---
title: "When evidence puts on sunglasses"
author: "Miraia S. Chiou"
creator: "Miraia Chiou"
published: "2026-09-07"
identifier: "epics-polarization-001"
source: "https://epistemicphysics.com/articles/when-evidence-puts-on-sunglasses/"
citation: "https://epistemicphysics.com/cite/?work=epics-polarization-001"
---

# When evidence puts on sunglasses

Follow a two-by-two matrix from an evidence ledger into polarizing sunglasses, quantum qubits, and wave interference.

[Read the illustrated article](https://epistemicphysics.com/articles/when-evidence-puts-on-sunglasses/)

Open a drawer on a physicist’s workbench, and you might find three curious objects:

1. A pair of polarized sunglasses designed to cut glare from the ocean.
2. A notebook recording the outcomes of Bernoulli coin flips.
3. A cryogenic chip hosting superconducting quantum qubits.

At first glance, these three items seem to belong to entirely separate domains of science. The sunglasses belong to classical electromagnetic optics. The notebook belongs to Bayesian statistics. The chip belongs to quantum information.

Yet when [I](https://raia.fun/) wrote down the mathematical matrices governing all three systems, the walls between them dissolved: they are governed by the exact same geometric equations.

In all three systems, the state is represented by a positive semidefinite two-by-two Hermitian matrix. The trace records total intensity or evidence, or is fixed to 1 for a normalized qubit, while four times the determinant gives the squared mass coordinate of the shared cone. And an invertible physical filter, with its overall intensity scale factored out, corresponds to an exact Lorentz transformation in special relativity.

Let us put on our mathematical sunglasses and watch the unified geometry of light, belief, and quantum information come into focus.

## The Hermitian matrix: Pauli spins and the light cone

Let us begin with our familiar two-channel Beta evidence notebook.

Suppose an observer records $Y$ positive trials and $N$ negative trials. Adding the standard neutral prior of half a count per channel yields the parameters $\alpha = Y + 1/2$ and $\beta = N + 1/2$.

Define the total concentration $\tau$ and the signed balance $x$:

$$
\tau = \alpha + \beta, \qquad x = \alpha - \beta.
$$

In our earlier studies, we found that the invariant epistemic rest mass squared is given by the difference of squares:

$$
m^2 = \tau^2 - x^2 = 4\alpha\beta.
$$

Now arrange these two positive counts along the diagonal of a two-by-two matrix:

$$
M = \begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix} = \frac{1}{2} \begin{pmatrix} \tau + x & 0 \\ 0 & \tau - x \end{pmatrix}.
$$

Let us compute the fundamental matrix invariants:

- **The trace** (the sum of the diagonal entries):
  $$
  \operatorname{tr}(M) = \alpha + \beta = \tau.
  $$
- **The determinant**:
  $$
  \det(M) = \alpha\beta - 0 = \alpha\beta.
  $$

Multiplying the determinant by 4 reveals our familiar relativistic interval:

$$
4\det(M) = 4\alpha\beta = \tau^2 - x^2 = m^2.
$$

The squared epistemic rest mass of a belief is four times the determinant of its evidence matrix!

Now give the notebook three spatial dimensions. Instead of a single balance coordinate $x$, let the belief point along a three-dimensional spatial vector $\mathbf{x} = (x_1, x_2, x_3)^{\mathsf T} \in \mathbb{R}^3$.

We can package this full three-dimensional state into a $2 \times 2$ complex matrix using the Pauli spin matrices:

$$
\sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad
\sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad
\sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},
$$

where $i^2 = -1$.

Define the general evidence matrix $M$ as:

$$
M = \frac{1}{2} \bigl(\tau I + \mathbf{x} \cdot \boldsymbol{\sigma}\bigr) = \frac{1}{2} \begin{pmatrix} \tau + x_3 & x_1 - i x_2 \\ x_1 + i x_2 & \tau - x_3 \end{pmatrix}.
$$

Notice that $M$ is Hermitian: it equals its own conjugate transpose ($M^\dagger = M$).

Let us calculate the trace and determinant of this general matrix:

$$
\operatorname{tr}(M) = \frac{(\tau + x_3) + (\tau - x_3)}{2} = \tau.
$$

$$
\det(M) = \frac{1}{4}\bigl[(\tau + x_3)(\tau - x_3) - (x_1 - i x_2)(x_1 + i x_2)\bigr] = \frac{1}{4}\bigl[\tau^2 - x_3^2 - (x_1^2 + x_2^2)\bigr] = \frac{\tau^2 - \|\mathbf{x}\|^2}{4}.
$$

Multiplying by 4 yields:

$$
\boxed{4\det(M) = \tau^2 - \|\mathbf{x}\|^2 = m^2.}
$$

The eigenvalues of $M$ are:

$$
\lambda_\pm = \frac{\tau \pm \|\mathbf{x}\|}{2}.
$$

A matrix is positive semidefinite ($M \ge 0$) if and only if both its eigenvalues are non-negative.

Because $\lambda_+ \ge \lambda_-$, this requires $\lambda_- \ge 0$, which means:

$$
\tau \ge \|\mathbf{x}\|.
$$

The condition that an evidence matrix is positive semidefinite is mathematically identical to the condition that a point lies inside the future light cone of Minkowski spacetime!

## The physics of sunglasses: Stokes parameters and polarization

Now take off the mathematician’s hat and step out into the sunlight.

Light is a transverse electromagnetic wave. In a beam propagating along the $z$-axis, the electric field vector oscillates within the two-dimensional $(x, y)$ plane.

Write the complex electric field components as a two-entry Jones vector:

$$
\mathbf{E}(t) = \begin{pmatrix} E_x(t) \\ E_y(t) \end{pmatrix}.
$$

To describe real, partially polarized sunlight, we take the time average of the outer product, forming the optical coherency matrix $J$:

$$
J = \langle \mathbf{E} \mathbf{E}^\dagger \rangle = \begin{pmatrix} \langle |E_x|^2 \rangle & \langle E_x E_y^* \rangle \\ \langle E_x^* E_y \rangle & \langle |E_y|^2 \rangle \end{pmatrix}.
$$

Here the angle brackets denote a time average, and the asterisk denotes complex conjugation.

Now decompose this coherency matrix into four real numbers: the famous Stokes parameters $(S_0, S_1, S_2, S_3)$:

$$
J = \frac{1}{2} \begin{pmatrix} S_0 + S_3 & S_1 - i S_2 \\ S_1 + i S_2 & S_0 - S_3 \end{pmatrix}.
$$

Compare this with our evidence matrix $M$:

- $S_0 = \tau$ is the total optical intensity of the beam.
- $S_3 = x_3$ measures horizontal versus vertical linear polarization.
- $S_1 = x_1$ measures diagonal linear polarization at $+45^\circ$ versus $-45^\circ$.
- $S_2 = x_2$ measures right versus left circular polarization.

The degree of polarization $P \in [0, 1]$ is the normalized length of the spatial polarization vector:

$$
P = \frac{\sqrt{S_1^2 + S_2^2 + S_3^2}}{S_0} = \frac{\|\mathbf{x}\|}{\tau}.
$$

Look at what this correspondence reveals:

1. **A completely unpolarized beam ($P = 0$)**: Light waves oscillate randomly in all directions. The spatial vector vanishes ($\mathbf{x} = 0$). The coherency matrix is proportional to the identity: $J = \frac{\tau}{2} I$. The invariant mass is maximized: $m = \tau$. An unpolarized beam is a balanced, high-mass state of maximal entropy!
2. **A fully polarized beam ($P = 1$)**: Light oscillates in a single pure state (such as horizontal linear polarization). The spatial length matches the intensity: $\|\mathbf{x}\| = \tau$. The determinant vanishes: $\det(J) = 0$. The invariant mass is zero: $m = 0$. A fully polarized beam travels on the lightlike boundary of the cone!

[Same brightness. A different kind of beam.](https://epistemicphysics.com/articles/when-evidence-puts-on-sunglasses/#polarization-lab)

In the interactive laboratory above, adjust the optical intensities and polarization angles. Watch how independent bundles of horizontal and vertical light mix: their diagonal elements sum, their phase fluctuations cancel, and the state settles into the interior of the light cone.

## The quantum qubit: Impurity as epistemic rest mass

Now move from classical optics to a quantum computer.

A qubit is a two-level quantum system spanned by the computational basis states $|0\rangle$ and $|1\rangle$.

In quantum mechanics, a general mixed state is described by a density matrix $\rho$. A density matrix is defined by two mathematical conditions:

1. It is positive semidefinite: $\rho \ge 0$.
2. It has unit trace: $\operatorname{tr}(\rho) = 1$.

We can obtain a valid quantum density matrix simply by dividing our evidence matrix $M$ by its trace $\tau$:

$$
\rho = \frac{M}{\tau} = \frac{1}{2} (I + \mathbf{v} \cdot \boldsymbol{\sigma}),
$$

where $\mathbf{v} = \mathbf{x}/\tau$ is the Bloch vector, living inside the unit Bloch ball ($\|\mathbf{v}\| \le 1$).

Now compute the quantum purity of this state:

$$
\gamma_{\mathrm{pure}} = \operatorname{tr}(\rho^2).
$$

Using the Pauli matrix identity $(\mathbf{v} \cdot \boldsymbol{\sigma})^2 = \|\mathbf{v}\|^2 I$:

$$
\rho^2 = \frac{1}{4} \bigl(I + 2\mathbf{v}\cdot\boldsymbol{\sigma} + \|\mathbf{v}\|^2 I\bigr) \implies \operatorname{tr}(\rho^2) = \frac{1 + \|\mathbf{v}\|^2}{2}.
$$

Now compare quantum purity with our relativistic rest mass:

$$
\frac{m^2}{\tau^2} = 1 - \|\mathbf{v}\|^2 = 1 - \bigl(2\operatorname{tr}(\rho^2) - 1\bigr) = 2\bigl(1 - \operatorname{tr}(\rho^2)\bigr).
$$

Look at that identity:

$$
\boxed{\frac{m^2}{\tau^2} = 2\bigl(1 - \operatorname{tr}(\rho^2)\bigr).}
$$

The fractional epistemic rest mass squared is exactly twice the loss of quantum purity!

In quantum information:

- A pure state ($\operatorname{tr}(\rho^2) = 1$) sits on the surface of the Bloch sphere ($\|\mathbf{v}\| = 1$). Its rest mass vanishes: $m = 0$.
- A maximally mixed state ($\operatorname{tr}(\rho^2) = 1/2$) sits at the exact center of the Bloch sphere ($\mathbf{v} = 0$). Its rest mass reaches its upper bound: $m = \tau$. When the qubit and its environment form a pure joint state, this is also maximum entanglement between them.

Fractional epistemic rest mass measures quantum impurity exactly! For a subsystem of a pure joint state, that same mixedness also measures entanglement; a classical mixture can have the same impurity without entanglement.

This suggests an evidence-aware extension of the usual qubit picture: retain both $\rho$ and $\tau$. The density matrix tells us the normalized state; the extra scale remembers how much supported that description. Whether that scale records detected photons, accumulated trials, or a deliberately designed learning quantity depends on the experiment. Our matrix has room for all three stories.

## Step inside the Bloch ball

Imagine the pure states as spots painted on a globe. Mixing states lets us travel inward, all the way to the balanced center. Choose an analyzer direction $\mathbf n$, a unit vector, and ask whether the state passes that directional test. The quantum probability is

$$
p_+ = \frac{1+\mathbf v\cdot\mathbf n}{2}.
$$

The evidence matrix gives directional weights $q_\pm=(\tau\pm\mathbf x\cdot\mathbf n)/2$. Dividing by their total $\tau$ produces exactly the same pair of probabilities. In the laboratory below, move the rose state and the teal analyzer, then compare the two columns. The [Bloch-ball account in IBM’s quantum information course](https://quantum.cloud.ibm.com/learning/en/courses/general-formulation-of-quantum-information/density-matrices/bloch-sphere) develops this state geometry from density matrices.

[Aim for an outcome inside the Bloch ball](https://epistemicphysics.com/articles/when-evidence-puts-on-sunglasses/#bloch-ball-lab)

Try the center first: every analyzer returns equal probabilities. Then move to the surface and point the analyzer along the state. What must happen when you point it in the opposite direction? Finally, increase $\tau$ while leaving the point still. The evidence matrix grows, while its normalized quantum state stays in place.

## Ask the light a directional question

Return to the optical bench with linearly polarized light at angle $\theta$ and an ideal polarizer at angle $a$. [Malus’s law](https://openstax.org/books/university-physics-volume-3/pages/1-7-polarization) says that the transmitted intensity is multiplied by $\cos^2(a-\theta)$. If only a fraction $P$ of the input is in that linear polarization and the remaining fraction is unpolarized, the transmitted fraction becomes

$$
\frac{I_{\mathrm{out}}}{I_{\mathrm{in}}} = \frac{1-P}{2}+P\cos^2(a-\theta).
$$

Here is the surprising change of coordinates: a physical polarization line has no arrowhead, so rotating it by $180^\circ$ brings it back to the same state. On the Bloch disk, that makes a full $360^\circ$ turn. In our Stokes convention the vectors are $\mathbf v=P(\sin 2\theta,0,\cos 2\theta)$ and $\mathbf n=(\sin 2a,0,\cos 2a)$. Their dot product is $P\cos[2(a-\theta)]$. The identity $\cos^2 u=(1+\cos 2u)/2$ now puts Malus’s law and the qubit probability on the very same curve.

[Turn the analyzer, watch both answers.](https://epistemicphysics.com/articles/when-evidence-puts-on-sunglasses/#analyzer-lab)

Cross the physical axes and watch the Bloch directions become opposites. With fully polarized input, no light passes; with unpolarized input, half passes at every orientation. An ideal polarizer also prepares any transmitted light along its own axis. That state change comes from the projector used in the optical model, an additional physical rule beyond the shared matrix geometry.

## Optical filters as Lorentz boosts: The SL(2, C) isomorphism

How does a beam of light change when it passes through an optical device, such as a polarizing filter?

In wave optics, a linear optical element (a polarizer, waveplate, or attenuator) acts on the electric field vector through a complex $2 \times 2$ Jones matrix $L$:

$$
\mathbf{E}' = L \mathbf{E}.
$$

The coherency matrix transforms according to:

$$
J' = \langle \mathbf{E}' (\mathbf{E}')^\dagger \rangle = \langle L \mathbf{E} \mathbf{E}^\dagger L^\dagger \rangle = L \langle \mathbf{E} \mathbf{E}^\dagger \rangle L^\dagger = L J L^\dagger.
$$

Now take the determinant of both sides:

$$
\det(J') = \det(L J L^\dagger) = \det(L) \det(J) \det(L^\dagger) = |\det L|^2 \det(J).
$$

Notice what happens when the optical element has a determinant of unit magnitude: $|\det L| = 1$.

The determinant of the coherency matrix is invariant:

$$
\det(J') = \det(J) \implies m' = m!
$$

Removing an irrelevant overall phase sets $\det L = 1$. The group of complex $2 \times 2$ matrices with this unit determinant is $\mathrm{SL}(2, \mathbb{C})$.

In mathematical physics, $\mathrm{SL}(2, \mathbb{C})$ is the double cover of the restricted Lorentz group $\mathrm{SO}^+(1, 3)$! Every unimodular optical Jones matrix corresponds to an exact Lorentz transformation in spacetime.

To see this explicitly, factor out the overall attenuation of an optical partial polarizer (a dichroic filter) that favors horizontal over vertical polarization:

$$
L = \begin{pmatrix} e^{r/2} & 0 \\ 0 & e^{-r/2} \end{pmatrix},
$$

where $r > 0$ is the filter strength. Notice that $\det L = e^{r/2} e^{-r/2} = 1$.

Apply this filter to an evidence matrix $M$:

$$
M' = L M L^\dagger = \begin{pmatrix} e^{r/2} & 0 \\ 0 & e^{-r/2} \end{pmatrix} \frac{1}{2}\begin{pmatrix} \tau + x_3 & 0 \\ 0 & \tau - x_3 \end{pmatrix} \begin{pmatrix} e^{r/2} & 0 \\ 0 & e^{-r/2} \end{pmatrix} = \frac{1}{2}\begin{pmatrix} e^r(\tau + x_3) & 0 \\ 0 & e^{-r}(\tau - x_3) \end{pmatrix}.
$$

Now compute the new total intensity $\tau'$ and new balance $x_3'$:

$$
\tau' = \frac{e^r(\tau + x_3) + e^{-r}(\tau - x_3)}{2} = \tau\left(\frac{e^r + e^{-r}}{2}\right) + x_3\left(\frac{e^r - e^{-r}}{2}\right) = \tau\cosh r + x_3\sinh r.
$$

$$
x_3' = \frac{e^r(\tau + x_3) - e^{-r}(\tau - x_3)}{2} = x_3\left(\frac{e^r + e^{-r}}{2}\right) + \tau\left(\frac{e^r - e^{-r}}{2}\right) = x_3\cosh r + \tau\sinh r.
$$

Look at those two equations:

$$
\boxed{
\begin{aligned}
\tau' &= \tau\cosh r + x_3\sinh r, \\
x_3' &= x_3\cosh r + \tau\sinh r.
\end{aligned}
}
$$

These are the exact equations for a relativistic Lorentz boost along the $x_3$ axis!

The determinant-normalized part of this partial polarizer literally performs a Lorentz velocity boost on the beam’s coherency state! An ideal polarizer lies at the singular limit, beyond any finite boost.

For a passive optical device, we can multiply $L$ by an overall attenuation factor so neither field component is amplified. That factor changes the total transmitted intensity. Likewise, a quantum filter is represented by a measurement operator $K$ satisfying $K^\dagger K \le I$, where $I$ is the identity matrix. Its selected output is $K\rho K^\dagger / p_{\mathrm{pass}}$, with passage probability $p_{\mathrm{pass}} = \operatorname{tr}(K\rho K^\dagger)$. Keeping the passage count beside the normalized output gives the geometry a physical experimental record.

Could attention work like this? A system amplifies distinctions in one direction, changes its normalized belief, and records how much input passed through. That is a concrete design for an attention mechanism: learn a matrix filter, carry its transmitted amount, and watch its trajectory through the cone.

## Complex amplitudes: When evidence learns to interfere

Let us take one final leap into the quantum world.

In classical Bayesian statistics, evidence is always real and additive: you count positive trials $\alpha$ and negative trials $\beta$.

What happens if we take the square roots of these counts and combine them into a single complex number?

Define the complex epistemic amplitude:

$$
\psi = \sqrt{\alpha} + i \sqrt{\beta}.
$$

Let us compute the squared modulus of this amplitude:

$$
|\psi|^2 = (\sqrt{\alpha})^2 + (\sqrt{\beta})^2 = \alpha + \beta = \tau.
$$

The squared modulus is the total evidence $\tau$!

Now compute the square of $\psi$ (without taking the absolute value):

$$
\psi^2 = (\sqrt{\alpha} + i \sqrt{\beta})^2 = (\sqrt{\alpha})^2 - (\sqrt{\beta})^2 + 2i\sqrt{\alpha}\sqrt{\beta} = (\alpha - \beta) + 2i\sqrt{\alpha\beta}.
$$

Substitute our coordinates $x = \alpha - \beta$ and $m = 2\sqrt{\alpha\beta}$:

$$
\boxed{\psi^2 = x + i m.}
$$

Look at that identity:

Squaring the complex amplitude $\psi$ automatically places the signed balance $x$ on the real axis and the epistemic rest mass $m$ on the imaginary axis!

Its modulus is:

$$
|\psi^2| = \sqrt{x^2 + m^2} = \sqrt{x^2 + (\tau^2 - x^2)} = \sqrt{\tau^2} = \tau.
$$

The entire geometry of the light cone is encoded in the complex square of a single number!

## The Bhattacharyya overlap: Quantum interference between beliefs

Now consider what happens when two observers combine their amplitudes.

In classical probability, if Observer 1 has parameters $(\alpha_1, \beta_1)$ and Observer 2 has parameters $(\alpha_2, \beta_2)$, pooling independent data adds their observed counts and retains their shared prior once: $\tau_{\mathrm{classical}} = \tau_1 + \tau_2 - n_0$, with $n_0 = 1$ here.

Now suppose they combine their quantum amplitudes first:

$$
\psi_{\mathrm{total}} = \psi_1 + \psi_2 = (\sqrt{\alpha_1} + \sqrt{\alpha_2}) + i (\sqrt{\beta_1} + \sqrt{\beta_2}).
$$

Now calculate the total effective evidence $|\psi_{\mathrm{total}}|^2$:

$$
|\psi_{\mathrm{total}}|^2 = (\sqrt{\alpha_1} + \sqrt{\alpha_2})^2 + (\sqrt{\beta_1} + \sqrt{\beta_2})^2.
$$

Expanding the squares:

$$
|\psi_{\mathrm{total}}|^2 = (\alpha_1 + \beta_1) + (\alpha_2 + \beta_2) + 2\sqrt{\alpha_1\alpha_2} + 2\sqrt{\beta_1\beta_2} = \tau_1 + \tau_2 + 2\bigl(\sqrt{\alpha_1\alpha_2} + \sqrt{\beta_1\beta_2}\bigr).
$$

Substitute the predicted probabilities $p_1 = \alpha_1/\tau_1$ and $p_2 = \alpha_2/\tau_2$:

$$
\boxed{|\psi_1 + \psi_2|^2 = \tau_1 + \tau_2 + 2\sqrt{\tau_1 \tau_2} \left[\sqrt{p_1 p_2} + \sqrt{(1 - p_1)(1 - p_2)}\right].}
$$

Look at the bracketed expression:

$$
B(p_1, p_2) = \sqrt{p_1 p_2} + \sqrt{(1 - p_1)(1 - p_2)}.
$$

In statistics and information theory, this is the celebrated Bhattacharyya coefficient, which measures the geometric overlap between two probability distributions!

- If the two observers agree completely ($p_1 = p_2$), the overlap is $B = 1$. The interference is fully constructive:
  $$
  |\psi_1 + \psi_2|^2 = \tau_1 + \tau_2 + 2\sqrt{\tau_1 \tau_2} = (\sqrt{\tau_1} + \sqrt{\tau_2})^2.
  $$
- If the two observers hold mutually contradictory, dogmatic beliefs ($p_1 = 1$ and $p_2 = 0$), the overlap vanishes: $B = 0$. The cross-term disappears, leaving only classical addition: $\tau_1 + \tau_2$.
- If we introduce an arbitrary quantum relative phase $e^{i\phi}$ between the two arguments ($\psi_{\mathrm{total}} = \psi_1 + e^{i\phi}\psi_2$), the cross-term can even become negative, producing destructive interference!

Destructive interference between arguments allows an autonomous intelligence to cancel out correlated biases and eliminate redundant noise, exactly like noise-canceling headphones.

Picture an inference engine with two ledgers: one for accumulated support, another for phase relationships between routes through an argument. Contradictory routes might cancel in one output channel and reinforce in another. A repeated explanation could change timing without manufacturing a new independent observation. The experiment would be to build such an engine and compare its predictions with ordinary count pooling.

## Looking through the lens

We began with three seemingly unrelated objects on a workbench: sunglasses, coin tosses, and qubits.

By following the mathematics without compromise, we discovered that:

- An evidence ledger forms a positive semidefinite Hermitian matrix $M$, with invariant Minkowski interval $m^2 = 4\det M$.
- Optical polarization parameters are the exact physical counterparts of epistemic evidence coordinates.
- Qubit purity is directly linked to spacetime geometry: fractional epistemic rest mass measures quantum impurity.
- Invertible partial polarizers, after factoring out overall attenuation, act via $\mathrm{SL}(2, \mathbb{C})$, performing exact relativistic Lorentz boosts on transmitted beams.
- Complex epistemic amplitudes $\psi = \sqrt{\alpha} + i\sqrt{\beta}$ generate the light cone through their square ($\psi^2 = x + i m$), while their sum reproduces quantum interference governed by the Bhattacharyya overlap.

The sunglasses, the notebook, and the qubit have given us a shared workbench: positive matrices for states, determinants for cone geometry, filters for selective transformation, and amplitudes for possible interactions.

The universe does not store its physical laws in separate compartments. The same light cone that bounds the cosmic speed of light stabilizes our beliefs, polarizes the sunset over the sea, and guides the quantum computer into the future. Next time a polarizer turns a patch of sky dark, there is a small mathematical universe worth imagining behind the lens.

## Interactive experiments

- [Aim for an outcome inside the Bloch ball](https://epistemicphysics.com/lab/?demo=bloch-ball#experiment): Turn an analyzer toward a target outcome, then discover which targets a mixed state can never reach.
- [Turn the analyzer, watch both answers](https://epistemicphysics.com/lab/?demo=analyzer#experiment): Turn a polarizer and overlay mixed-light Malus transmission with the matching qubit pass-probability curve.
- [Mix two beams of light](https://epistemicphysics.com/lab/?demo=polarization#experiment): Vary polarization and intensity, then follow their combined cone geometry.

## Citation

Chiou, M. S. (2026). When evidence puts on sunglasses. Epistemic Physics. https://epistemicphysics.com/articles/when-evidence-puts-on-sunglasses/

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