Quantum

Quantum Foundations

Start

Use the photoelectric effect and single-particle interference to motivate quantum ideas.

Wavefunctions

Interpret amplitudes, probabilities, and measurement in a one-dimensional quantum state.

Quantization

Here

Use bound-state potentials to build intuition for discrete energies in quantum systems.

Time Evolution

Finish

Evolve wavefunctions through free spreading, finite barriers, and two-dimensional interference.

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

-\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 $V(x)$ , to the allowed states $\psi$ and their energies $E$ .

Rigid Box

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

\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

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

with energies

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 $n=0$ bound state in this model, so even the lowest state carries nonzero energy.

Solving the Infinite Well

In the ideal infinite well,

V(x) = \begin{cases} 0, & 0 < x < L \\ \infty, & x \le 0 \text{ or } x \ge L \end{cases}

so the wavefunction must vanish outside the box and at the walls:

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

Inside the box, $V=0$ , so the time-independent Schrodinger equation becomes

-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi.

Rearrange it as

\frac{d^2\psi}{dx^2}+k^2\psi=0, \qquad k^2=\frac{2mE}{\hbar^2}.

The general interior solution is

\psi(x)=A\sin(kx)+B\cos(kx).

The first boundary condition gives

\psi(0)=B=0,

leaving $\psi(x)=A\sin(kx)$ . The second boundary condition then gives

\psi(L)=A\sin(kL)=0.

A nonzero wavefunction needs

kL=n\pi, \qquad n=1,2,3,\dots

k_n=\frac{n\pi}{L}.

Using $E=\hbar^2 k^2/(2m)$ ,

E_n=\frac{\hbar^2}{2m}\left(\frac{n\pi}{L}\right)^2 =\frac{n^2\pi^2\hbar^2}{2mL^2} =\frac{n^2 h^2}{8mL^2}.

Finally normalize:

1=\int_0^L |\psi_n(x)|^2\,dx =|A|^2\int_0^L \sin^2\!\left(\frac{n\pi x}{L}\right)\,dx =|A|^2\frac{L}{2}.

Choosing a real positive constant gives

A=\sqrt{\frac{2}{L}}, \qquad \psi_n(x)=\sqrt{\frac{2}{L}}\sin\!\left(\frac{n\pi x}{L}\right).

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) = \frac{1}{2}m\omega^2 x^2.

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

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

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

Solving the QHO

The 1D quantum harmonic oscillator is defined by the potential function

V(x)=\frac{1}{2}m\omega^2x^2,

so the time-independent Schrodinger equation is

-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} +\frac{1}{2}m\omega^2x^2\psi =E\psi.

Introduce the dimensionless coordinate

\xi=\sqrt{\frac{m\omega}{\hbar}}x, \qquad \epsilon=\frac{2E}{\hbar\omega}.

Since $d/dx=\sqrt{m\omega/\hbar}\,d/d\xi$ , the equation becomes

\frac{d^2\psi}{d\xi^2}+(\epsilon-\xi^2)\psi=0.

For large $|\xi|$ , the $\xi^2$ term dominates, and a normalizable solution must behave like a decaying Gaussian. Factor that part out:

\psi(\xi)=e^{-\xi^2/2}H(\xi).

Substitution gives

\frac{d^2H}{d\xi^2} -2\xi\frac{dH}{d\xi} +(\epsilon-1)H=0.

Now try a power series,

H(\xi)=\sum_{j=0}^{\infty}a_j\xi^j.

Matching equal powers of $\xi$ gives the recurrence

a_{j+2} = \frac{2j+1-\epsilon}{(j+2)(j+1)}a_j.

If the series never terminates, the large- $\xi$ behavior eventually ruins normalizability. The acceptable solutions occur when the numerator becomes zero for some nonnegative integer $n$ :

2n+1-\epsilon=0.

Therefore

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

The terminating polynomials are the Hermite polynomials $H_n(\xi)$ , so the stationary states have the form

\psi_n(x)=N_nH_n(\xi)e^{-\xi^2/2}.

Normalization fixes

N_n= \frac{1}{\sqrt{2^n n!}} \left(\frac{m\omega}{\pi\hbar}\right)^{1/4},

which gives the normalized oscillator wavefunctions.

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

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 $n$ to describe the allowed states. The hydrogen atom states are labeled with three quantum numbers $n$ , $l$ , and $m$ . The primary quantization pattern is determined by the principal quantum number $n$ :

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

Starting the Hydrogen Atom

The full hydrogen problem starts with the three-dimensional time-independent Schrodinger equation,

\left[ -\frac{\hbar^2}{2\mu}\nabla^2 -\frac{e^2}{4\pi\varepsilon_0r} \right]\psi(r,\theta,\phi) =E\psi(r,\theta,\phi),

where $\mu$ is the electron-proton reduced mass. Because the potential depends only on $r$ , spherical coordinates are the natural choice.

Separate the wavefunction as

\psi(r,\theta,\phi)=R(r)Y_l^m(\theta,\phi).

The angular part gives spherical harmonics, which introduce the quantum numbers

l=0,1,2,\dots, \qquad m=-l,-l+1,\dots,l.

For the radial part, it is useful to write

u(r)=rR(r).

The radial equation becomes a one-dimensional-looking bound-state equation:

-\frac{\hbar^2}{2\mu}\frac{d^2u}{dr^2} + \left[ \frac{l(l+1)\hbar^2}{2\mu r^2} -\frac{e^2}{4\pi\varepsilon_0r} \right]u =Eu.

The new term

\frac{l(l+1)\hbar^2}{2\mu r^2}

acts like an angular-momentum barrier, while the Coulomb term attracts the electron inward. The remaining work is to solve this radial equation with two physical requirements:

$u(r)$ must stay finite near $r=0$ .
$u(r)$ must be normalizable as $r\to\infty$ .

Those requirements make the radial series terminate, producing associated Laguerre polynomials and the principal quantum number

n=1,2,3,\dots, \qquad l=0,1,\dots,n-1.

Carrying that longer series solution through gives

E_n = -\frac{\mu e^4}{2(4\pi\varepsilon_0)^2\hbar^2} \frac{1}{n^2} \approx -\frac{13.6\ \text{eV}}{n^2}.

The radial part of the hydrogen wavefunction depends on both $n$ and $l$ , 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:

In infinite-well mode, raise the quantum number and count the added nodes.
Switch to the harmonic oscillator and compare the equal energy spacing with the box’s $n^2$ growth.
In the hydrogen panel, keep $n$ fixed and change $l$ to see the radial shape change even when the simple energy formula still depends only on $n$ .
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 $\psi(x)$ 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 $E_2/E_1$ and $E_3/E_1$ .

Given

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 $n^2$ .

Solve

\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

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

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

Given

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

Use

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

Solve

E_2 = -\frac{13.6}{2^2}\ \text{eV} = -3.4\ \text{eV}

E_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 $n=3$ state lies closer to $0$ , 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?

Choose an answer to get instant feedback.