How far apart are two quantum messages?

Imagine sending a secret message with two tiny arrows.

An arrow leaning left means A. An arrow leaning right means B. Your friend owns a receiver that can ask the arrow one directional question and record a plus or minus answer. The more differently your arrows answer that question, the easier the message is to read.

Now make the arrows quantum states. An unfamiliar piece of Epistemic Physics walks into the communications room: the mass coordinate of the two arrows’ average tells us exactly how well the best receiver can distinguish them.

We will build that equation, turn the receiver ourselves, then perform a small magic trick. We will keep the average state fixed while changing the readable information from one whole bit to zero. The missing ingredient will be the connection between the message labels and the states we prepared.

A qubit can be an arrow inside a ball

A qubit is a quantum system with two basis states. Its state can be represented by a density matrix, a positive two-by-two matrix whose diagonal entries sum to one. A convenient form is

Here $i^2=-1$, and the real vector $\mathbf{r}=(r_x,r_y,r_z)$ is the Bloch vector. It lives inside a unit ball:

$$|\mathbf{r}|\le 1.$$

Its direction says which questions favor which answers. Its length distinguishes pure states on the surface from mixed states inside. A pure state has a fully specified quantum preparation. A mixed state can arise by randomly choosing among preparations and averaging over that choice.

The receiver chooses a unit direction $\mathbf{n}$. The probability of a plus answer is

$$Pr(+\mid \mathbf{r},\mathbf{n})=\frac{1+\mathbf{n}\cdot \mathbf{r}}{2}.$$

The dot product $\mathbf{n}\cdot \mathbf{r}$ measures how strongly the state points along the receiver. Facing a pure arrow gives plus every time. Facing the opposite direction gives minus every time. Asking sideways gives equal chances.

When evidence puts on sunglasses develops this matrix geometry through polarization and light. Here we will use it to carry a message.

The same matrix has the cone identity

$$4det\rho =1-|\mathbf{r}{|}^2.$$

In Epistemic Physics, an unnormalized positive matrix $R$ has total amount ${\tau }_{\mathrm{E}}=\mathrm{tr}R$ and mass $m_{\mathrm{E}}=2\sqrt{detR}$. Normalizing to $\rho =R/{\tau }_{\mathrm{E}}$ leaves

$$z=\frac{m_{\mathrm{E}}}{{\tau }_{\mathrm{E}}}=\sqrt{1-|\mathbf{r}{|}^2}.$$

Thus the center of the ball has normalized mass one, and the surface has normalized mass zero. We can carry this geometric quantity straight into a decoding experiment.

The midpoint knows how far apart the messages are

Let the sender choose A or B equally often. Prepare them as pure states with unit Bloch vectors ${\mathbf{r}}_A,{\mathbf{r}}_B$. Call the angle between the arrows $\theta$, from zero to $\pi$ radians.

Before we learn the label, the average state has vector

$$\overline{\mathbf{r}}=\frac{{\mathbf{r}}_A+{\mathbf{r}}_B}{2}.$$

Draw the two arrows on a circle and join their tips. Their average is the midpoint of that chord. A little geometry gives

The average state’s normalized mass is therefore

$$\boxed{\overline{z}=\sin (\theta /2)}.$$

As the messages spread apart, the midpoint moves inward and its mass grows. Picture a bridge between two points on a circular shore: the farther apart its endpoints, the deeper its midpoint reaches into the lake.

Now measure the distance between the messages themselves. Quantum information uses trace distance

$$D({\rho }_A,{\rho }_B)=\frac{1}{2}\Vert {\rho }_A-{\rho }_B{\Vert }_1.$$

For a Hermitian matrix, the trace norm $\Vert \cdot {\Vert }_1$ is the sum of the absolute values of its eigenvalues. Subtracting our two qubit matrices gives eigenvalues $+|{\mathbf{r}}_A-{\mathbf{r}}_B|/2$ and its negative. Consequently,

The half-chord between the messages equals the mass at their midpoint. Two familiar geometric constructions have become the same communication quantity.

Turn geometry into a decoding error

Suppose the receiver asks along $\mathbf{n}$. Write $a=Pr(+\mid A)$ and $b=Pr(+\mid B)$. After each outcome, guess whichever label made it more likely.

For equal input probabilities, plus errors contribute $min(a,b)/2$ and minus errors contribute $min(1-a,1-b)/2$. Adding them gives the receiver’s error rate:

$$P_{\mathrm{e}\mathrm{r}\mathrm{r}\mathrm{o}\mathrm{r}}(\mathbf{n})=\frac{1-|a-b|}{2}.$$

The difference in plus probabilities follows directly from the Bloch rule:

$$a-b=\frac{\mathbf{n}\cdot ({\mathbf{r}}_A-{\mathbf{r}}_B)}{2}.$$

To make that difference as large as possible, point the receiver along the chord between the messages. This is the optimal Helstrom measurement for our pair. It reaches the minimum error allowed by quantum measurement:

$$\boxed{P_{\mathrm{e}\mathrm{r}\mathrm{r}\mathrm{o}\mathrm{r}}^{\mathrm{b}\mathrm{e}\mathrm{s}\mathrm{t}}=\frac{1-D}{2}=\frac{1-\overline{z}}{2}.}$$

The general distinguishability and measurement framework is developed in Christopher Fuchs’s dissertation, Chapter 3.

Try a prediction before touching the receiver. If the arrows are separated by ${90}^{\circ }$ on the Bloch sphere, should aiming directly at one arrow be best? Or should the receiver face along the line joining them?

A mass becomes a decoding advantage

Turn the receiver, catch the message

Restore the experiment

A sender chooses A or B with equal probability. Drag the teal ring to spread the two message arrows apart. Turn the outer gold handle to choose what the receiver measures. Arrow keys and the buttons below work too.

Message AMessage B, drag to spreadAverage stateReceiver, drag to turn

Message separation: 90°

Bring closer Spread apart Opposite poles

Receiver: 45°

Turn −15° Turn +15° Best receiver Aim between messages

A was sent

+ outcome: 50.0%

B was sent

+ outcome: 100.0%

Error with this receiver

25.0%

Best possible error

14.6%

Information this receiver reads

0.311 bits

Holevo information bound

0.601 bits

Average mass = 0.707; message distinguishability = 0.707. After +, guess B; after −, guess A. The difference between the two outcome probabilities is what the receiver can use.

$$D=\overline{z}=\sin (\theta /2)$$ $$P_{\mathrm{e}\mathrm{r}\mathrm{r}\mathrm{o}\mathrm{r}}^{\mathrm{b}\mathrm{e}\mathrm{s}\mathrm{t}}=(1-\overline{z})/2$$

Each trial uses one freshly prepared qubit. The two messages are pure states, chosen equally often. Values are exact expected probabilities, so you can compare directions without sampling noise. These are angles on the Bloch sphere.

Open the quantum receiver in the playground.

At ${90}^{\circ }$ separation,

$$\begin{array}{c}\overline{z}=\frac{1}{\sqrt{2}},\\ P_{\mathrm{e}\mathrm{r}\mathrm{r}\mathrm{o}\mathrm{r}}^{\mathrm{b}\mathrm{e}\mathrm{s}\mathrm{t}}=\frac{1-1/\sqrt{2}}{2}\\ \approx 14.64\mathrm{\%}.\end{array}$$

Point along the bisector instead, halfway between the arrows, and both messages produce identical outcome probabilities. The receiver then makes errors half the time. The geometry contains a good question and an entirely uninformative one.

At opposite poles, $\theta =\pi$, the states are orthogonal and the optimum error reaches zero. At the same point, $\theta =0$, no question can distinguish their labels.

Rapidity is hiding inside the receiver

The midpoint also has a radial rapidity,

$${\varphi }_{\mathrm{E}}=\mathrm{artanh}|\overline{\mathbf{r}}|,$$

where $\mathrm{artanh}$ is the inverse hyperbolic tangent. The cone equation becomes

$$\overline{z}=\mathrm{sech}{\varphi }_{\mathrm{E}},$$

with $\mathrm{sech}\varphi =1/\cosh \varphi$. Our optimal error can therefore be written

$$P_{\mathrm{e}\mathrm{r}\mathrm{r}\mathrm{o}\mathrm{r}}^{\mathrm{b}\mathrm{e}\mathrm{s}\mathrm{t}}=\frac{1-\mathrm{sech}{\varphi }_{\mathrm{E}}}{2}.$$

A midpoint near the surface has large radial rapidity and comes from two nearly coincident pure messages. Moving the midpoint toward rest at the center corresponds to spreading the messages toward perfect distinguishability. The rapidity belongs to their average, so it tracks this particular geometry of the preparation.

How many bits does the receiver learn?

Error rate counts wrong guesses. Mutual information measures how much observing the receiver reduces uncertainty about the label.

We use binary entropy in bits:

with $0{\log }_{2}0=0$. The equal A/B choice starts with one bit of label uncertainty.

For plus probabilities $a,b$, let $q=(a+b)/2$ be the receiver’s overall plus probability. Its mutual information with the label is

$$\begin{array}{c}I(\text{label};\text{outcome})\\ =h_2(q)-\frac{h_2(a)+h_2(b)}{2}.\end{array}$$

The first term is the entropy of the pooled readout. The second is the average readout entropy after the label is known. Their difference measures the part of the receiver’s variation that reveals which message was sent.

For the best receiver in our symmetric pure-state experiment, the two directions of error have the same probability $e=P_{\mathrm{e}\mathrm{r}\mathrm{r}\mathrm{o}\mathrm{r}}^{\mathrm{b}\mathrm{e}\mathrm{s}\mathrm{t}}$. That gives

$$I_{\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{e}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{r}}=1-h_2(e).$$

At ${90}^{\circ }$ separation, this is about $0.399$ bits per received qubit.

There is another quantity worth plotting beside it. The von Neumann entropy of a quantum state is the entropy of its eigenvalues. For a qubit,

$$S(\rho )=h_2\left(\frac{1+|\mathbf{r}|}{2}\right).$$

The Holevo information, written $\chi$, subtracts the average entropy within the prepared messages from the entropy of their average:

$$\chi =S(\overline{\rho })-\frac{S({\rho }_A)+S({\rho }_B)}{2}.$$

It bounds the information obtainable from the message label by any measurement. Mary Beth Ruskai’s treatment, Section 7, presents the entropy inequality and the conditions under which the bound is attained.

Our two messages are pure, so each individual entropy is zero. Expressing the midpoint length through its mass gives

$$\boxed{\chi =h_2\left(\frac{1+\sqrt{1-{\overline{z}}^2}}{2}\right).}$$

At ${90}^{\circ }$, $\chi \approx 0.601$ bits. The receiver’s $0.399$ bits sit below that bound. Both values are useful: one describes what the selected measurement reads; the other places a ceiling on the information a measurement could extract.

Keep the average, change the preparation

Time for the magic trick.

In the first recipe, A is the north-pole state and B is the south-pole state. Send each equally often. Their average is the center:

$$\overline{\mathbf{r}}=\mathbf{0},\overline{\rho }=\frac{\mathbf{1}}{2},$$

where $\mathbf{1}$ is the identity matrix. A vertical receiver reads a perfect bit.

In the second recipe, prepare north or south with a fair coin regardless of whether the requested label was A or B. Both labels now produce the same mixed state $\mathbf{1}/2$. The average is unchanged. Every receiver has lost all information about the requested label.

We can connect the recipes continuously. Choose a label contrast $r$ between zero and one and prepare

For A, this means choosing the upper pole with probability $(1+r)/2$. For B, choose it with probability $(1-r)/2$. The two recipes become identical as $r$ reaches zero.

Keep the preparation labels in view

One average state, many possible messages

Restore distinct messages

The sender still chooses A or B equally often. Start with opposite pure states, then blur the preparation labels. Drag the rose ring toward the center, or choose a recipe. The sky average stays fixed while the receiver's information changes.

Distinct messagesHalfway blurredForget the labels

A, drag to mixBFixed averageReceiver

Label contrast: 1.00

For A, prepare the upper pole 100.0% of the time. For B, prepare it 0.0% of the time. Pool both labels and it is always 50%.

Mix further Separate labels

Choose the receiver

Along the messagesAt 45 degreesAcross the messages

The receiver's + probability is 100.0% for A and 0.0% for B. Without the labels, the + probability is 50% for every receiver direction.

Average state's entropy1 bit

How that entropy divides

Within each message: 0.000 bitsHolevo information: 1.000 bits

Information this receiver reads1.000 bits

Average mass, always

1.000

Message distinguishability

1.000

Best possible error

0.0%

Error with this receiver

0.0%

The average sits at the center, yet the preparation labels are perfectly readable. Each qubit carries one full bit.

$$\chi =1-h_2\left(\frac{1+r}{2}\right)$$ $$\overline{z}=1,D=r$$

Here r is label contrast, the length of each message's Bloch vector. Mixing means randomly choosing between the upper and lower poles with the displayed label-dependent recipe. The preparation record supplies the relationship the pooled state leaves behind.

Open the preparation experiment in the playground.

Watch the two arrows shrink toward the center. Their pooled average stays at the center throughout, so

$$\overline{z}=1,S(\overline{\rho })=1.$$

The individual message entropies grow as the labels blur:

$$\begin{array}{c}S({\rho }_A)=S({\rho }_B)\\ =h_2\left(\frac{1+r}{2}\right).\end{array}$$

Therefore

A receiver along the messages reaches this $\chi$ exactly. At $r=1/2$, it reads about $0.189$ bits; at $r=0$, it reads zero; at $r=1$, it reads one.

Here the mass of the average stays one while distinguishability changes. We have given the geometry an extra degree of freedom: each prepared message may itself be mixed. Retaining those individual states lets the entropy subtraction recover the information the average alone leaves unspecified.

Mass asks what you averaged over

There is a delightful reversal when we compare this trip with When a channel carries mass.

In a classical decoding problem, a cone state can encode the two likelihoods of a received mark. Its normalized mass is large when that mark fits both possible inputs. In our pure quantum-message experiment, the mass belongs to the average of two prepared states. It grows as those preparations become easier to distinguish.

The same cone can organize both experiments because the coordinates record different parts of their stories.

This is a useful habit for evidence systems too. A pile of tracelets can preserve a total, a balance, and a geometric mass. Their lineage can additionally preserve which preparation, question, or message label each group belongs to. A later question may need exactly the relation that pooling discarded.

A small separation suggests a sharp prediction

Let the two pure messages be almost aligned, with a small Bloch angle $\theta$ measured in radians. Then

$$\overline{z}\approx \frac{\theta }{2},P_{\mathrm{e}\mathrm{r}\mathrm{r}\mathrm{o}\mathrm{r}}^{\mathrm{b}\mathrm{e}\mathrm{s}\mathrm{t}}\approx \frac{1}{2}-\frac{\theta }{4}.$$

The advantage over a random guess grows linearly with separation. The information read by the best receiver begins quadratically:

$$I_{\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{e}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{r}}\approx \frac{{\theta }^2}{8\ln 2}.$$

So doubling a very small message separation approximately doubles the guessing advantage and quadruples the readable information. The two measures respond differently to the same geometric change.

Imagine designing a learning system that represents alternative explanations by directional states. We could ask how its questions should turn to separate nearby possibilities, which preparation labels should survive compression, and whether a mass coordinate predicts the rate at which its answers become distinguishable. The quantum receiver supplies a complete little model in which to try those ideas.

The next experiment is yours: bring the arrows close together, find the best receiver, then double their separation. Watch the error and the bits move at different speeds.

Reading map

• When evidence puts on sunglasses introduces positive matrices, polarization, and the Bloch ball.
• When a channel carries mass follows cone geometry through classical channel overlap and information loss.
• Fuchs, Distinguishability and Accessible Information in Quantum Theory develops quantum discrimination and the optimization of measurements.
• Ruskai, Inequalities for Quantum Entropy develops the entropy inequalities behind the Holevo bound.