Stationary States and Expansions

Stationary states solve the time-independent Schrödinger equation. Expansions in stationary states solve the time-dependent problem when the Hamiltonian is time independent. Once the energy eigenstates are known, time evolution is mostly bookkeeping: each energy component receives its own phase.

For a discrete nondegenerate spectrum, the core formula is

$$\psi(x,0)=\sum_n c_n\psi_n(x), \qquad \psi(x,t)=\sum_n c_n\psi_n(x)e^{-iE_nt/\hbar}.$$

This page explains what the coefficients mean, when the formula is valid, how continuum states change it, and why interference between energy components produces time-dependent physics.

Stationary States

For a time-independent Hamiltonian,

$$\hat H\psi_n=E_n\psi_n$$

defines a stationary spatial wavefunction. The full time-dependent solution is

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

The density is time independent:

$$\lvert\Psi_n(x,t)\rvert^2 = \lvert\psi_n(x)\rvert^2.$$

The phase is not absent; it is physically global for a single energy component. The abstract ray-level statement is Stationary States.

Discrete Expansions

Suppose the bound-state eigenfunctions satisfy orthonormality

$$\int \psi_m^*(x)\psi_n(x)\,dx = \delta_{mn},$$

and are complete for the class of states being considered. Then an initial wavefunction can be expanded as

$$\psi(x,0)=\sum_n c_n\psi_n(x).$$

The coefficients are projections:

$$c_n = \int \psi_n^*(x)\psi(x,0)\,dx.$$

For a normalized initial state,

$$\sum_n\lvert c_n\rvert^2=1.$$

The probability of measuring energy $E_n$ is $\lvert c_n\rvert^2$ in the nondegenerate case. If the energy is degenerate, the probability for an energy value is the sum of squared amplitudes over a basis of that eigenspace, or equivalently the norm of the projected component.

Time Evolution From Coefficients

Once the coefficients are known, time evolution is

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

The coefficients $c_n$ are constant for a time-independent Hamiltonian. The phases change. This is why the energy basis is the diagonal basis for dynamics.

If the initial state is one eigenstate, only a global phase appears and probabilities are stationary. If two or more different energies are present, relative phases can affect densities and expectation values.

For a two-component state,

$$\psi(x,t) = c_1\psi_1(x)e^{-iE_1t/\hbar} +c_2\psi_2(x)e^{-iE_2t/\hbar},$$

the density contains an interference term oscillating at angular frequency

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

This is one of the simplest ways time-independent Hamiltonians produce time-dependent probability densities.

Orthonormality And Completeness

Orthonormality makes coefficients easy to compute. Completeness is the stronger statement that the eigenfunctions span the states under discussion.

For a purely discrete basis, a formal completeness relation is

$$\sum_n \psi_n(x)\psi_n^*(x') = \delta(x-x').$$

This identity means that any suitable wavefunction can be reconstructed from its coefficients. It is a coordinate-space version of a resolution of the identity.

Completeness is model dependent. The infinite square well has a discrete sine basis on the interval. The free particle has continuum plane waves. A finite well has both bound states and continuum states. The general mathematical background is Completeness and Orthonormal Bases and Spectral Theorem, Practical Version.

Continuous Spectra

For continuum eigenfunctions labeled by $k$ with delta normalization,

$$\int \psi_k^*(x)\psi_{k'}(x)\,dx = \delta(k-k'),$$

an expansion has the form

$$\psi(x,0) = \int c(k)\psi_k(x)\,dk,$$

with coefficient function

$$c(k) = \int \psi_k^*(x)\psi(x,0)\,dx.$$

The evolved state is

$$\psi(x,t) = \int c(k)\psi_k(x)e^{-iE(k)t/\hbar}\,dk.$$

The normalization condition becomes

$$\int \lvert c(k)\rvert^2\,dk=1$$

when the continuum convention is $k$-normalized. If the continuum is labeled by momentum $p$ instead, the measure and coefficient function change. The convention must be stated.

Mixed Spectra

Many physical one-dimensional Hamiltonians have both bound states and scattering states. A finite attractive well, for example, has discrete bound states below the continuum threshold and continuum states above it.

The schematic expansion is

$$\psi(x,0) = \sum_n c_n\psi_n(x) + \int c(k)\psi_k(x)\,dk.$$

The time-evolved state is

$$\psi(x,t) = \sum_n c_n\psi_n(x)e^{-iE_nt/\hbar} + \int c(k)\psi_k(x)e^{-iE(k)t/\hbar}\,dk.$$

The normalization splits into discrete and continuous contributions:

$$\sum_n\lvert c_n\rvert^2 + \int\lvert c(k)\rvert^2\,dk =1,$$

for the chosen continuum normalization.

This is why bound-state tables are not a complete basis for systems with continua. They describe only part of the Hilbert space.

Degeneracy

If the energy $E_n$ is degenerate, write eigenfunctions as $\psi_{n\alpha}$, where $\alpha$ labels independent states with the same energy:

$$\hat H\psi_{n\alpha} = E_n\psi_{n\alpha}.$$

The expansion becomes

$$\psi(x,0) = \sum_{n,\alpha} c_{n\alpha}\psi_{n\alpha}(x),$$

and time evolution attaches the same phase to all states within one degenerate eigenspace:

$$\psi(x,t) = \sum_{n,\alpha} c_{n\alpha}\psi_{n\alpha}(x)e^{-iE_nt/\hbar}.$$

Superpositions inside one degenerate eigenspace are stationary because they share one energy phase. Superpositions across different energies generally are not.

Revivals: First Encounter

A wave packet can spread and later reconstruct if its energy phases re-align. This is called a revival. The infinite square well gives a clean first example because

$$E_n=E_1n^2.$$

At the revival time

$$T_{\mathrm{rev}} = \frac{2\pi\hbar}{E_1} = \frac{4mL^2}{\pi\hbar},$$

each phase satisfies

$$e^{-iE_nT_{\mathrm{rev}}/\hbar} = e^{-i2\pi n^2} =1.$$

Thus the full wavefunction returns to its initial shape. Fractional revivals and imperfect revivals are richer topics, but the basic mechanism is already visible: stationary-state phases carry the long-time dynamics.

Practical Workflow

For a time-independent Hamiltonian:

1. Solve the stationary eigenvalue problem with the correct boundary conditions.
2. Normalize bound states or state the continuum normalization.
3. Check which part of the spectrum is needed: discrete, continuous, or mixed.
4. Compute expansion coefficients from inner products.
5. Attach phases $e^{-iEt/\hbar}$ to each energy component.
6. Use the resulting $\psi(x,t)$ to compute densities, currents, and expectation values.
7. Check limiting cases, conserved norm, and energy probabilities.

This is the basic strategy behind many exactly solvable model pages.

Common Mistakes

• Treating a stationary-state list as complete when continuum states are also present.
• Forgetting that expansion coefficients depend on the normalization convention.
• Attaching the same phase to all terms in a superposition of different energies.
• Dropping degeneracy labels and losing states.
• Thinking a time-independent Hamiltonian makes every probability density time independent.
• Normalizing continuum coefficients as if they were discrete amplitudes.
• Using model eigenfunctions without checking their boundary conditions match the problem.

Where This Is Used

• Time-Independent Schrödinger Equation derives the stationary eigenvalue problem.
• Normalization Conventions explains the bound-state and continuum normalizations used by coefficients.
• Expectation Values in Wave Mechanics uses evolved wavefunctions to compute averages and variances.
• Infinite Square Well is the first complete discrete-basis example.
• Free Particle and Gaussian Wave Packets use continuum expansions.
• Finite Square Well illustrates mixed bound and continuum structure.
• Energy Eigenstates gives the abstract Hamiltonian-eigenbasis viewpoint.

References

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

Exercises

1. Let $\{\psi_n\}$ be an orthonormal discrete energy basis. Derive $c_m=\int\psi_m^*(x)\psi(x,0)\,dx$ from $\psi(x,0)=\sum_n c_n\psi_n(x)$.

Solution

Multiply the expansion by $\psi_m^*(x)$ and integrate:

$$\int \psi_m^*(x)\psi(x,0)\,dx = \sum_n c_n \int \psi_m^*(x)\psi_n(x)\,dx.$$

Using orthonormality,

$$\int \psi_m^*(x)\psi_n(x)\,dx=\delta_{mn},$$

so the sum reduces to $c_m$.

2. A state is $\psi(x,0)=(\psi_1(x)+\psi_2(x))/\sqrt2$, where $\psi_1$ and $\psi_2$ have energies $E_1$ and $E_2$. Write $\psi(x,t)$.

Solution

Each energy component gets its own phase:

$$\psi(x,t) = \frac{1}{\sqrt2} \left( \psi_1(x)e^{-iE_1t/\hbar} + \psi_2(x)e^{-iE_2t/\hbar} \right).$$

Only if $E_1=E_2$ is this just a common global phase.

3. For a continuum expansion $\psi(x,0)=\int c(k)\psi_k(x)\,dk$ with $\int\psi_k^*\psi_{k'}dx=\delta(k-k')$, what is the normalization condition on $c(k)$?

Solution

For a normalized state and this $k$-normalization convention,

$$\int \lvert c(k)\rvert^2\,dk=1.$$

If a different continuum label or normalization is used, the measure and coefficient function must be translated accordingly.

4. Show that the infinite square well revival time $T_{\mathrm{rev}}=2\pi\hbar/E_1$ rephases every stationary component.

Solution

The energies are $E_n=E_1n^2$. Therefore

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

Every component has its original phase, so any superposition reconstructs.