The Harmonic Oscillator Has Evenly Spaced Levels
Put a particle not against hard walls but inside a soft, bowl-shaped potential V(x) = ½ m ω² x². In the box the energy rungs crowded together upward like 1, 4, 9, 16, but here it is the opposite: the spacing stays the same from bottom to top at ΔE = ℏω. The levels are Eₙ = (n + ½)ℏω, so the ladder rungs line up at a constant interval. In the first figure we look at the bowl V(x) and the evenly spaced energy ladder sitting on it together, and watch how both change when we tune the stiffness ω.
Use the slider to change the stiffness ω of the bowl V(x) = ½ m ω² x². This ω is the oscillator angular frequency, the same quantity as the ω from classical simple harmonic motion (SHM). Turn ω up and the bowl gets narrow and steep, and the energy ladder sitting on it stretches up and down with it. The rungs always sit at a constant spacing ΔE = ℏω, shown as a readout beside them. Unlike the box with 1, 4, 9, 16, the gaps never crowd together.
Now look at which wavefunction ψₙ(x) each rung of the ladder carries. Here Hₙ is a Hermite polynomial (a polynomial with a subscript), a different quantity from the Hamiltonian H that shares the same letter. ψₙ is a Gaussian hump multiplied by Hₙ(x), so the n-th state has exactly n nodes. Step n from 0 up to 4 and you see the hump ripple inside the bowl, gaining one more node each time.
Climb to high rungs and watch how the quantum and classical pictures meet. A classical oscillator slows down near its endpoints (turning points) and lingers there, so the probability of finding it there is largest, going as ∝ 1/√(xmax² − x²). Raise n with the slider and the rippling humps of the quantum |ψₙ(x)|² increasingly trace this bathtub-shaped classical curve. For small n the two differ greatly, but for large n the probability piles up near both turning points and they nearly overlap. This is the correspondence principle.
Same particle, different bowl. Put the two ladders side by side and compare. The box has hard walls so Eₙ ∝ n², and the rungs spread fast upward like 1, 4, 9, 16. The harmonic oscillator has a soft parabolic bowl so Eₙ = (n + ½)ℏω, and the spacing stays the same all the way at ΔE = ℏω. Toggle between the two ladders and you see at a glance how the shape of the potential sets the energy spacing.
Look closely at the ground state n=0. It is a simple Gaussian hump with no nodes at all, yet its energy is not zero but E₀ = ½ℏω. This is the zero-point energy: even the calmest state can never fully come to rest. Change ω with the slider and the hump grows wider or narrower, but E₀ = ½ℏω always stays at half the first rung height and never touches zero. As long as it is confined, the ground energy cannot be zero.