Information and Distinguishability

Alice wants to send Bob a yes-or-no answer. She could send a light pulse, turn a dial, or put a card in an envelope. In each case, she needs two preparations that Bob can tell apart. They agree that one means “yes” and the other means “no.”

This agreement is an encoding: a rule associating a message with a physical preparation. Bob also needs a decoder, a rule for turning what he observes into a guess about the message. If he can identify either preparation with certainty using the observations available to him, the preparations are perfectly distinguishable.

Let us give Bob a less reliable apparatus. Alice chooses a card marked A or B and puts it into a detector. Bob cannot see the card. He sees only a lamp, which either lights red or stays dark. Both cards can produce either report. Reading the lamp without error therefore does not guarantee that he can identify the card.

So the question becomes practical. After seeing the lamp, which card should Bob guess, and how often will he be right?

Suppose Alice has put card A into the detector. She knows which preparation she made; Bob does not. Their knowledge differs even though they are talking about the same physical carrier.

A state description tells us what outcomes to expect from possible observations. It must give more than a name such as “card A”: to make predictions, we also need to know how that preparation responds to the detector. Bob, uncertain about Alice’s choice, must combine the predictions for A and B with the probabilities of those choices.

For the cards, this uncertainty can come from not knowing which label Alice chose. Is that the only possible source of uncertainty? For cards it is: if Bob knew the card, he would know how likely each report is, and nothing more could be known. Quantum systems will add a second source. Alice can know her preparation completely, and the outcome of an observation can still be uncertain.

A classical value with two alternatives, written 0 and 1, is a bit. Two preparations that Bob can reliably distinguish can carry such a value. Merely having two different preparation labels does not establish that he can read it. What matters is how the preparations respond to the observations he can make.

Before using the detector, Bob already knows something: Alice chooses A on three quarters of trials and B on one quarter. We describe these choices by probabilities, \(p(A)=3/4\) and \(p(B)=1/4\). These are his prior probabilities, the probabilities assigned before he sees the report.

A probability lies between zero and one. Zero means an event never occurs in the model; one means it is certain. Because Alice chooses exactly one of A and B, their probabilities must add to one.

Bob could ignore the detector and always guess A. He would then be right on three quarters of trials. Any rule using the lamp should be compared with this \(3/4\) success rate. A fair guess would be a weaker starting point because it would discard information Bob already has.

The detector lights red on one fifth of A trials and four fifths of B trials. Red therefore occurs more often for B. But A is the more common card. To decide what a red report means, Bob must account for both facts.

Imagine 100 trials in the proportions specified by our model. Of the 75 A cards, one fifth produce red: 15 trials. The other 60 produce a dark report. Of the 25 B cards, four fifths produce red: 20 trials. The remaining 5 produce a dark report.

Now suppose Bob sees red. He knows that his trial belongs to the group of 35 red reports. Within that group, 20 came from B and 15 from A. His probability for B becomes

\[p(B\mid R)=\frac{20}{35}=\frac47.\]

The vertical bar means “given”: \(p(B\mid R)\) is the probability of B given a red report, written \(R\). Selecting the trials consistent with an observation and finding the proportions within that group is called conditioning. B was less likely before the observation; it is more likely within the red group.

We can obtain the same answer directly from the probabilities. The probability of both B and red is \(p(B,R)=(1/4)(4/5)=1/5\). The probability of red is \(p(R)=(3/4)(1/5)+(1/4)(4/5)=7/20\). Dividing the first by the second gives

\[p(B\mid R)=\frac{p(B,R)}{p(R)}=\frac{1/5}{7/20}=\frac47.\]

The comma means “both.” Notice the difference between \(p(R\mid B)=4/5\) and \(p(B\mid R)=4/7\). The first asks how the detector responds when the card is known. The second asks which card was sent when the report is known. The table makes the different denominators visible: 20 out of 25 B trials, compared with 20 out of 35 red trials.

These counts describe the model’s proportions. In an actual set of 100 trials the counts can fluctuate; they need not be exactly 15, 60, 20, and 5.

So far Bob has tried to learn which card Alice chose. Suppose instead that his job is to pass her card, unopened, to another receiver. He can succeed at that task without identifying it himself.

These are different requirements. Learning asks what Bob can infer from an observation. Preserving asks whether a physical state survives a procedure. Transmitting asks whether the required information or state reaches another place.

Consider a card sealed in an envelope. Passing the envelope on may preserve it. Opening it may reveal the card while damaging the seal. Whether opening counts as success depends on what was meant to survive: the printed letter, the intact envelope, or both. We have to state the task before judging the procedure.

Quantum information makes the distinction especially important. Measuring a carrier may disturb its state, and a single specimen need not reveal which quantum state was prepared. Faithfully transferring that state and identifying it are therefore separate goals.

Return to Bob’s lamp. His task is to guess the card correctly as often as possible, using one detector report. The table gives a decision rule: guess B on red, because 20 exceeds 15; guess A on dark, because 60 exceeds 5.

He succeeds on the 20 red B trials and the 60 dark A trials. His success probability is

\[p(\text{correct})=\frac{20+60}{100}=\frac45.\]

Always guessing A gave \(3/4\). Using the report raises the success rate from 75 to 80 correct decisions per 100 model trials.

Could a different decoder do better with this same report? On red trials, choosing A would replace 20 correct guesses by 15. On dark trials, choosing B would replace 60 correct guesses by 5. Randomly switching between the two guesses only mixes a better choice with a worse one. Our rule is therefore optimal among decoders that receive this detector’s report. A different detector might offer better information.

This comparison depends on the allowed preparations, their prior probabilities, the available observation, and the required output. Those details define the task. Other tasks add resources to count: transmitted systems, uses of a channel, classical messages, and correlations shared in advance.

The same report can be useful in one task and insufficient in another. Bob’s detector supports a guess with success \(4/5\), but it does not let him read every card without error. To say how much information a carrier provides, we must say what the receiver is trying to do with it.

One question remains open. Bob's uncertainty about the card has been described by two probabilities, but we have not yet measured how much uncertainty that is, or how many bits it would cost to remove it.