Qubits, bases and measurement
A qubit is a two-level quantum system with state $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$, $|\alpha|^2+|\beta|^2=1$. Measuring in the basis $\{|0\rangle,|1\rangle\}$ gives 0 with probability $|\alpha|^2$ and 1 with probability $|\beta|^2$ (the Born rule), and the state collapses to the result. For photons the qubit is usually a polarization:
| Basis | State 0 | State 1 | Photon polarization |
|---|---|---|---|
| Rectilinear (Z, “+”) | $|0\rangle$ | $|1 angle$ | horizontal ↔ / vertical ↕ |
| Diagonal (X, “×”) | $|+\rangle=\tfrac{|0\rangle+|1\rangle}{\sqrt2}$ | $|-\rangle=\tfrac{|0\rangle-|1\rangle}{\sqrt2}$ | 45° / 135° |
| Circular (Y) | $\tfrac{|0\rangle+i|1\rangle}{\sqrt2}$ | $\tfrac{|0\rangle-i|1\rangle}{\sqrt2}$ | right / left circular |
The Z and X bases are mutually unbiased: a state prepared in one gives a perfectly random outcome in the other, $|\langle +|0\rangle|^2=\tfrac12$. This single fact is the engine of BB84: a spy who guesses the wrong basis learns nothing and leaves a disturbed state behind.
No-cloning, uncertainty and information–disturbance
No-cloning theorem. No physical process maps every unknown state $|\psi\rangle|0\rangle\mapsto|\psi\rangle|\psi\rangle$. Proof sketch: a unitary $U$ preserves inner products. If it cloned two states $|\psi\rangle,|\phi\rangle$, then
which forces $\langle\psi|\phi\rangle\in\{0,1\}$: states must be orthogonal or identical. Non-orthogonal states, which BB84 deliberately uses, cannot be copied. (Wootters–Zurek, Dieks, 1982.) Related results: no-deleting, no-broadcasting (mixed states) and no-signalling (entanglement cannot carry messages).
Uncertainty. For non-commuting observables $\Delta A\,\Delta B\ge\tfrac12|\langle[A,B]\rangle|$. Modern QKD proofs use entropic uncertainty relations: if Bob can predict the outcome of a $Z$ measurement well, Eve cannot know the outcome of an $X$ measurement, and conversely.
Information–disturbance principle. Any measurement that extracts information about non-orthogonal states must disturb them. Quantitatively, the more Eve learns, the larger the error rate Bob sees. QKD turns this tradeoff into a security statement.
Entanglement, Bell states and CHSH
A two-qubit state is entangled if it cannot be written $|\psi_A\rangle\otimes|\psi_B\rangle$. The four Bell states are maximally entangled, e.g.
Measuring both halves in the same basis gives perfectly correlated results, though each result alone is uniformly random. The CHSH quantity
obeys $|S|\le2$ for any local hidden-variable model, but quantum mechanics reaches $|S|=2\sqrt2\approx2.83$ (Tsirelson’s bound). Violating CHSH certifies that the correlations are not pre-arranged and that no third party can hold a complete copy of the outcomes. This is the security principle of Ekert’s E91 protocol and of device-independent QKD.
Density operators, entropy and distinguishability
Imperfect or partially known states are described by a density operator $\rho\ge0$, $\mathrm{Tr}\rho=1$. Eve’s view of Alice’s qubits, when Eve does not know the basis, is the maximally mixed state $\rho=I/2$. Key quantities:
- Von Neumann entropy $S(\rho)=-\mathrm{Tr}\,\rho\log_2\rho$: 0 for pure states, 1 bit for $I/2$.
- Binary entropy $h(p)=-p\log_2p-(1-p)\log_2(1-p)$, the cost of an error rate $p$.
- Trace distance $D(\rho,\sigma)=\tfrac12\|\rho-\sigma\|_1$: the maximal probability advantage in telling two states apart. Secrecy of a key $K$ against Eve’s system $E$ is measured by $D(\rho_{KE},\,\tau_K\otimes\rho_E)\le\varepsilon$, where $\tau_K$ is the uniform key distribution.
- Holevo bound $\chi\le S(\rho)-\sum_x p_xS(\rho_x)$: the most classical information that can be extracted from an ensemble of quantum states.
The quantum one-time pad
To encrypt an unknown quantum state with a classical key $(a,b)\in\{0,1\}^2$, apply $X^aZ^b$. Averaging over the four keys,
for every $\rho$: the ciphertext carries no information. Thus 2n bits of key encrypt n qubits perfectly (Ambainis–Mosca–Tapp–de Wolf 2000), and with fewer key bits perfect secrecy is impossible. Decryption applies the Paulis in reverse. This is the quantum counterpart of Shannon’s classical one-time pad, which requires a key at least as long as the message.