Stationary States and Phases

Stationary-state language is the simplest way to understand exact time evolution under a time-independent Hamiltonian. Energy eigenstates acquire phases. A single energy eigenspace acquires only a global phase. Superpositions of different energies acquire relative phases, and those relative phases produce time-dependent interference, expectation values, beats, and revivals.

Energy Eigenstates

Let $H$ be time independent. An energy eigenstate satisfies

$$H\lvert E_n,\alpha\rangle = E_n\lvert E_n,\alpha\rangle,$$

where $\alpha$ labels possible degeneracy. The time-evolution operator gives

$$U(t,t_0)\lvert E_n,\alpha\rangle = e^{-iE_n(t-t_0)/\hbar} \lvert E_n,\alpha\rangle.$$

For one nondegenerate eigenstate, the state vector changes by a phase, but the ray does not. Since pure states are rays, physical predictions for time-independent measurements are unchanged.

In coordinate representation this is the familiar separated form

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

where $\psi_n(x)$ solves the time-independent Schrödinger equation. The detailed wave-mechanics expansion conventions are given in Stationary States and Expansions.

Global Phase of a Single Stationary State

If

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

then for any time-independent observable $A$,

$$\langle A\rangle_t = \langle E\rvert e^{iEt/\hbar} A e^{-iEt/\hbar} \lvert E\rangle = \langle E\rvert A\lvert E\rangle.$$

The phase cancels between bra and ket. This is why “stationary” does not mean “no symbol depends on time.” It means the time dependence is a physically invisible global phase for that state.

For the ray-level interpretation, see Rays and Global Phase.

Superpositions and Relative Phases

For an initial state expanded in an energy eigenbasis,

$$\lvert\psi(t_0)\rangle = \sum_n c_n\lvert E_n\rangle,$$

time evolution gives

$$\lvert\psi(t)\rangle = \sum_n c_n e^{-iE_n(t-t_0)/\hbar} \lvert E_n\rangle.$$

Only relative phases are observable. Between two components,

$$\frac{ e^{-iE_n(t-t_0)/\hbar} }{ e^{-iE_m(t-t_0)/\hbar} } = e^{-i(E_n-E_m)(t-t_0)/\hbar}.$$

The angular frequency

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

is a Bohr frequency. It controls oscillations in interference terms and in observables that connect the two energy components.

Time-Dependent Expectation Values

For a time-independent observable $A$, write

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

Then

$$\langle A\rangle_t = \sum_{m,n} c_m^*c_n e^{i(E_m-E_n)(t-t_0)/\hbar} A_{mn}.$$

The diagonal terms $m=n$ are constant. Off-diagonal terms can oscillate at energy-difference frequencies. If $A$ is diagonal in the energy basis, its expectation value is constant even for a superposition of energies. That does not make the state stationary; it means that particular observable does not see the evolving relative phases.

Energy probabilities are constant for a time-independent Hamiltonian:

$$p(E_n) = \lvert c_n\rvert^2$$

in the nondegenerate discrete case. Other measurement probabilities can still change.

Degeneracy

Degeneracy is the main reason “superposition” and “nonstationary” should not be identified. If all components lie in the same eigenspace,

$$\lvert\psi\rangle = \sum_{\alpha=1}^{g_n} c_\alpha\lvert E_n,\alpha\rangle,$$

then

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

The whole superposition evolves by one common phase and is stationary.

Density operators make the criterion especially clean. For a time-independent Hamiltonian,

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

In finite dimensions, the state is stationary precisely when

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

For an unbounded $H$, the domain-free statement is that $\rho$ commutes with every spectral projector of $H$, equivalently $U(t)\rho U(t)^\dagger=\rho$ for all times. Coherences inside one degenerate eigenspace may be stationary. Coherences between different energy eigenspaces generally rotate.

Beats and Revivals

For two energy components, the relative phase has a single beat frequency:

$$\omega = \frac{E_2-E_1}{\hbar}.$$

Observables with off-diagonal matrix elements between the two components can oscillate with period

$$T = \frac{2\pi}{\lvert\omega\rvert} = \frac{2\pi\hbar}{\lvert E_2-E_1\rvert}.$$

For a discrete superposition, exact recurrence requires all occupied energy components to acquire the same phase. If all relevant energy differences are commensurate, a state may revive exactly up to a global phase. A finite superposition with incommensurate gaps has arbitrarily close approximate recurrences but no exact common period. Continuous spectral components need not recur; their spreading cannot be inferred from a finite set of phases.

The infinite square well is a clean example. Its energies satisfy

$$E_n=E_1 n^2.$$

All phases return to one at

$$T_{\mathrm{rev}} = \frac{2\pi\hbar}{E_1},$$

because $e^{-i2\pi n^2}=1$ for integer $n$. Detailed wave-packet reconstruction is developed in Infinite Square Well.

Worked Example: Two-Level Superposition

Let

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

with eigenstates $\lvert 0\rangle$ and $\lvert 1\rangle$:

$$H\lvert 0\rangle = \frac{\hbar\omega}{2}\lvert 0\rangle, \qquad H\lvert 1\rangle = - \frac{\hbar\omega}{2}\lvert 1\rangle.$$

The $z$-basis eigenstates are stationary. The $x$-basis state

$$\lvert +x\rangle = \frac{1}{\sqrt2} \left( \lvert 0\rangle+\lvert 1\rangle \right)$$

is not stationary. It evolves as

$$\lvert\psi(t)\rangle = \frac{1}{\sqrt2} \left( e^{-i\omega t/2}\lvert 0\rangle + e^{i\omega t/2}\lvert 1\rangle \right).$$

The probability of finding $\lvert +x\rangle$ again is

$$\left\lvert \langle +x\rvert\psi(t)\rangle \right\rvert^2 = \cos^2\frac{\omega t}{2}.$$

Equivalently,

$$\langle\sigma_x\rangle_t=\cos\omega t, \qquad \langle\sigma_y\rangle_t=\sin\omega t, \qquad \langle\sigma_z\rangle_t=0.$$

The state remains normalized, but its relative phase rotates around the Bloch sphere.

Common Mistakes

• Saying an energy eigenstate has “no time dependence” instead of “only a global phase.”
• Treating every superposition as nonstationary; superpositions inside a degenerate eigenspace are stationary.
• Forgetting that energy-basis coefficients are constant only when the Hamiltonian is time independent.
• Assuming a constant expectation value for one observable means the whole state is stationary.
• Confusing energy measurement probabilities, which are constant here, with all measurement probabilities.
• Ignoring continuum components when using stationary-state expansions for systems with scattering states.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

Stationary criterion for pure states

Show that any normalized state inside one degenerate eigenspace is stationary under a time-independent Hamiltonian.

Solution

Let

$$\lvert\psi\rangle = \sum_{\alpha} c_\alpha\lvert E,\alpha\rangle.$$

Since every basis vector in the sum has the same energy,

$$H\lvert\psi\rangle = \sum_\alpha c_\alpha E\lvert E,\alpha\rangle = E\lvert\psi\rangle.$$

Therefore

$$\lvert\psi(t)\rangle = e^{-iE(t-t_0)/\hbar} \lvert\psi(t_0)\rangle.$$

The ray is unchanged, so time-independent physical predictions are stationary.

Oscillation from two energies

For

$$\lvert\psi(t)\rangle = \frac{1}{\sqrt2} \left( e^{-iE_1t/\hbar}\lvert E_1\rangle + e^{-iE_2t/\hbar}\lvert E_2\rangle \right),$$

compute the probability of projecting onto $\lvert\chi\rangle=(\lvert E_1\rangle+\lvert E_2\rangle)/\sqrt2$.

Solution

The amplitude is

$$\langle\chi\rvert\psi(t)\rangle = \frac12 \left( e^{-iE_1t/\hbar} + e^{-iE_2t/\hbar} \right).$$

Factor out an unobservable common phase and define $\Delta E=E_2-E_1$:

$$\langle\chi\rvert\psi(t)\rangle = e^{-i(E_1+E_2)t/(2\hbar)} \cos \left( \frac{\Delta E\,t}{2\hbar} \right).$$

Thus

$$P_\chi(t) = \cos^2 \left( \frac{\Delta E\,t}{2\hbar} \right).$$

Infinite-well revival time

For an infinite square well with $E_n=E_1n^2$, show that all components revive at $T_{\mathrm{rev}}=2\pi\hbar/E_1$.

Solution

At $T_{\mathrm{rev}}$,

$$e^{-iE_nT_{\mathrm{rev}}/\hbar} = e^{-iE_1n^2(2\pi\hbar/E_1)/\hbar} = e^{-i2\pi n^2} =1$$

for every integer $n$. Therefore every energy component has the same phase it had initially, and any superposition built from these components is reconstructed.