Lecture 2.3: Polarization and Measurement

Polarization as a Qubit¶

Light is an electromagnetic wave. For light traveling along the $z$-axis, the electric field oscillates in the transverse $x$-$y$ plane:

The amplitudes $E_x$ and $E_y$ and the relative phase $\phi$ together determine the polarization — the pattern traced by the electric field vector as the wave passes. Linear polarization, circular polarization, and everything in between are all encoded in these parameters.

The Jones Vector¶

We can factor out the propagation $e^{i(kx - \omega t)}$ and focus on what makes different polarizations different. What remains is the Jones vector:

This is a 2D complex vector — exactly like our qubit state vectors! For normalized states (total intensity = 1):

The Six Cardinal Polarization States¶

We identify horizontal and vertical polarization with our computational basis:

From these we can build the diagonal and anti-diagonal linear polarizations:

And the circular polarizations, where the relative phase between $x$ and $y$ components is $\pm \pi/2$:

For circular polarization, the $E_x$ and $E_y$ components have equal amplitude but are 90° out of phase. The electric field vector traces out a circle as the wave propagates — rotating clockwise (R) or counterclockwise (L) when viewed head-on.

The Polarization–Qubit Dictionary¶

These six states map directly onto the Bloch sphere:

Notice what distinguishes these states. $|D\rangle$ and $|A\rangle$ have the same amplitudes — both have $|E_x| = |E_y| = 1/\sqrt{2}$ — but differ by a relative phase of $\pi$. Similarly, $|R\rangle$ and $|L\rangle$ have the same amplitudes but differ by a relative phase of $\pi$. Relative phase is physical: it produces completely different polarizations.

The six states form three pairs of orthogonal states, corresponding to three measurement bases:

Polarizers and Projective Measurement¶

A polarizer transmits light polarized along one direction and blocks the orthogonal component. Let’s figure out two things: how much power comes through, and what is the state of the light afterward.

How Much Power Passes?¶

Consider a general input state $|\psi\rangle = c_0|H\rangle + c_1|V\rangle$ hitting a vertical polarizer. The polarizer only passes the $|V\rangle$ component.

The amplitude to pass is the overlap with $|V\rangle$:

The power (intensity) detected is the magnitude squared of the amplitude:

This is Malus’s Law in Dirac notation: the fraction of power transmitted through a polarizer $|\chi\rangle$ is $|\langle \chi|\psi\rangle|^2$.

What Is the State Afterward?¶

After the vertical polarizer, the output state is:

The polarizer doesn’t just filter — it projects the state onto $|V\rangle$. Whatever the input was, what comes out is always proportional to $|V\rangle$.

The Projection Operator¶

We can write this as the action of a projection operator:

Acting on an arbitrary state:

The $H$ component is gone; only the $V$ component survives.

Similarly, the horizontal projector is:

Projectors Are Idempotent¶

A key property of projectors: applying the same projection twice gives the same result as applying it once.

And more generally:

This makes physical sense: once you’ve projected onto $|V\rangle$, the state is already $|V\rangle$. Running it through the same polarizer again changes nothing.

From Waves to Single Photons¶

Everything so far has been classical wave optics. Malus’s Law tells us what fraction of the intensity passes through a polarizer. Now let’s ask: what happens when we turn the power way down?

Photons as Discrete Energy Packets¶

From the blackbody radiation problem (one of the founding puzzles of quantum mechanics), we know that the electromagnetic field carries energy in discrete packets — photons — each with energy:

where $h = 6.626 \times 10^{-34}$ J·s is Planck’s constant and $\nu$ is the frequency of the light.

Let’s calculate what this energy actually is. For a helium-neon (HeNe) laser at $\lambda = 633$ nm:

That is an incredibly small amount of energy.

How Many Photons in a Laser Pointer?¶

A typical laser pointer has a power of about $P = 1$ mW $= 10^{-3}$ W. Power is energy per unit time, so the number of photons emitted per second is:

That’s about 3 quadrillion photons per second! At this power level, the photons arrive so densely packed that the light appears perfectly continuous. Your detector sees a smooth signal proportional to $|E|^2$.

Turning Down the Power¶

Now imagine we gradually reduce the power. What do we expect to see on a detector?

At 1 mW, we have $\sim 10^{15}$ photons/s — a smooth, continuous signal.

At 1 nW (10^-9 W), we have $\sim 10^{9}$ photons/s — still essentially continuous.

But at extremely low power — say, a few photons per second — something qualitatively different happens.

How Do You Actually Detect Single Photons?¶

A standard photodetector works like this: an incoming photon hits a semiconductor and promotes an electron from the valence band to the conduction band. That electron flows as a current. The current passes through a resistor, and by Ohm’s law ($V = IR$), you measure a voltage drop.

The problem is that one photon produces one electron, which is a fantastically tiny current — far too small to measure directly.

The solution is an Avalanche Photodetector (APD). In an APD, a strong reverse bias voltage accelerates the initial photoelectron so that it knocks out additional electrons through impact ionization. Those electrons are also accelerated, knock out more, and so on — creating a cascade, an avalanche. A single absorbed photon triggers a macroscopic current pulse that you can easily measure.

A more modern alternative is the Superconducting Nanowire Single-Photon Detector (SNSPD). An SNSPD consists of a thin superconducting nanowire, cooled well below its critical temperature and biased with a current just below the critical current. When a single photon is absorbed, it breaks Cooper pairs in a small region of the wire, creating a resistive hotspot. The bias current is forced to divert around this hotspot, which pushes the current density in the remaining superconducting cross-section above the critical value. The entire wire cross-section goes normal (resistive), and the sudden appearance of resistance in the circuit produces a measurable voltage pulse. The wire then cools back to superconducting on a timescale of nanoseconds, ready for the next photon.

Regardless of the detection method, the end result is the same: a voltage spike on our oscilloscope, roughly 1 ps to 1 ns wide, that tells us exactly one photon has arrived.

Single Photons in the Mach-Zehnder Interferometer¶

Now put a single-photon detector at each output port of a Mach-Zehnder interferometer, and send in photons one at a time — on average, about one per second.

Classically, we found that the output intensities depend on the phase difference $\Delta\phi$ between the two arms:

With single photons, you don’t see continuous intensities. Instead, each photon produces a click at exactly one detector — never both, never neither. The photon is indivisible.

But if you accumulate many clicks, the statistics reproduce the classical interference pattern. If $\Delta\phi = 0$ (constructive interference at D1), every photon clicks at D1. If $\Delta\phi = \pi/2$, each photon has a 50/50 chance of clicking at either detector.

Which Path Did the Photon Take?¶

This is the central question — the MZI version of the double-slit argument.

After the first beam splitter, the quantum state of the photon is:

The photon is in a superposition of being in both arms. It accumulates phase in each arm, arrives at the second beam splitter, and interference determines which detector clicks.

How can we be confident the photon really went both ways? Because we see interference. If we block one arm, the interference disappears — both detectors click with equal probability regardless of $\Delta\phi$. The fact that $\Delta\phi$ controls the outcome means both arms must be contributing amplitudes. The only consistent picture is that the single photon traveled both paths simultaneously, as a wave with complex amplitudes along each arm. This is exactly the quantum description we’ve been building: multiple configurations, each carrying a complex amplitude.

The Wave Picture Works — Until Detection¶

So let’s trace the photon through the interferometer:

• At the first beam splitter: the photon is a wave, split into two amplitudes. Fine.

• Traveling through the arms: still a wave, accumulating phase. Fine.

• At the second beam splitter: still a wave, interfering. Fine.

• Right before the detectors: still a wave, with amplitudes at both output ports. Fine.

At every stage, the wave picture is perfectly logical and consistent. But then the photon hits a detector — and we get a click at exactly one location. The wave, which had amplitude at both detectors, seems to “choose” one.

What happened?

The Measurement Problem¶

This has been a source of a great deal of headache in physics. It’s called the measurement problem: how does a quantum superposition become a definite classical outcome?

Our goal here is not to resolve the full philosophical debate, but to understand the physical process that occurs during detection. The key framework is called open quantum system theory — a formalism that describes what happens to a quantum system when information or energy leaks out into an environment and can never be recovered. We won’t develop the full formalism here, but we can understand the essential physics by following the quantum state step by step through the detection process.

Step 1: Superposition at the Detectors¶

Just before detection, the photon is in a superposition of arriving at the two detectors:

where $|\gamma_{\text{D1}}\rangle$ means “photon heading toward detector D1” and likewise for D2.

Step 2: The Photon Interacts with One Electron¶

The photon hits the semiconductor in one of the detectors and is absorbed, promoting an electron from the valence band to the conduction band. But the photon was in a superposition of which detector it was at, so now the superposition transfers to the electron:

This is still a perfectly valid quantum superposition. The photon is gone, but its “which-detector” superposition has been handed off to the electron. In principle, if we could carefully isolate just these two electrons — one at D1, one at D2 — we could interfere them and still find evidence that the photon went both ways.

Step 3: The Avalanche¶

But we don’t isolate the electron. In an APD, that electron is accelerated and collides with other atoms, liberating more electrons. Those electrons hit more atoms. Within nanoseconds, the initial single-electron excitation has spread into an avalanche involving billions of electrons, atoms, and phonons.

Now the state looks something like:

The superposition hasn’t disappeared — it has spread to an enormous number of particles.

Step 4: The Two Branches Become Orthogonal¶

Here is the crucial point. As the entanglement spreads to more and more particles, the two terms in the superposition diverge. By “diverge” we mean they become orthogonal:

When the overlap between two branches of a superposition reaches zero, they can no longer interfere with each other. For all practical purposes, they have become two completely independent macroscopic events — unaware of each other, unable to ever communicate or recombine. You might picture it as two independent simulations running in parallel: one universe where the photon clicked at D1, and another where it clicked at D2.

The Schrödinger Equation Is Enough¶

This process is called decoherence, and it is described entirely by the Schrödinger equation. We did not need to add any special “measurement postulate” or “wavefunction collapse” rule to quantum mechanics. The Schrödinger equation, applied to the photon and the detector and the environment together, naturally produces the branching into orthogonal, non-interfering outcomes.

What the Schrödinger equation does not do is select one branch over the other. From the perspective of the equation, both branches continue to exist. But from our perspective — trapped inside one branch — it appears that nature “chose” one outcome. We experience a click at D1 or D2, never both.

This is admittedly unsatisfying. But it is the current state of the art: the Schrödinger equation fully describes the measurement process, including the apparent randomness, without requiring any additional postulates. The “randomness” we experience emerges from our inability to access the other branch once decoherence has made the two branches orthogonal.

Homework¶

Problem 1: Changing Basis¶

A photon is prepared in the state $|H\rangle$.

(a) Write $|H\rangle$ in the circular polarization basis $\{|R\rangle, |L\rangle\}$. That is, find coefficients $\alpha$ and $\beta$ such that $|H\rangle = \alpha|R\rangle + \beta|L\rangle$.

Hint: Use the definitions $|R\rangle = \frac{1}{\sqrt{2}}(|H\rangle + i|V\rangle)$ and $|L\rangle = \frac{1}{\sqrt{2}}(|H\rangle - i|V\rangle)$, and solve for $|H\rangle$.

(b) If this $|H\rangle$-polarized photon passes through a right-circular polarizer, what is the probability it is transmitted? What about a left-circular polarizer?

Problem 2: Sequential Polarizers¶

(a) A photon is prepared in the state $|H\rangle$ and sent through a $|V\rangle$ polarizer. What fraction of the light is transmitted?

(b) Now insert a $|D\rangle$ polarizer between the $|H\rangle$ source and the $|V\rangle$ polarizer. Walk through the sequence step by step: what is the state after each polarizer, and what is the total fraction of light transmitted?

(c) Instead of a $|D\rangle$ polarizer, insert a polarizer oriented at angle $\theta$ from horizontal. The corresponding state is $|\theta\rangle = \cos\theta\,|H\rangle + \sin\theta\,|V\rangle$. Show that the total transmission through the sequence $|H\rangle \to |\theta\rangle\text{ polarizer} \to |V\rangle\text{ polarizer}$ is:

What angle $\theta$ maximizes the transmission, and what is the maximum?

(d) Now insert $n$ polarizers evenly spaced in angle between 0° and 90°, at angles $\theta_k = k \cdot \frac{90°}{n+1}$ for $k = 1, 2, \ldots, n$. Each consecutive pair of polarizers differs by $\Delta\theta = \frac{90°}{n+1}$. Show that the total transmission is:

(e) Compute $T(n)$ numerically for $n = 1, 2, 3, 5, 10, 50, 100$. What happens as $n \to \infty$?

(f) Explain physically why adding more polarizers — each of which can only absorb light — increases the total transmission. What does this have to do with measurement changing the state?

Problem 3: Quantum Measurement Problem¶

In the Mach-Zehnder interferometer, a single photon enters the first beamsplitter and — based on the interference pattern — appears to travel through both arms simultaneously. Yet when it reaches a detector, it always produces exactly one click at one detector. It is never split between two detectors, and it never fails to arrive.

(a) In your own words, describe the quantum measurement problem. At what point does the photon stop behaving like a wave (traveling both paths, producing interference) and start behaving like a particle (clicking at one detector)? Be specific about what happens at each stage of the experiment. (b) What experimental evidence forces us to conclude the photon travels both paths? What happens to the interference pattern if you block one arm?

(c) Propose a resolution to the measurement problem. There is no single accepted answer — physicists have debated this for a century. Some approaches include: the Copenhagen interpretation, the many-worlds interpretation, decoherence/entanglement with the environment, or objective collapse theories. Pick one (or propose your own) and argue for it. Your answer should address: at what point (if any) does the wave description break down, and why do we only ever experience one outcome?

(d) Is the “collapse of the wavefunction” a physical process, or a mathematical update to our description? Defend your position.

You are welcome (and encouraged) to discuss with classmates, consult textbooks, or use an LLM. If you use an LLM, cite it and briefly note what you used it for. Your final answer must be in your own words and reflect your own understanding — we are looking for evidence that you’ve genuinely thought about this, not for a “correct” answer.