Stationary States and the Time-Independent Equation
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You should be comfortable with the time-dependent Schrödinger equation and the concept of probability density. Familiarity with basic calculus, specifically separation of variables in differential equations, is assumed.
Why this matters
Stationary states provide the bridge between the dynamic quantum world and the stable, measurable energy levels observed in nature.
In the previous lecture, we established that probability is conserved locally, leading to the continuity equation. However, solving the full time-dependent Schrödinger equation for arbitrary potentials is often intractable. By identifying a special class of solutions where the spatial and temporal dependencies decouple, we reduce a complex partial differential equation to a simpler ordinary differential equation. This allows us to determine the allowed energy states of a system, which are the fundamental observables in atomic and molecular physics.
The Idea of Stationarity
Imagine a wave on a string that is plucked and allowed to vibrate freely. While the string moves up and down, the shape of the wave pattern remains fixed in space; it does not travel along the string. In quantum mechanics, a "stationary state" is the analogous concept. It is a solution to the Schrödinger equation where the probability of finding the particle at a specific location does not change with time. The particle is not "static" in the sense that it has zero momentum or is frozen in place; rather, the statistical distribution of its position is time-independent. This stability arises because the phase of the wavefunction evolves uniformly in time, causing the interference terms that would otherwise create a time-dependent probability density to cancel out. This is the quantum equivalent of a normal mode of vibration.
Separation of Variables
To find these stationary states, we assume the potential energy \( V(\mathbf{r}) \) depends only on position and not on time. We propose a solution where the total wavefunction \( \Psi(\mathbf{r}, t) \) is the product of a spatial function \( \psi(\mathbf{r}) \) and a temporal function \( T(t) \): \[ \Psi(\mathbf{r}, t) = \psi(\mathbf{r}) T(t) \] Substituting this ansatz into the time-dependent Schrödinger equation: \[ i\hbar \frac{\partial \Psi}{\partial t} = \left[ -\frac{\hbar^2}{2m} \nabla^2 + V(\mathbf{r}) \right] \Psi \] yields: \[ i\hbar \psi(\mathbf{r}) \frac{dT}{dt} = T(t) \left[ -\frac{\hbar^2}{2m} \nabla^2 \psi(\mathbf{r}) + V(\mathbf{r}) \psi(\mathbf{r}) \right] \] Dividing both sides by \( \psi(\mathbf{r}) T(t) \) separates the variables: \[ \frac{i\hbar}{T(t)} \frac{dT}{dt} = \frac{-\frac{\hbar^2}{2m} \nabla^2 \psi(\mathbf{r}) + V(\mathbf{r}) \psi(\mathbf{r})}{\psi(\mathbf{r})} \] The left-hand side depends only on time, while the right-hand side depends only on position. For this equality to hold for all \( t \) and \( \mathbf{r} \), both sides must equal a constant. This constant is identified as the total energy \( E \).
The Time-Independent Schrödinger Equation
Setting the spatial part equal to \( E \) gives the time-independent Schrödinger equation (TISE): \[ \left[ -\frac{\hbar^2}{2m} \nabla^2 + V(\mathbf{r}) \right] \psi(\mathbf{r}) = E \psi(\mathbf{r}) \] This is an eigenvalue equation where the Hamiltonian operator \( \hat{H} \) acts on the spatial wavefunction \( \psi(\mathbf{r}) \) to yield the energy eigenvalue \( E \). The temporal part yields: \[ \frac{i\hbar}{T(t)} \frac{dT}{dt} = E \implies \frac{dT}{dt} = -\frac{iE}{\hbar} T(t) \] Solving this first-order differential equation gives the universal phase factor: \[ T(t) = e^{-iEt/\hbar} \] Thus, the full stationary state is: \[ \Psi(\mathbf{r}, t) = \psi(\mathbf{r}) e^{-iEt/\hbar} \] The probability density is: \[ |\Psi(\mathbf{r}, t)|^2 = \Psi^* \Psi = \psi^*(\mathbf{r}) e^{iEt/\hbar} \psi(\mathbf{r}) e^{-iEt/\hbar} = |\psi(\mathbf{r})|^2 \] The time-dependent phase factors cancel exactly, confirming that the probability distribution is stationary.
Physical Interpretation and Superposition
It is crucial to distinguish between a single stationary state and a superposition of states. A single stationary state has a time-independent probability density. It may carry a steady probability current, as a free-particle plane wave does (uniform density, definite momentum), but that current does not change in time either; for a bound state in one dimension the wavefunction can be chosen real and the current is zero. If we form a superposition of two stationary states with different energies \( E_1 \) and \( E_2 \): \[ \Psi(\mathbf{r}, t) = c_1 \psi_1(\mathbf{r}) e^{-iE_1 t/\hbar} + c_2 \psi_2(\mathbf{r}) e^{-iE_2 t/\hbar} \] The probability density \( |\Psi|^2 \) will contain cross-terms involving \( e^{-i(E_1 - E_2)t/\hbar} \). These terms oscillate in time, leading to a time-dependent probability distribution. Therefore, only pure energy eigenstates are truly stationary in the sense of having a static probability cloud. This distinction is vital for understanding phenomena such as quantum beats and the time evolution of wave packets.
The derivation
Separation of Variables and Stationary States
For a potential that does not depend on time, V = V(r), the time-dependent Schrödinger equation admits separable solutions Ψ(r,t) = ψ(r) T(t). Substituting and separating forces both sides to equal a single real constant E, yielding the time-independent Schrödinger equation Ĥψ = Eψ for the spatial factor and the universal phase evolution T(t) = e−iEt/ℏ for the temporal factor. The product is a stationary state: its probability density |Ψ|2 is independent of time.
A common misconception is that a stationary state implies the particle is at rest. In fact, for a bound state like the ground state of the hydrogen atom, the expectation value of momentum is zero, but the expectation value of kinetic energy is non-zero. The "stationarity" refers strictly to the probability density \( |\Psi|^2 \), not to the mechanical motion of the particle. The particle exhibits zero-point motion, a direct consequence of the uncertainty principle. C
Worked example
Problem. Consider a particle of mass \( m = 9.11 \times 10^{-31} \ \mathrm{kg} \) in a one-dimensional infinite square well of width \( L = 1.0 \times 10^{-9} \ \mathrm{m} \). The potential is \( V(x) = 0 \) for \( 0 < x < L \) and \( V(x) = \infty \) otherwise. Determine the energy of the ground state (\( n=1 \)) and verify that the probability density is time-independent.
- Identify the time-independent Schrödinger equation for the region \( 0 < x < L \) where \( V(x) = 0 \): \[ -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi \]
- Apply boundary conditions \( \psi(0) = 0 \) and \( \psi(L) = 0 \). The general solution is \( \psi(x) = A \sin(kx) \). The condition \( \psi(L) = 0 \) requires \( kL = n\pi \), so \( k_n = \frac{n\pi}{L} \).
- Calculate the energy eigenvalues using \( E_n = \frac{\hbar^2 k_n^2}{2m} = \frac{n^2 \pi^2 \hbar^2}{2mL^2} \).
- Substitute values for \( n=1 \): \[ E_1 = \frac{\pi^2 (1.055 \times 10^{-34} \ \mathrm{J \cdot s})^2}{2 (9.11 \times 10^{-31} \ \mathrm{kg}) (1.0 \times 10^{-9} \ \mathrm{m})^2} \]
- Compute the numerator: \( \pi^2 \approx 9.87 \), \( \hbar^2 \approx 1.113 \times 10^{-68} \). Numerator \( \approx 1.10 \times 10^{-67} \).
- Compute the denominator: \( 2 \times 9.11 \times 10^{-31} \times 10^{-18} = 1.822 \times 10^{-48} \).
- Divide: \( E_1 \approx \frac{1.10 \times 10^{-67}}{1.822 \times 10^{-48}} \approx 6.04 \times 10^{-20} \ \mathrm{J} \).
- Convert to electron-volts: \( E_1 \approx \frac{6.04 \times 10^{-20}}{1.602 \times 10^{-19}} \approx 0.377 \ \mathrm{eV} \).
- Verify stationarity: \( \Psi(x,t) = \sqrt{\frac{2}{L}} \sin(\frac{\pi x}{L}) e^{-iE_1 t/\hbar} \). \[ |\Psi(x,t)|^2 = \frac{2}{L} \sin^2\left(\frac{\pi x}{L}\right) \] The time factor cancels, confirming stationarity.
Units check. \( \frac{\mathrm{J^2 \cdot s^2}}{\mathrm{kg \cdot m^2}} = \frac{(\mathrm{kg \cdot m^2/s^2})^2 \cdot s^2}{\mathrm{kg \cdot m^2}} = \frac{\mathrm{kg^2 \cdot m^4}}{\mathrm{s^2}} \cdot \frac{s^2}{\mathrm{kg \cdot m^2}} = \mathrm{kg \cdot m^2/s^2} = \mathrm{J} \). Correct.
Three quick tests
- Why must the separation constant be real?
Answer
If the constant were complex, the time evolution would involve exponential growth or decay, violating probability conservation (unitarity) for a time-independent potential. - Does a stationary state have zero kinetic energy?
Answer
No. The expectation value of kinetic energy is generally non-zero, even if the expectation value of momentum is zero, due to the uncertainty principle. - What happens to the probability density if you superpose two stationary states with different energies?
Answer
The probability density becomes time-dependent due to the interference terms oscillating at the frequency proportional to the energy difference.
Problem set for PU-202 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.