Two-qubit system is a quantum system of two qubits. The Hilbert space is $\mathbb{C}^4$, meaning four complex amplitudes to describe it, not two.

A single qubit needs two amplitudes; two qubits need four; three qubits need eight. The state space doubles with every qubit you add. That exponential growth is what makes quantum computers interesting.

A general two-qubit state is a superposition over the four computational basis states, with the normalization condition saying the total probability must be 1:

$$\lvert\psi\rangle = \begin{pmatrix}c_{00}\\c_{01}\\c_{10}\\c_{11}\end{pmatrix} \qquad \lvert\psi\rangle = c_{00}\underbrace{\begin{pmatrix}1\\0\\0\\0\end{pmatrix}}_{\lvert 00\rangle} + c_{01}\underbrace{\begin{pmatrix}0\\1\\0\\0\end{pmatrix}}_{\lvert 01\rangle} + c_{10}\underbrace{\begin{pmatrix}0\\0\\1\\0\end{pmatrix}}_{\lvert 10\rangle} + c_{11}\underbrace{\begin{pmatrix}0\\0\\0\\1\end{pmatrix}}_{\lvert 11\rangle}, \qquad |c_{00}|^2 + |c_{01}|^2 + |c_{10}|^2 + |c_{11}|^2 = 1$$

The simplest two-qubit states are product states: states where the two qubits are independent of each other. If qubit 1 is in $a\lvert 0\rangle + b\lvert 1\rangle$ and qubit 0 is in $c\lvert 0\rangle + d\lvert 1\rangle$, the combined state is just their tensor product:

$$\lvert\psi_1\rangle\otimes\lvert\psi_0\rangle = ac\lvert 00\rangle + ad\lvert 01\rangle + bc\lvert 10\rangle + bd\lvert 11\rangle$$

The shorthand $\lvert\psi_1\psi_0\rangle$ means $\lvert\psi_1\rangle\otimes\lvert\psi_0\rangle$. There are infinitely many product states. Any pair of single-qubit states gives you one. The ones worth naming come from picking a standard single-qubit basis and using it for both qubits.

Pick $\lvert 0\rangle$ and $\lvert 1\rangle$ for each qubit and you get four product states that look exactly like a classical 2-bit register. Each qubit is definitively 0 or 1 with no superposition. These four states span all of $\mathbb{C}^4$, so every two-qubit state can be written in terms of them.

$$\lvert 00\rangle = \begin{pmatrix} 1\\0\\0\\0 \end{pmatrix} \qquad \lvert 01\rangle = \begin{pmatrix} 0\\1\\0\\0 \end{pmatrix} \qquad \lvert 10\rangle = \begin{pmatrix} 0\\0\\1\\0 \end{pmatrix} \qquad \lvert 11\rangle = \begin{pmatrix} 0\\0\\0\\1 \end{pmatrix} $$

Pick $\lvert +\rangle$ and $\lvert -\rangle$ for each qubit instead. These states look “spread out” in the computational basis, because $\lvert +\rangle$ $= (\lvert 0\rangle + \lvert 1\rangle)/\sqrt{2}$ mixes both. But in the X basis they are perfectly sharp, just like $\lvert 00\rangle$ is sharp in the Z basis. You get from the computational basis to this one by applying a Hadamard to every qubit.

$$\lvert ++\rangle = \frac{1}{2}\begin{pmatrix}1\\1\\1\\1\end{pmatrix} \qquad \lvert +-\rangle = \frac{1}{2}\begin{pmatrix}1\\-1\\1\\-1\end{pmatrix} \qquad \lvert -+\rangle = \frac{1}{2}\begin{pmatrix}1\\1\\-1\\-1\end{pmatrix} \qquad \lvert --\rangle = \frac{1}{2}\begin{pmatrix}1\\-1\\-1\\1\end{pmatrix} $$

Pick $\lvert +i\rangle$ and $\lvert -i\rangle$ for each qubit. These are the eigenstates of the Y gate. The coefficients are complex, so the Y-basis states are the first ones on this page with imaginary entries. You get from the computational basis to the Y basis by applying $SH$ to every qubit ($H$ first, then $S$).

$$\lvert{+i,+i}\rangle = \frac{1}{2}\begin{pmatrix}1\\i\\i\\-1\end{pmatrix} \qquad \lvert{+i,-i}\rangle = \frac{1}{2}\begin{pmatrix}1\\-i\\i\\1\end{pmatrix} \qquad \lvert{-i,+i}\rangle = \frac{1}{2}\begin{pmatrix}1\\i\\-i\\1\end{pmatrix} \qquad \lvert{-i,-i}\rangle = \frac{1}{2}\begin{pmatrix}1\\-i\\-i\\-1\end{pmatrix} $$

Some two-qubit states cannot be written as $\lvert\psi_1\rangle\otimes\lvert\psi_0\rangle$ for any choice of single-qubit states. These are entangled states. The give-away is that the four amplitudes $c_{00}, c_{01}, c_{10}, c_{11}$ cannot be factored into two pairs: entanglement means $c_{00}c_{11} \neq c_{01}c_{10}$.

What does that mean physically? Measuring one qubit of an entangled state instantly determines the outcome of measuring the other, even though neither qubit has a definite value before measurement. The correlations are stronger than anything a classical system can produce.

The four Bell states are the canonical maximally entangled two-qubit states. Each is an equal superposition of two computational basis states that are not factorable. Together they form a complete orthonormal basis for $\mathbb{C}^4$.

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\0\\0\\1\end{pmatrix} \qquad \lvert\Phi^-\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\0\\0\\-1\end{pmatrix} \qquad \lvert\Psi^+\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}0\\1\\1\\0\end{pmatrix} \qquad \lvert\Psi^-\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}0\\1\\-1\\0\end{pmatrix} $$

Any single-qubit Pauli gate on one qubit of a Bell state maps it to another Bell state: $X$ switches $\Phi\leftrightarrow\Psi$, $Z$ toggles the sign, $Y$ does both.

Measuring one qubit of an entangled state collapses the joint state. Take $\lvert\Phi^+\rangle = \tfrac{1}{\sqrt{2}}(\lvert 00\rangle + \lvert 11\rangle)$ and measure qubit 0 in the Z basis:

• result 0 with probability $\tfrac{1}{2}$: the state collapses to $\lvert 00\rangle$. Qubit 1 is now certain to be $\lvert 0\rangle$.
• result 1 with probability $\tfrac{1}{2}$: the state collapses to $\lvert 11\rangle$. Qubit 1 is now certain to be $\lvert 1\rangle$.

Before the measurement neither qubit has a definite value. After a measurement on qubit 0, qubit 1's outcome is fixed without touching it. The same pattern holds for any maximally entangled state and any measurement basis.

Two-qubit gates are $4\times 4$ unitary matrices acting on $\mathbb{C}^4$. The three most common ones are CX (CX), CZ, and SWAP. Row and column ordering follows the computational basis: $\lvert 00\rangle, \lvert 01\rangle, \lvert 10\rangle, \lvert 11\rangle$.

$$CX = \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix} \qquad CZ = \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&-1\end{pmatrix} \qquad SWAP = \begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}$$

CX flips qubit 0 when qubit 1 is $\lvert 1\rangle$ (qubit 1 is the control; Qiskit: cx(1, 0)). CZ applies a $-1$ phase when both qubits are $\lvert 1\rangle$; it is symmetric so either qubit can serve as the control. SWAP exchanges the two qubits entirely. For a full list see Two-qubit gates.

# Requires: pip install qiskit qiskit-aer
# Run: python bell.py
# Prepares |Φ+⟩ and samples 1000 shots; expect roughly equal counts of '00' and '11'.
from qiskit import QuantumCircuit
from qiskit_aer import AerSimulator
 
qc = QuantumCircuit(2, 2)
qc.h(0)       # Hadamard on qubit 0 → superposition
qc.cx(0, 1)   # CX: control=qubit 0, target=qubit 1 → entanglement
qc.measure([0, 1], [0, 1])
 
counts = AerSimulator().run(qc, shots=1000).result().get_counts()
print(counts)  # {'00': ~500, '11': ~500}