Liouville–von Neumann Equation

The Liouville–von Neumann equation is the closed-system equation of motion for a density operator:

$$i\hbar\frac{d\rho}{dt} = [H,\rho].$$

Equivalently,

$$\frac{d\rho}{dt} = -\frac{i}{\hbar}[H,\rho].$$

It is the density-operator version of the Schrödinger equation. It applies to a closed system evolving unitarily under a Hamiltonian $H(t)$. It does not describe dissipative dynamics, measurement conditioning, decoherence from an unmodeled environment, or phenomenological relaxation; those require an open-system map or master equation.

Scope and Assumptions

The equation assumes that the system state at time $t$ is represented by a density operator $\rho(t)$ and that all time evolution is generated by a self-adjoint Hamiltonian. For a time-dependent Hamiltonian, the same local equation holds:

$$i\hbar\frac{d\rho(t)}{dt} = [H(t),\rho(t)].$$

The Hamiltonian may include externally controlled fields, but the evolution is still closed-system unitary evolution. If the system is coupled to an environment and the environment is traced out, the reduced state usually does not obey this equation by itself.

The canonical density-operator definitions live in Density Operators. The expectation-value trace rule used below is reviewed in Trace Rule for Expectation Values.

Derivation from Pure-State Evolution

For a pure state,

$$\rho(t) = \lvert\psi(t)\rangle\langle\psi(t)\rvert.$$

The Schrödinger equation and its adjoint are

$$i\hbar\frac{d}{dt}\lvert\psi(t)\rangle = H\lvert\psi(t)\rangle, \qquad -i\hbar\frac{d}{dt}\langle\psi(t)\rvert = \langle\psi(t)\rvert H.$$

Differentiate $\rho(t)$:

$$\begin{aligned} \frac{d\rho}{dt} &= \frac{d\lvert\psi\rangle}{dt}\langle\psi\rvert + \lvert\psi\rangle\frac{d\langle\psi\rvert}{dt} \\ &= -\frac{i}{\hbar}H\lvert\psi\rangle\langle\psi\rvert + \frac{i}{\hbar}\lvert\psi\rangle\langle\psi\rvert H \\ &= -\frac{i}{\hbar}(H\rho-\rho H). \end{aligned}$$

Therefore

$$i\hbar\frac{d\rho}{dt} = [H,\rho].$$

This derivation is short, but it is often the best sign check: if $H$ is on the left of $\rho$ with the wrong sign, the equation will disagree with pure-state Schrödinger evolution.

Mixed-State Evolution

For an ensemble of pure states with fixed preparation probabilities,

$$\rho(t) = \sum_k p_k \lvert\psi_k(t)\rangle\langle\psi_k(t)\rvert, \qquad \sum_k p_k=1,$$

linearity gives the same equation:

$$\frac{d\rho}{dt} = \sum_k p_k\frac{d}{dt} \left( \lvert\psi_k\rangle\langle\psi_k\rvert \right) = -\frac{i}{\hbar}[H,\rho].$$

The result is not limited to a particular ensemble decomposition. If an initial density operator is given abstractly, the unitary evolution operator gives

$$\rho(t) = U(t,t_0)\rho(t_0)U^\dagger(t,t_0),$$

where $U(t,t_0)$ satisfies the operator Schrödinger equation described in Time-Evolution Operator.

Time-Independent Hamiltonians

For a time-independent Hamiltonian,

$$U(t,t_0) = e^{-iH(t-t_0)/\hbar},$$

so

$$\rho(t) = e^{-iH(t-t_0)/\hbar} \rho(t_0) e^{iH(t-t_0)/\hbar}.$$

In an energy eigenbasis $H\lvert n\rangle=E_n\lvert n\rangle$, the density-matrix elements evolve as

$$\rho_{mn}(t) = e^{-i(E_m-E_n)(t-t_0)/\hbar} \rho_{mn}(t_0).$$

The diagonal populations $\rho_{nn}$ are constant. The off-diagonal coherences rotate at Bohr frequencies $(E_m-E_n)/\hbar$.

Trace, Positivity, and Purity

Closed-system density-matrix dynamics preserves the defining properties of a density operator.

Trace is conserved because the trace of a commutator vanishes:

$$\frac{d}{dt}\operatorname{Tr}\rho = -\frac{i}{\hbar}\operatorname{Tr}[H,\rho] = 0.$$

Hermiticity is preserved because $\rho(t)=U\rho(t_0)U^\dagger$ is Hermitian whenever $\rho(t_0)$ is Hermitian. Positivity is also preserved:

$$\langle\phi|\rho(t)|\phi\rangle = \langle U^\dagger\phi|\rho(t_0)|U^\dagger\phi\rangle \ge0.$$

Purity is conserved:

$$\begin{aligned} \frac{d}{dt}\operatorname{Tr}\rho^2 &= 2\operatorname{Tr}\left(\rho\frac{d\rho}{dt}\right) \\ &= -\frac{2i}{\hbar} \operatorname{Tr}\bigl(\rho[H,\rho]\bigr) \\ &= -\frac{2i}{\hbar} \left[ \operatorname{Tr}(\rho H\rho) - \operatorname{Tr}(\rho^2H) \right] =0. \end{aligned}$$

More generally, unitary conjugation preserves the spectrum of $\rho$. Therefore the von Neumann entropy $S(\rho)=-\operatorname{Tr}(\rho\log\rho)$ is constant for a closed system. Entropy growth in a subsystem is not a violation of this statement; it reflects that the subsystem is not closed.

Expectation Values

If $A$ is a Schrödinger-picture observable with possible explicit time dependence, then

$$\langle A\rangle = \operatorname{Tr}(\rho A).$$

Using the Liouville–von Neumann equation and trace cyclicity,

$$\frac{d}{dt}\langle A\rangle = \frac{i}{\hbar} \operatorname{Tr}\bigl(\rho[H,A]\bigr) + \operatorname{Tr}\left( \rho\frac{\partial A}{\partial t} \right).$$

For a time-independent observable with $[H,A]=0$, the expectation value is conserved. This is the density-operator form of the same conservation-law criterion used in Heisenberg-picture dynamics.

Stationary Density Operators

For a time-independent Hamiltonian, a stationary density operator satisfies

$$\frac{d\rho}{dt}=0 \qquad\Longleftrightarrow\qquad [H,\rho]=0.$$

If the spectrum of $H$ is nondegenerate, stationarity implies that $\rho$ is diagonal in the energy basis. If $H$ has degeneracies, $\rho$ may have coherences inside a degenerate energy subspace while still commuting with $H$.

Any sufficiently well-defined function of the Hamiltonian is stationary. The canonical thermal example is

$$\rho_\beta = \frac{e^{-\beta H}}{Z}, \qquad Z=\operatorname{Tr}e^{-\beta H},$$

because $[H,e^{-\beta H}]=0$. This statement is only about closed-system stationarity once the thermal state is prepared; deriving thermal equilibrium requires statistical-mechanical assumptions beyond the Liouville–von Neumann equation itself.

Worked Example: Spin One-Half in a Static Field

Let

$$H=\frac{\hbar\omega}{2}\sigma_z, \qquad \rho=\frac12(I+\mathbf r\cdot\boldsymbol\sigma).$$

Using

$$[\sigma_z,\sigma_x]=2i\sigma_y, \qquad [\sigma_z,\sigma_y]=-2i\sigma_x,$$

the Liouville–von Neumann equation gives

$$\frac{d\rho}{dt} = \frac{\omega}{2} \left(r_x\sigma_y-r_y\sigma_x\right).$$

Comparing coefficients in $d\rho/dt=\frac12\dot{\mathbf r}\cdot\boldsymbol\sigma$ gives

$$\dot r_x=-\omega r_y, \qquad \dot r_y=\omega r_x, \qquad \dot r_z=0.$$

Thus

$$\dot{\mathbf r} = \omega\,\hat{\mathbf z}\times\mathbf r.$$

The Bloch vector precesses about the $z$ axis. Its length is constant, so a pure spin state stays pure and a mixed spin state stays equally mixed under the closed-system evolution.

Relation to Open-System Master Equations

The Liouville–von Neumann equation can be written as

$$\dot\rho = \mathcal L_H(\rho), \qquad \mathcal L_H(\rho) = -\frac{i}{\hbar}[H,\rho].$$

Here $\mathcal L_H$ is the Hamiltonian Liouvillian. It is trace-preserving, positivity-preserving, and entropy-preserving because it generates unitary conjugation. The operator-space and vectorized-generator viewpoint is developed in Liouvillian Superoperators.

Open-system master equations add non-Hamiltonian terms. A standard Markovian form is the Lindblad equation,

$$\dot\rho = -\frac{i}{\hbar}[H,\rho] + \sum_\mu \left( L_\mu\rho L_\mu^\dagger - \frac12\{L_\mu^\dagger L_\mu,\rho\} \right).$$

The first term is the Liouville–von Neumann term. The remaining dissipative terms describe information or energy exchanged with degrees of freedom not retained in the system state. For a compact reference, see Lindblad Equation.

Common Mistakes

• Using the equation for a subsystem while silently ignoring its environment.
• Forgetting that unitary evolution preserves the eigenvalues and entropy of the full density operator.
• Treating a time-dependent thermal expression $e^{-\beta H(t)}/Z(t)$ as a solution without checking the actual equation of motion.
• Reversing the commutator sign when translating from Schrödinger-picture states to density operators.
• Confusing ensemble probabilities $p_k$ changing by a classical update with unitary time evolution of the density operator.
• Calling every equation for $\rho$ a Liouville–von Neumann equation; dissipative terms belong to open-system master equations.

Cross-Links

• Operator Dynamics
• Liouvillian Superoperators
• Density Operators
• Trace Rule for Expectation Values
• Time-Evolution Operator
• Time-Dependent Hamiltonians
• Density Operators in Different Pictures
• Heisenberg Equations of Motion
• Formula Sheet
• Lindblad Equation

References

• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
• 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.
• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press, 2002.

Exercises

1. Verify directly that $\rho(t)=U(t,t_0)\rho(t_0)U^\dagger(t,t_0)$ satisfies the Liouville–von Neumann equation when $i\hbar\,\partial_t U=HU$.

Solution

Differentiate:

$$\frac{d\rho}{dt} = \frac{dU}{dt}\rho_0U^\dagger + U\rho_0\frac{dU^\dagger}{dt}.$$

Using $\dot U=-(i/\hbar)HU$ and $\dot U^\dagger=(i/\hbar)U^\dagger H$ gives

$$\frac{d\rho}{dt} = -\frac{i}{\hbar}H\rho + \frac{i}{\hbar}\rho H = -\frac{i}{\hbar}[H,\rho].$$

2. Let $H\lvert n\rangle=E_n\lvert n\rangle$. Derive the equation of motion for $\rho_{mn}(t)=\langle m|\rho(t)|n\rangle$.

Solution

Take the $mn$ matrix element of $i\hbar\dot\rho=[H,\rho]$:

$$i\hbar\dot\rho_{mn} = \langle m|H\rho-\rho H|n\rangle = (E_m-E_n)\rho_{mn}.$$

Therefore

$$\dot\rho_{mn} = -\frac{i}{\hbar}(E_m-E_n)\rho_{mn},$$

and

$$\rho_{mn}(t) = e^{-i(E_m-E_n)(t-t_0)/\hbar} \rho_{mn}(t_0).$$

3. Show that $\operatorname{Tr}\rho^n$ is conserved for any positive integer $n$ under closed-system Liouville–von Neumann evolution.

Solution

Using $\dot\rho=-(i/\hbar)[H,\rho]$,

$$\frac{d}{dt}\operatorname{Tr}\rho^n = n\operatorname{Tr}(\rho^{n-1}\dot\rho) = -\frac{in}{\hbar} \operatorname{Tr}\left(\rho^{n-1}[H,\rho]\right).$$

The last trace is

$$\operatorname{Tr}(\rho^{n-1}H\rho) - \operatorname{Tr}(\rho^nH).$$

By cyclicity,

$$\operatorname{Tr}(\rho^{n-1}H\rho) = \operatorname{Tr}(\rho^nH),$$

so the derivative vanishes.

4. Explain why $\rho_\beta=e^{-\beta H}/Z$ is stationary for time-independent $H$.

Solution

Since $e^{-\beta H}$ is a function of $H$, it commutes with $H$:

$$[H,e^{-\beta H}]=0.$$

The scalar normalization $Z$ also commutes with $H$, so $[H,\rho_\beta]=0$. The Liouville–von Neumann equation then gives $\dot\rho_\beta=0$.