Two-qubit system is a quantum system of two qubits. The state space is $\mathbb{C}^2\otimes\mathbb{C}^2 \cong \mathbb{C}^4$: you need 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:

$$\lvert\psi\rangle = c_{00}\lvert 00\rangle + c_{01}\lvert 01\rangle + c_{10}\lvert 10\rangle + c_{11}\lvert 11\rangle, \qquad |c_{00}|^2 + |c_{01}|^2 + |c_{10}|^2 + |c_{11}|^2 = 1$$

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

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

The shorthand $\lvert\psi_0\psi_1\rangle$ means $\lvert\psi_0\rangle\otimes\lvert\psi_1\rangle$. 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.

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

Notice the pattern: applying $H\otimes I$ then CNOT to each computational basis state gives a different Bell state. The same circuit run in reverse (CNOT then $H\otimes I$) is the Bell measurement, which maps every Bell state back to its corresponding computational basis state. This makes the two bases inverses of each other.

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 $\lvert\psi_0\rangle\otimes\lvert\psi_1\rangle$ 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. This is why the four Bell states map one-to-one to the four Pauli corrections in quantum teleportation.