Detection and the Double Slit

Physics / Photon

Detection and the Double Slit

We've seen that the electromagnetic field of a photon expands as a spherical shell with a structure that extends throughout space. Now we confront the deepest puzzle: how that extended field interacts with matter.

Detection is Localized and Quantized

The field is spread across space. But detection is the opposite — it happens at a single point, transferring a discrete quantum of energy:

: Energy transferred in one absorption event — always this exact amount: Planck's constant: $6.626 \times 10^{-34}$ J·s: Reduced Planck's constant: $\hbar = h/(2\pi) = 1.055 \times 10^{-34}$ J·s: Frequency of the electromagnetic wave: Angular frequency: $\omega = 2\pi\nu$

When detection occurs, the entire spatially-extended field vanishes. Not gradually — instantly. This is the "collapse" of quantum mechanics. The field was on a sphere 16 light-minutes across, and now it's gone, its energy deposited in one atom.

The probability of detection at any point follows the Born rule:[1]

: Probability density for a detection event at position $\mathbf{r}$: amplitude: Classical electromagnetic field — the classical limit of the quantum amplitude

For an electromagnetic field, the quantum probability amplitude is directly proportional to the classical field strength. This is why classical optics works: the interference patterns predicted by Maxwell's equations correctly predict the probability distribution of quantum detection events.

The Double Slit Experiment

Watch this in the double-slit experiment: the field passes through both slits, interferes with itself, then collapses to one detection event. Over many photons, the random detections build up into an interference pattern.

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Fraunhofer Diffraction

The detection probability on a screen at distance $L$ from two slits of width $a$ separated by distance $d$:[2]

: Double-slit interference: maxima where path difference is a whole number of wavelengths: Single-slit diffraction envelope: the finite slit width limits the pattern: Centre-to-centre distance between the two slits: Width of each individual slit: Wavelength of the photon: Distance from the slits to the detection screen: Position on the screen, measured from the central axis

where $\operatorname{sinc}(x) = \sin(x)/x$.

The $\cos^2$ term comes from the interference between the two slits — it creates the fringes. The $\operatorname{sinc}^2$ term is the diffraction envelope from each individual slit — it modulates the overall brightness. Together they predict the exact positions of every bright and dark band.

This formula is the Fraunhofer (far-field) limit, valid when $L \gg d^2/\lambda$. It assumes the path difference can be approximated as $d \sin\theta \approx dy/L$.

What the double slit proves
Each photon's field passes through **both** slits simultaneously. The interference pattern is not built from photons "choosing" one slit — it emerges because the field amplitude at each point on the screen is the **sum** of contributions from both slits. The detection probability $P \propto |E_1 + E_2|^2$ includes the cross term $2 \operatorname{Re}(E_1^* E_2)$ that creates the fringes.

The Field Itself is Not Quantized

Key Insight
**The electromagnetic field itself is not quantized.** It can take on any value. What *is* quantized are the **energy transfer processes** between the field and matter — absorption and emission always occur in discrete units of $\hbar\omega$.[3]

This distinction is crucial and widely misunderstood. Consider a laser beam passing through neutral density filters:

The beam starts as a continuous electromagnetic wave — a coherent field produced by stimulated emission. Each filter absorbs discrete quanta of energy from the field ($\hbar\omega$ per absorption event), but these absorption events happen randomly in time and are spread throughout the filter volume. After the filter, the remaining field is still continuous and coherent, just weaker.[3]

Single-Photon Intensity Calculation
For a HeNe laser (632.8 nm) with a path length of $L = 1300$ mm: The maximum lifetime of a photon in the setup: $$\tau = \frac{L}{c} = \frac{1.3}{3 \times 10^8} = 4.3 \text{ ns}$$ For an average of one photon in the beam at any time: $$\Phi = \frac{1}{\tau} = 2.3 \times 10^8 \text{ photons/s}$$ $$P = \Phi \cdot h\nu = \Phi \cdot \frac{hc}{\lambda} = 2.3 \times 10^8 \times \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{632.8 \times 10^{-9}} = 74 \text{ pW}$$ With optical density 9 filters (transmission $10^{-9}$), the beam is attenuated from 0.3 mW to 0.3 pW — about **250 times below** single-photon-equivalent intensity. Yet interference is still observed.[3] The field is continuous — the filters reduced its amplitude, but it is still a coherent wave capable of self-interference.
Light Source Matters
Had the experiment used a fluorescent lamp (spontaneous emission) instead of a laser, the single-photon calculation would have been valid — and interference would **not** have been observed in this setup.[4] Spontaneous emission produces a field with short coherence length ($L_c \sim \lambda^2/\Delta\lambda$, typically microns for broadband sources), so the two paths cannot interfere when their length difference exceeds $L_c$.

Shot Noise: The Randomness of Detection

At very low light levels, individual pixels light up at seemingly random positions. This is shot noise — the inherent randomness of quantum detection events.[4]

The spatial distribution differs between measurements even under identical conditions. Each detection is a random sample from the probability distribution $P \propto |E|^2$. Only after many detections does the pattern emerge.

Interference is never observed from a single detection. It requires the statistics of many events. What we call an "interference pattern" is really a probability distribution — the $|E|^2$ pattern made visible by accumulating thousands of individual, random, localized detection events.

Energy of a Visible Photon
A green photon with wavelength $\lambda = 532$ nm: $$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{532 \times 10^{-9}} = 5.64 \times 10^{14} \text{ Hz}$$ $$E = h\nu = 6.626 \times 10^{-34} \times 5.64 \times 10^{14} = 3.74 \times 10^{-19} \text{ J} \approx 2.33 \text{ eV}$$ This is the exact amount of energy transferred when this photon is absorbed — no more, no less, regardless of how spread out the field was before collapse.

The shot noise is not a limitation of the detector. It is fundamental. Even a perfect detector would show the same randomness, because the randomness is in the quantum detection process itself — the field's probability distribution determines where detections are likely, but each individual detection is irreducibly random.

Sources

[1]Mandel, L. & Wolf, E. — Optical Coherence and Quantum Optics (Cambridge, 1995), Ch. 12
[2]Saleh, B.E.A. & Teich, M.C. — Fundamentals of Photonics (Wiley, 2007), Ch. 2
[3]Huygens Optics — How big is a visible photon?
[4]Fox, M. — Quantum Optics: An Introduction (Oxford, 2006), Ch. 5–7