Phase Evolution e^{-iEt/ħ}

When a quantum system is in a stationary state — an energy eigenstate of its Hamiltonian — the time-dependent Schrödinger equation tells us exactly how the wave function evolves. If $\psi_n(x)$ is a spatial eigenfunction with energy eigenvalue $E_n$ , the full time-dependent solution is

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

The exponential $e^{-iE_n t/\hbar}$ is the phase factor, and it deserves close attention.

The phase factor is a unit complex number

Because $E_n$ , $t$ , and $\hbar$ are all real, the exponent $-iE_n t/\hbar$ is purely imaginary. A complex exponential of the form $e^{i\theta}$ (with $\theta \in \mathbb{R}$ ) lies on the unit circle in the complex plane: its real part is $\cos\theta$ and its imaginary part is $\sin\theta$ , so

$\left|e^{i\theta}\right| = \sqrt{\cos^2\theta + \sin^2\theta} = 1.$

Setting $\theta = -E_n t/\hbar$ confirms that $|e^{-iE_n t/\hbar}| = 1$ for every energy, every time, and every value of $\hbar$ .

Why the magnitude of the wave function does not change

Because the phase factor has unit magnitude, multiplying $\psi_n(x)$ by it leaves the spatial probability density unchanged:

$|\Psi_n(x,t)|^2 = \left|\psi_n(x)\right|^2 \left|e^{-iE_n t/\hbar}\right|^2 = |\psi_n(x)|^2.$

This is precisely what earns these solutions the name stationary states: the probability distribution $|\Psi_n(x,t)|^2$ is independent of time. The phase rotates continuously at angular frequency $\omega_n = E_n/\hbar$ , but nothing measurable oscillates — at least not until two stationary states are superposed.

The phase rotates at angular frequency E/ħ

Writing the phase factor in polar form makes the rotation explicit:

$e^{-iE_n t/\hbar} = \cos\!\left(\frac{E_n t}{\hbar}\right) - i\sin\!\left(\frac{E_n t}{\hbar}\right).$

The argument decreases linearly in time, so the phase vector sweeps clockwise around the unit circle at a rate $\omega_n = E_n/\hbar$ radians per second. A higher energy eigenstate rotates faster. In natural units where $\hbar \approx 6.582 \times 10^{-16}\,\text{eV}\cdot\text{s}$ , a state with $E = 1\,\text{eV}$ completes a full $2\pi$ rotation in a period

$T = \frac{2\pi\hbar}{E} = \frac{2\pi \times 6.582 \times 10^{-16}\,\text{eV}\cdot\text{s}}{1\,\text{eV}} \approx 4.13 \times 10^{-15}\,\text{s},$

which is on the scale of a few femtoseconds — an astonishingly fast oscillation that is physically meaningful when two states of different energy interfere.

Try it

This exercise focuses on the key insight above. Your code should return a number.

Run your code to see the quantum state.