Modern Physics

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The semiclassical picture

A full treatment of quantum mechanics uses the Schrödinger equation, beyond the scope of these USAPhO notes. Earlier semiclassical models, including Bohr’s, explain some quantum behavior with simpler methods. Here, we use the particle’s wave properties and require its phase to match after a complete orbit or round trip.

The phase $\phi$ of a wave varies in space and time according to its wavenumber and angular frequency,

k = \frac{d\phi}{dx}, \qquad \omega = \frac{d\phi}{dt},

and the group velocity (the speed at which a wavepacket travels) is

v_g = \frac{d\omega}{dk}.

A standing wave can only form if the phase lines back up with itself after one round trip:

\oint k\,dx = 2\pi n, \qquad n \in \mathbb{Z}.

For a string of length $L$ with two fixed ends this gives $2kL = 2\pi n$ , so $k_n = \pi n / L$ and $\omega_n = \pi v n / L$ , exactly the familiar result.

Reflection phase shifts

The naive condition above is not quite complete: a wave can pick up an extra phase when it reflects off a boundary. A reflection off a fixed (hard) end adds a phase shift of $\pi$ . For a string with one fixed and one free end, this single extra $\pi$ modifies the round-trip condition to

\oint k\,dx = 2\pi\left(n + \tfrac{1}{2}\right),

which yields

\omega_n = \frac{\pi v}{L}\left(n + \tfrac{1}{2}\right).

For two fixed ends, the two $\pi$ shifts add to $2\pi$ and have no net effect — which is why we never noticed them above.

The WKB approximation

In quantum mechanics a particle is described by a wavefunction $\psi(x,t)$ .

Theorem (de Broglie relations). Momentum and energy obey

p = \hbar k, \qquad E = \hbar\omega,

where $\hbar$ is the reduced Planck constant (the exact value, $\hbar = \frac{h}{2\pi} = 1.05 \cdot 10^{-34} J \cdot s$ is usually given if necessary). For a nonrelativistic particle in a potential $V(x)$ ,

E = \frac{p^2}{2m} + V(x).

The group velocity is then $v_g = d\omega/dk = dE/dp = p/m$ , which just the ordinary classical velocity. (The same relations hold relativistically if $E$ is the relativistic energy and $p = \gamma m v_g$ .)

Since $\omega$ (hence $E$ ) is uniform for a standing wave, these quantum standing waves are states of definite energy. The quantization condition becomes the WKB / Bohr–Sommerfeld rule:

\oint p\,dx = \left(n + \frac{\alpha}{2\pi}\right) h, \qquad n = 0, 1, 2, \dots

where the integral runs over one full classical period of the motion, and the extra phase $\alpha$ accounts for what happens at the turning points.

The rule for $\alpha$ at the two ends of the motion:

A hard wall (potential jumps to infinity) contributes $\pi$ to $\alpha$ .
A soft turning point (potential rises smoothly through $E$ ) contributes $\pi/2$ to $\alpha$ .

So a box with two hard walls has $\alpha = 2\pi$ (equivalent to $\alpha = 0$ ), while a smooth potential well like the harmonic oscillator has $\alpha = \pi$ .

The left-hand side $\oint p\,dx$ is equivalent to the adiabatic invariant of classical mechanics, which is conserved when system parameters change slowly. This guarantees the quantization condition stays self-consistent over time.

Example. For a particle of mass $m$ in a box of length $L$ with hard walls, $\alpha = 0$ and $p$ is constant inside, find the energy of the particle inside the box.

Inside the box $V = 0$ , so the particle moves freely at constant speed and constant momentum magnitude $p$ , reflecting off each wall. One full period of the motion is a round trip: it crosses the box once moving right (momentum $+p$ ) and once moving left (momentum $-p$ ). The loop integral therefore picks up the same positive contribution on each leg,

\oint p\,dx = \int_0^L (+p)\,dx + \int_L^0 (-p)\,dx = pL + pL = 2pL.

Two hard walls give $\alpha = 2\pi$ , equivalent to $\alpha = 0$ , so the quantization rule is simply $\oint p\,dx = nh$ :

2pL = nh \quad\Longrightarrow\quad p = \frac{nh}{2L}.

Since all the energy is kinetic,

E_n = \frac{p^2}{2m} = \frac{n^2 h^2}{8 m L^2}, \qquad n = 1, 2, 3, \dots

The levels grow as $n^2$ , and the spacing widens with increasing $n$ — the opposite of the evenly spaced oscillator levels below. (Note $n=0$ is excluded: it would mean $p=0$ , a particle at rest spread over the whole box, which violates uncertainty.)

If instead the “particle” is a photon with $E = pc$ rather than $p^2/2m$ , the same momenta $p = nh/2L$ give the standing-wave (cavity) frequencies

E_n = pc = \frac{nhc}{2L},

the electromagnetic modes of a box with conducting walls. These are exactly the modes whose zero-point energies are summed in the Casimir and blackbody discussions later.

Example. For a particle inside a harmonic oscillator with $V(x) = \tfrac{1}{2}kx^2$ , find the energy of the particle.

A particle of energy $E$ satisfies $E = \dfrac{p^2}{2m} + \tfrac{1}{2}kx^2$ . Rearranging,

\frac{p^2}{2mE} + \frac{x^2}{2E/k} = 1,

which is an ellipse in the $(x,p)$ phase plane with semi-axes

x_{\max} = \sqrt{\frac{2E}{k}}, \qquad p_{\max} = \sqrt{2mE}.

The loop integral $\oint p\,dx$ is just the area enclosed by this ellipse, $\pi\, x_{\max} p_{\max}$ :

\oint p\,dx = \pi\sqrt{\frac{2E}{k}}\,\sqrt{2mE} = 2\pi E\sqrt{\frac{m}{k}} = \frac{2\pi E}{\omega_0}, \qquad \omega_0 = \sqrt{\frac{k}{m}}.

The particle turns around at two soft turning points (the potential rises smoothly through $E$ ), each contributing $\pi/2$ , so $\alpha = \pi$ and the quantization rule reads $\oint p\,dx = \left(n + \tfrac{1}{2}\right)h$ . Setting the two equal,

\frac{2\pi E}{\omega_0} = \left(n + \tfrac{1}{2}\right)h \quad\Longrightarrow\quad E_n = \left(n + \tfrac{1}{2}\right)\frac{h\,\omega_0}{2\pi} = \hbar\omega_0\left(n + \tfrac{1}{2}\right).

Unlike the box, these levels are evenly spaced by $\hbar\omega_0$ . Remarkably this is the exact answer, even though WKB is an approximation. The leftover $\tfrac{1}{2}\hbar\omega_0$ at $n=0$ is the zero-point energy — the oscillator can never sit perfectly still, consistent with the uncertainty-principle estimate at the end of this page.

Bohr quantization

The integral $\oint p\,dx$ can use any conjugate momentum–coordinate pair. For rotational motion,

\oint L\,d\theta = nh.

When angular momentum is conserved the integrand is constant, so the left side is $2\pi L$ , giving Bohr’s condition

L = n\hbar.

Rotation differs from back-and-forth motion in two ways: $n$ can be positive or negative (clockwise vs. counterclockwise), and there is no $\alpha$ phase, because the particle circulates freely without ever reflecting.

Example. Find the energy of an electron at energy level $n$ of a hydrogen atom. Assume circular orbit.

Two equations govern a circular orbit. First, the Coulomb attraction supplies the centripetal force,

\frac{mv^2}{r} = \frac{e^2}{4\pi\epsilon_0 r^2} \quad\Longrightarrow\quad mv^2 = \frac{e^2}{4\pi\epsilon_0 r}. \tag{1}

Second, the Bohr condition quantizes angular momentum,

L = mvr = n\hbar \quad\Longrightarrow\quad v = \frac{n\hbar}{mr}. \tag{2}

Substitute $(2)$ into $(1)$ to eliminate $v$ :

m\left(\frac{n\hbar}{mr}\right)^2 = \frac{e^2}{4\pi\epsilon_0 r} \;\Longrightarrow\; \frac{n^2\hbar^2}{m r^2} = \frac{e^2}{4\pi\epsilon_0 r},

and solving for $r$ gives the allowed radii

r_n = \frac{4\pi\epsilon_0 \hbar^2}{m e^2}\, n^2 = a_0 n^2, \qquad a_0 \approx 5.3\times10^{-11}\,\text{m},

where $a_0$ is the Bohr radius. For the energy, note the kinetic energy from $(1)$ is $\tfrac{1}{2}mv^2 = \tfrac{1}{2}\dfrac{e^2}{4\pi\epsilon_0 r}$ , while the potential energy is $U = -\dfrac{e^2}{4\pi\epsilon_0 r}$ . Their sum is

E = \frac{1}{2}\frac{e^2}{4\pi\epsilon_0 r} - \frac{e^2}{4\pi\epsilon_0 r} = -\frac{e^2}{8\pi\epsilon_0 r},

the general fact that for an inverse-square force the total energy is half the potential energy (the virial theorem). Inserting $r = r_n$ ,

E_n = -\frac{e^2}{8\pi\epsilon_0 r_n} = -\frac{m e^4}{2(4\pi\epsilon_0)^2 \hbar^2}\,\frac{1}{n^2}.

The constant prefactor is the Rydberg energy, $13.6\,\text{eV}$ , so $E_n = -13.6\,\text{eV}/n^2$ . A jump from level $n_i$ to $n_f$ emits a photon of energy $13.6\,\text{eV}\,(1/n_f^2 - 1/n_i^2)$ — the Rydberg formula for the hydrogen spectral lines.

The same method handles hydrogen-like systems: for positronium (electron + positron), replace $m$ with the reduced mass $\mu = m_e/2$ , halving all the binding energies.

The correspondence principle

The correspondence principle says quantum results must smoothly match classical ones in the limit $\hbar \to 0$ , i.e. for large quantum numbers $n \to \infty$ . For high $n$ you can superpose nearby orbitals into a sharply peaked wavepacket that orbits just like a classical particle, which is why the Bohr model still describes highly excited Rydberg atoms. In fact, demanding that the classical orbital frequency match the quantum transition frequency $\Delta E = \hbar\omega$ as $n\to\infty$ is exactly how Bohr originally derived his quantization rule.

Higher dimensions and degeneracy

For a system with several degrees of freedom, the WKB condition holds for each one independently:

\oint p_i\,dx_i = \left(n_i + \frac{\alpha_i}{2\pi}\right) h.

Several distinct sets of quantum numbers can give the same energy; the number of states at one energy is the degeneracy of that level.

For a particle in a 2D or 3D box of side $L$ (hard walls, all $\alpha_i = 0$ ),

E = \frac{h^2}{8mL^2}\left(n_1^2 + n_2^2 + \cdots\right),

and degeneracies arise whenever different integer combinations give the same sum of squares.

Density of states

For a large box it is easier to count states than to list them. Working in momentum space with axes $p_x, p_y, p_z$ :

Hard walls: $k_i = (\pi/L)n_i$ with $n_i$ a positive integer. States live in the first octant, one per volume $(\pi\hbar/L)^3$ .
Periodic boundaries: $k_i = (2\pi/L)n_i$ with $n_i$ any integer. States fill all of momentum space, one per volume $(2\pi\hbar/L)^3$ .

Both give the same density of states. The number of states with energy at most $E_0$ (a sphere of radius $p = \sqrt{2mE_0}$ ) is

N \approx \frac{4}{3}\pi (2mE_0)^{3/2}\left(\frac{2\pi\hbar}{L}\right)^{-3} = \frac{V}{6\pi^2}\left(\frac{2mE_0}{\hbar^2}\right)^{3/2}.

The boundary conditions don’t matter for bulk statistical properties — a fact worth remembering, since Rayleigh originally botched it by using hard walls and allowing negative $n_i$ , overcounting by a factor of 8 (later fixed by Jeans).

Heisenberg’s Uncertainty Principle

Theorem (Heisenberg uncertainty principle). The standard deviations of position and momentum obey

\Delta x\,\Delta p \ge \frac{\hbar}{2}.

The semiclassical limit is just the regime where the required uncertainty is small compared to the scales involved, reached for $n \gg 1$ .

Example. Approximate the energy of an oscillator at ground state.

In the ground state the particle is localized to within $\Delta x$ of the origin, with a spread of momentum $\Delta p$ . Its typical kinetic energy is $\sim (\Delta p)^2/2m$ and its typical potential energy is $\sim \tfrac{1}{2}k(\Delta x)^2$ . Uncertainty ties the two scales together, $\Delta p \sim \hbar/\Delta x$ , so writing everything in terms of $\Delta x$ (and dropping order-unity factors),

E \sim \frac{\hbar^2}{m(\Delta x)^2} + k(\Delta x)^2.

Squeezing the particle (small $\Delta x$ ) raises the kinetic term; spreading it out raises the potential term — the ground state balances the two. Minimize: set $dE/d(\Delta x) = 0$ ,

-\frac{2\hbar^2}{m(\Delta x)^3} + 2k(\Delta x) = 0 \quad\Longrightarrow\quad (\Delta x)^4 \sim \frac{\hbar^2}{mk} \quad\Longrightarrow\quad (\Delta x)^2 \sim \frac{\hbar}{\sqrt{km}}.

Substituting back, both terms become $\sim \hbar\sqrt{k/m}$ , so

E \sim \hbar\sqrt{\frac{k}{m}} \sim \hbar\omega_0,

reproducing the zero-point energy $\tfrac12\hbar\omega_0$ up to the numerical factor. The same trick on hydrogen — balancing $\hbar^2/2m(\Delta x)^2$ against $-e^2/4\pi\epsilon_0\Delta x$ — yields the Bohr radius and Rydberg.

Example. Find the width of a diffraction of a ray with wavelength $\lambda$ going through a single slit with width $a$ from a distance $D$ away.

Passing through a slit of width $a$ confines the photon’s transverse position to $\Delta y \sim a$ . By uncertainty it therefore picks up a transverse momentum

\Delta p_y \sim \frac{\hbar}{a} \sim \frac{h}{a}.

Meanwhile its forward momentum is $p_x = h/\lambda$ . The beam spreads by an angle equal to the ratio of transverse to forward momentum,

\Delta\theta \sim \frac{\Delta p_y}{p_x} \sim \frac{h/a}{h/\lambda} = \frac{\lambda}{a},

so on a screen a distance $D$ away the pattern has width

\Delta y_{\text{screen}} \sim D\,\Delta\theta \sim \frac{D\lambda}{a}.

The narrower the slit, the wider the spread — the hallmark of diffraction. This quantum derivation reproduces the classical wave-optics result, and now applies to matter waves too, with the de Broglie wavelength $\lambda = h/p$ .

Energy–time uncertainty

If a system is observed for only a finite time $\Delta t$ , or changes its state significantly in time $\Delta t$ , its energy is uncertain by

\Delta E\,\Delta t \ge \frac{\hbar}{2}.

A common phrasing is that “energy conservation can be violated by $\Delta E$ for a time $\Delta t$ .” This is technically wrong — quantum systems always conserve energy — but it gives the right answers by dimensional analysis. A typical application: an unstable particle with lifetime $\tau$ has an unavoidable energy (mass) width $\Delta E \sim \hbar/\tau$ .

Bosons and fermions

When many non-interacting identical particles are put together, their behavior splits into two types:

Fermions obey the Pauli exclusion principle — no two can occupy the same quantum state. The ground state of $N$ fermions fills the $N$ lowest single-particle states, one per state.
Bosons have no such restriction — any number can pile into the same state.

Bose–Einstein distribution (photons)

For a mode that can hold any number of photons of energy $E$ , the Boltzmann weights give occupation probability $p_n \propto e^{-nE/k_B T}$ . Summing the geometric series, the expected occupation is

\langle n\rangle = \frac{1}{e^{E/k_B T} - 1}.

Applied to the electromagnetic modes of a box (two polarizations per mode), this yields the total energy

U = \frac{L^3 \hbar}{\pi^2 c^3}\int_0^\infty \frac{\omega^3\,d\omega}{e^{\hbar\omega/k_B T}-1},

whose integrand is Planck’s law for blackbody radiation. Substituting $x = \hbar\omega/k_B T$ and using $\int_0^\infty x^3/(e^x-1)\,dx = \pi^4/15$ gives the Stefan–Boltzmann form

U = \frac{\pi^2}{15}\,\frac{(k_B T)^4}{(\hbar c)^3}\,L^3.

This is the modern “quantum field theory” route: find the classical modes of a field, then let bosons (photons) occupy them. The analogous quantization of a displacement field gives phonons.

Fermi–Dirac distribution (electrons)

For fermions, each state is either empty or singly occupied. Introducing the chemical potential $\mu$ (the energy cost to add one fermion), the occupation probability is

\langle n\rangle = \frac{1}{e^{(E-\mu)/k_B T} + 1}.

At $T = 0$ this is a step function: every state below $\mu$ is filled, every state above is empty. The cutoff energy is the Fermi energy $E_F = \mu(T=0)$ .

Exanoke. We fill the lowest states with $N$ electrons in a box of volume $V$ . Find the energy $E_F$ of the highest occupied state.

The number of spatial states with energy at most $E$ was found above from the density of states,

N_{\text{spatial}}(E) = \frac{V}{6\pi^2}\left(\frac{2mE}{\hbar^2}\right)^{3/2}.

Each spatial state holds two electrons (spin up and spin down), so at $T=0$ all states up to $E_F$ are filled and

N = 2\,N_{\text{spatial}}(E_F) = \frac{V}{3\pi^2}\left(\frac{2mE_F}{\hbar^2}\right)^{3/2}.

Solve for $E_F$ . First isolate the bracket,

\left(\frac{2mE_F}{\hbar^2}\right)^{3/2} = \frac{3\pi^2 N}{V} \quad\Longrightarrow\quad \frac{2mE_F}{\hbar^2} = \left(\frac{3\pi^2 N}{V}\right)^{2/3},

so that

E_F = \frac{\hbar^2}{2m_e}\left(\frac{3\pi^2 N}{V}\right)^{2/3} = \frac{h^2}{2m_e}\left(\frac{3N}{8\pi V}\right)^{2/3},

where the second form just uses $\hbar = h/2\pi$ to absorb the $\pi$ ‘s. The Fermi energy depends only on the number density $N/V$ , not on $N$ and $V$ separately — it is an intensive property of the gas.

Degeneracy pressure

Because fermions are forced into ever-higher momentum states, a cold Fermi gas exerts pressure even at $T=0$ . For $N$ fermions in volume $V$ the ground-state energy scales as $E \sim N^{5/3}/(mV^{2/3})$ , so

P = -\frac{\partial E}{\partial V} \propto \frac{\hbar^2}{m}\,n^{5/3}, \qquad n = N/V.

This degeneracy pressure is what supports compact objects like white dwarfs and neutron stars against gravity.

Zero-point energy and the Casimir effect

Each standing-wave mode of frequency $\omega_n$ carries a zero-point energy $\tfrac{1}{2}\hbar\omega_n$ . Summing over all modes between two plates (or pins on a string) naively diverges, but the difference between the plated configuration and empty space is finite. Regulating the sums with an exponential cutoff and using the famous

1 + 2 + 3 + \cdots \;\to\; -\frac{1}{12}

(which physically means: the regulated sum minus the corresponding integral is $-1/12$ , independent of the regulator) yields an attractive Casimir force. For a string of wave speed $v$ and pin separation $L$ ,

F = \frac{\pi\hbar v}{24 L^2}.

For light ( $v = c$ ) between conductors this has been measured precisely and confirmed.

Nuclear notation and decay

A nucleus is written $^A_Z X$ , where $X$ is the element, $A$ is the mass number (protons + neutrons), and $Z$ is the atomic number (protons). Since $Z$ is fixed by $X$ , it is often omitted.

The common decay channels:

\begin{aligned} \text{alpha:}\quad & ^A_Z X \to\; ^{A-4}_{Z-2}X' + {}^4_2\text{He} \\[4pt] \beta^- \text{ decay:}\quad & ^A_Z X \to\; ^{A}_{Z+1}X' + e^- + \bar\nu_e \\[4pt] \beta^+ \text{ decay:}\quad & ^A_Z X \to\; ^{A}_{Z-1}X' + e^+ + \nu_e \\[4pt] \text{electron capture:}\quad & ^A_Z X + e^- \to\; ^{A}_{Z-1}X' + \nu_e \\[4pt] \text{gamma:}\quad & ^A_Z X^* \to\; ^A_Z X + \gamma \end{aligned}

Conservation laws

Which decays are allowed is governed by three conserved quantities. In the nuclear setting:

\begin{aligned} \text{baryon number} &= (\text{protons}) + (\text{neutrons}) \\ \text{electric charge} &= (\text{protons} + \text{positrons}) - (\text{electrons}) \\ \text{electron number} &= (\text{electrons} + \nu_e) - (\text{positrons} + \bar\nu_e) \end{aligned}

These are what tell you the identity of the missing particle in a reaction. The (anti)neutrinos exist precisely to balance electron number in beta decay.

Energetics

The energy released equals the drop in rest-mass energy,

\Delta E = (\Delta m)\,c^2,

and a decay can occur spontaneously only if it lowers the total energy of the nucleus. At the level of individual nucleons,

n \to p + e^- + \bar\nu_e, \qquad p \to n + e^+ + \nu_e,

and either can be favorable inside a nucleus depending on its composition. But a free proton never decays, because the proton is lighter than the neutron. Useful energy scales:

Process	Scale
$k_B T$ at room temperature	$\sim 10^{-2}\,\text{eV}$
chemical bonds	$\sim 0.1\text{–}1\,\text{eV}$
$k_B T$ at the Sun’s core	$\sim 10^{3}\,\text{eV}$
electron rest energy	$\sim 0.5\,\text{MeV}$
energy released in nuclear reactions	$\sim 0.1\text{–}100\,\text{MeV}$
nucleon rest energy	$\sim 1\,\text{GeV}$

Since nuclear energies dwarf chemical ones, decay rates are usually insensitive to the chemical environment. (Rare exceptions exist, like electron-capture isotopes such as $^7$ Be.)

The radioactive decay law

Theorem (Radioactive decay law). Radioactive decay is memoryless: in any interval $dt$ a nucleus has probability $\lambda\,dt$ of decaying, regardless of history. The survival probability decays exponentially,

p(t) = e^{-\lambda t}, \qquad \tau = \frac{1}{\lambda} \;(\text{mean lifetime}).

With $N_0 \gg 1$ nuclei initially, the number remaining and the activity (decay rate) are

N(t) = N_0 e^{-\lambda t}, \qquad A(t) = \left|\frac{dN}{dt}\right| = \lambda N_0 e^{-\lambda t}.

The half-life is $t_{1/2} = \tau\ln 2$ . Note that isotopes rarely decay in isolation: long decay chains mean a short-lived isotope is continually replenished by its longer-lived parents, reaching a steady state (secular equilibrium).

Tunneling and alpha decay

An alpha particle is held in the nucleus by the short-range strong force but must escape through a Coulomb barrier. WKB describes this: in the classically forbidden region $p$ is imaginary, so the wavefunction picks up an imaginary phase

\theta = \frac{1}{\hbar}\int p\,dx,

meaning it exponentially decays across the barrier. The escape probability per collision is $\lvert e^{i\theta} \rvert^2$ , and the decay timescale comes out as

\tau \sim \tau_{\text{collision}}\, e^{+2\lvert\mathrm{Im}\,\theta\rvert}, \qquad \text{with tunneling probability} \;\sim e^{-\sqrt{E_g/E}}.

The strong exponential dependence of lifetime on energy (the Geiger–Nuttall law) is the key feature, and matches experiment. Run in reverse, the same barrier governs fusion in stars: averaging the tunneling rate $e^{-\sqrt{E_g/E}}$ over a Boltzmann distribution $e^{-E/k_B T}$ gives a sharply peaked Gamow window of optimal energies.

Nuclear processes

Critical mass

A neutron-induced chain reaction runs away when each fission triggers, on average, more than one further fission. The probability a neutron collides before escaping a sample of radius $r$ is $p \sim n\sigma r$ , so criticality occurs at fixed $r_{\text{crit}} \propto 1/n \propto 1/\rho$ . The critical mass is then

m_{\text{crit}} \propto \rho\, r_{\text{crit}}^3 \propto \frac{1}{\rho^2}.

Compressing the material lowers the critical mass — the principle behind implosion-type weapons.

The liquid-drop model

Model the nucleus as an incompressible drop of uniform density, so volume $\propto A$ , surface area $\propto A^{2/3}$ , and radius $\propto A^{1/3}$ . The binding energy has competing contributions:

E_B = a_V A - a_S A^{2/3} - a_C \frac{Z^2}{A^{1/3}} - \cdots

Volume term $+a_V A$ : the strong force is short-ranged, so each nucleon binds only to its neighbors — energy proportional to the number of nucleons.
Surface term $-a_S A^{2/3}$ : nucleons at the surface have fewer neighbors, costing energy proportional to surface area.
Coulomb term $-a_C Z^2/A^{1/3}$ : the long-ranged electromagnetic repulsion has every proton push on every other, scaling as $Z^2$ over the radius.

Further terms (asymmetry, pairing) require quantum mechanics to motivate. This model explains the peak of the binding-energy-per-nucleon curve near iron, and hence why both fusion of light nuclei and fission of heavy nuclei release energy.

Fusion in stars

The Sun runs on the proton–proton chain, net $4\,{}^1\text{H} \to {}^4\text{He} + 2e^+ + 2\nu_e + \gamma$ . Heavier stars also use the CNO cycle, in which carbon acts as a catalyst:

^{12}\text{C} \xrightarrow{p} {}^{13}\text{N} \xrightarrow{\beta^+} {}^{13}\text{C} \xrightarrow{p} {}^{14}\text{N} \xrightarrow{p} {}^{15}\text{O} \xrightarrow{\beta^+} {}^{15}\text{N} \xrightarrow{p} {}^{12}\text{C} + {}^4\text{He}.

The $^{12}$ C is regenerated, so the net reaction is again four protons fusing into one helium nucleus.

Basic particle physics

The Standard Model fundamental particles, worth knowing roughly:

Quarks (six flavors: up, down, charm, strange, top, bottom) — charges $+\tfrac{2}{3}$ or $-\tfrac{1}{3}$ ; they feel the strong force and combine into protons, neutrons, and other hadrons.
Leptons: the electron, muon, tau (charge $-1$ ) and their neutrinos (neutral); they do not feel the strong force.
Gauge bosons: photon (electromagnetism), $W^\pm$ and $Z$ (weak force), gluons (strong force).
Higgs boson: gives mass to the others.

Almost everything in the everyday world is made of up quarks, down quarks, and electrons. Quarks and gluons feel the strong interaction; all the fermions feel the weak interaction. The exotic particles (muons, neutrinos, pions, quarks, vector bosons) play no role in chemistry because they are either too short-lived, too weakly interacting, or confined inside nucleons — chemistry only sees the stable electrons and the nuclear charge.

A forbidden process

A free electron cannot absorb a single photon, $e^- + \gamma \to e^-$ : energy and momentum conservation cannot be satisfied simultaneously (in the electron’s rest frame the photon brings momentum but the final electron would need energy without the right momentum). By time reversal, a free electron cannot emit a single photon either. Atoms can absorb photons because the recoil is taken up by the rest of the atom (or nucleus); free electrons interact with light only through processes like Thomson scattering, $e^- + \gamma \to e^- + \gamma$ .

Atomic physics

Beyond the Bohr model, most quantitative atomic physics needs full quantum mechanics — but given the energy levels, transitions are straightforward.

An electron falling from level $E_1$ to $E_0$ emits a photon of angular frequency

\omega = \frac{E_1 - E_0}{\hbar}.

Because levels are discrete, the emitted light has a sharply peaked spectrum, and each element’s characteristic spectral lines identify it. Conversely, an atom can absorb a photon to climb from $E_0$ to $E_1$ ; a field also drives stimulated emission from $E_1$ down to $E_0$ (the basis of lasers). If a photon has more than enough energy, it can eject the electron entirely — the photoelectric effect, with final kinetic energy $\hbar\omega - E$ (where $-E$ was the binding energy).

Spectral line broadening

Real lines have nonzero width, from two main effects:

Lifetime (natural) broadening. An excited state living for time $\Delta t$ has an energy spread $\Delta E \sim \hbar/\Delta t$ by the energy–time uncertainty principle, giving a wavelength spread

\frac{\Delta\lambda}{\lambda} \sim \frac{\Delta E}{E} \sim \frac{\hbar}{E\,\Delta t}.

Doppler broadening. Thermal motion at temperature $T$ shifts wavelengths by the Doppler effect. With typical speed $v \sim \sqrt{k_B T/m}$ ,

\frac{\Delta\lambda}{\lambda} \sim \frac{v}{c} \sim \frac{1}{c}\sqrt{\frac{k_B T}{m}}.

The Sun’s spectrum is the reverse situation: a hot continuous (blackbody) source seen through cooler gas, which absorbs at its characteristic wavelengths, producing dark absorption lines instead of bright emission lines.

Problem-solving strategy

Modern-physics problems usually reduce to “which quantization or conservation rule applies.” Match the situation to the right tool:

Problem-solving strategy

A particle is trapped and you want its energy levels: apply the WKB / Bohr–Sommerfeld rule $\oint p\,dx = \left(n+\tfrac{\alpha}{2\pi}\right)h$ . Count turning points to fix $\alpha$ — $\pi$ per hard wall, $\pi/2$ per soft turning point — then do the loop integral (box gives $E_n = n^2h^2/8mL^2$ , oscillator gives $E_n = \hbar\omega_0(n+\tfrac12)$ ).
Rotational / circular-orbit quantization: use Bohr’s condition $L = n\hbar$ . For hydrogen, pair it with the Coulomb-centripetal balance $mv^2/r = e^2/4\pi\epsilon_0 r^2$ to get $r_n = a_0 n^2$ and $E_n = -13.6\,\text{eV}/n^2$ ; transitions emit $\hbar\omega = E_i - E_f$ .
An order-of-magnitude estimate of a ground-state energy or size: use the uncertainty principle $\Delta x\,\Delta p \ge \hbar/2$ , set $\Delta p \sim \hbar/\Delta x$ , write $E(\Delta x)$ and minimize. The same trick gives spectral linewidths and lifetimes via $\Delta E\,\Delta t \ge \hbar/2$ .
Counting states / thermodynamics of a gas: go to momentum space, use the density of states $N = \frac{V}{6\pi^2}(2mE/\hbar^2)^{3/2}$ , then occupy modes with Bose–Einstein $\langle n\rangle = (e^{E/k_BT}-1)^{-1}$ (photons, blackbody) or Fermi–Dirac $\langle n\rangle = (e^{(E-\mu)/k_BT}+1)^{-1}$ (electrons, $E_F$ , degeneracy pressure).
Identifying the missing particle in a decay: balance baryon number, electric charge, and electron number simultaneously. Then check it is energetically allowed via $\Delta E = (\Delta m)c^2 > 0$ .
How much is left / how fast it decays: the decay law $N(t) = N_0 e^{-\lambda t}$ , with $\tau = 1/\lambda$ and $t_{1/2} = \tau\ln 2$ ; activity is $A = \lambda N$ .
Barrier penetration (alpha decay, fusion): WKB tunneling, $\tau \sim \tau_{\text{coll}}\,e^{+2\lvert\mathrm{Im}\,\theta\rvert}$ with probability $\sim e^{-\sqrt{E_g/E}}$ — note the steep dependence of lifetime on energy.
Nuclear stability / energy release: the liquid-drop binding energy $E_B = a_V A - a_S A^{2/3} - a_C Z^2/A^{1/3} - \cdots$ ; the peak near iron explains why both fusion (light) and fission (heavy) release energy.

Common mistakes

The photoelectric kinetic energy is $\hbar\omega - E_{\text{bind}}$ , not the full photon energy — the work function (binding energy) must be subtracted, and below threshold no electron escapes regardless of intensity.
Don’t conflate energy and momentum for a photon: $E = \hbar\omega = pc$ , so $p = E/c$ — never $p^2/2m$ .
For the particle in a box $n$ starts at $1$ (an $n=0$ “state at rest” violates uncertainty), while the WKB integer starts at $0$ ; keep the $\alpha/2\pi$ offset straight.
“Energy is borrowed for time $\Delta t$ ” from $\Delta E\,\Delta t \ge \hbar/2$ is a heuristic, not real non-conservation — use it only for estimates.
A free electron cannot absorb or emit a single photon ( $e^-+\gamma\to e^-$ is forbidden); energy and momentum cannot both be conserved without a recoiling atom or nucleus.
A free proton does not decay (it is lighter than the neutron); $n\to p$ and $p\to n$ channels are only favorable inside a nucleus, set by the rest-mass balance.
Hydrogen-like systems need the reduced mass: positronium uses $\mu = m_e/2$ , halving the binding energies.