The Ground State

The ground state $|0\rangle$ is the bottom rung of the ladder — the state that the lowering operator sends to zero. That single condition, $a|0\rangle = 0$, determines the entire ground-state wavefunction without solving the Schrödinger equation.

The defining condition

The ladder terminates at the bottom: there is no $|{-1}\rangle$, so

$a|0\rangle = 0.$

This is the algebraic fingerprint of the ground state. Everything about $|0\rangle$ follows from it. Its energy, for instance, is immediate:

So $E_0 = \tfrac{1}{2}\hbar\omega$, the zero-point energy, with no integration required.

Recovering the Gaussian wavefunction

To find $\psi_0(x) = \langle x|0\rangle$, write the condition $a|0\rangle = 0$ in position space. In natural units $a = (\hat{x} + i\hat{p})/\sqrt{2}$ with $\hat{p} = -i\,d/dx$, so

This is a first-order differential equation — far simpler than the original second-order Schrödinger equation — and it integrates at once to a Gaussian:

where the prefactor is fixed by normalization $\int |\psi_0|^2\,dx = 1$. The algebraic method turns finding the ground state into solving one first-order equation.

Spread of the ground state

Because $\psi_0$ is a Gaussian centred at the origin, $\langle\hat{x}\rangle = \langle\hat{p}\rangle = 0$. Its variances follow from the ladder operators. Using $\hat{x} = (a + a^\dagger)/\sqrt{2}$ and $a|0\rangle = 0$,

An identical calculation gives $\langle\hat{p}^2\rangle = \tfrac{1}{2}$. Hence the uncertainty product is

which saturates the Heisenberg bound $\Delta x\,\Delta p \ge \hbar/2$. The ground state is a minimum-uncertainty state — the tightest a quantum particle can be squeezed into a parabolic well.

Try it

This is a numerical exercise — your code should return a number. Using $\hat{x} = (a + a^\dagger)/\sqrt{2}$ and the truncated number-basis matrices, compute the ground-state spread $\langle 0|\hat{x}^2|0\rangle$ (which equals $\tfrac{1}{2}$).

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