Part 5 · Chapter 38

Photons and Matter Waves

Relativity stretched our intuitions about space and time at high speed; the subatomic world now breaks them again. Quantum physics opens with a startling claim: light, which we spent four chapters treating as a smooth wave, also arrives in indivisible grains of energy called photons. From that one idea — energy comes in lumps of size $hf$ — flow the photoelectric effect Einstein used to launch the quantum era, the Compton scattering that hands a photon momentum, and a deep puzzle about how anything can be both wave and particle. De Broglie then turned the idea inside out: if waves act like particles, particles must act like waves — and electrons fired one at a time really do build an interference pattern. We follow that logic to its core tools: the de Broglie wavelength, Schrödinger's equation and its probability waves, Heisenberg's uncertainty principle, and the uncanny way a particle can tunnel through a wall it could never classically cross.

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 90 min

i What you'll learn

Light is quantized into photons, each of energy $E = hf$ and momentum $p = \dfrac{hf}{c} = \dfrac{h}{\lambda}$, where $h$ is the Planck constant.
The photoelectric effect obeys Einstein's equation $hf = K_{\max} + \Phi$, with a work function $\Phi$ and a cutoff frequency: intensity sets the number of photons, not the energy each delivers.
Compton scattering proves a photon carries momentum; the scattered wavelength grows by the Compton shift $\Delta\lambda = \dfrac{h}{mc}(1 - \cos\phi)$.
Light is a probability wave: the chance of detecting a photon at a point is proportional to the square of the wave's amplitude there.
De Broglie: every particle has a wavelength $\lambda = \dfrac{h}{p}$, and electrons, neutrons, even molecules, produce interference patterns.
Matter waves obey Schrödinger's equation for the wave function $\psi$, with probability density $|\psi|^{2}$; Heisenberg's principle sets $\Delta x\cdot\Delta p_x \ge \hbar$; and particles can tunnel through barriers with $T \approx e^{-2bL}$.

Section 38-1

What Is Physics?

Relativity carried us into the world of the very fast; now we enter the world of the very small — the subatomic realm governed by quantum physics. This is the theory that answers the deepest "why" questions of ordinary matter: why the stars shine, why the periodic table has the structure it does, why copper conducts and glass does not, how transistors and every microelectronic device function. Because quantum physics underlies all of chemistry, and thus all of biochemistry, it is in a real sense the physics of life itself. Its predictions can seem bizarre even to the physicists who use them daily, yet experiment after experiment has confirmed the theory and, often, revealed it to be stranger still. We begin our tour of this quantum amusement park where Einstein began — with the photon.

Section 38-2

The Photon, the Quantum of Light

A quantity is quantized when it comes only in a smallest indivisible amount — its quantum — or in whole-number multiples of it. (U.S. currency is loosely quantized: every amount is a whole number of pennies; you cannot pay $\$0.755$.) In 1905 Einstein proposed that light itself is quantized, existing in elementary packets we now call photons. This should feel jarring after several chapters of treating light as a smooth electromagnetic wave with $f = c/\lambda$ — and indeed, what a photon "is" remains subtle. But the proposal is precise about energy.

Photon energy and the Planck constant

\[ E = hf, \qquad h = 6.63\times10^{-34}\,\mathrm{J\cdot s} = 4.14\times10^{-15}\,\mathrm{eV\cdot s} \]

The smallest energy a light wave of frequency $f$ can carry is one photon's worth, $hf$; any greater energy is a whole-number multiple of it. The wave cannot have, say, $0.6\,hf$. When an atom absorbs light it swallows one photon whole (the photon vanishes, its energy $hf$ transferred); when it emits, a photon of energy $hf$ appears.

Section 38-3

The Photoelectric Effect

Shine light of short enough wavelength on a clean metal and electrons fly off — the photoelectric effect, the basis of night-vision viewers and many detectors. Two experimental facts wreck the classical wave picture. First, the maximum kinetic energy $K_{\max}$ of the ejected electrons (found from the stopping potential, $K_{\max} = eV_{\text{stop}}$) does not depend on the light's intensity — a blinding beam and a feeble one of the same color give the same maximum kick. Second, below a cutoff frequency $f_0$ no electrons emerge at all, however bright the light. Classically, a brighter wave should always deliver a bigger kick, and any frequency should work if intense enough. Neither is true.

Einstein's photoelectric equation

\[ hf = K_{\max} + \Phi \]

This is energy conservation for a single photon absorbed by one electron. Of the photon's energy $hf$, an amount $\Phi$ — the work function, a property of the metal — is spent escaping the surface; the rest emerges as kinetic energy, at most $K_{\max}$. Below the cutoff ($hf < \Phi$) the electron cannot escape at all, no matter how many photons arrive.

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Why intensity changes nothing per electron

Brighter light means more photons, not more energetic ones. Each electron still interacts with a single photon of energy $hf$, fixed by the frequency alone.

Rewriting the equation as $V_{\text{stop}} = (h/e)f - \Phi/e$ predicts a straight line of stopping potential versus frequency, with slope $h/e$. Millikan measured exactly that line in 1916 and read off $h$, in agreement with every other method — a triumph for the photon idea. (Curiously, in 1969 the photoelectric effect was shown to be explainable with quantum physics but without photons; light is quantized, but this experiment is no longer its best proof.)

Section 38-4

Photons Have Momentum

In 1916 Einstein extended the idea: a photon also carries linear momentum. So when light meets matter, both energy and momentum can transfer, exactly as in a mechanical collision.

Photon momentum

\[ p = \frac{hf}{c} = \frac{h}{\lambda} \]

Smaller wavelength means larger momentum. This is the relation de Broglie will soon borrow for matter; here it sets the scale of the photon's mechanical punch.

Arthur Compton tested this in 1923 by firing X-rays of a single wavelength $\lambda$ at carbon and measuring the scattered rays. Classically the scattered light should keep the same wavelength; instead a second peak appears at a longer wavelength $\lambda'$, shifted by an amount that grows with scattering angle. Treating the encounter as a photon colliding with a loosely bound electron — conserving relativistic energy and momentum, then eliminating the recoiling electron's variables — yields the shift exactly.

The Compton shift

\[ \Delta\lambda = \lambda' - \lambda = \frac{h}{mc}\,(1 - \cos\phi) \]

Here $\phi$ is the scattering angle and $m$ the electron mass. The constant $h/mc$ is the Compton wavelength (about $2.43\,\mathrm{pm}$ for an electron). The shift is zero for forward scatter ($\phi = 0$) and largest for backscatter ($\phi = 180^\circ$). The small leftover peak at the original wavelength comes from photons scattering off whole tightly-bound atoms, whose huge mass makes $\Delta\lambda$ undetectably tiny.

Section 38-5

Light as a Probability Wave

How can light be a spread-out wave classically yet be emitted and absorbed as point-like photons? The double-slit experiment lies at the heart of the mystery. Put a tiny photon detector in the interference pattern and it clicks at random instants — but the click rate rises and falls exactly with the bright and dark fringes.

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Light is also a probability wave

The probability per unit time of detecting a photon at a point is proportional to the square of the wave's electric-field amplitude there — the same quantity that sets the classical intensity.

Run the source so feebly that only one photon is in flight at a time (as G. I. Taylor did in 1909) and the fringes still build up, photon by photon, over months. We cannot say which slit a given photon takes; between source and screen the photon travels as a probability wave filling the apparatus, then vanishes in a single absorption. Three consistent ingredients survive every version of the experiment: light is born as photons, detected as photons, and travels as a probability wave.

Section 38-6

Electrons and Matter Waves

In 1924 Louis de Broglie made a daring symmetry argument: if a light wave transfers energy and momentum to matter at points, via photons, why shouldn't a beam of particles behave as a wave? He proposed that the photon relation $p = h/\lambda$ works in reverse for matter.

The de Broglie wavelength

\[ \lambda = \frac{h}{p} \]

Any particle of momentum $p$ has this wavelength. It is tiny for everyday objects (a thrown baseball's wavelength is absurdly small), but for an electron it is about the size of an atom — large enough to matter.

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Particles interfere with themselves

Electrons sent one at a time through a double slit land at seemingly random points — yet after many thousands, an interference pattern emerges, exactly as for light.

De Broglie's prediction was confirmed by Davisson and Germer (and G. P. Thomson) in 1927, and the self-interference has since been shown for protons, neutrons, whole atoms, iodine molecules, and even 60-atom carbon "buckyballs." Each particle travels as a matter wave whose two slit-portions interfere, fixing the probability of where it materializes. Electron and neutron diffraction are now everyday tools for probing the atomic structure of solids — though a cat, mercifully, is far too large and complex to interfere with itself.

Section 38-7

Schrödinger's Equation

A light wave is described by its oscillating electric field; a matter wave is described by its wave function $\Psi(x,y,z,t)$, which is generally a complex quantity. In the cases we meet, the time part separates as $\Psi = \psi\,e^{-i\omega t}$, leaving the space part $\psi$ to carry the physics. The wave function itself is not directly observable; what has meaning is its squared magnitude.

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The probability density

The probability per unit time of detecting the particle in a small volume around a point is proportional to $|\psi|^{2}$ at that point — a quantity that is always real and non-negative even though $\psi$ is complex.

Where light waves obey Maxwell's equations, matter waves obey Schrödinger's, the founding principle of quantum mechanics — it cannot be derived from anything deeper. For one-dimensional motion through a region of potential energy $U(x)$, it reads as below, with $E$ the particle's total (non-relativistic) energy.

Schrödinger's equation (one dimension)

\[ \frac{d^{2}\psi}{dx^{2}} + \frac{8\pi^{2}m}{h^{2}}\,\bigl[E - U(x)\bigr]\psi = 0 \]

For a free particle ($U = 0$) the solution is a traveling wave $\psi(x) = \psi_0 e^{ikx}$ with $k = 2\pi/\lambda$, giving $|\psi|^{2} = \psi_0^{2}$, a constant. The particle is then equally likely to be found anywhere along $x$ — its position is completely undetermined, a hint of what comes next.

Section 38-8

Heisenberg's Uncertainty Principle

That a free particle of perfectly known momentum has a completely unknown position is no accident. It is the first glimpse of Heisenberg's 1927 principle: position and momentum cannot both be pinned down with unlimited precision at once.

Heisenberg's uncertainty principle

\[ \Delta x\cdot\Delta p_x \ge \hbar, \qquad \Delta y\cdot\Delta p_y \ge \hbar, \qquad \Delta z\cdot\Delta p_z \ge \hbar, \qquad \hbar = \frac{h}{2\pi} \]

Each product of a position uncertainty and the corresponding momentum uncertainty is at least $\hbar$ ("h-bar"), never less. Sharpen the momentum and the position blurs, and vice versa. This is not a defect of our instruments: if a particle's momentum is exactly known, the phrase "position of the particle" simply has no meaning.

Section 38-9

Barrier Tunneling

Slide a puck at an icy hill: if its energy $E$ is less than the hilltop's potential energy $U_b$, it rolls back, every time. Send an electron of energy $E < U_b$ at a potential-energy barrier and classically the same should hold. But the electron is a matter wave, and a wave does something a puck cannot.

Transmission through a barrier (tunneling)

\[ T \approx e^{-2bL}, \qquad b = \sqrt{\frac{8\pi^{2}m(U_b - E)}{h^{2}}} \]

Inside the barrier the wave function does not stop dead — it decays exponentially, and if the barrier of thickness $L$ is thin enough, a small but nonzero amplitude emerges on the far side. The transmission coefficient $T$ is the probability that an incident particle leaks through. Because $b$ sits in an exponent, $T$ is fiercely sensitive to the particle's mass $m$, the barrier thickness $L$, and the energy shortfall $U_b - E$.

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Tunneling is real — and built into technology

For an electron with $E = 5.1\,\mathrm{eV}$ facing a barrier $U_b = 6.8\,\mathrm{eV}$ high and $L = 750\,\mathrm{pm}$ thick, $T \approx 45\times10^{-6}$ — about 45 of every million electrons pass straight through.

Swap in a proton (far heavier) and $T$ collapses to roughly $10^{-186}$: tunneling is overwhelmingly a feature of the lightest particles. The effect powers the tunnel diode and the Josephson junction, and it is the working principle of the scanning tunneling microscope, whose fine tip reads a surface atom-by-atom from the tunneling current that flows across the vacuum gap — sharpened by the same exponential sensitivity to distance.

Worked Examples

Putting It to Work

1 Counting photons from a lamp

Problem. A sodium lamp emits $P = 100\,\mathrm{W}$, all at wavelength $\lambda = 590\,\mathrm{nm}$, into a sphere that absorbs every photon. At what rate are photons absorbed?

Solution. The rate is the power divided by the energy per photon, $E = hf = hc/\lambda$.

R = P / (hc/λ) = Pλ / hc

\[ R = \frac{P\lambda}{hc} = \frac{(100)(590\times10^{-9})}{(6.63\times10^{-34})(3.00\times10^{8})} \approx 2.97\times10^{20}\,\text{photons/s} \]

Nearly $3\times10^{20}$ photons a second — so vast a torrent that the graininess of light is utterly hidden, which is exactly why classical wave optics served us so well until single-photon experiments became possible.

2 Work function from the cutoff

Problem. A photoelectric plot of $V_{\text{stop}}$ versus $f$ for sodium crosses the frequency axis at a cutoff $f_0 \approx 5.5\times10^{14}\,\mathrm{Hz}$. Find sodium's work function.

Solution. At the cutoff $K_{\max} = 0$, so the entire photon energy goes into escape: $\Phi = hf_0$.

Φ = h f₀

\[ \Phi = hf_0 = (6.63\times10^{-34})(5.5\times10^{14}) \approx 3.6\times10^{-19}\,\mathrm{J} \approx 2.3\,\mathrm{eV} \]

About $2.3\,\mathrm{eV}$ — the minimum energy one photon must deliver to free an electron from sodium. Light of any lower frequency, however intense, leaves the metal undisturbed.

3 Compton shift and energy loss

Problem. X-rays of $\lambda = 22\,\mathrm{pm}$ scatter from carbon and are detected at $\phi = 85^\circ$. Find (a) the Compton shift and (b) the fraction of photon energy lost.

Solution. Use the Compton shift with the electron mass, then express the fractional loss in wavelengths.

Δλ = (h/mc)(1 − cos φ)

\[ \Delta\lambda = \frac{6.63\times10^{-34}}{(9.11\times10^{-31})(3.00\times10^{8})}(1 - \cos 85^\circ) \approx 2.2\,\mathrm{pm} \]

fractional loss = Δλ / (λ + Δλ)

\[ \frac{\Delta E}{E} = \frac{\Delta\lambda}{\lambda + \Delta\lambda} = \frac{2.21}{22 + 2.21} \approx 0.091 = 9.1\% \]

The shift itself is independent of the incident wavelength, but the fractional energy loss is not — it grows as $\lambda$ shrinks, which is why the Compton effect is glaring for X-rays and gamma rays yet hopelessly tiny for visible light.

4 De Broglie wavelength of an electron

Problem. What is the de Broglie wavelength of an electron with kinetic energy $K = 120\,\mathrm{eV}$?

Solution. Since $K$ is far below the electron's rest energy ($0.511\,\mathrm{MeV}$), use the classical link $p = \sqrt{2mK}$, then $\lambda = h/p$.

p = √(2mK), then λ = h/p

\[ p = \sqrt{2(9.11\times10^{-31})(120)(1.60\times10^{-19})} \approx 5.91\times10^{-24}\,\mathrm{kg\cdot m/s}, \qquad \lambda = \frac{h}{p} \approx 112\,\mathrm{pm} \]

About $112\,\mathrm{pm}$ — roughly the size of an atom, which is precisely why low-energy electrons diffract from crystals and make electron microscopy possible. Give the electron more energy and its wavelength shrinks, letting it resolve ever finer detail.

Review

Chapter Summary

✓ Photons

Light is quantized: each photon has energy $E = hf$ and momentum $p = hf/c = h/\lambda$.

✓ Photoelectric effect

$hf = K_{\max} + \Phi$; no emission below the cutoff; intensity sets photon number, not $K_{\max}$.

✓ Compton shift

$\Delta\lambda = \dfrac{h}{mc}(1 - \cos\phi)$; photons carry momentum into the collision.

✓ Probability wave

Detection probability per unit time $\propto$ the square of the field amplitude; light is born and absorbed as photons but travels as a probability wave.

✓ Matter waves

Every particle has $\lambda = h/p$; electrons, neutrons, atoms, and molecules all interfere.

✓ Wave function

Schrödinger's equation governs $\psi$; $|\psi|^{2}$ is the probability density. A free particle has uniform $|\psi|^{2}$.

✓ Uncertainty

$\Delta x\cdot\Delta p_x \ge \hbar$; position and momentum cannot both be exact.

✓ Tunneling

$T \approx e^{-2bL}$, $b = \sqrt{8\pi^{2}m(U_b-E)/h^{2}}$; basis of the STM.

Practice

Problems

Take $h = 6.63\times10^{-34}\,\mathrm{J\cdot s} = 4.14\times10^{-15}\,\mathrm{eV\cdot s}$, the electron mass $m = 9.11\times10^{-31}\,\mathrm{kg}$ (rest energy $0.511\,\mathrm{MeV}$), and $c = 3.00\times10^{8}\,\mathrm{m/s}$. For photon problems work with $E = hf = hc/\lambda$; for matter waves use $\lambda = h/p$ and, when the kinetic energy is well below the rest energy, the classical $p = \sqrt{2mK}$.

What is the photon energy (in eV) of the yellow sodium-lamp light at $589\,\mathrm{nm}$?
Photographic film records light only if each photon carries at least $0.6\,\mathrm{eV}$. (a) What is the greatest wavelength that can be recorded? (b) In what part of the spectrum does it lie?
Assuming the Sun radiates $3.9\times10^{26}\,\mathrm{W}$ entirely at $550\,\mathrm{nm}$, at what rate does it emit photons?
A helium–neon laser emits $5.0\,\mathrm{mW}$ at $633\,\mathrm{nm}$ in a beam $3.5\,\mathrm{mm}$ across, fully absorbed by a detector. At what rate per unit area are photons absorbed?
Light strikes a sodium surface (work function $2.2\,\mathrm{eV}$) and the stopping potential is $5.0\,\mathrm{V}$. What is the wavelength of the light?
The work function of tungsten is $4.50\,\mathrm{eV}$. Find the speed of the fastest electrons ejected when light of photon energy $5.80\,\mathrm{eV}$ shines on it.
For a photocell to work with visible light via the photoelectric effect, which of these are suitable (work functions in parentheses): tungsten ($4.5\,\mathrm{eV}$), barium ($2.5\,\mathrm{eV}$), lithium ($2.3\,\mathrm{eV}$)? Explain.
Light of wavelength $200\,\mathrm{nm}$ strikes aluminum, which requires $4.20\,\mathrm{eV}$ to eject an electron. Find (a) the kinetic energy of the fastest ejected electron, (b) the stopping potential, and (c) the cutoff wavelength.
Light of wavelength $2.40\,\mathrm{pm}$ is scattered from free electrons. Find the scattered wavelength at (a) $30.0^\circ$ and (b) $120^\circ$ from the incident direction.
What is the maximum Compton wavelength shift for a photon scattering from a free proton?
X-rays of $0.0100\,\mathrm{nm}$ scatter from loosely bound electrons. For backscatter ($180^\circ$), find (a) the Compton shift and (b) the kinetic energy of the recoiling electron.
Calculate the Compton wavelength $h/mc$ for (a) an electron and (b) a proton.
Calculate the de Broglie wavelength of (a) a $1.00\,\mathrm{keV}$ electron, (b) a $1.00\,\mathrm{keV}$ photon, and (c) a $1.00\,\mathrm{keV}$ neutron.
In an old television tube, electrons are accelerated through $25.0\,\mathrm{kV}$. What is their de Broglie wavelength? (Non-relativistic.)
At what kinetic energy would an electron's de Broglie wavelength equal the $590\,\mathrm{nm}$ sodium emission wavelength?
For an electron and a proton with the same (a) kinetic energy, (b) momentum, and (c) speed, which has the shorter de Broglie wavelength?
The uncertainty in an electron's position along the $x$-axis is $50\,\mathrm{pm}$ (about a hydrogen-atom radius). What is the least uncertainty in its momentum component $p_x$?
Consider an electron approaching a barrier of height $U_b = 6.0\,\mathrm{eV}$ and thickness $L = 0.70\,\mathrm{nm}$. What incident energy $E$ gives a transmission coefficient of $0.0010$?

Tip: three habits carry this whole chapter. First, decide early whether you are counting quanta or measuring a wave: a "rate of photons" or "energy per photon" question is just $E = hf = hc/\lambda$ with a power divided through, while a photoelectric question is the single bookkeeping line $hf = K_{\max} + \Phi$ — at the cutoff $K_{\max} = 0$ so $\Phi = hf_0$, and intensity never changes $K_{\max}$, only the number of electrons. Second, keep the two "wavelength" relations distinct: the Compton shift $\Delta\lambda = (h/mc)(1 - \cos\phi)$ is a change in a photon's wavelength on scattering (largest at $180^\circ$, independent of the incident $\lambda$), whereas the de Broglie wavelength $\lambda = h/p$ is the whole wavelength a massive particle carries — and for sub-relativistic particles get $p$ from $p = \sqrt{2mK}$ before dividing. Third, treat $|\psi|^{2}$, not $\psi$, as the physical thing (it is the probability density), remember Heisenberg's floor $\Delta x\cdot\Delta p_x \ge \hbar$ with $\hbar = h/2\pi$, and recall that tunneling $T \approx e^{-2bL}$ is exponentially sensitive to mass, thickness, and the energy gap $U_b - E$ — double the thickness and the probability plummets.