Energy Eigenstates Evolve Simply

The last lesson built the general solution as a sum of phase-rotated eigenstates. Now we zoom in on the simplest possible case — a state that is an energy eigenstate — and see why such states barely change at all.

A single phase factor

Let $|E_n\rangle$ be an eigenstate of the Hamiltonian, $\hat H|E_n\rangle = E_n|E_n\rangle$. Applying the evolution operator is trivial because $\hat H$ acts as a number on this state:

The only effect of time is to multiply the state by the unit-modulus complex number $e^{-iE_n t/\hbar}$. The "direction" of the state vector in Hilbert space never moves; only its phase winds around.

Why these are called stationary states

Because the change is a pure overall phase, every observable prediction is time-independent. For any observable $\hat A$,

since the two scalar phases cancel. Measurement probabilities, expectation values — all frozen. An energy eigenstate prepared at $t=0$ looks experimentally identical at every later time. This is why $|E_n\rangle$ is called a stationary state: nothing measurable evolves.

The phase still matters — eventually

Saying the phase is "invisible" is true only as long as the eigenstate stays alone. The moment you add a second component or interfere two copies of the system, the accumulated phase $e^{-iE_n t/\hbar}$ becomes physical. So energy eigenstates are not boring — they are the carriers of phase information. The clock they carry, ticking at angular frequency $E_n/\hbar$, is exactly what makes superpositions oscillate.

Try it

This is a numerical exercise — your code should return a number. The evolved amplitude of the eigenstate is $e^{+i\omega t/2} = \cos(\omega t/2) + i\sin(\omega t/2)$. Confirm stationarity by computing its modulus squared (the measurement probability), which must equal $1$ for the arbitrary value $\omega t = 1.234$.

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