In short

A photon is the quantum of the electromagnetic field — a lump of light energy that cannot be split. For light of frequency f (and vacuum wavelength \lambda = c/f) each photon carries

\boxed{\;E = hf = \dfrac{hc}{\lambda}\;} \qquad\text{and}\qquad \boxed{\;p = \dfrac{h}{\lambda} = \dfrac{E}{c}\;}

where h = 6.626 \times 10^{-34} J·s is Planck's constant. A green photon (\lambda = 550 nm) carries about 3.6 \times 10^{-19} J, which is 2.25 eV. The momentum is tiny but real: the same green photon carries p \approx 1.2 \times 10^{-27} kg·m/s.

Compton scattering. Shine X-rays on a block of graphite and the electrons inside recoil as if each X-ray were a billiard ball of energy hf and momentum h/\lambda. Conservation of energy and momentum, applied to a photon-electron collision, predicts a wavelength shift

\boxed{\;\Delta\lambda = \dfrac{h}{m_e c}\,(1 - \cos\theta)\;}

where \theta is the scattering angle and h/m_e c \approx 2.43 pm is the Compton wavelength of the electron. This shift was measured by Compton in 1923 and settled the question — light carries momentum in discrete parcels.

Wave-particle duality. Light propagates like a wave (it diffracts, interferes, polarises) and interacts like a particle (it arrives one photon at a time, kicks electrons individually, exchanges momentum in collisions). These are not two separate things light does — they are two faces of one quantum object. In a double-slit experiment run at extremely low intensity, you can see the duality: each photon lands as a single dot on the screen, but the accumulated dots, over millions of photons, form the exact wave interference pattern.

Complementarity (Bohr). Any experiment that reveals the wave nature of light (an unobstructed double slit) simultaneously conceals its particle nature (you cannot tell which slit each photon went through). Any experiment that reveals the particle nature (a detector at each slit) destroys the interference pattern. Wave and particle descriptions are not contradictory; they are complementary — each is needed, neither alone is complete.

Point a camera at the night sky from a hillside outside Ladakh. Leave the shutter open for ten minutes. A faint star appears in the image — one whose light takes a hundred years to reach the Earth and is so dim, by the time it arrives, that your camera's sensor receives photons from it only a few per second. Those photons arrive one at a time and are counted one at a time by the silicon pixels. Yet the same light, coming from a bright star, passes through a prism and diffracts neatly — it is a wave with a definite wavelength.

This is the strangest and most important fact about light that physics discovered in the twentieth century. Light is not a wave that sometimes behaves like a particle, and it is not a particle that sometimes behaves like a wave. It is a quantum object — something whose behaviour in any particular experiment is captured either by a wave picture or a particle picture, but never both at once. The article on the photoelectric effect showed that light delivers energy in discrete parcels. This article takes that idea and completes it: those parcels — photons — also carry momentum, they collide with electrons like billiard balls, and they do all of this while, in a separate experiment on the same desk, interfering with themselves like waves. Everything about this sounds wrong. It is the truth.

Photons carry energy — a quick recap and the generalisation

In the photoelectric effect you saw that a beam of light of frequency f behaves, for the purpose of knocking electrons out of a metal, as if it were a stream of particles each carrying energy

E = hf

Why: this was Einstein's 1905 move. Only a particle-like delivery of energy explains the threshold frequency (below some f_0 no electrons are emitted, no matter how bright the light) and the linear rise of maximum kinetic energy with frequency.

For a wave in vacuum, frequency and wavelength are tied by c = f\lambda, so you can equally write

E = hf = \frac{hc}{\lambda}.

Three numerical facts you should keep in your head:

Photon energy across the electromagnetic spectrumA horizontal spectrum bar labelled from radio at the left through microwave, infrared, visible (shown in colour), ultraviolet, X-ray and gamma. A second row gives approximate photon energies in electronvolts, rising from around 10⁻⁶ eV for radio to 10⁶ eV for gamma.radiomicroIRvisibleUVX-rayγ10⁻⁶ eV10⁻³ eV0.1 eV2–3 eV10 eV10³–10⁵ eV>10⁶ eVlonger λ, smaller Eshorter λ, larger Eone photon of visible light ≈ 2–3 eV ≈ the energy to break a weak chemical bond
Photon energy runs over twelve orders of magnitude from radio through gamma. Visible-light photons sit in the chemistry-relevant band — a few electronvolts, the scale of atomic and molecular transitions.

This order-of-magnitude feel is useful everywhere. An FM radio wave at 100 MHz has a photon energy of h \times 10^8 \approx 4 \times 10^{-7} eV — so tiny that a tuned receiver picks up 10^{15} photons per second without ever noticing them individually. An X-ray photon at 10^{18} Hz has E \approx 4000 eV, enough to ionise an inner-shell electron. Gamma rays go higher still. The same formula covers all of it.

Photons carry momentum — deriving p = h/\lambda

Here is the step that is not obvious. A photon has no rest mass. Yet it carries momentum. Where does this come from?

Start from Einstein's relativistic relation between energy, momentum, and mass:

E^2 = (pc)^2 + (m c^2)^2.

Why: this is the relativistic Pythagorean theorem for a particle, derived in any special-relativity text from the four-momentum p^\mu p_\mu = m^2 c^2. It reduces to E = mc^2 for a stationary particle and to E = pc for a massless one. You do not need to derive relativity here — but you should recognise this relation as the natural generalisation of E = \tfrac12 mv^2 once you include rest-mass energy.

A photon has m = 0. Substitute:

E^2 = (pc)^2 \quad\Rightarrow\quad E = pc \quad\Rightarrow\quad p = \frac{E}{c}.

Why: the photon sits on the "null" line of the mass relation — energy purely kinetic, no rest energy.

Now substitute E = hf = hc/\lambda:

p = \frac{E}{c} = \frac{hf}{c} = \frac{h}{\lambda}.
\boxed{\;p = \frac{h}{\lambda}\;}

Why: momentum is inversely proportional to wavelength. Longer waves, softer kicks. An X-ray photon at 0.1 nm carries p = 6.626\times10^{-34}/10^{-10} = 6.6\times10^{-24} kg·m/s — comparable to the momentum an electron picks up after falling through 100 V.

A green photon (\lambda = 550 nm) carries p = h/\lambda = 6.626\times10^{-34}/5.5\times10^{-7} = 1.2\times 10^{-27} kg·m/s. A laser pointer delivering 10^{17} photons per second exerts a force on whatever absorbs it of roughly 10^{17} \times 1.2\times10^{-27} = 1.2 \times 10^{-10} N — ten nanonewtons, the mass of a single bacterium. Small, but measured and measurable. This is radiation pressure. It is what makes a comet's tail point away from the Sun (the Sun's photons push gas and dust outwards). It is what ISRO's solar-sail concepts would use for deep-space propulsion. It is the force that "optical tweezers" use to hold a single bacterium suspended in a laser beam.

Compton scattering — photons as billiard balls

If photons really carry momentum h/\lambda, then a collision between a photon and a free electron should obey conservation of momentum and energy exactly like any other two-body collision — the kind you derived in elastic collisions. Arthur Compton did this experiment in 1923 and the result was decisive.

The setup. A beam of monochromatic X-rays of wavelength \lambda_0 is directed at a thin block of graphite. The graphite's loosely bound electrons act, to good approximation, as free stationary particles. X-rays that scatter off these electrons emerge at an angle \theta to the original beam; a moveable detector measures the wavelength \lambda' of the scattered X-rays as a function of \theta.

The result. The scattered X-rays have a longer wavelength than the incident ones, and the shift depends only on the angle:

\Delta\lambda = \lambda' - \lambda_0 = \frac{h}{m_e c}\,(1 - \cos\theta).

At \theta = 90° the shift is h/m_e c = 2.43 pm; at \theta = 180° (backscatter) it is twice that. A classical wave picture predicts no wavelength shift at all — the electron oscillates at the wave's frequency and re-radiates at the same frequency. The Compton shift is a particle effect.

Compton scattering geometryAn incoming photon of wavelength lambda-nought approaches a stationary electron at the origin from the left. After the collision, a scattered photon of longer wavelength lambda-prime goes out at angle theta above the original direction, and the electron recoils at angle phi below.incident directione⁻ (at rest)incoming photon, λ₀, p₀ = h/λ₀scattered photonλ′ > λ₀, p′ = h/λ′recoiling electron, p_eθφ
Compton geometry. A photon of momentum $p_0 = h/\lambda_0$ scatters off a stationary electron. The photon emerges at angle $\theta$ with longer wavelength $\lambda' > \lambda_0$. The electron recoils at angle $\phi$ carrying the missing energy and momentum.

Deriving the Compton shift

Treat the photon–electron encounter as a two-body collision with the photon's four-momentum known on both sides. Use non-relativistic notation for clarity, keeping the photon fully relativistic and the electron relativistic where it matters.

Step 1. Write conservation of momentum. Let the incoming photon momentum be p_0 = h/\lambda_0 along \hat x, the scattered photon momentum be p' = h/\lambda' at angle \theta above \hat x, and the electron recoil momentum be p_e at angle \phi below \hat x.

x:\quad p_0 = p' \cos\theta + p_e \cos\phi
y:\quad 0 = p' \sin\theta - p_e \sin\phi

Why: the electron was at rest, so the total momentum before is purely the incoming photon's. Both components must match after.

Step 2. Eliminate the recoil angle \phi. Rearrange the x and y equations:

p_e \cos\phi = p_0 - p'\cos\theta, \qquad p_e \sin\phi = p'\sin\theta.

Square and add:

p_e^2 = (p_0 - p'\cos\theta)^2 + (p'\sin\theta)^2 = p_0^2 - 2 p_0 p' \cos\theta + p'^2.

Why: the identity \cos^2\phi + \sin^2\phi = 1 kills the unknown angle \phi, leaving one equation in the three magnitudes p_0, p', p_e.

Step 3. Write conservation of energy. The incoming photon has E_0 = p_0 c. The scattered photon has E' = p' c. The electron starts at rest (rest energy m_e c^2) and ends with total energy \sqrt{(p_e c)^2 + (m_e c^2)^2}. So

p_0 c + m_e c^2 = p' c + \sqrt{(p_e c)^2 + (m_e c^2)^2}.

Why: relativistic energy conservation. The photon is always relativistic; so is the electron after absorbing the X-ray's kick, since the kick can give it speeds comparable to c.

Step 4. Rearrange and square.

\sqrt{(p_e c)^2 + (m_e c^2)^2} = (p_0 - p')c + m_e c^2.

Squaring:

(p_e c)^2 + (m_e c^2)^2 = (p_0 - p')^2 c^2 + 2(p_0 - p') c \cdot m_e c^2 + (m_e c^2)^2.

Cancel (m_e c^2)^2 and divide through by c^2:

p_e^2 = (p_0 - p')^2 + 2(p_0 - p') m_e c.

Why: squaring removes the square root and the electron's rest-energy term cancels, leaving a clean relation between p_e^2, (p_0 - p'), and m_e c.

Step 5. Substitute the momentum-conservation result for p_e^2 from Step 2 and expand.

p_0^2 - 2 p_0 p' \cos\theta + p'^2 = p_0^2 - 2 p_0 p' + p'^2 + 2(p_0 - p') m_e c.

The p_0^2 and p'^2 terms cancel. Rearranging:

2 p_0 p'(1 - \cos\theta) = 2(p_0 - p') m_e c.

Divide by 2 p_0 p' m_e c:

\frac{1 - \cos\theta}{m_e c} = \frac{1}{p'} - \frac{1}{p_0}.

Why: this is the step that turns the momentum bookkeeping into a wavelength statement. The left side contains the scattering angle and the right side the momentum difference — and photon momentum is h/\lambda.

Step 6. Substitute 1/p = \lambda/h on the right.

\frac{1 - \cos\theta}{m_e c} = \frac{\lambda'}{h} - \frac{\lambda_0}{h} = \frac{\lambda' - \lambda_0}{h}.

Multiplying through by h:

\boxed{\;\Delta\lambda = \lambda' - \lambda_0 = \frac{h}{m_e c}\,(1 - \cos\theta)\;}

Why: the final result, known as the Compton wavelength shift. The quantity \lambda_C \equiv h/m_e c = 2.426 \times 10^{-12} m is the Compton wavelength of the electron. It is a universal constant — the shift depends only on the angle, never on the initial wavelength or the intensity of the beam.

What this means. The shift is measurable only when the incident wavelength is itself of order picometres — hence X-rays, not visible light. For a visible photon (\lambda_0 \sim 500 nm) a shift of 2.4 pm is a fractional change of 5 \times 10^{-6}, far below spectroscopic resolution. For a 70 pm X-ray (typical of a copper K-alpha source) it is a fractional change of 3 percent — easily seen. This explains why Compton's experiment needed X-rays and why centuries of visible-light experiments never hinted at this effect.

Compton's measurements agreed with the formula to within experimental error and earned him the 1927 Nobel Prize. The key point is not the number — it is that the formula was derived assuming the photon behaves as a particle with energy hf and momentum h/\lambda, colliding with the electron using ordinary conservation laws. No wave argument could reproduce it. After Compton, the particle nature of light was not a hypothesis; it was a measurement.

The double slit, one photon at a time

You already know, from Young's double slit experiment, that a coherent beam of light passing through two slits produces a pattern of bright and dark fringes on a screen. The fringe spacing \beta = \lambda D/d is a wave formula — it comes from superposition of two waves with a path difference.

Now turn the intensity down. Way down. Turn it down far enough that, on average, only one photon is in flight between the source and the screen at any given time. What happens?

If light were truly a classical wave, you would expect the screen to light up dimly but continuously, with the full fringe pattern faintly visible everywhere at once. If light were truly a classical particle, you would expect each photon to pass through one slit or the other like a bullet, and you would see two bright stripes on the screen behind the slits — no fringes, because there is no interference between particles that went through different slits.

The experiment has been done, many times, in many labs. Here is what happens. Each photon arrives at the screen as a single dot, at a single place — a single silver grain exposed, or a single photomultiplier click. The arrivals look random: if you take a photograph after a handful of photons, you see a scatter of dots with no obvious pattern.

Leave the experiment running. After a million photons, the scatter of dots has filled in the exact interference pattern predicted by the wave theory — bright bands, dark bands, the same \beta = \lambda D/d spacing, everything.

Interactive build-up of a two-slit interference pattern, photon by photonA two-slit intensity distribution on a screen. A draggable red point controls the number of photons N that have arrived, from a few up to a million. At low N the pattern is flat with fluctuations; at high N the classical cos-squared fringes emerge sharply. Readouts display N and the fringe visibility.position on screendetection ratedrag the red point
Drag the red point to increase $\log_{10} N$, the log of the photon count. At $N \sim 10$ the distribution looks flat (the fringes are invisible in the statistical noise). By $N \sim 10^6$ the clean $\cos^2$ pattern has emerged. Each individual photon still arrives as a single random dot — the fringes are only a statement about how those dots are distributed.

This is what "wave-particle duality" actually says. The photon arrives like a particle — one discrete interaction with the screen at one point. But the probability that it lands at any given point is computed from a wave — the amplitude of a wave that passed through both slits, interfering with itself. The single photon does not "split" and go through both slits as a classical wave would; nor does it "choose" one slit as a classical particle would. It exists as a quantum object whose propagation is wave-like and whose detection is particle-like.

Here is the punchline. If you put a detector at each slit to see which one the photon went through, the fringe pattern disappears. The moment you extract which-slit information, the wave behaviour dies. The photon now truly chooses a slit, and what appears on the screen is two bright bands, no fringes. This is not a technical limitation of the detectors; it is a statement about what "existing in two places at once" can mean for a quantum object. Observation changes the outcome — not because observation is invasive (though it can be), but because the wave description and the particle description are mutually exclusive in the same experiment.

The complementarity principle

Niels Bohr made this principle the philosophical centrepiece of quantum mechanics: for a quantum system, wave and particle descriptions are complementary. Each one applies in some experiments and not others, each one is needed for a full description, and they cannot both be applied at the same time to the same event.

Three equivalent ways to say this:

Complementarity is not "light is a wave on Mondays and a particle on Wednesdays". It is "any complete description of light involves both languages, and the two languages each answer different questions about the same underlying quantum object".

Worked examples

Example 1: Energy and momentum of a photon in Compton scattering

A beam of X-rays of wavelength \lambda_0 = 70.8 pm (from the copper K-alpha line) is scattered off a graphite block. A detector is placed at \theta = 90°. Find (a) the wavelength of the scattered X-rays, (b) the energy of one incident photon, (c) the energy carried away by the recoiling electron, and (d) the speed of the electron in non-relativistic approximation.

Step 1. Compton shift at 90°.

\Delta\lambda = \frac{h}{m_e c}(1 - \cos 90°) = \lambda_C \cdot 1 = 2.43\ \text{pm}.

Why: \cos 90° = 0, so the shift is exactly one Compton wavelength — the universal unit.

So \lambda' = \lambda_0 + 2.43 = 73.23 pm.

Step 2. Energy of the incident photon.

E_0 = \frac{hc}{\lambda_0} = \frac{1240\ \text{eV}\cdot\text{nm}}{70.8\times10^{-3}\ \text{nm}} = 17\,514\ \text{eV} \approx 17.5\ \text{keV}.

Why: hc in eV·nm makes this a single division. A 70.8 pm photon is a typical hard X-ray.

Step 3. Energy of the scattered photon.

E' = \frac{hc}{\lambda'} = \frac{1240}{73.23\times10^{-3}} = 16\,933\ \text{eV} \approx 16.93\ \text{keV}.

Step 4. Energy given to the electron by conservation.

K_e = E_0 - E' = 17\,514 - 16\,933 = 581\ \text{eV}.

Why: energy conservation. The photon lost this much; the electron (starting at rest) gained exactly this much as kinetic energy.

Step 5. Electron speed, assuming non-relativistic.

K_e = \tfrac12 m_e v^2 \quad\Rightarrow\quad v = \sqrt{\frac{2 K_e}{m_e}}.

Convert K_e = 581 eV = 9.3\times10^{-17} J and use m_e = 9.11\times10^{-31} kg:

v = \sqrt{\frac{2 \times 9.3\times10^{-17}}{9.11\times10^{-31}}} \approx \sqrt{2.04\times10^{14}} \approx 1.43\times10^{7}\ \text{m/s}.

Why: v/c \approx 0.05, so non-relativistic kinematics (K = \tfrac12 mv^2) is accurate to better than 1%. For much higher-energy photons the full relativistic formula would be needed.

Result. \lambda' = 73.23 pm, E_0 = 17.5 keV, E' = 16.93 keV, K_e = 581 eV, v \approx 1.43 \times 10^7 m/s.

Compton scattering at 90° — energy bookkeepingTwo horizontal energy bars: the top bar shows 17.5 keV for the incident photon as a single wide block. The bottom bar shows the same total length split into a scattered-photon segment of 16.93 keV and a recoil-electron segment of 0.58 keV, labelled.energy bookkeeping at θ = 90°incident photon: 17.514 keVscattered photon: 16.933 keVrecoil e⁻0.581 keV
The scattered photon keeps 96.7% of the incident energy; the electron carries away 3.3%. The small fraction makes sense — the electron is much heavier than the photon's "effective mass" $E/c^2$.

What this shows. Most of the incident energy survives the scattering; only a few percent goes to the electron at 90°. The fraction rises as \theta \to 180° (backscatter), where the full Compton shift is doubled and the electron can take away tens of percent of the energy at keV scales. You have the tools to recompute any of these cases.

Example 2: Photon count from a bulb

A 40 W white-light bulb radiates uniformly in all directions, with an average photon wavelength of 550 nm. (a) How many photons per second does it emit? (b) How many photons per second enter the pupil of a student whose eye (pupil radius 3 mm) is 2 m from the bulb?

Step 1. Energy per photon.

E = \frac{hc}{\lambda} = \frac{1240}{550}\ \text{eV} = 2.25\ \text{eV} = 3.61\times10^{-19}\ \text{J}.

Why: the 1240 eV·nm trick again. Every visible photon has energy of order a few eV — this is why chemical bond energies and photon energies overlap.

Step 2. Assume the bulb converts 100% of its electrical power to light (an overestimate — real bulbs are more like 5% for incandescent, 25% for LED, but it keeps the arithmetic clean). Number of photons per second:

N/t = \frac{P}{E} = \frac{40}{3.61\times10^{-19}} \approx 1.1 \times 10^{20}\ \text{photons/second}.

Why: power is energy per second; dividing by energy per photon gives photons per second.

Step 3. Fraction of those photons entering the pupil. The bulb radiates uniformly over a sphere of radius r = 2 m, area 4\pi r^2 = 4\pi(2)^2 = 50.3\ \text{m}^2. The pupil has area \pi (3\times10^{-3})^2 = 2.83\times10^{-5}\ \text{m}^2. Fraction:

f = \frac{2.83\times10^{-5}}{50.3} = 5.6\times10^{-7}.

Why: geometric dilution. Only the photons heading in the direction of the pupil are caught; the rest fly past.

Step 4. Photon rate into the eye.

N_\text{eye}/t = f \cdot N/t = 5.6\times10^{-7} \times 1.1\times10^{20} \approx 6.2 \times 10^{13}\ \text{photons/s}.

Result. The bulb emits \sim 10^{20} photons/s; the eye catches \sim 6 \times 10^{13} of them each second.

Photon flux from a bulb to the eyeA 40 W bulb on the left radiates outward in all directions, shown as rays into a sphere of radius 2 m. A small disc representing the pupil of the eye is on the right; only the rays heading into the disc are caught. Labels give the photon rates at each stage.40 W bulb1.1×10²⁰ γ/sr = 2 mpupil (3 mm)6×10¹³ γ/s
Even a modest bulb fires out $10^{20}$ photons per second. The eye, catching just a millionth of a millionth of the sphere, still registers tens of trillions of photons per second — which is why we never notice the discrete nature of light in daily life.

What this shows. The photon flux from even a dim source is staggering. The quantum nature of light is invisible at ordinary intensities because the number of photons is so large that statistical fluctuations average out. The discrete, particle-like behaviour shows up only in ultra-low-light experiments (astronomy, single-photon optics, photon-counting detectors of the kind ISRO flies on astronomy payloads) or in interactions with individual atoms (photoelectric, Compton, atomic spectra).

Common confusions

Where Indian physics sits in this story

The particle nature of light is braided with three Indian contributions, two of which are direct and one through the statistics that govern every photon ever emitted.

Satyendra Nath Bose (1924). While teaching at Dhaka University, Bose wrote a short paper deriving Planck's law of blackbody radiation by treating photons as indistinguishable particles with a new counting rule. He sent it to Einstein, who recognised it as exactly the right foundation for quantum statistics, translated it into German, and sent it to be published. The particles that obey "Bose statistics" — photons, gluons, the Higgs — are called bosons in his honour. Every equation that describes how photons populate the modes of a laser, a blackbody, or the cosmic microwave background traces back to Bose's Dhaka calculation.

C.V. Raman (1928). Scattered light that exchanges a quantum of vibrational energy with a molecule is wavelength-shifted (as in Compton scattering, but with molecular vibration rather than electron recoil as the partner). Raman discovered this effect in liquid benzene and won the 1930 Nobel Prize for it — the first Asian scientist to do so. Raman scattering is a direct consequence of the photon picture: a classical wave theory predicts elastic scattering only.

Photon-counting instruments in Indian astronomy. The Astrosat satellite launched by ISRO in 2015 carries an Ultraviolet Imaging Telescope (UVIT) that images the sky in the UV band by counting individual photons — at the low intensities of distant galaxies, every photon matters and the discrete-particle nature of light is not a curiosity but an engineering reality. The same is true for the X-ray instrument on Astrosat and for the gamma-ray observatories at Ooty and Hanle.

Going deeper

If what you came for was the photon energy and momentum formulas and a sense of how they relate, you have them. What follows is the rigorous version — the quantised-field language in which a photon is really defined, the full relativistic Compton formula, single-photon interference, and a precise statement of complementarity.

The photon as a quantum of the electromagnetic field

Classically, the electromagnetic field is a continuous function (\vec E, \vec B) that can have any amplitude. When you quantise this field — the subject of quantum electrodynamics — you find that each mode of the field (each allowed wave pattern in the cavity or in free space) behaves like a quantum harmonic oscillator with spacing \hbar\omega = hf. The n-th energy level of the mode corresponds to n photons of frequency f. A photon is not a little object travelling along; it is an excitation of a particular mode of the field, carrying one quantum of that mode's energy and one quantum of its momentum \hbar \vec k. Two-slit interference is then a statement about the mode: a single-photon state that is a superposition of "went through slit 1" and "went through slit 2" is a single photon living in a two-slit mode, and its detection probability at the screen is the squared amplitude of that mode at each point.

The full relativistic Compton formula, with the electron at rest

The derivation in the body took the electron as initially at rest. Nothing about the steps required the electron to be slow; the squared-four-momentum algebra handles arbitrary final electron velocities. You can now read the derivation as exact to all orders in v/c, and the recoil kinematics at high photon energies (where the electron becomes relativistic) follow from the same \Delta\lambda = \lambda_C(1-\cos\theta) without correction. For extremely high-energy photons (E_0 \gtrsim m_e c^2 \approx 511 keV) the scattered photon loses a large fraction of its energy and the electron recoils relativistically — this is the regime exploited by gamma-ray detectors and by the Klein-Nishina formula, which gives the full angular cross-section.

Single-photon interference — the Mandel-Pfleegor experiment and what it settled

In 1967 Mandel and Pfleegor performed an interference experiment between two independent lasers at intensities so low that at most one photon from either source was in the apparatus at any time. Fringes appeared. The result proved that interference is not between two photons of the same source — it is a property of the quantum state of the electromagnetic field itself, which can carry superpositions across independent sources when the geometry is right. The conclusion: a photon "interferes only with itself" (Dirac's phrase), where "itself" means the full quantum state of the field, not the individual classical-emission event. For JEE, the relevant statement is simpler — interference works one photon at a time and the pattern is built statistically.

Heisenberg uncertainty and the sharpness of complementarity

Complementarity is quantified by an uncertainty relation. If V is the visibility of the fringes on the screen (how clean the interference pattern is) and D is the "distinguishability" of the two paths (how well you know which slit the photon went through), Englert proved in 1996 that

V^2 + D^2 \le 1.

Perfect fringes (V=1) demand no path information (D=0). Full path information (D=1) forces flat fringes (V=0). Any partial measurement (a weak which-path monitor) sits on the curve — you get some of both, bounded by a circle. This is complementarity as a sharp theorem, not a slogan.

Why photons are bosons

The statistics a particle obeys is determined by its spin: spin-integer particles are bosons, spin-half-integer particles are fermions. Photons have spin 1, so they are bosons — they obey Bose-Einstein statistics, the same rule Bose derived in 1924 for the Planck law. One consequence is stimulated emission: a photon of frequency f passing an excited atom encourages the atom to emit a second identical photon into the same mode. This is the operating principle of every laser in the world. Lasers would not work if photons were fermions; Pauli exclusion would forbid two in the same mode.

Photon momentum at work — laser cooling

If an atom moving towards you at speed v absorbs a photon of frequency just below its natural absorption line, the Doppler shift brings the photon into resonance; the atom absorbs it and picks up a kick of momentum h/\lambda in the opposite direction to its motion. A laser beam tuned this way systematically slows atoms moving towards it and does nothing to atoms moving away. Three pairs of counter-propagating laser beams in orthogonal directions can cool a gas of atoms to microkelvin temperatures — the "optical molasses" technique that underlies atomic clocks, Bose-Einstein condensates in the lab, and precision quantum-metrology experiments currently being set up at the Raman Research Institute in Bengaluru.

Where this leads next