Energy and Frequency

Stationary states oscillate in time

When a quantum system has a definite energy $E$ — an energy eigenstate $\psi(x)$ satisfying $\hat{H}\psi = E\psi$ — the full, time-dependent wave function takes the form

\Psi(x, t) = \psi(x)\,e^{-iEt/\hbar}.

This result follows directly from separating variables in the time-dependent Schrödinger equation. The spatial part $\psi(x)$ is fixed; only the phase rotates. The complex exponential $e^{-iEt/\hbar}$ completes one full cycle whenever $Et/\hbar$ advances by $2\pi$ , so the angular frequency of this phase rotation is

\omega = \frac{E}{\hbar}.

Higher energy means faster oscillation. This is the quantum analogue of the classical statement that a higher frequency oscillator carries more energy, and it is a direct descendant of Planck's and Einstein's relation $E = h\nu = \hbar\omega$ .

Why the oscillation is "stationary"

Despite the time-varying phase, all observable quantities are frozen. The probability density is

|\Psi(x, t)|^2 = |\psi(x)|^2 \left|e^{-iEt/\hbar}\right|^2 = |\psi(x)|^2,

because $|e^{-iEt/\hbar}| = 1$ for any real $E$ and $t$ . The expectation value of every time-independent observable likewise carries no time dependence for a pure energy eigenstate. This is why such states are called stationary even though the wave function itself is not constant.

Superpositions: the relative phase matters

Consider two energy eigenstates $\psi_1$ and $\psi_2$ with energies $E_1$ and $E_2$ . Their superposition

\Psi(x, t) = c_1\,\psi_1(x)\,e^{-iE_1 t/\hbar} + c_2\,\psi_2(x)\,e^{-iE_2 t/\hbar}

is no longer stationary. The relative phase between the two terms advances at the beat frequency

\omega_{12} = \frac{E_2 - E_1}{\hbar},

so the probability density oscillates with angular frequency $\omega_{12}$ . The period of this oscillation is

T = \frac{2\pi}{\omega_{12}} = \frac{h}{E_2 - E_1}.

This is the quantum version of a classical beating pattern: two sinusoids of slightly different frequency combine into a slowly modulated envelope. In quantum mechanics the envelope is the time-varying probability density, and the "frequency" is set by the energy gap.

Connecting to Planck and de Broglie

The relation $\omega = E/\hbar$ is equivalent to $E = \hbar\omega$ , which Planck wrote as $E = h\nu$ (with $h = 2\pi\hbar$ and $\nu = \omega/2\pi$ ). For a free particle of momentum $p$ , the energy is $E = p^2/2m$ , and the wave function is a plane wave $e^{ipx/\hbar - iEt/\hbar}$ . The spatial frequency is $k = p/\hbar$ (de Broglie) and the temporal frequency is $\omega = E/\hbar$ , both encoded in the same wave function.

Numerical example

For an electron in an infinite square well of width $a = 0.1\,\text{nm}$ , the ground-state energy is

E_1 = \frac{\pi^2 \hbar^2}{2 m_e a^2}.

Substituting $\hbar = 1.055 \times 10^{-34}\,\text{J·s}$ , $m_e = 9.109 \times 10^{-31}\,\text{kg}$ , and $a = 10^{-10}\,\text{m}$ :

E_1 = \frac{\pi^2 (1.055 \times 10^{-34})^2}{2 \times (9.109 \times 10^{-31}) \times (10^{-10})^2} \approx 6.02 \times 10^{-18}\,\text{J} \approx 37.6\,\text{eV}.

The wave function therefore oscillates at

\omega_1 = \frac{E_1}{\hbar} = \frac{6.02 \times 10^{-18}}{1.055 \times 10^{-34}} \approx 5.7 \times 10^{16}\,\text{rad/s}.

This is an ultraviolet-range frequency, consistent with the energy scale of atomic physics.

Try it

This is a numerical exercise — your code should return a number. An energy eigenstate has energy $E = 2.0\,\text{eV}$ . Use $1\,\text{eV} = 1.602 \times 10^{-19}\,\text{J}$ and $\hbar = 1.055 \times 10^{-34}\,\text{J·s}$ . Return the angular frequency $\omega = E/\hbar$ in rad/s.

Run your code to see the quantum state.