integration of the commutator differential equation. Agreement between two methods is stronger evidence than norm preservation alone.

Common mistakes

• Reversing the sign without also reversing commutator order.
• Forgetting the explicit derivative term.
• Mixing a Schrödinger-picture Hamiltonian with a Heisenberg-picture observable without the matching transformation.
• Treating pictures as different theories or confusing them with basis representations.
• Assuming a time-independent Schrödinger operator is automatically conserved.
• Inferring an operator constant of motion from one constant expectation value in one state.
• Replacing $\langle V'(X)\rangle$ by $V'(\langle X\rangle)$ without a justified approximation.
• Solving operator equations as though all quantities commute.
• Using finite nested-commutator series outside their domain or convergence regime.
• Applying unitary conjugation to a reduced open system.
• Ignoring boundary conditions and domains for unbounded operators.

Cross-links

• Canonical Heisenberg Derivation
• Heisenberg Picture
• Pictures of Motion Overview
• Commutator Dynamics
• Explicitly Time-Dependent Operators
• Conservation Laws
• Ehrenfest Theorem
• Commutator Identities
• Baker–Campbell–Hausdorff
• Liouville–von Neumann Equation
• Larmor Precession

References

• W. Heisenberg, “Über quantentheoretische Umdeutung kinematischer und mechanischer Beziehungen,” Zeitschrift für Physik 33, 879–893 (1925).
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.

Exercises

1. Use $[X,P]=i\hbar I$ to derive $\dot X_H=P_H/m$ for a free particle.

Solution

For $H=P^2/(2m)$,

$$\dot X_H = \frac{i}{2m\hbar}[P_H^2,X_H].$$

The product rule gives

$$\begin{aligned} [P_H^2,X_H] &= P_H[P_H,X_H] +[P_H,X_H]P_H \\ &= -2i\hbar P_H. \end{aligned}$$

Therefore

$$\dot X_H = \frac{i}{2m\hbar} (-2i\hbar P_H) = \frac{P_H}{m}.$$

2. For the harmonic oscillator, use the Heisenberg equations for $a_H$ and $a_H^\dagger$ to prove that $N_H=a_H^\dagger a_H$ is constant.

Solution

Use the product rule:

$$\begin{aligned} \dot N_H &= \dot a_H^\dagger a_H +a_H^\dagger\dot a_H \\ &= (i\omega a_H^\dagger)a_H +a_H^\dagger(-i\omega a_H) \\ &= 0. \end{aligned}$$

Equivalently, $[H,N]=0$.

3. Suppose $A_S(t)=tB_S$, with $[H,B_S]=0$ and time-independent $H$. Find $A_H(t)$ and its derivative. Is it conserved?

Solution

Because $B_S$ commutes with $H$,

$$(B_S)_H=B_S.$$

Therefore

$$A_H(t)=tB_S.$$

The commutator contribution vanishes, but the explicit derivative is

$$\dot A_H=B_S.$$

The operator is not conserved. Commutation with $H$ is sufficient only when there is no explicit time dependence.

4. Show that unitary Heisenberg evolution preserves the canonical commutator.

Solution

Using $X_H=U^\dagger X_SU$ and $P_H=U^\dagger P_SU$,

$$\begin{aligned} [X_H,P_H] &= U^\dagger X_SU U^\dagger P_SU \\ &\quad- U^\dagger P_SU U^\dagger X_SU \\ &= U^\dagger[X_S,P_S]U. \end{aligned}$$

Since $[X_S,P_S]=i\hbar I$,

$$[X_H,P_H] = i\hbar U^\dagger U = i\hbar I.$$