A switch holds one bit. A dial with sixteen positions holds four. A qubit looks like a dial with infinitely many positions. So does it hold infinitely many bits? This page answers that question honestly, one step at a time, starting from binary and ending at the sphere.
Two positions, one bit. A row of switches is a register. This is the whole of classical computing, and it is worth one minute before we change it.
n switches hold 2n patterns, and that is n bits. Not more, not less.
Flip the switches. The register below shows the pattern and the number it stands for. A gate is a rule for turning one pattern into another. NOT flips a switch. XOR compares two.
Why stop at two? A symbol that can sit in any one of k positions carries log2k bits. Communications engineers do this every day.
More positions, more bits. The count grows as the logarithm of the number of positions.
A switch is a two-position dial. Give the dial four positions and it holds two bits. Sixteen positions, four bits. This is exactly what a 16-QAM symbol does on a Wi-Fi link. Slide k and watch the ticks and the bit count.
A pointer that can rest anywhere on the circle has infinitely many positions. It does not hold infinitely many bits, and the reason is one every engineer already knows.
A dial holds only as many bits as you can read back. Noise sets that number, not the dial.
Every real pointer wobbles. If the reader cannot tell positions apart that are closer than some angle δ, then the dial has only 360°/δ usable positions, and holds log2(360°/δ) bits. Turn up the noise and watch the bit count fall. This is Shannon's capacity idea in its simplest form.
A qubit is a dial too. Its pointer can rest anywhere. The difference is the reading rule, which is assumption three from the top of the page.
The pointer can be anywhere. The reading is always a single 0 or 1.
Twelve o'clock is |0⟩. Six o'clock is |1⟩. Everything in between is a superposition, and the label means nothing more than "the pointer is there". When you measure, you do not get the angle. You get one bit, with the odds set by the angle.
Start discrete. Snap the dial to 2, 4, or 8 positions and measure. Then let it go anywhere. Notice what does not change.
Now ask the storage question properly. Alice sets the dial to one of k positions. Bob measures once and guesses. How often is he right, and how many bits did he get?
With a classical dial Bob is always right. With a quantum dial his best odds are 2/k, and he never gets more than one bit.
Play the game. For each round the page picks a position at random, prepares the qubit, measures it, and lets Bob make his best guess from the one bit he saw. The chart on the right tracks how his success rate and his information fall as k grows.
You might object that Bob measured the lazy way, always asking "0 or 1?". He can rotate his detector. Below, two positions only. Turn the detector and find the best angle. You will discover a limit that has a name.
The angle is real. You can recover it, but not from one copy. Statistics gives it back one square root at a time.
Measure N identical copies and the angle comes back, with an error that shrinks as 1/√N.
The page hides a pointer angle. Choose how many copies to measure. It counts the zeros, works out cos2(θ/2), and reports an estimate with an error bar. Four times the copies, half the error. The continuum is there, and you pay for it in copies.
A real qubit has two angles, not one. The circle becomes a sphere. The count of what you can read does not change at all.
Two real numbers to write a qubit down. One bit to read it. That gap is the entire story.
The second angle is called the phase. It moves the pointer around the vertical axis and changes nothing about the odds of 0 and 1. It is invisible to the measurement we have been using. It is not useless. It is what quantum computers compute with, and the companion page on this site is about exactly that.
This is the count from section one, finished. Compare how many numbers it takes to write a register down against how many bits you can read out of it.
n qubits need 2n amplitudes to describe and give back n bits when read.
If every qubit kept to its own dial, n qubits would need 2n numbers, which is a lot but not alarming. Real qubits can be correlated in ways no separate dials can capture. The word for that is entanglement, and it means a general state of n qubits needs about 2n+1 real numbers. Thirty for four qubits. Over a hundred thousand for sixteen. The readout stays at n bits the whole way.
There is one honest way to get more than a bit out of a sent qubit. Share an entangled pair first. Then one qubit carries two bits, and a lookup table shows how.
Sending one qubit delivers two bits, if Alice and Bob already share one entangled pair.
Alice and Bob each hold half of a pair. Alice wants to send two bits. She applies one of four operations to her half, chosen by the two bits, and sends that one qubit to Bob. Bob measures the pair together and reads both bits, every time. No probabilities anywhere in this table.
Three answers, and all three are true at once.
The gap between the first number and the second is not a flaw. It is the resource. Everything a quantum computer or a quantum sensor does is a way of making that hidden continuum act before it collapses to one bit. How it acts, through phase and interference, is the subject of the companion page, Bits, Qubits, and Phase.
