Classical vs Quantum Oscillator

The quantum harmonic oscillator has discrete energy levels $E_n = \hbar\omega(n + \tfrac{1}{2})$ and wave functions built from Hermite polynomials. But how does the position of the particle compare to what a classical pendulum or spring would predict? The answer reveals one of the deepest contrasts between the two theories.

Where does a classical oscillator spend its time?

Consider a classical particle of mass $m$ oscillating in the potential $V(x) = \tfrac{1}{2}m\omega^2 x^2$ with total energy $E$ . The turning points are at $x = \pm A$ , where the kinetic energy is zero and $E = \tfrac{1}{2}m\omega^2 A^2$ , giving the amplitude

$A = \sqrt{\frac{2E}{m\omega^2}}.$

At position $x$ the speed is found from energy conservation, $\tfrac{1}{2}mv^2 = E - \tfrac{1}{2}m\omega^2 x^2$ , so

$v(x) = \omega\sqrt{A^2 - x^2}.$

The probability of finding the particle in an interval $dx$ is proportional to the time it spends there, which is inversely proportional to its speed. Normalising over a full half-period (the particle crosses every $x$ twice per cycle) gives the classical probability density

$P_{\text{cl}}(x) = \frac{1}{\pi\sqrt{A^2 - x^2}}, \qquad |x| \leq A.$

This density diverges at the turning points $x = \pm A$ : a classical oscillator moves slowest — and therefore lingers longest — at the extremes of its motion, exactly where it momentarily stops and reverses direction. Near the centre $x = 0$ it moves fastest and the probability density is at its minimum. The distribution is U-shaped, concentrating probability at the edges of the classically allowed region.

Quantum probability densities

For the quantum oscillator in energy eigenstate $|n\rangle$ the probability density is

$P_n(x) = |\psi_n(x)|^2,$

where $\psi_n(x)$ is the $n$ -th energy eigenfunction. These eigenfunctions involve Hermite polynomials and a Gaussian envelope. The key structural property is that $P_n(x)$ has exactly $n$ nodes (zeros) inside the classically allowed region and decays exponentially beyond the classical turning points.

The ground state: peaked at the centre

For $n = 0$ the wave function is a pure Gaussian,

\exp\!\left(-\frac{m\omega}{2\hbar}x^2\right),$$ so $P_0(x) = |\psi_0(x)|^2 \propto e^{-m\omega x^2/\hbar}$ is also a Gaussian. It has a single peak **at the centre** $x = 0$ — exactly where the classical particle moves fastest and its density is lowest. A measurement of a ground-state oscillator is most likely to find the particle near equilibrium — precisely the location where the classical probability density is *smallest*. The two pictures are therefore most sharply in conflict at the centre. ### Excited states: oscillations approaching the classical limit For large $n$ the quantum density $P_n(x)$ develops rapid oscillations, with peaks concentrating more and more toward the turning points. The local maxima and minima of $|\psi_n|^2$ average out over many oscillations to a smooth envelope that closely follows the classical $P_{\text{cl}}(x)$ derived above. This convergence is an instance of the **correspondence principle**: quantum mechanics reduces to classical mechanics in the limit of large quantum numbers. To make the comparison concrete, the classical turning point at energy $E_n$ corresponds to amplitude $A_n = \sqrt{2E_n/(m\omega^2)}$. Substituting $E_n = \hbar\omega(n+\tfrac{1}{2})$: $$A_n = \sqrt{\frac{2\hbar(n+\tfrac{1}{2})}{m\omega}} = \sqrt{(2n+1)\,\frac{\hbar}{m\omega}}.$$ The characteristic length of the oscillator is $x_0 = \sqrt{\hbar/(m\omega)}$, so $A_n = \sqrt{2n+1}\,x_0$. For $n = 0$ the turning point is at $A_0 = x_0$, just one characteristic length from the origin. ## Quantum tunnelling beyond the turning points One of the most striking differences is that $P_n(x)$ is **nonzero** for $|x| > A_n$. Classically, the particle cannot exist beyond the turning point because its kinetic energy would be negative. Quantum mechanically the wave function decays exponentially into this forbidden zone but does not vanish. The probability of finding the particle in the classically forbidden region for the ground state is $$\int_{|x|>A_0} P_0(x)\,dx \approx 0.157,$$ meaning that about $16\%$ of measurements would find the ground-state particle outside the region a classical particle of the same energy could reach. This probability decreases as $n$ increases — another facet of the correspondence principle. ## Summary of contrasts | Feature | Classical oscillator | Quantum oscillator | |---|---|---| | Probability distribution | U-shaped; highest at turning points | Oscillatory; Gaussian at $n=0$, U-shaped average at large $n$ | | Most probable location ($n=0$) | Turning points $\pm A$ | Centre $x=0$ | | Beyond turning points | Forbidden (zero probability) | Exponential decay (nonzero probability) | | Large $n$ limit | Fixed smooth distribution | Rapid oscillations that average to classical | <Callout type="tip"> A useful mnemonic: classically, the particle is most often found where it moves most slowly; quantum mechanically (at low $n$), it is most often found where the restoring force is smallest — near the equilibrium. Only at large $n$ do the two pictures agree on average. </Callout>