The Qubit & Superposition
The quantum bit, and what it means to be '0 and 1 at once'
By the end of this topic you'll be able to
A two-state system is the simplest non-trivial quantum system there is: something with exactly two mutually exclusive outcomes when measured along a given axis, the way a spin-half particle's spin along z is always found to be either +½ or −½. Binney and Skinner introduce the states |+⟩ and |−⟩ for these two outcomes; in the language of computing we instead write |0⟩ and |1⟩ and call the system a qubit.
By the state postulate, any state of the qubit is some linear combination of the two: |ψ⟩ = α|0⟩ + β|1⟩, with complex amplitudes α, β satisfying |α|² + |β|² = 1. This is exactly the same construction Binney and Skinner use for a spin-half particle's orientation, just relabeled for a computational bit.
Superposition is nothing more than this: a state with nonzero amplitude on more than one basis vector. It's the same fact about vectors that lets any point in a plane be written as a combination of two basis directions — nothing mystical is added by calling the vector a qubit instead of an arrow.
The subtlety is entirely in what happens on measurement. You never observe α and β directly; a measurement returns one classical outcome, 0 or 1, with probability |α|² and |β|² respectively, and — as the measurement postulate demands — the state collapses to whichever outcome was found.
This makes 'a qubit is both 0 and 1 at once' a slippery phrase. It is true that the state is neither definitely 0 nor definitely 1 before measurement; it is a distinct third thing, a vector with two nonzero components. It is false, or at least unjustified, if it is taken to mean that two separate realities coexist, each secretly holding a value, waiting to be uncovered.
There is one qubit and one vector. A measurement is not an act of discovery of a pre-existing fact, in the way reading a hidden classical bit would be — it is, in Binney and Skinner's phrase, a real physical disturbance of the system, and its outcome is genuinely undetermined beforehand.
The symmetric case α = β = 1/√2 is the state a Hadamard gate produces from |0⟩: a fair coin between 0 and 1. It is one particular, highly symmetric superposition among an entire continuum of possible qubit states allowed by the normalization constraint.
That continuum of states, parameterized by every complex α and β obeying |α|²+|β|²=1, has a natural geometric picture: every pure qubit state corresponds to exactly one point on the surface of a sphere, the subject of the next topic.
Try It Yourself
A qubit is prepared in |ψ⟩ = (2/3)|0⟩ + (√5/3)|1⟩. Is this a valid quantum state? If so, what's the probability of each measurement outcome?
- 1Check normalization first: |2/3|² + |√5/3|² = 4/9 + 5/9 = 9/9 = 1. ✓ — this is a valid single-qubit state.
- 2By the Born rule, P(0) = |α|² = |2/3|² = 4/9.
- 3P(1) = |β|² = |√5/3|² = 5/9.
- 4Sanity check: 4/9 + 5/9 = 1, confirming the two probabilities account for the entire outcome space, since 0 and 1 are the only possible results.
Yes, it's valid. P(measure 0) = 4/9 ≈ 0.44, and P(measure 1) = 5/9 ≈ 0.56.
Reference
| Qubit state | General single-qubit state | |
| Normalization | Total probability must equal 1 | |
| P(measure 0) | Born-rule probability of the |0⟩ outcome | |
| P(measure 1) | Born-rule probability of the |1⟩ outcome |
Quick Check
A qubit is in state (√3/2)|0⟩ + (1/2)|1⟩. What's the probability of measuring 1?
Which statement about superposition is most accurate?