Stationary States

A stationary state is a state whose physical predictions remain unchanged under a time-independent Hamiltonian. A pure-state vector may still acquire a phase:

$$\lvert\psi(t)\rangle = e^{-iEt/\hbar}\lvert\psi(0)\rangle.$$

Because quantum states are rays, this time-dependent vector represents the same physical pure state at every time. In density-operator language, stationarity is literal:

$$\rho(t)=\rho(0).$$

For pure states, stationary states are precisely the normalizable energy eigenstates, including arbitrary superpositions inside one degenerate energy eigenspace. For mixed states, the broader criterion is

$$[H,\rho]=0.$$

This distinction explains why a mixture spanning several energies can be stationary even though a coherent pure superposition of those energies generally is not.

Physical Definition

Let

$$U(t)=e^{-iHt/\hbar}$$

for a time-independent self-adjoint Hamiltonian $H$. A state is stationary when its density operator is invariant:

$$U(t)\rho U^\dagger(t)=\rho \qquad \text{for every }t.$$

Consequently, every time-independent measurement has a constant outcome distribution. If $\{M_a\}$ is a POVM, then

$$\begin{aligned} p(a,t) &= \operatorname{Tr}\!\left[\rho(t)M_a\right]\\ &= \operatorname{Tr}\!\left[\rho M_a\right]. \end{aligned}$$

This is stronger than saying that one selected expectation value happens to be constant. Stationarity is a property of the state under a specified Hamiltonian; it fixes the statistics of all time-independent observables and measurements.

For a pure state,

$$\rho_\psi = \lvert\psi\rangle\langle\psi\rvert.$$

The density operator is unchanged exactly when the evolving vector stays on the same ray:

$$U(t)\lvert\psi\rangle = e^{i\chi(t)}\lvert\psi\rangle$$

for some real phase $\chi(t)$.

Pure-State Criterion

Suppose

$$H\lvert E,\lambda\rangle = E\lvert E,\lambda\rangle,$$

where $\lambda$ labels possible degeneracy. Functional calculus gives

$$U(t)\lvert E,\lambda\rangle = e^{-iEt/\hbar}\lvert E,\lambda\rangle.$$

The vector changes, but only by a global phase. Its projector is constant:

$$\begin{aligned} \rho_E(t) &= e^{-iEt/\hbar} \lvert E,\lambda\rangle \langle E,\lambda\rvert e^{iEt/\hbar}\\ &= \rho_E(0). \end{aligned}$$

The converse is also true for a differentiable pure-state evolution. If

$$U(t)\lvert\psi\rangle = e^{i\chi(t)}\lvert\psi\rangle$$

for all $t$, differentiate at $t=0$. The Schrödinger equation gives

$$H\lvert\psi\rangle = -\hbar\dot\chi(0)\lvert\psi\rangle.$$

Thus $\lvert\psi\rangle$ is an energy eigenstate, with

$$E=-\hbar\dot\chi(0).$$

For an unbounded Hamiltonian, this argument assumes that $\lvert\psi\rangle$ lies in the domain of $H$. The Hamiltonian-specific measurement interpretation belongs to Energy Eigenstates.

Why the Phase Is Invisible

Let $\lvert\psi(t)\rangle=e^{-iEt/\hbar}\lvert E\rangle$. For any fixed outcome vector $\lvert\phi\rangle$,

$$\begin{aligned} \left| \langle\phi\vert\psi(t)\rangle \right|^2 &= \left| e^{-iEt/\hbar} \langle\phi\vert E\rangle \right|^2\\ &= \left| \langle\phi\vert E\rangle \right|^2. \end{aligned}$$

For any time-independent observable $A$,

$$\langle A\rangle_t = \langle E\vert A\vert E\rangle.$$

More directly, the phase cancels in every spectral projector of $A$, so its full outcome distribution is unchanged; all moments that exist are therefore constant. The phase is not omitted from the Schrödinger solution; it is present but common to the entire state vector.

In coordinate representation,

$$\psi_E(x,t) = e^{-iEt/\hbar}\psi_E(x),$$

so

$$\left|\psi_E(x,t)\right|^2 = \left|\psi_E(x)\right|^2.$$

The spatial wavefunction $\psi_E(x)$ solves a boundary-value problem for the time-independent Schrödinger equation. That problem is treated in Time-Independent Schrödinger Equation.

Density-Operator Criterion

A density operator evolves according to

$$\rho(t) = e^{-iHt/\hbar} \rho(0) e^{iHt/\hbar}.$$

Equivalently, it obeys the von Neumann equation

$$i\hbar\dot\rho=[H,\rho].$$

Therefore

$$\rho(t)=\rho(0) \quad\Longleftrightarrow\quad [H,\rho]=0$$

in the usual finite-dimensional setting, and under the corresponding domain assumptions in infinite dimensions.

Spectral-block form

For a discrete spectral resolution

$$H=\sum_E E P_E,$$

decompose the density operator into energy blocks:

$$\rho = \sum_{E,E'} P_E\rho P_{E'}.$$

Evolution gives

$$\rho(t) = \sum_{E,E'} e^{-i(E-E')t/\hbar} P_E\rho P_{E'}.$$

The state is stationary exactly when blocks connecting distinct energies vanish:

$$P_E\rho P_{E'}=0 \qquad \text{whenever }E\ne E'.$$

Equivalently,

$$\rho = \sum_E P_E\rho P_E.$$

This form displays both key facts:

• populations in several different energy eigenspaces may coexist in a stationary mixture;
• coherences within one degenerate energy eigenspace may also be stationary.

It is not necessary for a stationary density operator to be a function of $H$. When an eigenspace is degenerate, operators acting nontrivially within that eigenspace can still commute with $H$.

Important examples

Any normalized function of the Hamiltonian is stationary:

$$\rho = \frac{f(H)}{\operatorname{Tr}f(H)}$$

when the denominator exists and $f(H)$ is positive. The canonical thermal state is the familiar case

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

Stationarity alone does not imply thermal equilibrium. Many nonthermal density operators commute with $H$.

Superpositions and Relative Phases

Let

$$\lvert\psi(0)\rangle = \sum_n c_n\lvert E_n\rangle.$$

Then

$$\lvert\psi(t)\rangle = \sum_n c_n e^{-iE_nt/\hbar} \lvert E_n\rangle.$$

A pair of components accumulates the relative phase

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

For a time-independent observable $A$,

$$\langle A\rangle_t = \sum_{m,n} c_m^*c_n e^{-i(E_n-E_m)t/\hbar} A_{mn},$$

where

$$A_{mn} = \langle E_m\vert A\vert E_n\rangle.$$

The diagonal terms are constant. Off-diagonal terms can oscillate at Bohr angular frequencies

$$\omega_{nm} = \frac{E_n-E_m}{\hbar}.$$

A particular observable may be insensitive to these coherences if its relevant off-diagonal matrix elements vanish. That does not make the state stationary. Another measurement can reveal the changing relative phase.

A state that returns to its initial ray after a finite recurrence time is also not necessarily stationary. Stationarity requires invariance at every time, not only at isolated revivals.

Degeneracy

Suppose the energy-$E$ eigenspace has basis

$$\left\{ \lvert E,\lambda\rangle \right\}_{\lambda=1}^{g_E}.$$

Any vector in that eigenspace,

$$\lvert\psi\rangle = \sum_{\lambda=1}^{g_E} c_\lambda\lvert E,\lambda\rangle,$$

satisfies

$$H\lvert\psi\rangle = E\lvert\psi\rangle.$$

It therefore evolves by one common phase and is stationary. The coefficients $c_\lambda$ may contain arbitrary coherences because the Hamiltonian does not distinguish directions within the degenerate eigenspace.

For a mixed state, a block

$$P_E\rho P_E$$

may contain arbitrary positive matrix structure within that same eigenspace. By contrast, $P_E\rho P_{E'}$ with $E\ne E'$ rotates with a nonzero relative phase and prevents stationarity.

If a perturbation lifts the degeneracy, a former superposition within the old eigenspace need no longer be stationary. Stationarity is always relative to the Hamiltonian actually generating the evolution.

Continuous Spectra

For continuous spectra, formal eigenkets such as momentum kets can evolve by a single phase and are often called stationary states. They are generalized eigenvectors, however, and are not normalizable Hilbert-space vectors.

For example, a free-particle plane wave has

$$\psi_p(x,t) \propto e^{ipx/\hbar} e^{-ip^2t/(2m\hbar)}.$$

Its formal probability density is time independent, but the state is delta-normalized. A normalizable wave packet superposes a range of energies and generally spreads, so it is not stationary.

The rigorous mixed-state criterion remains invariance under $U(t)$, or equivalently commutation with the spectral projections of $H$. The discrete phrase “diagonal in the energy basis” must be used with care when no countable normalizable energy basis exists. See Discrete and Continuous Spectra for the underlying distinction.

Example: Qubit Criterion

Take

$$H = \frac{\hbar\omega}{2}\sigma_z.$$

In the $\sigma_z$ basis, write

$$\rho(0) = \begin{pmatrix} \rho_{00} & \rho_{01}\\ \rho_{10} & \rho_{11} \end{pmatrix}.$$

Evolution gives

$$\rho(t) = \begin{pmatrix} \rho_{00} & e^{-i\omega t}\rho_{01}\\ e^{i\omega t}\rho_{10} & \rho_{11} \end{pmatrix}.$$

The state is stationary exactly when

$$\rho_{01}=\rho_{10}=0.$$

Thus every energy-basis mixture

$$\rho = p\lvert0\rangle\langle0\rvert + (1-p)\lvert1\rangle\langle1\rvert$$

is stationary. The pure stationary states are only the two energy eigenstates. A coherent state such as

$$\lvert+x\rangle = \frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2}$$

is not stationary: its off-diagonal density-matrix elements rotate, and its $\sigma_x$ measurement probabilities oscillate.

Example: Infinite Square Well

For a well of width $L$, each normalized energy eigenfunction

$$\psi_n(x) = \sqrt{\frac{2}{L}} \sin\!\left(\frac{n\pi x}{L}\right)$$

produces the stationary solution

$$\psi_n(x,t) = \psi_n(x)e^{-iE_nt/\hbar}.$$

Its probability density is constant:

$$\left|\psi_n(x,t)\right|^2 = \frac{2}{L} \sin^2\!\left(\frac{n\pi x}{L}\right).$$

A coherent superposition of levels $n=1$ and $n=2$ contains an interference term oscillating at

$$\omega_{21} = \frac{E_2-E_1}{\hbar}.$$

The complete boundary-value problem and spectrum are developed in Infinite Square Well.

Example: Stationary Does Not Mean No Current

For a particle of mass $M$ on a ring of radius $R$, a normalized angular-momentum eigenfunction is

$$\psi_m(\varphi,t) = \frac{1}{\sqrt{2\pi}} e^{im\varphi} e^{-iE_mt/\hbar}.$$

Its angular probability density is uniform and time independent:

$$\left|\psi_m(\varphi,t)\right|^2 = \frac{1}{2\pi}.$$

Nevertheless, its probability current in the angular coordinate is

$$\begin{aligned} j_\varphi &= \frac{\hbar}{MR^2} \operatorname{Im} \left( \psi_m^* \frac{\partial\psi_m}{\partial\varphi} \right)\\ &= \frac{\hbar m}{MR^2} \left|\psi_m\right|^2. \end{aligned}$$

For $m\ne0$, the current is nonzero and constant. A stationary state can carry persistent flow; “stationary” means time-independent statistics, not classical rest. See Particle on a Ring for the full system.

Time-Dependent Hamiltonians

The standard stationary-state concept assumes a time-independent Hamiltonian. If

$$H(t)\lvert n(t)\rangle = E_n(t)\lvert n(t)\rangle,$$

an instantaneous eigenvector $\lvert n(t)\rangle$ does not generally solve the time-dependent Schrödinger equation by acquiring only a dynamical phase. Differentiation also produces $\lvert\dot n(t)\rangle$, which can couple different instantaneous eigenspaces.

Adiabatic following, Floquet states, and invariant subspaces provide related but distinct ideas. They should not be labeled stationary without specifying the intended sense. The Core entry point is Time-Dependent Hamiltonians.

Nearby Concepts

Stationary state versus conserved observable. A stationary state has constant statistics for every fixed observable. A conserved observable has constant statistics for every evolving state, under the appropriate commutator and explicit-time conditions. These are different claims.

Stationary state versus equilibrium. Thermal equilibrium states are stationary, but stationarity alone imposes no temperature, maximum-entropy principle, or thermodynamic interpretation.

Stationary state versus a fixed vector. In the Schrödinger picture, a stationary pure-state vector normally carries the phase $e^{-iEt/\hbar}$. In the Heisenberg picture all state vectors are fixed by convention, so fixed coordinates alone do not diagnose physical stationarity.

Stationary density versus one constant probability. A nonstationary state may give a constant distribution for one measurement, such as energy. Testing one observable is insufficient.

Common Mistakes

• Thinking “stationary” means the Schrödinger-picture vector is literally constant.
• Omitting the phase $e^{-iEt/\hbar}$ from an energy-eigenstate solution.
• Calling every superposition nonstationary, including superpositions within one degenerate eigenspace.
• Calling every mixture of energies nonstationary; an energy-block-diagonal mixture is stationary.
• Checking only one expectation value and concluding that the entire state is stationary.
• Assuming a state that revives periodically is stationary at intermediate times.
• Treating generalized continuum eigenkets as normalizable states.
• Equating stationarity with zero probability current or classical rest.
• Assuming an instantaneous eigenstate of $H(t)$ is automatically stationary.
• Equating any stationary density operator with thermal equilibrium.

Cross-Links

• Energy Eigenstates
• Time-Evolution Operator
• Superposition and Relative Phase
• Density Operators
• Conservation Laws
• Discrete and Continuous Spectra
• Infinite Square Well
• Particle on a Ring

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958, secs. 26–28.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977, vol. 1, ch. 3.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994, chs. 4 and 12.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, ch. 2.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, World Scientific, 1998, chs. 3–4.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018, ch. 2.

Exercises

1. Let a normalized pure state be stationary under a time-independent Hamiltonian. Assuming differentiability, prove that it is an energy eigenstate.

Solution

Stationarity of a pure state means that its ray is fixed, so

$$U(t)\lvert\psi\rangle = e^{i\chi(t)}\lvert\psi\rangle.$$

Differentiate at $t=0$. The left side gives

$$\left. \frac{dU}{dt} \right|_{t=0} \lvert\psi\rangle = -\frac{i}{\hbar}H\lvert\psi\rangle,$$

while the right side gives

$$i\dot\chi(0)\lvert\psi\rangle.$$

Equating them yields

$$H\lvert\psi\rangle = -\hbar\dot\chi(0)\lvert\psi\rangle.$$

Thus $\lvert\psi\rangle$ is an eigenvector of $H$.

2. Show that every fixed POVM outcome probability is constant in an energy eigenstate.

Solution

Let $M_a$ be a POVM effect and

$$\lvert\psi(t)\rangle = e^{-iEt/\hbar}\lvert E\rangle.$$

Then

$$\begin{aligned} p(a,t) &= \langle\psi(t)\vert M_a\vert\psi(t)\rangle\\ &= e^{iEt/\hbar}e^{-iEt/\hbar} \langle E\vert M_a\vert E\rangle\\ &= \langle E\vert M_a\vert E\rangle. \end{aligned}$$

The result applies to arbitrary fixed measurements, not only projectors.

3. Suppose

$$\lvert\psi(0)\rangle = c_1\lvert E_1\rangle + c_2\lvert E_2\rangle, \qquad E_1\ne E_2.$$

Derive $\langle A\rangle_t$ for a time-independent observable $A$ and identify the terms that can oscillate.

Solution

The evolved state is

$$\lvert\psi(t)\rangle = c_1e^{-iE_1t/\hbar}\lvert E_1\rangle + c_2e^{-iE_2t/\hbar}\lvert E_2\rangle.$$

Writing $A_{mn}=\langle E_m\vert A\vert E_n\rangle$ gives

$$\begin{aligned} \langle A\rangle_t &= \lvert c_1\rvert^2A_{11} + \lvert c_2\rvert^2A_{22}\\ &\quad+ c_1^*c_2 e^{-i(E_2-E_1)t/\hbar}A_{12}\\ &\quad+ c_2^*c_1 e^{i(E_2-E_1)t/\hbar}A_{21}. \end{aligned}$$

The last two terms contain the changing relative phase. They oscillate when the coefficients and corresponding off-diagonal matrix elements are nonzero.

4. Let

$$H=E_1P_1+E_2P_2, \qquad E_1\ne E_2.$$

Characterize all stationary density operators for this Hamiltonian, allowing either eigenspace to be degenerate.

Solution

Decompose

$$\rho = P_1\rho P_1 + P_1\rho P_2 + P_2\rho P_1 + P_2\rho P_2.$$

The cross blocks evolve with phases

$$e^{-i(E_1-E_2)t/\hbar} \quad\text{and}\quad e^{-i(E_2-E_1)t/\hbar}.$$

They must vanish for stationarity. Hence every stationary density operator has the form

$$\rho = P_1\rho P_1 + P_2\rho P_2,$$

subject to positivity and unit trace. Each diagonal block may contain arbitrary populations and coherences within its degenerate eigenspace.

5. Two orthonormal states $\lvert E,1\rangle$ and $\lvert E,2\rangle$ have the same energy. Show that both

$$\lvert\psi\rangle = \alpha\lvert E,1\rangle + \beta\lvert E,2\rangle$$

and the coherence operator $\lvert E,1\rangle\langle E,2\rvert$ are stationary.

Solution

By linearity,

$$H\lvert\psi\rangle = E\lvert\psi\rangle,$$

so the vector acquires only the common phase $e^{-iEt/\hbar}$.

For the coherence operator $C=\lvert E,1\rangle\langle E,2\rvert$,

$$\begin{aligned} e^{-iHt/\hbar}Ce^{iHt/\hbar} &= e^{-iEt/\hbar} e^{iEt/\hbar} C\\ &= C. \end{aligned}$$

Degeneracy makes the phase difference zero.

6. For the qubit Hamiltonian

$$H=\frac{\hbar\omega}{2}\sigma_z,$$

show directly from the commutator that a density operator is stationary exactly when its Bloch vector points along the $z$ axis.

Solution

Write

$$\rho = \frac12 \left( I+r_x\sigma_x+r_y\sigma_y+r_z\sigma_z \right).$$

Using

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

gives

$$[H,\rho] = \frac{i\hbar\omega}{2} \left( r_x\sigma_y-r_y\sigma_x \right).$$

This vanishes exactly when $r_x=r_y=0$. The remaining Bloch vector is parallel or antiparallel to the $z$ axis, with any allowed length $\lvert r_z\rvert\le1$.

7. Verify that the ring eigenfunction

$$\psi_m(\varphi,t) = \frac{e^{im\varphi-iE_mt/\hbar}}{\sqrt{2\pi}}$$

has time-independent density but nonzero current when $m\ne0$.

Solution

Its density is

$$\left|\psi_m(\varphi,t)\right|^2 = \frac{1}{2\pi},$$

which is independent of $\varphi$ and $t$. Also,

$$\frac{\partial\psi_m}{\partial\varphi} = im\psi_m.$$

Therefore

$$\begin{aligned} j_\varphi &= \frac{\hbar}{MR^2} \operatorname{Im} \left( \psi_m^* \frac{\partial\psi_m}{\partial\varphi} \right)\\ &= \frac{\hbar m}{MR^2} \left|\psi_m\right|^2. \end{aligned}$$

This is nonzero for $m\ne0$, demonstrating that stationarity does not require zero flow.

8. Assume $e^{-\beta H}$ is trace class. Prove that the canonical state

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

is stationary. Does the converse hold: must every stationary state have this form?

Solution

The operator $e^{-\beta H}$ is a function of $H$, so

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

The partition function in the denominator is a scalar. Hence

$$[H,\rho_\beta]=0,$$

and the state is stationary.

The converse does not hold. Any positive, unit-trace density operator that is block diagonal in the energy eigenspaces is stationary. Its populations need not have Boltzmann weights, and degenerate energy blocks may contain coherences.