The need for quantization - Quantum Mechanics for Chemistry

What is the nature of light?¶

Figure 1:Electromagnetic radiation has perpendicular electric and magnetic components that propagate at the speed of light. Unlike other waves (water, sound), light needs no medium and can travel in vacuum.

According to classical wave theory, light is seen as a traveling wave consisting of electric and magnetic components.
We will soon see that this picture of light as an electromagnetic wave is not the whole story, and radically new ideas are needed to understand a wide variety of phenomena involving the interaction of light with atoms and molecules.

Spectrum of electromagnetic waves¶

Visible light occupies a narrow frequency region in between.
High-frequency waves carry much higher energy. This means X-rays or gamma rays can only be generated by heating “stuff” up to very high temperatures. This happens naturally at the core of the Sun!
Low-frequency waves carry less energy. They can be generated in a “microwave” or by broadcasting antennas.
So what is the relationship between the frequency of radiation $\nu$ and its energy $E$ ? This is not such a trivial question. In fact, this very question arose in connection with black body radiation, an experiment that forever changed the course of history by giving birth to quantum mechanics!

Relationship between frequency, wavelength and speed of light.¶

Figure 3:Definitions of wavelength $\lambda$ , amplitude, and frequency $\nu$ . At fixed speed $c$ a wave with twice the frequency has half the wavelength.

Black body as a model for heated objects.¶

Figure 4:A guide to black body radiation from PhD Comics.

Watch this beautiful animation; the first 3 minutes focus solely on blackbody radiation.

A light bulb problem: how a standards lab started quantum mechanics

In 1887 Werner von Siemens and Hermann von Helmholtz founded the Physikalisch-Technische Reichsanstalt (PTR) in Berlin, the world’s first national metrology institute, with the job of giving German industry reliable standards Arrieta, 2022. One pressing customer was the new electric lighting business: to rate a lamp you must know how much light a glowing filament gives per watt, and how that depends on its temperature. That is exactly the black body question.
Through the 1890s the PTR experimenters Lummer, Pringsheim, Rubens, and Kurlbaum built cavity radiators and measured the emitted spectrum with unprecedented precision, pushing far into the infrared Arrieta, 2022. By 1900 their data showed that the best available formula, Wien’s, failed at long wavelengths.
Max Planck, a theorist in Berlin who had been working on the problem for years, learned of the new infrared results from Rubens on a Sunday in October 1900 and found an interpolation formula that fit everything, the law we now call Planck’s law, within the day Hoffmann & Friedrich, 2026. He then spent weeks searching for a derivation, and in December 1900 presented the only one that worked: the oscillators exchange energy in packets $E = h\nu$ . He later called it “an act of desperation.”
Planck did not set out to overthrow classical physics; he wanted a formula for lamp engineers. Quantum mechanics is a by-product of the search for a better light bulb.
The whole story, with the people and the instruments, is told in this video.

Black body as an idealized model¶

This model is called a black body because it absorbs every wavelength that hits its surface, therefore appearing as a perfectly black object.
If an object has a color, it is because it reflects certain wavelengths of light, which are then detected by the retina of our eye. The distribution of wavelengths emitted by a black body is determined only by its temperature!

vs wavelength

vs frequency

Black body spectra versus wavelength for temperatures from 3000 K to 7000 K — Figure 6:Black body spectra versus wavelength at increasing temperature, colored red to blue. Heating a material does three things: (1) the radiated intensity rises steeply; (2) the peak shifts to shorter wavelengths, following the dotted Wien line; and (3) as the peak crosses the visible band the color changes from red to yellow to white to blue.

Ultraviolet catastrophe of classical mechanics¶

What is radiation in classical mechanics? Radiation is considered a wave with frequency $\nu$ . In a heated body, naturally vibrating springs (which represent atoms or molecules) generate waves with the same frequency.

The recipe for a radiation spectrum

\rho_{\nu}(T)\, d\nu = \underbrace{\langle E \rangle}_{\text{energy per mode}} \times \underbrace{dN_{\nu}}_{\text{number of modes}}

(2)

A mode is one standing wave that fits in the cavity, like one harmonic of a guitar string; each mode has its own frequency. The energy stored in radiation at frequency $\nu$ is the energy each mode of that frequency carries on average, times the number of such modes: $\langle E \rangle$ is the average energy of one wave mode at temperature $T$ , and $dN_{\nu}$ is the number of modes per volume with frequency in $[\nu, \nu + d\nu]$ .
The average mode energy $\langle E \rangle$ is thermodynamics: what the waves are and how much energy each holds. It is where classical and quantum physics part ways.
The number of modes $dN_{\nu}$ is pure geometry: how many standing waves fit in a box. It is the same in classical and quantum physics. Everything that follows is about these two ingredients, in that order.

Figure 9:Visualization of atomic vibrations in a solid body. These vibrational modes are called phonons, not to be confused with the photons introduced in the next section. Each mode is one of the dots counted above.

Average mode energy, the classical way: equipartition. From thermodynamics we know that in equilibrium each degree of freedom, or each oscillator, gets the same energy $k_BT$ , where $k_B$ is the Boltzmann constant.

\langle E\rangle = k_BT

(3)

Every vibrating spring in a heated body thus has the same energy regardless of frequency. Think about this assumption for a second!
Number of modes in one breath. Shorter waves fit into a box of length $L$ more easily than long ones: the number that fit grows as $L/\lambda \sim \nu$ in one dimension, and in three dimensions the wave has to fit along each direction independently, so the count grows as $\nu^3$ . The number of modes in a thin frequency slice is therefore $dN \sim d(\nu^3) \sim \nu^2 d\nu$ . The box below fills in the details on this calculation.

Counting the number of modes in a box

standing waves with nodes at the walls of a one-dimensional box, and the grid of modes of a two-dimensional box — A wave trapped between two walls must vanish at each wall, so only a whole number $n$ of half-wavelengths fits: $\lambda_n = 2L/n$ . Shorter waves fit more easily, and the number that fit up to a given frequency grows as $2L/\lambda = 2L\nu/c$ .

Where the integers come from. A wave trapped between two walls has to vanish at both walls, like a guitar string pinned at its ends. That is only possible if a whole number of half-wavelengths spans the box, $L = n\,\lambda/2$ , so the allowed wavelengths are $\lambda_n = 2L/n$ with $n = 1, 2, 3, \ldots$ The integer $n$ is nothing more than the number of half-waves that fit.
Shorter waves fit more easily. Since $\nu = c/\lambda$ , the allowed frequencies are $\nu_n = n\,c/2L$ . Counting all the waves that fit with frequency up to $\nu$ (wavelength down to $\lambda$ ) gives $N(\nu) = 2L\nu/c = 2L/\lambda$ : the count grows in proportion to $\nu$ , in inverse proportion to $\lambda$ , and in proportion to the size of the box.
Two and three dimensions. In a square box the wave must fit along $x$ and along $y$ separately, so a mode needs two integers $(n_x, n_y)$ ; in a cube it needs three. The frequency of a mode is set by the distance of the point $(n_x, n_y, n_z)$ from the origin, $\nu = \frac{c}{2L}\sqrt{n_x^2 + n_y^2 + n_z^2}$ , so the modes with frequency below $\nu$ are the grid points inside a sphere of radius $2L\nu/c$ . Their number grows like the size of that region: $N(\nu) \propto \nu$ in one dimension (a line), $\propto \nu^2$ in two (a quarter disk), $\propto \nu^3$ in three (an octant of a sphere).
A thin slice. The modes between $\nu$ and $\nu + d\nu$ form a thin spherical shell whose volume is surface times thickness, $dN \propto \nu^2\, d\nu$ . This is all that “more short waves than long waves fit in a box” means.

Result of the counting. Keeping track of the constants (two polarizations of light, only positive integers, so one octant of the sphere, and division by the box volume $L^3$ ) gives the number of modes per unit volume in $[\nu, \nu+d\nu]$ :

dN_{\nu} = \frac{8\pi}{c^3} \cdot \nu^2 d\nu

(4)

Radiation energy distribution. Putting the two factors of the recipe together, $k_BT$ per mode times modes per volume, gives the classical Rayleigh-Jeans law:

\rho({\nu}) = \frac{8\pi}{c^3}\nu^2 \cdot k_B T

(5)

Ultraviolet catastrophe. The energy distribution shoots to infinity at high $\nu$ (or low $\lambda$ ). This is known as the ultraviolet catastrophe! Integrating $\rho$ over all frequencies gives the total amount of radiation, which in this case is infinite. A light bulb could destroy the universe! Something is off with our classical prediction.

Max Planck and the trick of quantization¶

In 1900 Planck found that the theoretical curve can very closely match the experimental curve if one postulates that only discrete (quantized) values of energy are possible.

This means atoms and molecules absorb and emit radiation in discrete quantities, multiples of $h\nu$ , which are called quanta!
When light is emitted or absorbed, the atom or molecule jumps from one state to another and the energy difference $h\nu$ is either coming from light or is used to generate light.
Note how small $h$ is in the macroscopic units (such as J s). This is why quantization of energy is hardly noticeable and classical mechanics works so well at the macro scale. In the limit $h \rightarrow 0$ , $E$ becomes continuous, and an arbitrary real value of E is allowed. This is the classical limit.

The black body radiation distribution function¶

Deriving Black Body radiation formula

Planck hypothesized that the energy of oscillators in a black body is quantized and given by:

E_n = n h \nu

(7)

where $n = 0, 1, 2, \ldots$ is a non-negative integer, $h$ is Planck’s constant, and $\nu$ is the frequency. (Planck’s 1900 oscillators start at zero energy; the extra $\tfrac{1}{2}h\nu$ of zero-point energy is a later result of full quantum mechanics that we will meet with the harmonic oscillator.)

The average energy of an oscillator is found by summing over all possible energies, weighted by the Boltzmann factor:

\langle E \rangle = \frac{\sum_n E_n e^{-E_n/kT}}{\sum_n e^{-E_n/kT}}

(8)

Substituting $E_n = n h \nu$ , the sum becomes:

\langle E \rangle = \frac{\sum_n n h \nu e^{-n h \nu / kT}}{\sum_n e^{-n h \nu / kT}}

(9)

This sum is a geometric series. For the geometric series of the form:

S = \sum_{n=0}^{\infty} x^n

(10)

The sum is given by:

S = \frac{1}{1 - x} \quad \text{for} \quad |x| < 1

(11)

In the context of Planck’s derivation, we use the series:

\sum_{n=0}^{\infty} e^{-n h \nu / kT}

(12)

This series can be summed as:

\sum_{n=0}^{\infty} e^{-n h \nu / kT} = \frac{1}{1 - e^{-h \nu / kT}}

(13)

The series involving $n$ in the numerator is:

\sum_{n=0}^{\infty} n e^{-n h \nu / kT}

(14)

This can be evaluated using the derivative with respect to $x$ :

\sum_{n=0}^{\infty} n x^n = x \frac{d}{dx} \left( \frac{1}{1 - x} \right) = \frac{x}{(1 - x)^2}

(15)

Substituting $x = e^{-h \nu / kT}$ , we get:

\sum_{n=0}^{\infty} n e^{-n h \nu / kT} = \frac{e^{-h \nu / kT}}{(1 - e^{-h \nu / kT})^2}

(16)

Using these results, Planck’s formula for the average energy becomes:

\langle E \rangle = \frac{h \nu}{e^{h \nu / kT} - 1}

(17)

The energy density $\rho(\nu, T)$ is then obtained by multiplying the average energy by the density of states and the number of oscillators per unit volume:

\rho(\nu, T) = \frac{8 \pi \nu^2}{c^3} \cdot \frac{h \nu}{e^{h \nu / kT} - 1}

(18)

This is Planck’s law, which describes the spectral density of radiation emitted by a black body in thermal equilibrium at a temperature $T$ .

Planck kept the number of modes untouched. Assuming that the energy of an oscillator is quantized, he derived a new average mode energy which, unlike the classical $k_BT$ , depends on the frequency of oscillation:

\langle E \rangle = \Big[ \frac{h\nu}{e^{\frac{h\nu}{ kT}} - 1}\Big]

(19)

Modes per volume times this average energy gives Planck’s law, a distribution that tends to zero in the high-frequency limit because the exponential in the average energy wins over the $\nu^2$ of the mode count. The same law can be written per unit frequency or per unit wavelength (substitute $\nu = c/\lambda$ and $d\nu = c\,d\lambda/\lambda^2$ ):

per unit frequency

per unit wavelength

\rho_{\nu}(T) = \frac{8\pi \nu^2}{c^3} \cdot \Big[\frac{h\nu}{e^{\frac{h\nu}{kT}} - 1} \Big]

(20)

The expressions $\rho_{\lambda}(T)d\lambda$ and $\rho_{\nu}(T)d\nu$ have units of energy per volume, which is why they are often referred to as the energy density of radiation. By integrating over the entire spectrum (e.g., all frequencies or wavelengths) we obtain the total energy of radiation per volume!

\int^{\infty}_0 \rho_{\nu}(T)d\nu = \frac{4\sigma}{c} T^4

(22)

The power radiated per unit surface area of the black body is the more familiar Stefan-Boltzmann law, $P/A = \sigma T^4$ , where $\sigma=5.67 \cdot 10^{-8}\, \text{W m}^{-2} \text{K}^{-4}$ is the Stefan-Boltzmann constant. Doubling the temperature increases the radiated power sixteenfold. Try it: the shaded area below is the integral, and the readout compares it with the area at 3000 K.

Wien’s displacement law¶

Connecting the temperature of a black body with wavelength or frequency. The energy density peaks at a wavelength $\lambda_{max}$ that is inversely proportional to the temperature.
This relationship is described by Wien’s displacement law. You can derive it by setting the derivative $d\rho(\lambda)/d\lambda=0$ , which gives the wavelength at the peak of the distribution.

Explore black body radiation¶

Drag the temperature and watch three things at once: the Planck curve (solid) rises and its peak slides to shorter wavelengths, following Wien’s law; the visible band lights up only above a few thousand kelvin; and the classical Rayleigh-Jeans prediction (dashed) agrees with Planck at long wavelengths but shoots off the top of the plot at short ones. That divergence is the ultraviolet catastrophe.

Applications of Black Body radiation¶

planets — Figure 11:The black body is used as a standard against which the absorption of real bodies is compared. To a good approximation, stars radiate like black bodies, so we can use blackbody radiation as a model to infer the temperatures of stars from their colors. Find out more in this video on Visible Light Waves.

Rayleigh Scattering and the Color of the Sky¶

Problems¶

Problem 1: Why is the sky blue and not violet?¶

If hotter objects radiate more strongly at shorter wavelengths, why does the daytime sky appear blue instead of violet (or even ultraviolet)?

Solution

The Sun’s spectrum indeed peaks in the green/yellow (≈500 nm), and it emits significant violet and ultraviolet.
However:

Rayleigh scattering in the atmosphere is much stronger at shorter wavelengths (scattering $\sim 1/\lambda^4$ ).
Human eyes are more sensitive to blue than violet.
Most ultraviolet is absorbed by the ozone layer.

Thus, the scattered light that reaches us is predominantly blue.

Problem 2: Color of a 3000 K black body¶

A blackbody has temperature $T = 3000 \,\text{K}$ . According to Wien’s law, what is the approximate color of its peak emission?

Red/Orange
Green
Blue
Ultraviolet

Solution

Using $\lambda_{\max} = \dfrac{2.898 \times 10^{-3}}{3000} \approx 9.7 \times 10^{-7}\,\text{m} = 970 \,\text{nm}$ , the peak is in the infrared, but the visible portion is dominated by the red/orange end.
Correct choice: (1) Red/Orange.

Problem 3: Wavelength from photon energy¶

For a monochromatic (single wavelength) radiation with an energy of $3.5 \, \text{eV}$ calculate the wavelength. Use Planck’s equation to relate the energy of radiation to its wavelength. The values of constants are:

Planck’s constant, $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$
Speed of light, $c = 3.00 \times 10^8 \, \text{m/s}$
$1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$

Solution

First, convert the energy of the radiation from electron volts (eV) to joules (J):

E = 3.5 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 5.607 \times 10^{-19} \, \text{J}

(24)

Now, use Planck’s equation to relate the energy $E$ of the photon to its wavelength $\lambda$ :

E = \frac{hc}{\lambda}

(25)

Rearranging to solve for $\lambda$ :

\lambda = \frac{hc}{E}

(26)

Substitute the known values:

\lambda = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (3.00 \times 10^8 \, \text{m/s})}{5.607 \times 10^{-19} \, \text{J}}

(27)

\lambda = \frac{1.988 \times 10^{-25} \, \text{J} \cdot \text{m}}{5.607 \times 10^{-19} \, \text{J}} \approx 3.55 \times 10^{-7} \, \text{m} = 355 \, \text{nm}

(28)

The wavelength of the radiation is approximately $355 \, \text{nm}$ , which is in the ultraviolet range of the electromagnetic spectrum.

Problem 4: Peak of the solar spectrum¶

Using Wien’s displacement law, determine the wavelength $\lambda_{\text{max}}$ at which the spectral radiance of a blackbody is maximized. Calculate $\lambda_{\text{max}}$ for $T = 5800 \, \text{K}$ , approximately the temperature of the Sun’s surface.

Solution

Wien’s displacement law states that the wavelength at which the spectral radiance of a blackbody peaks is inversely proportional to the temperature:

\lambda_{\text{max}} = \frac{b}{T}

(29)

where $b = 2.897 \times 10^{-3} \, \text{m} \cdot \text{K}$ is Wien’s displacement constant.

For $T = 5800 \, \text{K}$ :

\lambda_{\text{max}} = \frac{2.897 \times 10^{-3} \, \text{m} \cdot \text{K}}{5800 \, \text{K}} = 5.0 \times 10^{-7} \, \text{m} = 500 \, \text{nm}

(30)

So, the peak wavelength $\lambda_{\text{max}}$ for a blackbody at $5800 \, \text{K}$ is $500 \, \text{nm}$ , which is in the visible range.

Problem 5: Star colors as thermometers¶

Betelgeuse looks distinctly red, Rigel blue-white. Their spectra peak near 830 nm and 240 nm respectively. Estimate the surface temperature of each star. Which of the two radiates more power per square meter of surface, and by what factor?

Problem 6: Counting photons¶

A red laser pointer emits 1.0 mW at 650 nm. How many photons leave it per second? Compare with the number of photons per second in a 1.0 mW beam of X-rays with wavelength 0.10 nm. In which beam is the “graininess” of light easier to detect?

Problem 7: Where the classical formula hides inside Planck’s¶

Expand $e^{h\nu/k_BT}$ for $h\nu \ll k_BT$ and show that Planck’s average oscillator energy $\langle E \rangle = h\nu/(e^{h\nu/k_BT}-1)$ reduces to the classical equipartition value $k_BT$ . Then examine the opposite limit, $h\nu \gg k_BT$ , and show that the average energy dies off exponentially. Explain in one sentence why this second limit is what cures the ultraviolet catastrophe.

Problem 8: The Sun’s power output¶

The Sun has radius $6.96 \times 10^8$ m and a surface temperature of about 5770 K. Treating it as a black body, use the Stefan-Boltzmann law $P/A = \sigma T^4$ to compute its total radiated power. The Earth is $1.50 \times 10^{11}$ m away; what power per square meter arrives at the top of our atmosphere? (The measured value, the solar constant, is about 1360 W/m $^2$ .)

Reference Table of Constants¶

Constant	Symbol	Value
Speed of light	$c$	$3.00 \cdot 10^8\, \text{m/s}$
Planck’s constant	$h$	$6.626 \cdot 10^{-34}\, \text{J s}$
Boltzmann constant	$k_B$	$1.381 \cdot 10^{-23}\, \text{J/K}$
Stefan-Boltzmann constant	$\sigma$	$5.67 \cdot 10^{-8}\, \text{W m}^{-2}\text{K}^{-4}$
Wien’s displacement constant	$b$	$2.898 \cdot 10^{-3}\, \text{m K}$

Extra: quantized oscillators explain heat capacities too¶

Einstein’s 1907 heat capacity model (optional reading)

Planck’s quantized oscillator did more than fix the radiation curve. Classical equipartition says every atom in a solid, vibrating in three directions, stores $3k_BT$ of energy, so the molar heat capacity should be a constant $C_V = 3R \approx 25$ J/(mol K). This is the Dulong-Petit law, and it works at room temperature for most metals.
Yet measurements showed $C_V$ falling toward zero as $T \to 0$ , and diamond fell far short of $3R$ even at room temperature. Classical physics had no answer.
In 1907 Einstein modeled each atom as a Planck oscillator of frequency $\nu$ and replaced $k_BT$ with Planck’s average energy. Differentiating with respect to $T$ gives

C_V = 3R \left(\frac{\theta_E}{T}\right)^2 \frac{e^{\theta_E/T}}{\left(e^{\theta_E/T}-1\right)^2}, \qquad \theta_E = \frac{h\nu}{k_B}

(31)

The Einstein temperature $\theta_E$ marks where quantization kicks in: for $T \gg \theta_E$ the formula returns Dulong-Petit, for $T \ll \theta_E$ the oscillators freeze out and $C_V \to 0$ . Diamond’s stiff bonds give a large $\nu$ , hence a high $\theta_E$ , which is why it looks “quantum” already at room temperature.

Einstein heat capacity of copper, aluminum, and diamond versus temperature — Einstein heat capacity of three solids. Vibrations freeze out below the Einstein temperature, and diamond is still far from the classical limit at room temperature.

References¶

González de Arrieta, I. (2022). Beyond the infrared: a centenary of Heinrich Rubens’s death. The European Physical Journal H, 47, 11. 10.1140/epjh/s13129-022-00044-x
Hoffmann, D., & Friedrich, B. (2026). Max Planck (1858–1947): A Revolutionary Against His Will. Natural Sciences, 6(3). 10.1002/ntls.70077