The Harmonic Oscillator

The quantum harmonic oscillator is the canonical bound-state problem: a particle in a quadratic potential

The energy levels are quantised:

The ground state has energy $E_0 = \tfrac{1}{2}\hbar\omega$ — the zero-point energy, a purely quantum effect.

Time evolution

Every state above is stationary. An eigenstate evolves as $\psi_n(x,t) = \psi_n(x)\,e^{-iE_n t/\hbar}$, so $|\psi_n(x,t)|^2$ never changes: nothing in that figure would move if we animated it.

Motion requires a superposition. And since the eigenstates form a complete basis, every state is one:

The motion comes entirely from the relative phases. Here they are unusually well behaved: since $E_n$ is evenly spaced, every phase difference is a whole multiple of $\omega$, so any superposition returns exactly to its initial shape after one period $T = 2\pi/\omega$. The harmonic oscillator never dephases.

A Gaussian wavepacket is one such superposition,

an envelope of width $\sigma$ that says where the particle is, times a plane wave that says how fast it is moving. It is not an alternative to the eigenbasis — it is a particular vector in it, with the $c_n$ fixed by $x_0$, $k_0$ and $\sigma$.

Coherent states

Among all these Gaussians, one width is singled out:

which is precisely the width of the ground state. A packet prepared at $\sigma = \sigma_0$ and displaced to $x_0$ is a coherent state: the ground state, translated. Its coefficients follow a Poisson distribution,

with mean occupation $\bar n = |\alpha|^2$ and energy $E = \hbar\omega\left(\bar n + \tfrac12\right)$ — the same formula as a single level, with the integer replaced by an average. For $\omega = 1$ and $x_0 = 2.5$ this gives $\bar n = 3.125$ and $E = 3.625$, the packet spread across roughly states $n = 0$ to $7$.

Three properties make it remarkable, and they all hold at once:

1. It does not spread. $\Delta x = \sigma_0$ for all time. Every other Gaussian breathes, its width oscillating at $2\omega$ — twice the frequency of the sloshing.
2. Its centre obeys Newton exactly. $\langle x\rangle(t) = x_0\cos\omega t + \dfrac{k_0}{m\omega}\sin\omega t$. Ehrenfest's theorem gives $m\,\mathrm{d}^2\langle x\rangle/\mathrm{d}t^2 = -\langle V'(x)\rangle$, and because the force is linear here, $\langle V'(x)\rangle = V'(\langle x\rangle)$ — the mean position follows the classical path with no approximation whatsoever.
3. It sits on the Heisenberg bound. $\Delta x\,\Delta p = \hbar/2$ at every instant, the minimum any state is allowed.

Taken together: a lump of probability that keeps its shape, tracks the classical trajectory, and is as sharply defined as the uncertainty principle permits. It is the closest quantum mechanics comes to a marble rolling in a bowl, and it is why coherent states are the natural bridge between the quantum and classical descriptions.

They also escape the textbook. A single mode of the electromagnetic field is a harmonic oscillator, and its coherent states are what an ideal laser emits — $|\alpha|^2$ becomes the mean photon number, and the Poisson statistics above become the shot noise of the beam. Glauber received the 2005 Nobel Prize for working this out.

All of this is special to the parabola. Coherent states exist because the levels are evenly spaced; in an anharmonic well the phase differences are no longer commensurate, the terms of the sum drift out of step, and the packet spreads and never rigidly recurs. There, the gap between $\langle x\rangle(t)$ and the classical trajectory is exactly the part of the motion with no classical counterpart.