The Birth of Quantum Mechanics: How Black-Body Radiation, the Photoelectric Effect and Matter Waves Broke the Classical Picture

0. Key points

• The spectrum of cavity radiation can never be obtained from classical electromagnetism together with the equipartition theorem. Combining the two gives $u(\nu,T) \propto \nu^2 T$, and the total energy density diverges (the ultraviolet catastrophe).
• Planck assumed that the energy of each mode is restricted to the discrete values $E_n = nh\nu$, and thereby derived $u(\nu,T) = \dfrac{8\pi h\nu^3}{c^3}\dfrac{1}{e^{h\nu/k_BT}-1}$, in complete agreement with experiment. Both the Stefan-Boltzmann law and Wien’s displacement law follow from this one formula.
• The four experimental facts about the photoelectric effect (independence of intensity, a threshold frequency, the straight line $K_{\max} = h\nu - W$, and instantaneity) are all explained at once if light is absorbed as lumps of energy $E = h\nu$. On the classical wave picture, emission should begin only after nearly an hour.
• De Broglie read the relation $p = h/\lambda$ for light backwards and assigned a wavelength $\lambda = h/p$ to every particle. The Davisson-Germer electron diffraction experiment confirms this to about one percent.
• Bohr’s quantization condition $L = n\hbar$ is exactly the condition that a de Broglie standing wave fit around a circular orbit. At that point what is missing is the equation the wave obeys.

1. Motivation: the two clouds left at the end of the nineteenth century

In 1900 Lord Kelvin gave a lecture entitled “Nineteenth-century clouds over the dynamical theory of heat and light”. The first cloud was the motion of the Earth through the ether (the Michelson-Morley experiment); the second was the failure of the equipartition theorem to give the correct specific heats of gases. The first led to special relativity, the second to quantum theory.

The tools physicists held at that moment were Newtonian mechanics, Maxwell’s electromagnetism and Boltzmann’s statistical mechanics. Together these three covered an astonishing range, from the motion of celestial bodies to the propagation of radio waves to the equation of state of a gas. And yet they could not answer a single one of the following three questions.

1. Why does a heated body glow with the colour it does? Where does the shape of that spectrum come from, the one that shifts from red to white as the temperature rises?
2. Why do electrons fly out of a metal illuminated by ultraviolet light, while no matter how intense the red light, not one electron emerges?
3. Why does the atom not collapse? According to Maxwell’s equations, an electron in circular motion radiates electromagnetic waves, loses energy, and should fall into the nucleus in $10^{-11}$ seconds.

This article follows the logic by which quantum theory arose, taking the first and second questions as its starting points. A partial answer to the third (Bohr’s quantization condition) is obtained in Proposition 6.5 as a consequence of de Broglie’s matter waves.

It is worth emphasizing that quantum theory was not introduced because classical physics fitted the data only approximately. Classical physics gave an answer that was wrong even qualitatively. The energy density of cavity radiation did not deviate from a finite value; it came out infinite. This qualitative breakdown, finite against infinite, is what forced a change in the underlying assumptions.

flowchart TD
A["Measured cavity-radiation spectrum (1890s)"] --> B["Classical electromagnetism + equipartition"]
B --> C["Rayleigh-Jeans law u ∝ ν²T"]
C --> D["Divergence at high frequency (ultraviolet catastrophe)"]
D --> E["Planck's quantum hypothesis E = nhν (1900)"]
E --> F["Planck's radiation law: exact agreement with experiment"]
F --> G["Photoelectric effect → light-quantum hypothesis E = hν (1905)"]
G --> H["De Broglie's matter waves λ = h/p (1924)"]
H --> I["Confirmed by electron diffraction (1927) → a wave equation is needed"]

The chain of reasoning followed in this article

2. Preliminaries: black bodies and cavity radiation

Definition 2.1（Black body）

A body that absorbs completely all incident electromagnetic radiation of every frequency, reflecting and transmitting none of it, is called a black body.

A black body is an idealization, but a good approximation can be built in the laboratory. Bore a small hole in a cavity whose walls are held at a fixed temperature $T$: light entering the hole is almost certain to be absorbed inside before it can escape. The hole therefore behaves as a black body, and conversely the radiation leaking out of the hole reproduces exactly the spectrum of the electromagnetic field that is in thermal equilibrium with the walls inside. This is called cavity radiation, or black-body radiation.

As Kirchhoff showed in 1859, the spectrum of the radiation in a cavity in thermal equilibrium depends neither on the material of the walls nor on the shape of the cavity, but only on the temperature. This universality matters. Were the spectrum material-dependent, it would be a property of a particular metal and nothing more; being a universal function, its shape must be fixed not by the details of matter but by general laws governing the electromagnetic field and thermal equilibrium themselves. That is precisely why a failure to explain its shape is fatal.

Definition 2.2（Spectral energy density）

Inside a cavity in thermal equilibrium at temperature $T$, write $u(\nu,T)\,d\nu$ for the energy per unit volume carried by the electromagnetic field with frequency between $\nu$ and $\nu + d\nu$. The function $u(\nu,T)$ is called the spectral energy density. The total energy density is $U(T) = \int_0^\infty u(\nu,T)\,d\nu$.

The classical computation proceeds in two stages. First one counts how many degrees of freedom (modes) there are near a frequency $\nu$; then one determines, by statistical mechanics, how much energy a single mode carries on average. The first stage is purely a matter of geometry and wave theory, and it survives unchanged in quantum theory.

Proposition 2.3（Mode density of a cavity）

Inside a cubical cavity of side $L$ with perfectly conducting walls, let $N(\nu)$ denote the number of eigenmodes (standing waves) of the electromagnetic field with frequency at most $\nu$. In the limit $L\nu/c \gg 1$,

$$N(\nu) = \frac{8\pi L^3 \nu^3}{3c^3}$$

and hence the mode density per unit volume at frequency $\nu$ is

$$g(\nu) = \frac{1}{L^3}\frac{dN}{d\nu} = \frac{8\pi \nu^2}{c^3},$$

where $c$ is the speed of light.

Proof（Proposition 2.3）

At a perfectly conducting wall the tangential component of the electric field vanishes. The standing waves in the cube satisfying this boundary condition are labelled by triples of positive integers $(n_x, n_y, n_z)$, with wave vector

$$\boldsymbol{k} = \frac{\pi}{L}(n_x, n_y, n_z), \qquad n_x, n_y, n_z \in \mathbb{N}.$$

The vacuum dispersion relation $\nu = c|\boldsymbol{k}|/(2\pi)$ gives

$$\nu = \frac{c}{2L}\sqrt{n_x^2 + n_y^2 + n_z^2}.$$

The condition that the frequency be at most $\nu$ is therefore equivalent to the lattice point $(n_x,n_y,n_z)$ lying inside the sphere of radius $R = 2L\nu/c$.

Each lattice point corresponds to one unit cube, so for $R \gg 1$ the number of admissible points is approximated by the volume of the octant with $n_x, n_y, n_z > 0$ (the error is of the order of the surface area, $O(R^2)$, which is $O(1/R)$ relative to the leading term $O(R^3)$). Moreover each wave vector carries two independent polarizations perpendicular to $\boldsymbol{k}$. Hence

$$N(\nu) = 2 \cdot \frac{1}{8}\cdot \frac{4\pi}{3}R^3 = \frac{\pi}{3}\left(\frac{2L\nu}{c}\right)^3 = \frac{8\pi L^3\nu^3}{3c^3}.$$

Differentiating with respect to $\nu$ and dividing by $L^3$ gives $g(\nu) = 8\pi\nu^2/c^3$. That this result does not depend on the shape of the cavity is checked in the Appendix.

∎

That $g(\nu)$ grows like $\nu^2$ is the key to everything that follows. The higher the frequency, the overwhelmingly more numerous the modes. Whether the mean energy per mode can hold this abundance in check is what separates the finite from the infinite.

3. The classical prediction and the ultraviolet catastrophe

Proposition 3.1（The Rayleigh-Jeans law and the divergence of the total energy）

Assume that each mode of the electromagnetic field in the cavity behaves as a classical harmonic oscillator and obeys the equipartition theorem in thermal equilibrium at temperature $T$. Then

$$u_{\mathrm{RJ}}(\nu,T) = \frac{8\pi\nu^2}{c^3}k_B T,$$

where $k_B$ is Boltzmann’s constant. For every $T > 0$ the total energy density then diverges:

$$U(T) = \int_0^\infty u_{\mathrm{RJ}}(\nu,T)\,d\nu = \infty.$$

Proof（Proposition 3.1）

Expanding the field in the cavity in its eigenmodes, the amplitude $q$ of each mode obeys the equation of motion of a harmonic oscillator of angular frequency $\omega = 2\pi\nu$ (general solution of the harmonic oscillator（Theorem 5.2）［Foundations of Newtonian Mechanics］), and its energy $E = \frac{1}{2}\dot q^2 + \frac{1}{2}\omega^2 q^2$ is a sum of quadratic forms in $\dot q$ and $q$. The classical equipartition theorem states that each quadratic term in the Hamiltonian carries a mean energy $\frac{1}{2}k_BT$, so the mean energy per mode is

$$\langle E\rangle_{\text{classical}} = \frac{1}{2}k_BT + \frac{1}{2}k_BT = k_BT,$$

independently of $\nu$. Multiplying by the mode density of Proposition 2.3 gives

$$u_{\mathrm{RJ}}(\nu,T) = g(\nu)\,\langle E\rangle_{\text{classical}} = \frac{8\pi\nu^2}{c^3}k_BT.$$

As for the integral, for every $\Omega > 0$

$$\int_0^{\Omega} \frac{8\pi\nu^2}{c^3}k_BT\,d\nu = \frac{8\pi k_BT}{3c^3}\Omega^3 \xrightarrow[\Omega\to\infty]{} \infty ,$$

and the integrand is non-negative, so the improper integral diverges to $+\infty$.

∎

This conclusion is physically unacceptable. A cavity in a room at ordinary temperature is in equilibrium with walls of finite heat capacity, and so cannot store an infinite amount of energy. Moreover the divergence comes from the high-frequency side. In this sense the name ultraviolet catastrophe (Ultraviolettkatastrophe), coined by Ehrenfest in 1911, is apt.

Planck's radiation law against the Rayleigh-Jeans law. The horizontal axis is the dimensionless frequency x = hν/(k_B T); the vertical axis is x³/(eˣ−1) (Planck) and x² (Rayleigh-Jeans)

Remark 3.2

Textbooks tend to say that Planck proposed the quantum hypothesis in order to resolve the ultraviolet catastrophe, but the historical record is more involved. Rayleigh wrote down $u \propto \nu^2 T$ in June 1900, and the complete form with the correct coefficient had to wait for Jeans’s correction of 1905. When Planck wrote his radiation formula in October 1900, what he was matching against was Wien’s law $u \propto \nu^3 e^{-a\nu/T}$, an empirical formula that works well at high frequencies, together with the newly discovered fact, found by Rubens and Kurlbaum in the far infrared, that $u$ is proportional to $T$. Planck interpolated the second derivative of the entropy between these two limits, and only afterwards, while trying to give the resulting formula a statistical-mechanical foundation, did he reluctantly introduce the energy element $\varepsilon = h\nu$. This history is set out in detail in T. S. Kuhn’s study Black-Body Theory and the Quantum Discontinuity, 1894-1912 (Oxford University Press, 1978).

4. Planck’s quantum hypothesis and the radiation law

Axiom 4.1（Planck's quantum hypothesis）

The energies available to a mode of the electromagnetic field of frequency $\nu$ (or to a wall oscillator in equilibrium with it) are not continuous but are restricted to

$$E_n = n h \nu, \qquad n = 0, 1, 2, \ldots$$

Here $h$ is a universal constant independent of both $\nu$ and $T$, called Planck’s constant. In the present SI it is a defined value, $h = 6.62607015 \times 10^{-34}\ \mathrm{J\,s}$.

Before the calculation, here is why this assumption cures the divergence. Equipartition says that every mode receives its equal share $k_BT$; under the quantum hypothesis, however, putting any energy at all into a mode requires paying an entrance fee of at least $h\nu$. In a mode with $h\nu \gg k_BT$, thermal fluctuations cannot afford this fee, and the mode is effectively frozen out. The occupation probability falls exponentially and overwhelms the number of modes, which grows only as $\nu^2$. That is what separates the finite from the infinite.

Theorem 4.2（Planck's radiation law）

Under Axiom 4.1, if a mode of frequency $\nu$ follows the canonical distribution $p_n \propto e^{-E_n/k_BT}$ at temperature $T$, then the mean energy of that mode is

$$\langle E\rangle_\nu = \frac{h\nu}{e^{h\nu/k_BT}-1}$$

and the spectral energy density is

$$u(\nu,T) = \frac{8\pi h \nu^3}{c^3}\,\frac{1}{e^{h\nu/k_BT}-1}.$$

Proof（Theorem 4.2）

Put $\beta = 1/(k_BT)$ and $z = e^{-\beta h\nu}$. Since $\nu > 0$ and $T > 0$ we have $0 < z < 1$. The partition function is a geometric series, so it has the closed form

$$Z = \sum_{n=0}^{\infty} e^{-\beta n h\nu} = \sum_{n=0}^{\infty} z^n = \frac{1}{1-z}.$$

The mean energy is

$$\langle E\rangle_\nu = \frac{\sum_{n=0}^\infty (nh\nu)z^n}{\sum_{n=0}^\infty z^n}.$$

The series in the numerator is obtained by differentiating $\sum_{n\ge 0} z^n = (1-z)^{-1}$ with respect to $z$ and multiplying by $z$:

$$\sum_{n=0}^\infty n z^n = z\frac{d}{dz}\frac{1}{1-z} = \frac{z}{(1-z)^2}$$

(term-by-term differentiation is legitimate since $|z| < 1$). Hence

$$\langle E\rangle_\nu = h\nu \cdot \frac{z/(1-z)^2}{1/(1-z)} = h\nu\,\frac{z}{1-z} = \frac{h\nu}{z^{-1}-1} = \frac{h\nu}{e^{h\nu/k_BT}-1}.$$

It remains to multiply by the mode density $g(\nu) = 8\pi\nu^2/c^3$ of Proposition 2.3:

$$u(\nu,T) = \frac{8\pi\nu^2}{c^3}\cdot\frac{h\nu}{e^{h\nu/k_BT}-1} = \frac{8\pi h\nu^3}{c^3}\frac{1}{e^{h\nu/k_BT}-1}.$$

The only difference from Proposition 3.1 is that $\langle E\rangle$ has changed from the constant $k_BT$ into a function of $\nu$.

∎

Corollary 4.3（The two limits）

The Planck distribution of Theorem 4.2 has the following two limits.

1. Low-frequency limit. For $h\nu \ll k_BT$, $u(\nu,T) = \dfrac{8\pi\nu^2}{c^3}k_BT\left(1 - \dfrac{h\nu}{2k_BT} + O\!\left((h\nu/k_BT)^2\right)\right)$. The leading term is the Rayleigh-Jeans law Proposition 3.1.
2. High-frequency limit. For $h\nu \gg k_BT$, $u(\nu,T) = \dfrac{8\pi h\nu^3}{c^3}e^{-h\nu/k_BT}\left(1 + O\!\left(e^{-h\nu/k_BT}\right)\right)$. This is the form of Wien’s empirical law.

Proof（Corollary 4.3）

Put $x = h\nu/(k_BT)$.

(1) As $x \to 0$, the Taylor expansion $e^x - 1 = x + \frac{x^2}{2} + O(x^3) = x\left(1 + \frac{x}{2}+O(x^2)\right)$ gives

$$\frac{h\nu}{e^x-1} = \frac{k_BT\,x}{x\left(1+\frac{x}{2}+O(x^2)\right)} = k_BT\left(1 - \frac{x}{2} + O(x^2)\right)$$

(the last equality uses $(1+u)^{-1} = 1-u+O(u^2)$). Multiplying by $g(\nu)$ gives the stated form.

(2) As $x \to \infty$ we have $e^{-x} \to 0$, so

$$\frac{1}{e^x - 1} = \frac{e^{-x}}{1-e^{-x}} = e^{-x}\left(1 + e^{-x} + e^{-2x}+\cdots\right) = e^{-x}\left(1+O(e^{-x})\right).$$

Multiplying by $8\pi h\nu^3/c^3$ completes the proof.

∎

Planck’s formula thus unifies in a single expression the two limiting behaviours that had been established experimentally. And it agreed with experiment throughout the intermediate region as well. That was decisive.

Corollary 4.4（The Stefan-Boltzmann law）

Under Theorem 4.2 the total energy density is finite, namely

$$U(T) = \int_0^\infty u(\nu,T)\,d\nu = \frac{8\pi^5 k_B^4}{15c^3h^3}T^4 .$$

Consequently the radiant exitance of a black body per unit area is $M = \dfrac{c}{4}U = \sigma T^4$ with $\sigma = \dfrac{2\pi^5k_B^4}{15c^2h^3} = 5.670\times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$.

Proof（Corollary 4.4）

Substitute $x = h\nu/(k_BT)$. Since $\nu = k_BTx/h$ and $d\nu = (k_BT/h)dx$,

$$U(T) = \frac{8\pi h}{c^3}\int_0^\infty \frac{\nu^3\,d\nu}{e^{h\nu/k_BT}-1} = \frac{8\pi h}{c^3}\left(\frac{k_BT}{h}\right)^4\int_0^\infty \frac{x^3}{e^x-1}\,dx.$$

We evaluate the remaining integral. For $x > 0$ we have $\dfrac{1}{e^x-1} = \dfrac{e^{-x}}{1-e^{-x}} = \sum_{n=1}^\infty e^{-nx}$, and every term of the integrand is non-negative, so the monotone convergence theorem permits term-by-term integration. Using $\int_0^\infty x^3 e^{-nx}dx = 3!/n^4 = 6/n^4$,

$$\int_0^\infty \frac{x^3}{e^x-1}dx = \sum_{n=1}^\infty \frac{6}{n^4} = 6\,\zeta(4) = 6\cdot\frac{\pi^4}{90} = \frac{\pi^4}{15}$$

(the value $\zeta(4) = \pi^4/90$ is Euler’s). Therefore

$$U(T) = \frac{8\pi h}{c^3}\cdot\frac{k_B^4T^4}{h^4}\cdot\frac{\pi^4}{15} = \frac{8\pi^5k_B^4}{15c^3h^3}T^4.$$

That the flux escaping through a hole of area $A$ from an isotropic radiation field is $\frac{c}{4}U$ follows from the solid-angle average $\langle c\cos\theta\rangle_{\text{hemisphere}} = c/4$. Inserting numbers,

$$\sigma = \frac{2\pi^5 (1.380649\times10^{-23})^4}{15(2.99792\times10^8)^2(6.62607\times10^{-34})^3} = 5.670\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}},$$

which agrees with the value Stefan obtained from experiment in 1879. It was precisely this agreement, together with Wien’s displacement law below, that first allowed Planck to determine numerical values for $h$ and $k_B$.

∎

Proposition 4.5（Wien's displacement law）

The spectral energy density per unit wavelength $u_\lambda(\lambda,T)$ (defined by $u_\lambda\,d\lambda = -u_\nu\,d\nu$) is

$$u_\lambda(\lambda,T) = \frac{8\pi hc}{\lambda^5}\frac{1}{e^{hc/\lambda k_BT}-1},$$

and the wavelength $\lambda_{\max}$ maximizing it satisfies

$$\lambda_{\max} T = \frac{hc}{x_0 k_B} = 2.898\times10^{-3}\ \mathrm{m\,K},$$

where $x_0 \approx 4.9651$ is the unique positive root of the equation $x = 5(1-e^{-x})$.

Proof（Proposition 4.5）

From $\nu = c/\lambda$ we get $|d\nu/d\lambda| = c/\lambda^2$, so $u_\lambda = u_\nu \cdot c/\lambda^2$. Substituting $\nu = c/\lambda$ into $u_\nu$ from Theorem 4.2,

$$u_\lambda = \frac{8\pi h (c/\lambda)^3}{c^3}\frac{1}{e^{hc/\lambda k_BT}-1}\cdot\frac{c}{\lambda^2} = \frac{8\pi hc}{\lambda^5}\frac{1}{e^{hc/\lambda k_BT}-1}.$$

Now put $x = hc/(\lambda k_BT)$. For fixed $T$ the map $\lambda \mapsto x$ is a strictly decreasing bijection of $(0,\infty)$ onto itself, so maximizing $u_\lambda$ over $\lambda$ is equivalent to maximizing

$$F(x) = \frac{x^5}{e^x-1}$$

over $x$ (indeed $u_\lambda = \dfrac{8\pi (k_BT)^5}{h^4c^4}F(x)$, and the constant of proportionality does not depend on $x$). Since $F(x) > 0$ we may differentiate $\log F$:

$$\frac{F'(x)}{F(x)} = \frac{5}{x} - \frac{e^x}{e^x-1} = 0 \;\Longleftrightarrow\; 5(e^x-1) = xe^x \;\Longleftrightarrow\; x = 5(1-e^{-x}),$$

the last equivalence coming from dividing both sides by $e^x > 0$. Setting $G(x) = 5(1-e^{-x}) - x$, we have $G(0)=0$ and $G'(x) = 5e^{-x}-1$, so $G$ increases for $x < \log 5$ and decreases for $x > \log 5$, with $G(\log 5) = 4-\log 5 > 0$ and $G(x)\to-\infty$ as $x\to\infty$. Hence there is exactly one positive root. Substituting $x=4.9$ gives $5(1-e^{-4.9}) = 4.9628$, and $x = 4.9651$ gives $5(1-e^{-4.9651}) = 4.96511$, so the iteration converges to $x_0 = 4.96511$. Therefore

$$\lambda_{\max}T = \frac{hc}{x_0k_B} = \frac{(6.62607\times10^{-34})(2.99792\times10^{8})}{4.96511\times1.380649\times10^{-23}} = 2.8978\times10^{-3}\ \mathrm{m\,K}.$$∎

Note

The peak of $u_\nu$ and the peak of $u_\lambda$ do not correspond to each other under $\lambda = c/\nu$. The maximum of $u_\nu$ occurs at $x = h\nu/k_BT \approx 2.821$, the root of $x = 3(1-e^{-x})$. A density function picks up a Jacobian factor under a change of variable, and the peak moves accordingly. Whenever the “peak wavelength of a black body” is mentioned, check which of the two densities is meant.

Example 4.6（The cosmic microwave background）

The cosmic microwave background (CMB) is the most nearly ideal black-body spectrum ever measured. The FIRAS instrument aboard the COBE satellite found deviations from a Planck distribution at temperature $T = 2.7255\ \mathrm{K}$ smaller than $10^{-4}$. By Proposition 4.5,

$$\lambda_{\max} = \frac{2.8978\times10^{-3}\ \mathrm{m\,K}}{2.7255\ \mathrm{K}} = 1.063\times10^{-3}\ \mathrm{m} = 1.06\ \mathrm{mm}.$$

The peak seen in frequency is $\nu_{\max} = 2.821\,k_BT/h = 2.821\times(1.3806\times10^{-23})(2.7255)/(6.6261\times10^{-34}) = 1.602\times10^{11}\ \mathrm{Hz} = 160.2\ \mathrm{GHz}$. The two do not match, since $c/\nu_{\max} = 1.87\ \mathrm{mm} \ne 1.06\ \mathrm{mm}$, exactly as warned above. The total energy density follows from Corollary 4.4:

$$U = \frac{4\sigma}{c}T^4 = \frac{4(5.670\times10^{-8})(2.7255)^4}{2.998\times10^8} = 4.17\times10^{-14}\ \mathrm{J\,m^{-3}},$$

which in electronvolts is about $0.26\ \mathrm{MeV\,m^{-3}}$.

5. The photoelectric effect and the light-quantum hypothesis

Planck himself regarded the quantum hypothesis as a convenient computational device for the exchange of energy with the wall oscillators, and never claimed that the electromagnetic field itself is corpuscular. It was Einstein, in a paper of 1905, who took that step. The stage was the photoelectric effect.

Discovered by Hertz in 1887 and quantified by Lenard in 1902, the experimental facts about the photoelectric effect can be summarized in four points. The middle column of the table gives the prediction of the picture in which light is a classical continuous wave and an electron in the metal gradually accumulates energy from its electric field.

Experimental fact	Classical wave prediction	What is observed
Maximum kinetic energy of the emitted electrons versus intensity	Grows in proportion to the intensity (the square of the field amplitude)	Independent of intensity. Intensity changes only the number of electrons
Maximum kinetic energy versus frequency	No dependence on frequency	A linear function of frequency, with a slope independent of the metal
Existence of a threshold	Emission occurs at any frequency if one waits long enough	Below a threshold frequency there is no emission at all
Delay between illumination and emission	With weak light the accumulation takes a long time	Within $10^{-9}$ seconds

Let us make the fourth point quantitative. Suppose light of intensity $10^{-2}\ \mathrm{W\,m^{-2}}$ falls on the cross-section of a single atom, about $(10^{-10}\ \mathrm{m})^2 = 10^{-20}\ \mathrm{m^2}$; then one atom receives power $10^{-22}\ \mathrm{W}$. The energy needed to liberate an electron is typically a few electronvolts, say $2.28\ \mathrm{eV} = 3.65\times10^{-19}\ \mathrm{J}$ for sodium, so the accumulation time would be $3.65\times10^{-19}/10^{-22} \approx 3.7\times10^{3}$ seconds, about an hour. In experiment the current flows immediately. The discrepancy is not three or four orders of magnitude but twelve.

Axiom 5.1（Einstein's light-quantum hypothesis）

Monochromatic light of frequency $\nu$ is created and absorbed as a collection of spatially localized, independent lumps of energy

$$E = h\nu$$

(light quanta, later called photons). The intensity of the light is proportional to the number of photons per unit time and has no effect on the energy of a single photon.

This is a stronger claim than Axiom 4.1. Planck said that the wall oscillators exchange energy only in units of $h\nu$; Einstein said that the electromagnetic field travelling through free space is itself made of lumps of $h\nu$. Almost nobody accepted this at the time, since interference and diffraction of light were perfectly accounted for by the wave theory. Even Planck, in the document recommending Einstein for the Prussian Academy of Sciences in 1913, added an apologetic proviso that the light-quantum hypothesis alone went too far.

Definition 5.2（Work function）

The minimum energy required to remove an electron from the interior of a metal and bring it to rest outside is called the work function $W$ of that metal. Typical values are about $2.1\ \mathrm{eV}$ for caesium, $2.28\ \mathrm{eV}$ for sodium, $4.7\ \mathrm{eV}$ for copper and $5.6\ \mathrm{eV}$ for platinum.

Proposition 5.3（Einstein's photoelectric equation）

Let monochromatic light of frequency $\nu$ fall on a metal of work function $W$. Under Axiom 5.1 the following hold.

1. If $h\nu < W$, no electrons are emitted, however far the intensity of the light is increased.
2. If $h\nu \ge W$, electrons are emitted, and the maximum of their kinetic energy is

$$K_{\max} = h\nu - W .$$

Consequently the stopping voltage $V_s$ that just halts the emitted electrons satisfies $eV_s = h\nu - W$, and plotting $V_s$ against $\nu$ gives a straight line of slope $h/e$. This slope does not depend on the metal.

Proof（Proposition 5.3）

By Axiom 5.1, photons are absorbed one at a time, independently. Under ordinary conditions of low intensity, the probability that a single electron absorbs two or more photons at once is negligibly small (it scales as the square of the intensity or higher), so the energy an electron gains per absorption event is exactly $h\nu$.

To leave the metal, the electron must expend at least the energy $W$, by Definition 5.2. Conservation of energy therefore gives, for the kinetic energy after emission,

$$K = h\nu - (\text{energy actually spent on escaping}) \le h\nu - W .$$

Hence if $h\nu < W$ then $K$ would have to be negative, and no electron is emitted at all. This is (1). Increasing the number of photons leaves the energy of each photon at $h\nu$, so raising the intensity changes nothing.

For (2), equality is attained when the most weakly bound electron escapes without suffering any other energy loss (collisions with the lattice, for instance). The actual electron distribution includes levels below the Fermi level, so the kinetic energies of the emitted electrons form a continuous distribution from $0$ up to $K_{\max}$, whose upper end is $h\nu - W$.

As for the stopping voltage: applying a potential difference $V$ between the electrodes makes each electron lose the work $eV$, so the current stops completely once $eV \ge K_{\max}$. The smallest such value is $V_s$, whence $eV_s = K_{\max} = h\nu - W$, that is, $V_s = (h/e)\nu - W/e$. No information about the metal enters the slope $h/e$; only the intercept depends on the metal, through $W$.

∎

The prediction that the slope is independent of the material is an extremely sharp one. Millikan, who did not believe Einstein’s hypothesis, spent ten years on precision measurements and in 1916 confirmed the linear relation and obtained $h = 6.57\times10^{-34}\ \mathrm{J\,s}$. This agrees to within 0.5 % with the value Planck had extracted from the radiation law. The appearance of the same constant in a completely different phenomenon is what finally won acceptance for the light-quantum hypothesis.

Example 5.4（The photoelectric effect in sodium, worked out numerically）

Consider a sodium surface with $W = 2.28\ \mathrm{eV}$. The combination $hc = 1239.84\ \mathrm{eV\,nm}$ makes the arithmetic easy.

Threshold wavelength. From $h\nu_0 = W$,

$$\lambda_0 = \frac{hc}{W} = \frac{1239.84\ \mathrm{eV\,nm}}{2.28\ \mathrm{eV}} = 543.8\ \mathrm{nm}.$$

This is green light. Yellow, orange and red light therefore eject no electrons, however intense.

At $\lambda = 400\ \mathrm{nm}$ (violet). The photon energy is

$$h\nu = \frac{1239.84}{400} = 3.100\ \mathrm{eV},$$

so by Proposition 5.3

$$K_{\max} = 3.100 - 2.28 = 0.82\ \mathrm{eV}, \qquad V_s = 0.82\ \mathrm{V}.$$

The maximum speed is $v = \sqrt{2K_{\max}/m_e} = \sqrt{2(0.82)(1.602\times10^{-19})/(9.109\times10^{-31})} = 5.4\times10^{5}\ \mathrm{m\,s^{-1}}$, which is $0.2\ \%$ of the speed of light, so a non-relativistic treatment is ample.

At $\lambda = 300\ \mathrm{nm}$ (ultraviolet). Here $h\nu = 1239.84/300 = 4.133\ \mathrm{eV}$ and $K_{\max} = 1.85\ \mathrm{eV}$. Changing the wavelength from $400$ to $300\ \mathrm{nm}$ multiplies $K_{\max}$ by 2.3, whereas multiplying the intensity by 2.3 leaves $K_{\max}$ at $0.82\ \mathrm{eV}$. This asymmetry is the fingerprint of the light-quantum hypothesis.

Remark 5.5

That photons also carry momentum was shown by Compton in 1923, through the scattering of X-rays. Applying conservation of energy and momentum to a two-body collision between a photon and an electron at rest (treating the electron relativistically) yields, for scattering angle $\theta$, the wavelength shift

$$\Delta\lambda = \frac{h}{m_ec}(1-\cos\theta), \qquad \frac{h}{m_ec} = 2.426\ \mathrm{pm},$$

in agreement with experiment. The derivation lies off the main line of this article, so we omit it; a complete calculation is in Chapter 2 of the textbook by Eisberg and Resnick. What it uses is the relation obtained by combining $E = pc$ and $E = h\nu$ for a photon:

$$p = \frac{h\nu}{c} = \frac{h}{\lambda}.$$

The next section begins by reading this formula in the opposite direction.

6. De Broglie’s matter waves

Nineteenth-century physics divided the world into particles and waves. Up to this point we have seen that light, which ought to be a wave, has a corpuscular side. In his doctoral thesis of 1924, Louis de Broglie crossed the dividing line in the other direction as well, and his reason was symmetry: if nature imposes a double aspect on light, why should it not do the same for the electron?

Axiom 6.1（The de Broglie relations）

To every particle with momentum $\boldsymbol{p}$ and energy $E$ there is associated a wave of wavelength

$$\lambda = \frac{h}{|\boldsymbol{p}|}$$

and frequency $\nu = E/h$. In terms of the wave vector $\boldsymbol{k}$ (with $|\boldsymbol{k}| = 2\pi/\lambda$) and the angular frequency $\omega = 2\pi\nu$, writing $\hbar = h/2\pi$,

$$\boldsymbol{p} = \hbar \boldsymbol{k}, \qquad E = \hbar\omega .$$

This $\lambda$ is called the de Broglie wavelength.

Remark 6.2

Once we say that a wave corresponds to a particle, does the speed of that wave match the speed of the particle? For a relativistic free particle, $E^2 = p^2c^2 + m^2c^4$, so the phase velocity is

$$v_{\text{phase}} = \frac{\omega}{k} = \frac{E}{p} = \frac{\gamma mc^2}{\gamma mv} = \frac{c^2}{v} > c ,$$

which exceeds the speed of light. But the phase velocity carries no information. What does carry information is the group velocity of a wave packet:

$$v_{\text{group}} = \frac{d\omega}{dk} = \frac{dE}{dp} = \frac{d}{dp}\sqrt{p^2c^2+m^2c^4} = \frac{pc^2}{\sqrt{p^2c^2+m^2c^4}} = \frac{pc^2}{E} = v .$$

The wave packet thus travels at exactly the speed of the particle. De Broglie took this coincidence as evidence that his hypothesis pointed in the right direction. Building a wave packet requires superposing waves of different wavelengths, and from that arises a limit on the simultaneous determination of position and momentum. This is the origin of the uncertainty relation, treated in The Schrödinger Equation and the Wave Function and Operators and Observables (in particular Heisenberg's uncertainty principle（Corollary 5.4）［Operators and Observables］).

Example 6.3（The scale of de Broglie wavelengths）

For a non-relativistic particle $p = \sqrt{2mK}$, so $\lambda = h/\sqrt{2mK}$.

An electron accelerated through a potential difference $V$. From $K = eV$,

$$\lambda = \frac{h}{\sqrt{2m_e eV}} = \frac{6.626\times10^{-34}}{\sqrt{2(9.109\times10^{-31})(1.602\times10^{-19})V}} = \frac{1.226\ \mathrm{nm}}{\sqrt{V/\mathrm{V}}}.$$

For $V = 100\ \mathrm{V}$ this gives $\lambda = 0.123\ \mathrm{nm}$, the same order as the spacing between atoms. That is why a crystal can serve as a diffraction grating.

A neutron at room temperature. In thermal equilibrium $K = \frac{3}{2}k_BT$, so $p = \sqrt{3mk_BT}$. Putting $T = 300\ \mathrm{K}$ and $m_n = 1.675\times10^{-27}\ \mathrm{kg}$,

$$p = \sqrt{3(1.675\times10^{-27})(1.381\times10^{-23})(300)} = 4.56\times10^{-24}\ \mathrm{kg\,m\,s^{-1}},$$$$\lambda = \frac{6.626\times10^{-34}}{4.56\times10^{-24}} = 1.45\times10^{-10}\ \mathrm{m} = 0.145\ \mathrm{nm}.$$

This too is comparable to the spacing of crystal lattice planes, which is why neutron diffraction has become a standard method for determining the structure of matter.

A ball of mass $1\ \mathrm{g}$ moving at $1\ \mathrm{m\,s^{-1}}$.

$$\lambda = \frac{6.626\times10^{-34}}{(10^{-3})(1)} = 6.6\times10^{-31}\ \mathrm{m}.$$

This is sixteen orders of magnitude smaller than the radius of a proton, about $10^{-15}\ \mathrm{m}$, so no experiment could ever detect the diffraction. Classical mechanics holds in the macroscopic world because the de Broglie wavelength is preposterously small compared with the scale of the system.

Example 6.4（The Davisson-Germer experiment）

In 1927 Davisson and Germer directed a beam of electrons at a single crystal of nickel and measured the angular distribution of the scattered electrons. For electrons accelerated through $54\ \mathrm{eV}$, a sharp peak in intensity appeared at $\phi = 50^\circ$ from the incident direction.

The spacing of the rows of atoms at the surface is known from X-ray diffraction to be $d = 0.215\ \mathrm{nm}$. Putting $n=1$ in the condition $d\sin\phi = n\lambda$ for constructive interference at the surface, the required wavelength is

$$\lambda = (0.215\ \mathrm{nm})\times\sin 50^\circ = 0.215\times0.7660 = 0.1647\ \mathrm{nm}.$$

On the other hand, from Axiom 6.1 and the formula in Example 6.3, the de Broglie wavelength of a $54\ \mathrm{eV}$ electron is

$$\lambda = \frac{1.226\ \mathrm{nm}}{\sqrt{54}} = \frac{1.226}{7.348} = 0.1669\ \mathrm{nm}.$$

The two differ by $1.3\ \%$, and correcting for refraction inside the crystal narrows the gap further. A “particle”, the electron, produced diffraction governed by an independently measured lattice constant. With that, de Broglie’s hypothesis ceased to be a hypothesis. In the same year G. P. Thomson obtained concentric diffraction rings from an electron beam transmitted through a thin film. His father, J. J. Thomson, had received the Nobel Prize for showing that the electron is a particle; the son received one for showing that the same electron is a wave.

Proposition 6.5（De Broglie's interpretation of Bohr's quantization condition）

Let a particle of mass $m$ move at speed $v$ on a circular orbit of radius $r$, and impose the condition that the de Broglie wave form a standing wave along that orbit, that is, that the phase return to its initial value after one turn:

$$2\pi r = n\lambda, \qquad n = 1,2,3,\ldots$$

Then, under Axiom 6.1, the magnitude of the angular momentum is quantized as

$$L = mvr = n\hbar .$$

This coincides with the condition Bohr postulated ad hoc in 1913 in order to explain the spectrum of hydrogen.

Proof（Proposition 6.5）

By Axiom 6.1, $\lambda = h/p = h/(mv)$. Substituting into the condition $2\pi r = n\lambda$,

$$2\pi r = \frac{nh}{mv}.$$

Multiplying both sides by $mv/(2\pi)$ and rearranging,

$$mvr = \frac{nh}{2\pi} = n\hbar .$$

The left-hand side is precisely the magnitude of the angular momentum on a circular orbit, $L = |\boldsymbol{r}\times\boldsymbol{p}| = mvr$ (on circular motion $\boldsymbol{r} \perp \boldsymbol{p}$).

∎

Why Bohr’s condition should involve an integer was a mystery from 1913 to 1924. De Broglie’s reading explains it in the same language as the resonance of an organ pipe: the wave must join up with itself after one turn. Here for the first time a quantum number inside the atom acquired a geometric meaning as the number of nodes of a wave.

Example 6.6（The hydrogen atom in the Bohr model）

Apply Proposition 6.5 to an electron of charge $-e$ in circular motion under the Coulomb attraction of a proton of charge $+e$. The equation of motion is

$$\frac{m_ev^2}{r} = \frac{e^2}{4\pi\varepsilon_0 r^2}.$$

Substituting $v = n\hbar/(m_er)$, which follows from $L = m_evr = n\hbar$, the left-hand side becomes $m_e\cdot\dfrac{n^2\hbar^2}{m_e^2r^2}\cdot\dfrac{1}{r} = \dfrac{n^2\hbar^2}{m_er^3}$, so

$$\frac{n^2\hbar^2}{m_er^3} = \frac{e^2}{4\pi\varepsilon_0r^2} \;\Longrightarrow\; r_n = \frac{4\pi\varepsilon_0\hbar^2}{m_ee^2}n^2 = a_0 n^2,$$$$a_0 = \frac{4\pi\varepsilon_0\hbar^2}{m_ee^2} = 5.29\times10^{-11}\ \mathrm{m} = 0.0529\ \mathrm{nm}.$$

The energy is $E = \frac{1}{2}m_ev^2 - \dfrac{e^2}{4\pi\varepsilon_0 r}$, and the equation of motion gives $\frac{1}{2}m_ev^2 = \dfrac{e^2}{8\pi\varepsilon_0 r}$, so

$$E_n = \frac{e^2}{8\pi\varepsilon_0 r_n} - \frac{e^2}{4\pi\varepsilon_0 r_n} = -\frac{e^2}{8\pi\varepsilon_0 a_0}\frac{1}{n^2} = -\frac{13.606\ \mathrm{eV}}{n^2}.$$

The photon emitted in the transition $n=2\to1$ has energy $13.606(1 - 1/4) = 10.20\ \mathrm{eV}$, corresponding to a wavelength $1239.84/10.20 = 121.6\ \mathrm{nm}$, which matches the observed Lyman $\alpha$ line. The model is nevertheless incorrect in treating the electron orbit as a classical circle: the angular momentum of the true ground state is $0$, not $\hbar$. The correct treatment is given in The Hydrogen Atom, where the levels $E_n = -13.606\ \mathrm{eV}/n^2$ obtained here are derived from the Schrödinger equation as the energy levels of the hydrogen atom（Theorem 5.2）［水素原子］.

7. What is needed next

We have now seen that both light and matter have a wave aspect and a particle aspect, and that the bridge between them is given by the two relations $E = h\nu$ and $p = h/\lambda$. But nothing yet deserves the name of a theory. What we hold is an assortment of rules that deliver answers in particular situations.

Two things are missing.

1. An equation for the wave. Just as electromagnetic waves have Maxwell’s equations, matter waves need an equation determining how the wave evolves in time in an arbitrary field of force. That is the time-dependent Schrödinger equation（Definition 3.1）［The Schrödinger Equation and the Wave Function］, which is obtained by substituting $E = \hbar\omega$ and $\boldsymbol{p} = \hbar\boldsymbol{k}$ into the classical energy relation $E = p^2/(2m)+V$. Here the Hamiltonian, a tool of classical mechanics (Hamiltonian Mechanics, the Hamiltonian（Definition 3.6）［ハミルトン形式の力学］), comes into its own.
2. The meaning of the wave. What is it that oscillates in the wave of an electron? The answer Born gave in 1926 was that the square of the amplitude is the probability density for finding the particle there (the Born rule（Definition 4.1）［The Schrödinger Equation and the Wave Function］). This is the deepest break with classical physics.

These two are treated in The Schrödinger Equation and the Wave Function, and the route to formulating physical quantities as linear operators is set out in Operators and Observables. There the linear algebra learned in Vector Spaces and Linear Maps (linear maps（Definition 6.1）［Vector Spaces and Linear Maps］) and in The Spectral Theorem (the spectral theorem for Hermitian matrices（Theorem 4.2）［スペクトル定理］) becomes, without alteration, the language of physics.

8. Exercises

Exercise 8.1Easy

For the mean energy per mode of the Planck distribution, $\langle E\rangle_\nu = h\nu/(e^{x}-1)$ with $x = h\nu/k_BT$, show the following.

1. $\langle E\rangle_\nu = k_BT\left(1 - \dfrac{x}{2} + \dfrac{x^2}{12} + O(x^4)\right)$.
2. For $x = 0.1$, find the relative error between this three-term approximation and the exact value.

Solution

(1) We have $\langle E\rangle_\nu = k_BT\cdot\dfrac{x}{e^x-1}$. From $e^x - 1 = x + \dfrac{x^2}{2}+\dfrac{x^3}{6}+O(x^4)$,

$$\frac{x}{e^x-1} = \frac{1}{1 + \frac{x}{2} + \frac{x^2}{6}+O(x^3)}.$$

Putting $u = \frac{x}{2}+\frac{x^2}{6}+O(x^3)$ and using $(1+u)^{-1} = 1 - u + u^2 - O(u^3)$, with $u^2 = \frac{x^2}{4}+O(x^3)$, we get

$$\frac{x}{e^x-1} = 1 - \left(\frac{x}{2}+\frac{x^2}{6}\right) + \frac{x^2}{4} + O(x^3) = 1 - \frac{x}{2} + \frac{x^2}{12}+O(x^3).$$

That the coefficient of $x^3$ vanishes, so that the error is in fact $O(x^4)$, follows from the fact that $\dfrac{x}{e^x-1}+\dfrac{x}{2}$ is an even function. Indeed

$$\frac{x}{e^x-1}+\frac{x}{2} = \frac{x}{2}\cdot\frac{2+e^x-1}{e^x-1} = \frac{x}{2}\coth\frac{x}{2},$$

and since $\coth$ is odd, $x\coth(x/2)$ is even.

(2) The approximate value is $1 - 0.05 + \dfrac{0.01}{12} = 0.9508333$. The exact value is $0.1/0.10517092 = 0.9508331$, using $e^{0.1} - 1 = 0.10517092$. The relative error is about $2\times10^{-7}$, that is $0.00002\ \%$. The rapidity of this convergence is why the Rayleigh-Jeans law agreed so well with experiment in the region $h\nu \ll k_BT$.

Exercise 8.2Standard

Two spectral lines of a mercury lamp are shone on a metal surface and the stopping voltage is measured: $\lambda_1 = 253.7\ \mathrm{nm}$ gives $V_{s1} = 2.60\ \mathrm{V}$, and $\lambda_2 = 365.0\ \mathrm{nm}$ gives $V_{s2} = 1.11\ \mathrm{V}$. From these two data points, determine Planck’s constant $h$, the work function $W$ of this metal, and the threshold wavelength $\lambda_0$. Take $c = 2.998\times10^{8}\ \mathrm{m\,s^{-1}}$ and $e = 1.602\times10^{-19}\ \mathrm{C}$.

Solution

By Proposition 5.3, $eV_s = h\nu - W$. Subtracting the two equations eliminates $W$:

$$h = \frac{e(V_{s1}-V_{s2})}{\nu_1 - \nu_2}.$$

The frequencies are

$$\nu_1 = \frac{2.998\times10^8}{253.7\times10^{-9}} = 1.1817\times10^{15}\ \mathrm{Hz},\qquad \nu_2 = \frac{2.998\times10^8}{365.0\times10^{-9}} = 8.213\times10^{14}\ \mathrm{Hz},$$

so $\nu_1-\nu_2 = 3.604\times10^{14}\ \mathrm{Hz}$ and $V_{s1}-V_{s2} = 1.49\ \mathrm{V}$. Hence

$$h = \frac{(1.602\times10^{-19})(1.49)}{3.604\times10^{14}} = 6.62\times10^{-34}\ \mathrm{J\,s}.$$

The work function is most easily computed in electronvolts. With $hc = 1239.84\ \mathrm{eV\,nm}$,

$$W = \frac{hc}{\lambda_1} - eV_{s1} = \frac{1239.84}{253.7} - 2.60 = 4.887 - 2.60 = 2.29\ \mathrm{eV}.$$

As a check, $\lambda_2$ gives $W = 1239.84/365.0 - 1.11 = 3.397-1.11 = 2.29\ \mathrm{eV}$, in agreement. The threshold wavelength is

$$\lambda_0 = \frac{hc}{W} = \frac{1239.84}{2.29} = 541\ \mathrm{nm}.$$

From these values the metal is presumably sodium. Note that the estimate of $h$ used only the difference of the stopping voltages, and no value of $W$ at all. This is the practical consequence of the claim in Proposition 5.3 that the slope is independent of the material.

Exercise 8.3Standard

The radiation spectrum of the Sun is well approximated by black-body radiation peaking at wavelength $\lambda_{\max} \approx 500\ \mathrm{nm}$.

1. Find the temperature $T$ of the solar surface.
2. Taking the solar radius to be $R_\odot = 6.96\times10^{8}\ \mathrm{m}$, find the total radiated power (the luminosity) $L_\odot$.
3. Find the radiative flux per unit area at the Sun-Earth distance $d = 1.496\times10^{11}\ \mathrm{m}$ (the solar constant).

Solution

(1) By Proposition 4.5,

$$T = \frac{2.898\times10^{-3}\ \mathrm{m\,K}}{500\times10^{-9}\ \mathrm{m}} = 5.80\times10^{3}\ \mathrm{K}.$$

(2) Integrate $M = \sigma T^4$ from Corollary 4.4 over the whole sphere. Since $T^4 = (5795)^4 = 1.128\times10^{15}\ \mathrm{K^4}$,

$$M = (5.670\times10^{-8})(1.128\times10^{15}) = 6.39\times10^{7}\ \mathrm{W\,m^{-2}},$$$$L_\odot = 4\pi R_\odot^2 M = 4\pi(6.96\times10^{8})^2(6.39\times10^{7}) = 3.9\times10^{26}\ \mathrm{W}.$$

The astronomically measured value is $3.828\times10^{26}\ \mathrm{W}$, agreeing to within $2\ \%$.

(3) The luminosity spreads over the whole sphere of radius $d$, so

$$S = \frac{L_\odot}{4\pi d^2} = \frac{3.9\times10^{26}}{4\pi(1.496\times10^{11})^2} = \frac{3.9\times10^{26}}{2.81\times10^{23}} = 1.4\times10^{3}\ \mathrm{W\,m^{-2}}.$$

The measured value is $1361\ \mathrm{W\,m^{-2}}$. It is worth savouring that a single formula containing Planck’s constant yields the temperature of the Sun, its luminosity, and the sunlight reaching the ground, without being off by even an order of magnitude.

Exercise 8.4Hard

Show that the number density of photons in black-body radiation at temperature $T$ is

$$n(T) = \frac{8\pi}{c^3}\left(\frac{k_BT}{h}\right)^3\cdot 2\zeta(3),$$

and compute its value for the CMB at $T = 2.7255\ \mathrm{K}$, where $\zeta(3) = 1.20206$. Also find the mean energy per photon in units of $k_BT$.

Solution

By Axiom 4.1 the energy of a mode of frequency $\nu$ is $nh\nu$, so the mean number of photons in that mode is $\langle n\rangle = \langle E\rangle_\nu/(h\nu)$. From Theorem 4.2,

$$\langle n\rangle_\nu = \frac{1}{e^{h\nu/k_BT}-1}$$

(this is the Bose-Einstein distribution). Multiplying by the mode density of Proposition 2.3 and integrating,

$$n(T) = \int_0^\infty \frac{8\pi\nu^2}{c^3}\frac{d\nu}{e^{h\nu/k_BT}-1} = \frac{8\pi}{c^3}\left(\frac{k_BT}{h}\right)^3\int_0^\infty\frac{x^2}{e^x-1}dx.$$

As in the proof of Corollary 4.4, expand $\dfrac{1}{e^x-1} = \sum_{n\ge1}e^{-nx}$ and use $\int_0^\infty x^2e^{-nx}dx = 2/n^3$:

$$\int_0^\infty\frac{x^2}{e^x-1}dx = 2\sum_{n=1}^\infty\frac{1}{n^3} = 2\zeta(3) = 2.4041.$$

Now the numbers. We have $k_BT/h = (1.3806\times10^{-23})(2.7255)/(6.6261\times10^{-34}) = 5.678\times10^{10}\ \mathrm{s^{-1}}$, whose cube is $1.831\times10^{32}$, and $8\pi/c^3 = 25.13/(2.694\times10^{25}) = 9.328\times10^{-25}$, so

$$n = (9.328\times10^{-25})(1.831\times10^{32})(2.4041) = 4.1\times10^{8}\ \mathrm{m^{-3}} = 411\ \mathrm{cm^{-3}}.$$

Everywhere in the universe, about 411 CMB photons per cubic centimetre are flying about. The baryon number density is about $2.5\times10^{-7}\ \mathrm{cm^{-3}}$, so photons outnumber nucleons by nine orders of magnitude.

For the mean energy, divide $U$ from Corollary 4.4 by $n$:

$$\frac{U}{n} = \frac{(8\pi h/c^3)(k_BT/h)^4\cdot\pi^4/15}{(8\pi/c^3)(k_BT/h)^3\cdot2\zeta(3)} = k_BT\cdot\frac{\pi^4/15}{2\zeta(3)} = k_BT\cdot\frac{6.4939}{2.4041} = 2.701\,k_BT.$$

The value $\langle E\rangle \approx 2.70\,k_BT$ is worth remembering.

References

• Sin-Itiro Tomonaga, Ryōshi Rikigaku I (Quantum Mechanics I, 2nd ed.), Misuzu Shobo, 1969 (in Japanese) — Chapter I, “The old quantum theory”. Follows the path from black-body radiation to the old quantum theory closely, staying near the original papers.
• R. Eisberg, R. Resnick, Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles, 2nd ed., Wiley, 1985 — Chapters 1-3 (black-body radiation, photons, de Broglie waves). The complete derivation of Compton scattering is here as well.
• M. Planck, “Zur Theorie des Gesetzes der Energieverteilung im Normalspectrum”, Verhandlungen der Deutschen Physikalischen Gesellschaft 2 (1900), 237-245.
• A. Einstein, “Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt”, Annalen der Physik 17 (1905), 132-148. DOI: 10.1002/andp.19053220607
• R. A. Millikan, “A Direct Photoelectric Determination of Planck’s ‘h’”, Physical Review 7 (1916), 355-388. DOI: 10.1103/PhysRev.7.355
• C. Davisson, L. H. Germer, “Diffraction of Electrons by a Crystal of Nickel”, Physical Review 30 (1927), 705-740. DOI: 10.1103/PhysRev.30.705

Appendix: Supplements to the mode count

Changing the boundary condition. In Proposition 2.3 we imposed perfectly conducting walls (vanishing tangential electric field), but periodic boundary conditions give the same answer. With periodic boundary conditions the wave vectors are

$$\boldsymbol{k} = \frac{2\pi}{L}(n_x,n_y,n_z), \qquad n_x,n_y,n_z\in\mathbb{Z},$$

so the spacing is doubled but negative integers are now allowed. The density of lattice points in $\boldsymbol{k}$ space is $(L/2\pi)^3$, and this time one uses the whole sphere rather than an octant, so the number of points satisfying $|\boldsymbol{k}| \le 2\pi\nu/c$ is

$$2 \cdot \left(\frac{L}{2\pi}\right)^3\cdot\frac{4\pi}{3}\left(\frac{2\pi\nu}{c}\right)^3 = \frac{8\pi L^3\nu^3}{3c^3}$$

(the leading 2 is for polarization), in exact agreement. This is no accident: when the cavity is large compared with the wavelength, the details of the boundary contribute only corrections of the order of the surface area.

Independence of shape. More generally, for the counting of eigenvalues of the Laplacian on a bounded region of volume $V$, Weyl’s asymptotic formula

$$N(\nu) \sim \frac{4\pi V \nu^3}{3c^3} \quad (\nu\to\infty)$$

holds (without the polarization factor). It guarantees that the leading term is determined by the volume alone and does not depend on the shape, consistently with Kirchhoff’s universality (Section 2).

Why the modes may be regarded as oscillators. Maxwell’s equations in vacuum make the vector potential $\boldsymbol{A}$ obey the wave equation $\nabla^2\boldsymbol{A} = c^{-2}\partial_t^2\boldsymbol{A}$. Expanding $\boldsymbol{A}$ in the eigenmodes of the cavity as $\boldsymbol{A}(\boldsymbol{r},t) = \sum_\alpha q_\alpha(t)\boldsymbol{u}_\alpha(\boldsymbol{r})$, orthogonality of the mode functions makes each coefficient satisfy $\ddot q_\alpha = -\omega_\alpha^2 q_\alpha$ independently, and the total field energy takes the form

$$\int \frac{\varepsilon_0 E^2 + B^2/\mu_0}{2}\,dV = \sum_\alpha \frac{1}{2}\left(\dot q_\alpha^2 + \omega_\alpha^2 q_\alpha^2\right)$$

(under a suitable normalization). This is the justification for the statement, in the proof of Proposition 3.1, that each mode is a harmonic oscillator. It also means that Axiom 4.1 anticipated the later quantum-mechanical conclusion that the energy levels of such an oscillator form an evenly spaced ladder $\left(n+\frac12\right)\hbar\omega$ (the spectrum of the harmonic oscillator（Theorem 5.3）［1次元の簡単な系］); the zero-point energy $\frac12\hbar\omega$ is a temperature-independent constant and so does not appear in the temperature-dependent part of $u(\nu,T)$.