Stationary State Time Evolution

In one sentence: A stationary state has one fixed energy, so the chance of finding the particle in each place stays the same over time, even though the wave keeps turning.

$\Psi(x, t) = \psi(x)\, e^{-i E t/\hbar}, \qquad |\Psi(x, t)|^2 = |\psi(x)|^2$

Read it out loud: “Psi of x and t equals psi of x, times e to the power of minus i times E times t over h-bar. Psi of x and t is the full wave function, which depends on the place x and the time t. Psi of x is its position part, which does not change with time. E is the fixed energy, t is time, and h-bar is the reduced Planck constant. The second rule says that the squared size of psi of x and t equals the squared size of psi of x, so the probability density does not change with time.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Wave function $\Psi$ and $\psi$ , and probability density: A wave function is a complex number attached to each place and each time. Its squared size, written $|\Psi|^2$ , tells how likely you are to find the particle near that place. The chance per unit of length is called the probability density. Capital psi, $\Psi$ , is the full wave function with time. Lowercase psi, $\psi$ , is its position part only, which does not change with time.
Powers of ten, like $10^{-34}$ and $10^{9}$: Very big and very small numbers are written as a short number times a power of ten. The exponent says how many times to multiply by 10, and a minus sign means divide. So $10^{3} = 1000$ and $10^{-3} = 0.001$ . The power $10^{-34}$ is a decimal point, then 33 zeros, then the digit 1. Multiplying powers adds the exponents, and dividing subtracts them, so $10^{-34} \times 10^{14} = 10^{-20}$ .
Complex numbers, $i$ and the phase $e^{i\theta}$: A complex number has an ordinary part and a part multiplied by $i$ , the imaginary unit with $i^2 = -1$ . The number $e$ , about 2.718, is a fixed number. The expression $e^{i\theta}$ is a complex number with size 1 that sits on a circle at the angle $\theta$ . That angle is the phase. Multiplying a wave by $e^{i\theta}$ turns the wave but does not change how big it is.
Joules, $\hbar$ and femtoseconds: Energy is measured in joules (J). One joule is about the energy needed to lift a small apple, of 100 grams, one meter straight up. The reduced Planck constant $\hbar$ , said h-bar, is $1.054571817 \times 10^{-34}$ J s, which means joule seconds. A femtosecond, written fs, is one quadrillionth of a second, $10^{-15}$ s, a tiny time that matters for atoms.
Radians, sine and square roots: An angle in radians is measured along a circle of radius 1. One radian is about 57.3 degrees, and $2\pi$ radians is a full turn. The sine of an angle, $\sin x$ , is the height of the matching point on that circle. For example $\sin 0 = 0$ and $\sin\frac{\pi}{2} = 1$ . A square root, $\sqrt{x}$ , is the number that multiplied by itself gives $x$ , so $\sqrt{9} = 3$ .
Hats on letters, $\hat{H}$ , and the Schrödinger equation: A hat over a letter, such as $\hat{H}$ , marks an operator: a rule that takes a wave function and returns a new one. The operator $\hat{H}$ is the energy rule, called the Hamiltonian. The Schrödinger equation is the basic equation that says how a wave function changes in time. Its time-independent form, $\hat{H}\psi = E\psi$ , says which energies are allowed.
Partial derivative, $\frac{\partial \Psi}{\partial t}$: A derivative measures a rate of change, like the speedometer of a car, which shows how fast the distance is changing. The curly symbol $\partial$ means a partial derivative: only one input is allowed to move. $\frac{\partial \Psi}{\partial t}$ is how fast $\Psi$ changes as the time $t$ moves on, while the place $x$ is held fixed.

What each symbol means

Symbol	Meaning
$\Psi(x, t)$	The full wave function, which depends on the place $x$ and the time $t$ . In one dimension its unit is $\text{m}^{-1/2}$ , one over the square root of a meter. That unit makes the squared size a chance per meter. The wave function is not a probability by itself, but its squared size is.
$\psi(x)$	The position part of the wave function. It depends only on the place $x$ , not on time, and it has the same unit, $\text{m}^{-1/2}$ . It solves the time-independent equation $\hat{H}\psi = E\psi$ , which sets the allowed energies.
$e^{-i E t/\hbar}$	The time part, a phase factor with size 1 at every time. It has no unit, because the exponent $Et/\hbar$ is a pure number: joule seconds divided by joule seconds. It only turns the wave, so the angle it has turned through grows steadily with time.
$E$	The fixed energy of the stationary state, measured in joules (J). It has the same value at every time. In the example $E = 6.02 \times 10^{-20}$ J, which is about 0.376 electron volts, a very small energy.
$t$	Time, measured in seconds (s). It is the clock that moves the phase forward. In the example we use $1.0 \times 10^{-15}$ s, which is one femtosecond, a very short time.
$\hbar$	The reduced Planck constant, $\hbar = 1.054571817 \times 10^{-34}$ J s, in joule seconds. The product $Et$ is in joule seconds, so dividing it by $\hbar$ leaves a pure number. The constant is tiny because quantum effects are small in everyday objects.
$\|\Psi(x, t)\|^2$	The probability density at the place $x$ and time $t$ : the chance of finding the particle per meter of length near that place. Its unit is $\text{m}^{-1}$ , one over a meter. For a stationary state it equals $\|\psi(x)\|^2$ at every time. In the example it is $1.00 \times 10^{9}\ \text{m}^{-1}$ .

The formula piece by piece

$\psi(x)$: Start with the position part of the wave function. It describes how the wave is spread over places, and it does not change with time. Its squared size gives the chance of finding the particle near each place.
$E$: The energy, a fixed number in joules. For a stationary state it is the same at every time, and the position part is what fixes it through the equation $\hat{H}\psi = E\psi$ .
$\frac{E t}{\hbar}$: The phase angle, a pure number. It grows steadily as time passes, because the energy times the time is divided by the tiny constant $\hbar$ . After enough time the angle completes full turns, but the size of the wave does not change.
$e^{-i E t/\hbar}$: The turning factor, built from the phase angle. It is a complex number with size 1, so it turns the wave around a circle without stretching or shrinking it at any time.
$|\Psi(x, t)|^2 = |\psi(x)|^2$: The probability density stays the same at every time. The turning factor has size 1, so it drops out when we take the squared size. Only the position part decides where the particle is likely to be found.
$\Psi(x, t) = \psi(x)\, e^{-i E t/\hbar}, \qquad |\Psi(x, t)|^2 = |\psi(x)|^2$: A stationary state is the position part of the wave multiplied by a turning factor. That turning changes the phase over time, while the chance of finding the particle in each place stays fixed.

What the formula is doing, and why

What a stationary state is. A stationary state is a solution with one definite energy $E$ . Its wave function splits into two parts: a position part $\psi(x)$ , which says where the particle tends to be, and a time part, the turning factor $e^{-iEt/\hbar}$ . The position part solves the time-independent equation $\hat{H}\psi = E\psi$ , and that equation sets the energy. The chance of finding the particle in each place does not change in time, which is why the name stationary, meaning not moving, is used.

Why this turning factor works. The time-dependent Schrödinger equation is $i\hbar\,\partial\Psi/\partial t = \hat{H}\Psi$ . Taking the time derivative of $e^{-iEt/\hbar}$ brings down the factor $-iE/\hbar$ . Multiplying by $i\hbar$ turns that into $E$ , because $i \times (-i) = 1$ . Both sides then equal $E\psi$ times the turning factor, so the equation holds. Why the minus sign? The $i$ on the left side of the equation forces the minus sign. A plus sign would bring down $+iE/\hbar$ , and multiplying by $i\hbar$ would give $-E$ , which does not match $\hat{H}\psi = E\psi$ .

Why the density does not move. The density is $|\Psi|^2 = |\psi|^2\,|e^{-iEt/\hbar}|^2$ . A complex number of the form $e^{i\theta}$ always has size 1. The squared size of $\cos\theta - i\sin\theta$ is $\cos^2\theta + \sin^2\theta = 1$ . So the time part drops out, and the chance of finding the particle at each place stays the same at every time.

The phase still turns. At a fixed spot, the complex value of $\Psi$ keeps rotating even though its size does not change. Any quantity built only from the density, such as the average position of the particle, is also fixed in time for a stationary state. The state is still changing in phase. What stays fixed is the probability and the energy.

Motion needs a mix. If the state is a mix of two stationary states with different energies $E_1$ and $E_2$ , the relative turning between them, $e^{-i(E_2 - E_1)t/\hbar}$ , changes with time. That changes the density, which can move, as in a wave packet, a bump of waves that travels. Real excited atoms also give off light and decay, which this model leaves out.

An everyday picture

Think of the hand of a clock, a fixed length, turning around the face. Its tip moves around a circle, so its direction keeps changing, but the length of the hand never changes. The turning is the phase factor. The length is the size of the wave, and the square of that length is the probability density, which stays put.

Worked example: A particle in a box at two times

An electron is in the lowest state of a box 1.0 nm wide, about the width of a few atoms. Its energy is $E = 6.02 \times 10^{-20}$ J, which is 0.376 electron volts. This value comes from the box model and is chosen as an example. The position part is $\psi(x) = \sqrt{2/L}\,\sin(\pi x / L)$ with $L = 1.0$ nm. We compare the probability density at $x = L/4$ , one quarter of the way across the box, at time $t = 0$ and at time $t = 1.0$ fs, which is $1.0 \times 10^{-15}$ s.

Find the phase angle after 1.0 fs. The phase angle is $\frac{Et}{\hbar}$ . Multiply $6.02 \times 10^{-20}$ J by $1.0 \times 10^{-15}$ s: the exponents add, $-20 + (-15) = -35$ , giving $6.02 \times 10^{-35}$ J s. Divide by $\hbar = 1.054571817 \times 10^{-34}$ J s. Dividing powers subtracts exponents, $-35 - (-34) = -1$ , and $6.02 / 1.0546 \approx 5.71$ , so the angle is $0.571$ radians. The joule seconds cancel, which leaves a pure number. In degrees, 0.571 radians is about 33 degrees.
Find the position part at $x = L/4$ . The box is $L = 1.0 \times 10^{-9}$ m long, so $L/4$ is a quarter of the way across. The factor $\sqrt{\frac{2}{L}}$ is $\sqrt{2.0 \times 10^{9}}$ , which is about $4.47 \times 10^{4}$ m $^{-1/2}$ , because $\sqrt{2}$ is about 1.414 and $\sqrt{10^{9}}$ is about $3.16 \times 10^{4}$ . The sine of $\frac{\pi}{4}$ is the sine of 45 degrees, which is $\frac{\sqrt{2}}{2} \approx 0.707$ . So $\psi \approx 4.47 \times 10^{4} \times 0.707 \approx 3.16 \times 10^{4}$ m $^{-1/2}$ .
Square it to get the density at $t = 0$ . The density is $|\psi|^2 = \frac{2}{L}\sin^2\frac{\pi}{4} = 2.0 \times 10^{9} \times 0.5 = 1.00 \times 10^{9}$ m $^{-1}$ . Squaring the square root removes it and leaves $\frac{2}{L}$ , and $\sin^2 45^\circ = 0.5$ . Multiplying $2.0 \times 10^{9}$ by 0.5 gives $1.0 \times 10^{9}$ .
Note the starting time. At $t = 0$ the phase angle is zero, so the turning factor is $e^{0} = 1$ . The full wave function equals its position part, and the density is $1.00 \times 10^{9}$ m $^{-1}$ . This is the starting picture that we compare against.
Find the density at $t = 1.0$ fs. The turning factor is now $e^{-0.571i}$ , which has size 1. Its squared size is $\cos^2 0.571 + \sin^2 0.571$ , and that sum is always 1, by the rule $\cos^2 + \sin^2 = 1$ . So the density is $\psi^2 \times 1 = 1.00 \times 10^{9}$ m $^{-1}$ , the same as before.
Compare the two times. The wave has turned by 0.571 radians, about 33 degrees, but the density at $x = L/4$ did not change. A one-femtosecond gap is very short, and the turn is the only change. For a stationary state the chance of finding the particle stays put at every time.

Answer: The density at $x = L/4$ is the same at both times, $1.00 \times 10^{9}$ m $^{-1}$ . The phase of $\Psi$ has turned by 0.571 radians, about 33 degrees, but the probability density has not changed.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Thinking the wave function stops changing. For a stationary state only the probability density is fixed. The complex wave function keeps turning in phase.
Using $e^{+iEt/\hbar}$ instead of $e^{-iEt/\hbar}$ . With the standard form of the Schrödinger equation, the minus sign is required, because the $i$ on the left side must match the time derivative.
Assuming every state has a fixed density. The result holds for a stationary state with one energy. A mix of two states with different energies has a density that changes in time.
Mixing units in the phase. $E$ must be in joules and $\hbar$ in J s. Putting in the energy as 0.376, the number of electron volts, instead of $6.02 \times 10^{-20}$ J makes the phase about $6 \times 10^{18}$ times too large.

Where you will meet it

Atomic and molecular physics, where, in the simplest model, the lowest orbitals of an atom are stationary states with fixed probability clouds.
Spectroscopy, where the energy difference $\Delta E$ between two stationary states sets the frequency $f$ of the light an atom emits or absorbs, through $\Delta E = hf$ , with $h$ Planck’s constant.
Wave packets in simulations, which are built from mixes of stationary states, so their motion comes from the changing relative turning between the pieces.
Quantum chemistry programs, which compute stationary states of molecules to predict their energy levels and the shapes of their orbitals.

Questions people ask

If the density never changes, is the state frozen?

No. The phase keeps turning, so the state is changing the whole time. What stays fixed is the probability of finding the particle at each place, and the energy.

When does the density change?

When the state is a mix of stationary states with different energies. The relative turning between the pieces changes with time, and the density moves with it, as in a wave packet.

Are stationary states realistic?

They are exact solutions of the model, but real atoms and molecules interact with light and with each other. An excited state in a real atom eventually decays, which this model does not include.

What is a phase, in plain words?

The phase is the angle that tells where a wave is in its cycle, like the position of a clock hand. Two waves of the same size but different phases can add up in different ways. The phase changes with time, but a phase alone does not change the chance of finding a particle.

Schrödinger Equation: The Schrödinger equation says how a quantum particle’s wave changes from moment to moment, the way Newton’s laws say how a moving ball changes its position.
Particle in a Box: A particle trapped between two walls can have only certain energies, like steps on a ladder, and the narrower the box, the higher those steps are.
Born Rule: The Born rule turns a quantum wave into chances: the probability of a result is the squared size of the wave for that result, so bigger waves mean more likely results.