Quantization

Once a quantum system is described by a wavefunction, the next question is: which states are actually allowed? The time-independent Schrodinger equation does not accept every imaginable shape. Boundary conditions, normalizability, and the form of the potential select only certain stationary states. That is what quantization means: a bound system admits a discrete set of energies rather than a continuous range.

Time-Independent Schrodinger Equation

For a particle in one dimension, the time-independent Schrodinger equation is

22md2ψdx2+V(x)ψ=Eψ.-\frac{\hbar^2}{2m}\frac{d^2 \psi}{dx^2} + V(x)\psi = E\psi.

This equation plays a role somewhat analogous to Newton’s second law in classical mechanics: it connects the system’s environment, represented by the potential energy function V(x)V(x), to the allowed states ψ\psi and their energies EE.

Rigid Box

Inside a rigid one-dimensional box (aka infinite potential well), the walls force the wavefunction to vanish at the boundaries:

ψ(0)=0,ψ(L)=0.\psi(0) = 0, \qquad \psi(L) = 0.

Those conditions mean only standing waves that fit exactly inside the box are allowed. The resulting stationary states are

ψn(x)=2Lsin ⁣(nπxL),n=1,2,3,\psi_n(x) = \sqrt{\frac{2}{L}} \sin\!\left(\frac{n\pi x}{L}\right), \qquad n = 1, 2, 3, \dots

with energies

En=n2h28mL2,n=1,2,3,E_n = \frac{n^2 h^2}{8mL^2}, \qquad n = 1, 2, 3, \dots

The allowed energies are therefore quantized, not continuous. There is no n=0n=0 bound state in this model, so even the lowest state carries nonzero energy.

Quantization Beyond Hard Walls

The rigid box is the cleanest example, but discrete energies are not unique to hard-wall boundaries.

For the harmonic oscillator, the potential is smooth rather than abrupt:

V(x)=12mω2x2.V(x) = \frac{1}{2}m\omega^2 x^2.

Here the wavefunction must stay normalizable as x|x| \to \infty. That requirement picks out the equally spaced energy ladder

En=(n+12)ω,n=0,1,2,E_n = \left(n + \frac{1}{2}\right)\hbar\omega, \qquad n = 0, 1, 2, \dots

The ground state already has the nonzero zero-point energy 12ω\frac{1}{2}\hbar\omega, and each excited state adds another node.

For the hydrogen atom, the potential comes from the Coulomb attraction between the proton and electron.

V(r)=e24πε0r,V(r) = -\frac{e^2}{4\pi\varepsilon_0 r},

As a more complex system, we can expect the wavefunction to require more than one integer nn to describe the allowed states. The hydrogen atom states are labeled with three quantum numbers nn, ll, and mm. We do not need the full derivation here to see the main quantization pattern:

En=13.6 eVn2,n=1,2,3,E_n = -\frac{13.6\ \text{eV}}{n^2}, \qquad n = 1, 2, 3, \dots

The radial part of the hydrogen wavefunction depends on both nn and ll, while the angular part is carried by spherical harmonics. In the explorer below, the hydrogen panel focuses on that radial component so you can compare its nodes and radial probability pattern with the box and oscillator states.

As you explore the diagram below, try this sequence:

  1. In infinite-well mode, raise the quantum number and count the added nodes.
  2. Switch to the harmonic oscillator and compare the equal energy spacing with the box’s n2n^2 growth.
  3. In the hydrogen panel, keep nn fixed and change ll to see the radial shape change even when the simple energy formula still depends only on nn.
  4. Compare the probability view in all three systems and look for how nodes in the amplitude create dips in the probability density.

Note that it’s convention to overlay the wavefunction ψ(x)\psi(x) on the same plot describing the allowed energies of a system. Below, the potential energy function is also drawn in black. It’s a lot at first, but helpful once you start to see the connection between things like frequency and energy.

Quantization Explorer

Compare stationary states in the infinite well, harmonic oscillator, and hydrogen radial problem.

Problem Solving

Compare Particle-In-A-Box Energy Levels

Problem

In a one-dimensional infinite square well, compare the energies of the first three states by finding E2/E1E_2/E_1 and E3/E1E_3/E_1.

Given

En=n2h28mL2.E_n = \frac{n^2 h^2}{8mL^2}.

Use

Since the prefactor is the same for every level in the same box, the ratios depend only on n2n^2.

Solve

E2E1=2212=4,E3E1=3212=9.\frac{E_2}{E_1} = \frac{2^2}{1^2} = 4, \qquad \frac{E_3}{E_1} = \frac{3^2}{1^2} = 9.

So the second state has four times the ground-state energy, and the third state has nine times the ground-state energy.

Check

The rapid growth reflects the fact that fitting shorter standing wavelengths into the same box requires stronger curvature in the wavefunction, which the Schrodinger equation associates with higher energy.

Compare Hydrogen Energy Levels

Problem

Use the hydrogen energy formula

En=13.6 eVn2E_n = -\frac{13.6\ \text{eV}}{n^2}

to find the energies of the n=2n=2 and n=3n=3 levels.

Given

En=13.6 eVn2E_n = -\frac{13.6\ \text{eV}}{n^2}

Use

Substitute n=2n=2 and n=3n=3.

Solve

E2=13.622 eV=3.4 eVE_2 = -\frac{13.6}{2^2}\ \text{eV} = -3.4\ \text{eV}E3=13.632 eV1.51 eVE_3 = -\frac{13.6}{3^2}\ \text{eV} \approx -1.51\ \text{eV}

Check

Both values are less negative than the ground-state energy, so they are less tightly bound. The n=3n=3 state lies closer to 00, which is the ionization limit in this simple model.

Quantization Checkpoint

Question 1 of 3

Why do energy levels become discrete for a particle trapped in an ideal one-dimensional box?

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