Energy Spectrum from Ladders

We now collect the payoff. Without ever solving a differential equation, the ladder algebra delivers the complete energy spectrum of the harmonic oscillator.

Reading the spectrum off the Hamiltonian

The factorized Hamiltonian and the number operator give, in physical units,

Acting on a number state,

so the energy eigenvalues are

$\boxed{\ E_n = \hbar\omega\left(n + \tfrac{1}{2}\right), \qquad n = 0, 1, 2, \ldots\ }$

This is identical to the result quoted from the Hermite-polynomial solution — but here it followed purely from $[a, a^\dagger] = 1$ and the requirement $n \ge 0$. The integer $n$ is simply the number of quanta, each carrying energy $\hbar\omega$.

The two defining features, explained

Equal spacing. Every gap is the same:

The spacing is constant because adding one quantum with $a^\dagger$ always costs exactly one $\hbar\omega$, regardless of the level. This uniformity is special to the oscillator — the infinite square well, by contrast, has gaps growing like $2n+1$.

Zero-point energy. The lowest level is

It traces back to the $\tfrac{1}{2}$ in $\hat{H} = a^\dagger a + \tfrac{1}{2}$, which itself came from $[a, a^\dagger] = 1$ — i.e. from $[\hat{x}, \hat{p}] = i\hbar$. The ground state cannot have zero energy because that would require simultaneously vanishing kinetic and potential energy, pinning down both $x$ and $p$ in violation of the uncertainty principle.

A quick consistency check

The energy of a number state should equal its expectation value $\langle n|\hat{H}|n\rangle = \hbar\omega(\langle n|\hat{N}|n\rangle + \tfrac{1}{2}) = \hbar\omega(n + \tfrac{1}{2})$, with no variance: $\langle\hat{H}^2\rangle - \langle\hat{H}\rangle^2 = 0$. Number states are exact energy eigenstates, so a measurement of energy always returns $\hbar\omega(n+\tfrac{1}{2})$ with certainty.

Try it

This is a numerical exercise — your code should return a number. Using $E_n = \hbar\omega(n + \tfrac{1}{2})$ with $\hbar\omega = 1$, compute the energy gap $E_4 - E_2$ (which equals $2$ because the spectrum is evenly spaced).

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