The Qubit
Flip a coin and, mid-air, it is neither heads nor tails — it is somehow on the way to both,
and only when it lands does it commit to one face. A classical bit is the coin
already on the table: definitely 0 or definitely
1. A qubit (quantum bit) is the coin in flight — a
genuine blend of both possibilities that only picks a value when you look. That blend is the whole
reason a quantum computer can do things a classical one cannot.
Formally, a qubit is a unit vector in a two-dimensional complex space. Two special
states stand in for the classical 0 and 1 —
the computational basis, written in
Dirac notation:
|0\rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \qquad |1\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.
The new power is superposition — a qubit may be a combination of both at once:
|\psi\rangle = \alpha\,|0\rangle + \beta\,|1\rangle = \begin{bmatrix} \alpha \\ \beta \end{bmatrix},
where the amplitudes \alpha, \beta are
complex numbers.
The normalisation rule
Not every pair of amplitudes is allowed. Because a measurement must yield some outcome,
with total probability 1, the amplitudes obey the
normalisation condition
|\alpha|^2 + |\beta|^2 = 1.
Geometrically, |\psi\rangle is a unit vector — length
exactly one. The squared magnitudes are the measurement probabilities:
|\alpha|^2 is the chance of finding the qubit in
|0\rangle and |\beta|^2 the chance of
|1\rangle. The single most important superposition is the perfectly
balanced one,
|{+}\rangle = \tfrac{1}{\sqrt{2}}\,|0\rangle + \tfrac{1}{\sqrt{2}}\,|1\rangle, \qquad |\alpha|^2 = |\beta|^2 = \tfrac12,
a coin caught exactly halfway — a 50/50 chance of either outcome.
Picturing a qubit
When the amplitudes are real we can draw the state as a unit arrow in the plane spanned by
|0\rangle and |1\rangle. Its tip rides on the
unit circle, and its two components are \alpha and
\beta. (Complex amplitudes need the extra room of the
Bloch sphere.)
Worked example: is it a valid state?
Is |\psi\rangle = 0.6\,|0\rangle + 0.8\,|1\rangle a legal qubit? Check
the norm:
|0.6|^2 + |0.8|^2 = 0.36 + 0.64 = 1. \quad\checkmark
Yes — it is normalised, so it is a perfectly good qubit, and measuring it gives
0 with probability 0.36 and
1 with probability 0.64. Now the reverse: if a
real qubit has \alpha = \tfrac12, what is \beta?
From |\beta|^2 = 1 - \tfrac14 = \tfrac34, we get
\beta = \tfrac{\sqrt3}{2}.
Worked example: the minus sign matters
Compare |{+}\rangle = \tfrac{1}{\sqrt2}(|0\rangle + |1\rangle) with
|{-}\rangle = \tfrac{1}{\sqrt2}(|0\rangle - |1\rangle). Both are
normalised, and both give a 50/50 measurement in the computational basis, because
|{+}\tfrac{1}{\sqrt2}|^2 = |{-}\tfrac{1}{\sqrt2}|^2 = \tfrac12. So they
look identical to a measurement — yet they are different states. That hidden minus sign is
a relative phase, and although it is invisible to a single measurement, it is
exactly the ingredient that lets quantum states interfere, cancelling wrong answers
and reinforcing right ones. Phase is where quantum algorithms do their real work.
- a qubit is a unit vector |\psi\rangle = \alpha|0\rangle + \beta|1\rangle in \mathbb{C}^2;
- the amplitudes are complex, and obey |\alpha|^2 + |\beta|^2 = 1;
- |\alpha|^2 and |\beta|^2 are the
probabilities of measuring |0\rangle and |1\rangle;
- unlike a bit, a qubit can be in a superposition of both — the source of its power.
In one sense, an enormous amount: \alpha and
\beta are continuous complex numbers, so specifying a qubit exactly would
take infinitely many digits. Yet when you measure it, out pops a single bit —
0 or 1 — and the rest is gone. Nature hands you
one bit per qubit and hides the amplitudes behind the randomness of measurement. Holevo's theorem
makes this precise: you cannot reliably store more than one classical bit of retrievable
information in one qubit. The magic of quantum computing is not in reading out more, but in
how those hidden amplitudes interfere while you compute.
A superposition \alpha|0\rangle + \beta|1\rangle does not
mean "the qubit is secretly 0 or 1 and we just
don't know which". That would be ordinary classical probability. A qubit in
|{+}\rangle is in a definite state — just not a definite
0-or-1 — and you can prove it: the amplitudes
can be negative or complex and cancel each other, something probabilities
(which are never negative) can never do. That interference is the experimental fingerprint that
tells a real superposition apart from mere lack of knowledge.