The bottom rung is the state the lowering operator kills, and in the position representation that condition is a first-order differential equation. Its solution is a Gaussian, the shape the chapter on position and momentum met as the state of least uncertainty. Its energy is half a quantum, the zero-point energy: the oscillator can never be still at the bottom, because a state with no spread in position and none in momentum does not exist.

A ball in a bowl can be put at rest at the very bottom. A quantum particle cannot: pinning it to the bottom would make its momentum wildly uncertain, and letting it spread costs potential energy. The ground state is the best compromise.

The ground state $\psi_{0}\propto e^{-\xi^{2}/2}$ drawn on its level $E_{0}=\tfrac12\hbar\omega$, with $\xi=x/x_{0}$. A classical particle with this energy would stay between the two dots, at $\xi=\pm1$. The quantum state reaches beyond them: the probability of finding it outside is about 0.16.

In the position representation the lowering operator is a first-order differential operator, so $\hat{a}\ket{0}=0$ is an equation with one solution up to a factor: a Gaussian of width about $x_{0}$.

• In $\xi=x/x_{0}$: $\psi_{0}(\xi)=\pi^{-1/4}e^{-\xi^{2}/2}$.

Write the lowering operator in the variable $\xi$: it is $\xi+d/d\xi$ up to a factor. Killing a state then means $\psi'=-\xi\psi$, which is separable and gives a Gaussian. Normalising fixes the constant, and the Hamiltonian written with ladder operators gives the energy at once.

The lowering operator in $\xi=x/x_{0}$, using $\hat{p}=-i\hbar\,d/dx$.

The condition $\hat{a}\psi_{0}=0$: a first-order equation.

Separate the variables and integrate: $\ln\psi_{0}=-\xi^{2}/2+\text{const}$.

The Gaussian integral, so $C=\pi^{-1/4}$.

Since $\hat{a}\psi_{0}=0$, only the half quantum survives.

The lowest energy of the oscillator is half a quantum, the zero-point energy. A particle resting at the bottom would have no spread in position and none in momentum, which the uncertainty relation forbids.

The lowest state solves a first-order equation, $\hat{a}\psi=0$, whose solution is a Gaussian centred at the bottom of the well.

A classical oscillator with energy $\hbar\omega/2$ would never go farther than $x_{0}$ from the centre. The ground state reaches beyond, with a probability of about 16 per cent.

In the ground state $\Delta x=x_{0}/\sqrt{2}$ and . Their product is , the smallest the uncertainty relation allows.

Great work! You now know how to:

• solve $\hat{a}\ket{0}=0$ for the Gaussian ground state
• explain the zero-point energy by the uncertainty relation
• compute spreads and probabilities in the ground state