Solving the Schrödinger Equation

We have the time-evolution operator $U(t) = e^{-i\hat H t/\hbar}$ . Now we put it to work: given a Hamiltonian and an initial state, find the state — and a measurable probability — at a later time.

The recipe

Solving the Schrödinger equation for a time-independent Hamiltonian is mechanical once you know the energy eigenbasis:

Diagonalise $\hat H$ : find eigenstates $|E_n\rangle$ with $\hat H|E_n\rangle = E_n|E_n\rangle$ .
Expand the initial state in that basis: $|\psi(0)\rangle = \sum_n c_n |E_n\rangle$ with $c_n = \langle E_n|\psi(0)\rangle$ .
Attach a phase to each component: each eigenstate evolves as $|E_n\rangle \mapsto e^{-iE_n t/\hbar}|E_n\rangle$ .
Reassemble: $|\psi(t)\rangle = \sum_n c_n\, e^{-iE_n t/\hbar}\,|E_n\rangle.$

This is just $U(t)$ written in its own eigenbasis, where it is diagonal.

A concrete qubit example

Take a spin- $\tfrac12$ with Hamiltonian

\hat H = \frac{\hbar\omega}{2}\,\sigma_z,

whose eigenstates are the computational-basis states with energies $E_0 = +\tfrac{\hbar\omega}{2}$ for $|0\rangle$ and $E_1 = -\tfrac{\hbar\omega}{2}$ for $|1\rangle$ . Start the system in

|\psi(0)\rangle = |+\rangle = \frac{1}{\sqrt 2}\big(|0\rangle + |1\rangle\big).

Following the recipe, each basis state picks up its phase:

|\psi(t)\rangle = \frac{1}{\sqrt 2}\Big(e^{-i\omega t/2}|0\rangle + e^{+i\omega t/2}|1\rangle\Big).

Up to an overall phase this is

|\psi(t)\rangle \simeq \frac{1}{\sqrt 2}\big(|0\rangle + e^{i\omega t}|1\rangle\big),

so a relative phase $\omega t$ accumulates between the two components.

The survival probability

How likely are we to still find the system in the original state $|+\rangle$ ? The amplitude is

\langle +|\psi(t)\rangle = \frac{1}{2}\Big(e^{-i\omega t/2} + e^{+i\omega t/2}\Big) = \cos\!\frac{\omega t}{2},

using $\cos\theta = \tfrac12(e^{i\theta} + e^{-i\theta})$ . The survival probability is therefore

P_{+}(t) = \big|\langle +|\psi(t)\rangle\big|^2 = \cos^2\!\frac{\omega t}{2}.

It starts at $1$ , drops to $0$ when $\omega t = \pi$ (the state has rotated to $|-\rangle$ ), and returns to $1$ at $\omega t = 2\pi$ . The phase that $\hat H$ stamps onto the eigenstates becomes a physical oscillation as soon as we measure in a basis that mixes them.

Try it

This is a numerical exercise — your code should return a number. Compute the survival probability $P_{+} = \cos^2(\omega t/2)$ at $\omega t = \pi/3$ .

Run your code to see the quantum state.