Quantum harmonic oscillator and hydrogen atom

Two of the most important bound systems in physics get the Schrödinger-equation treatment here: a particle in a parabolic well (the quantum harmonic oscillator) and an electron bound to a proton (the hydrogen atom). Both are classical, continuously-tunable systems in Newtonian mechanics, yet quantum mechanics forces their energies into discrete, evenly or unevenly spaced ladders. This article gives you the two energy formulas you must know and use for PH11001 — $E_n=(n+\tfrac12)\hbar\omega$ for the oscillator and $E_n=-13.6\ \text{eV}/n^2$ for hydrogen — along with the reasoning behind the SHO's most surprising feature, zero-point energy, derived directly from the uncertainty principle using nothing beyond calculus you already know. Full wavefunctions, ladder operators, and angular-momentum quantum numbers stay out of scope.

The core insight

Confining a particle in 1D by a parabolic potential (SHO) or in 3D by a Coulomb potential (hydrogen) both force the Schrödinger equation to admit only discrete energies, $E_n=(n+\tfrac12)\hbar\omega$ and $E_n=-13.6\ \text{eV}/n^2$ respectively, with the SHO's ground state pinned above zero by the uncertainty principle.

The core idea

The quantum SHO ladder is equally spaced. For the potential $V(x)=\tfrac12 m\omega^2x^2$ , solving the time-independent Schrödinger equation gives allowed energies

E_n = \left(n+\frac12\right)\hbar\omega, \qquad n = 0,1,2,\dots

where: $m$ is the particle's mass, $\omega$ is the classical angular frequency of the potential, $\hbar=h/2\pi$ is the reduced Planck constant, and $n$ is the vibrational quantum number. Every step up costs exactly $\hbar\omega$ : $E_{n+1}-E_n=\hbar\omega$ for any $n$ , a rigid, uniform ladder unlike the growing gaps of a square well.

Even "at rest," the SHO cannot have zero energy. The $n=0$ ground state sits at $E_0=\tfrac12\hbar\omega\neq0$ , the zero-point energy. A particle truly at rest at $x=0$ would need $\Delta x=0$ and $\Delta p=0$ simultaneously, which the uncertainty principle forbids; the derivation below turns that qualitative argument into the exact number $\tfrac12\hbar\omega$ .

Macroscopic springs hide this quantization. For everyday oscillators $\hbar\omega\ll k_BT$ at room temperature, so thousands of levels are packed within one unit of thermal energy and the spectrum looks continuous — the same Boltzmann-factor logic used elsewhere in this course for why classical mechanics "emerges" from quantum mechanics at large scale.

Hydrogen is a genuinely 3D bound-state problem. An electron bound to a proton by $V(r)=-e^2/4\pi\varepsilon_0 r$ requires spherical coordinates and angular-momentum operators to solve fully (outside this course's scope), but the resulting bound-state energies depend on a single principal quantum number:

E_n = -\frac{13.6\ \text{eV}}{n^2}, \qquad n = 1,2,3,\dots

where: $13.6\ \text{eV}$ is the hydrogen ground-state binding energy (the Rydberg energy) and $n=1,2,3,\dots$ indexes the shell. The negative sign marks a bound state below the $E=0$ ionization threshold; levels crowd together and approach zero as $n\to\infty$ , unlike the SHO's rigid, uniform spacing.

Figure

Two energy level ladders side by side, the harmonic oscillator with equal rung spacing and the hydrogen atom with rungs crowding together toward zero

Derivation AZero-point energy from the uncertainty principle

The goal is to show, using only the uncertainty relation $\Delta x\,\Delta p\gtrsim\hbar/2$ and ordinary calculus minimization, that a quantum particle held in a parabolic well can never have less than $\tfrac12\hbar\omega$ of energy — reproducing the exact $n=0$ result of the full Schrödinger solution without solving any differential equation.

Step 1. Write the classical energy of a particle in the SHO potential in terms of its momentum and position spreads. Since the particle's average position and momentum are both zero by symmetry, its typical momentum is of order $\Delta p$ and its typical displacement is of order $\Delta x$ , so the energy is roughly $E \approx \dfrac{(\Delta p)^2}{2m} + \tfrac12 m\omega^2(\Delta x)^2$ .

Step 2. Use the uncertainty principle to eliminate one variable. Taking the minimum-uncertainty case $\Delta p\,\Delta x = \hbar/2$ , substitute $\Delta p = \hbar/(2\Delta x)$ into the energy expression, giving $E(\Delta x) = \dfrac{\hbar^2}{8m(\Delta x)^2} + \tfrac12 m\omega^2(\Delta x)^2$ , a function of the single unknown $\Delta x$ alone.

Step 3. Minimize over $\Delta x$ using calculus. Setting $dE/d(\Delta x)=0$ : $-\dfrac{\hbar^2}{4m(\Delta x)^3} + m\omega^2(\Delta x) = 0$ , which rearranges to $(\Delta x)^4 = \dfrac{\hbar^2}{4m^2\omega^2}$ , so $(\Delta x)^2 = \dfrac{\hbar}{2m\omega}$ .

Step 4. Substitute back to find the minimum energy. $E_{\min} = \dfrac{\hbar^2}{8m}\cdot\dfrac{2m\omega}{\hbar} + \tfrac12 m\omega^2\cdot\dfrac{\hbar}{2m\omega} = \dfrac{\hbar\omega}{4} + \dfrac{\hbar\omega}{4} = \dfrac{\hbar\omega}{2}$ . This matches the exact quantum result $E_0=\tfrac12\hbar\omega$ exactly, confirming that zero-point energy is a direct consequence of the trade-off the uncertainty principle forces between localizing a particle (raising kinetic energy) and letting it spread out (raising potential energy) — a particle cannot minimize both at once, so the true minimum is strictly positive.

Worked example

An electron in a hydrogen atom sits in the $n=2$ state. Find (a) its energy in eV, (b) the wavelength of the photon emitted dropping to $n=1$ (Lyman- $\alpha$ ), and, as a stretch, (c) the wavelength emitted dropping from $n=3$ to $n=2$ (Balmer- $\alpha$ ).

Step 1. Energy at $n=2$ : $E_2 = -13.6/2^2 = -3.4\ \text{eV}$ .

Step 2. Photon energy for $2\to1$ : $E_1=-13.6\ \text{eV}$ , so $\Delta E = E_2-E_1 = -3.4-(-13.6) = 10.2\ \text{eV} = 1.632\times10^{-18}\ \text{J}$ .

Step 3. Wavelength using $\lambda = hc/\Delta E$ : $\lambda = \dfrac{6.63\times10^{-34}\times3\times10^8}{1.632\times10^{-18}} = 1.219\times10^{-7}\ \text{m} = 121.9\ \text{nm}$ , matching the known Lyman- $\alpha$ ultraviolet line.

Competitive stretch

Step 4. Energy at $n=3$ : $E_3 = -13.6/3^2 = -1.51\ \text{eV}$ . Photon energy for $3\to2$ : $\Delta E = E_3-E_2 = -1.51-(-3.4) = 1.89\ \text{eV} = 3.02\times10^{-19}\ \text{J}$ .

Step 5. Wavelength: $\lambda = \dfrac{6.63\times10^{-34}\times3\times10^8}{3.02\times10^{-19}} = 6.59\times10^{-7}\ \text{m} = 659\ \text{nm}$ — red light, matching the well-known Balmer- $\alpha$ line. Final answers: $E_2=-3.4\ \text{eV}$ , Lyman- $\alpha$ at $121.9\ \text{nm}$ , Balmer- $\alpha$ at $659\ \text{nm}$ .

Common traps

Writing the SHO ground state as $E_0=0$ — quantum zero-point energy is unavoidable and strictly positive, unlike the classical "no motion, no energy" intuition.
Confusing the equally-spaced SHO ladder ( $E_{n+1}-E_n=\hbar\omega$ , constant) with the square well's spectrum, whose spacing grows with $n$ .
Dropping the negative sign in $E_n=-13.6\ \text{eV}/n^2$ , or mixing up "more negative" with "higher energy" — the $n=1$ ground state is the most negative and thus lowest-energy state.
Trying to minimize $E(\Delta x)$ without first eliminating one variable via the uncertainty relation — both terms must be written in a single variable before calculus can find the minimum.

Ask the tutor: Want to see why the minimum-uncertainty trick gives the exact zero-point energy for the SHO but only an estimate for other wells, or need help converting an eV transition energy into a wavelength — ask the AI tutor to walk through it.

Why does the SHO have equally spaced levels but hydrogen does not?

The spacing pattern comes from how steeply each potential rises. The SHO potential $V(x)=\tfrac12m\omega^2x^2$ grows the same way in every direction and at every energy, so the Schrödinger equation returns a perfectly uniform ladder, $E_{n+1}-E_n=\hbar\omega$ always. The Coulomb potential $V(r)=-e^2/4\pi\varepsilon_0 r$ is steep near the proton and flattens out far away, so higher states feel a progressively gentler confinement and their energies crowd together as $n\to\infty$ , following $E_n\propto-1/n^2$ instead of a constant step.

Do I need to know the actual SHO or hydrogen wavefunctions for this course?

No. PH11001 requires only the energy-level formulas, $E_n=(n+\tfrac12)\hbar\omega$ and $E_n=-13.6\ \text{eV}/n^2$ , their spacing behaviour, and the zero-point-energy argument from the uncertainty principle. Constructing the Hermite-polynomial wavefunctions for the SHO, using ladder (creation/annihilation) operators, or solving the 3D radial equation with angular momentum quantum numbers for hydrogen are all outside this course's scope — save them for a dedicated quantum mechanics course.

Is the uncertainty-principle derivation of zero-point energy exact or just an estimate?

For the SHO specifically it happens to land exactly on $\tfrac12\hbar\omega$ , matching the full Schrödinger-equation result, because the ground-state wavefunction genuinely is the minimum-uncertainty state ( $\Delta x\,\Delta p=\hbar/2$ exactly). For other potentials the same technique only gives an order-of-magnitude estimate, since their ground states are not minimum-uncertainty states. Treat the method as a powerful shortcut that is exact here, not a universal exact formula.

Quantum harmonic oscillator and hydrogen atom

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads