Multi-Qubit States and Entanglement

Alice enters the twin observatory

One qubit taught Alice that a state is not a hidden bit. Two qubits teach a sharper lesson: the state of a pair is not always a pair of individual states. The new observatory has two wires, two dials, and one shared sky. Sometimes the whole sky has a pattern that neither wire owns alone.

Entanglement is essential groundwork for the protocol, algorithm, and hardware chapters that follow. The goal here is to replace vague language about "spooky connection" with a state-vector account of correlations and their limits.

Prerequisite recap

You need four ideas from earlier chapters:

1. A pure state is a normalized vector of amplitudes.
2. Ideal circuit gates are linear and unitary.
3. Measurement probabilities come from squared magnitudes in the chosen basis.
4. Relative phase can matter even when a single measurement basis hides it.

Joint state space

A two-qubit computational basis has four labels:

Throughout this course we use that order unless we say otherwise. A general two-qubit pure state is

with normalization

Tensor products

If the first qubit is $|\alpha\rangle = x|0\rangle + y|1\rangle$ and the second is $|\beta\rangle = u|0\rangle + v|1\rangle$, their independent joint state is the tensor product:

For example,

This is separable because it was built as a product of two one-qubit states. Not every four-amplitude vector has that form.

Controlled gates

A controlled gate acts on a target depending on a control. The most important first example is CNOT. With the first qubit as control and the second as target:

Linearity means CNOT applies that rule to every basis component. Starting from $|00\rangle$, apply H to the first qubit:

Then apply CNOT:

Bell states

The state

is a Bell state. It cannot be factored as $(x|0\rangle+y|1\rangle)\otimes(u|0\rangle+v|1\rangle)$. To see the tension, a product state would have amplitudes $xu, xv, yu, yv$. For $|\Phi^+\rangle$, the $|01\rangle$ and $|10\rangle$ amplitudes are zero, so $xv=0$ and $yu=0$, but the $|00\rangle$ and $|11\rangle$ amplitudes require $xu\neq0$ and $yv\neq0$. Those requirements cannot all hold at once.

For this pure two-qubit state, this is entanglement: the pair has a well-defined joint state that is not a product of one-qubit pure states.

Worked example: a separability test

For a pure two-qubit state $a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$, a useful test in our basis order is

The state is separable exactly when this equality holds. For $(|00\rangle+|01\rangle)/\sqrt{2}$, both sides are zero and the state factors as $|0\rangle|+\rangle$. For $|\Phi^+\rangle$, $ad=1/2$ while $bc=0$, so it is entangled. This shortcut is specific to pure states of two qubits; mixed-state entanglement needs more careful tools.

Common misconceptions

Guided practice

Write $|\Phi^+\rangle$ in the X basis using $|0\rangle=(|+\rangle+|-\rangle)/\sqrt{2}$ and $|1\rangle=(|+\rangle-|-\rangle)/\sqrt{2}$. The cross terms cancel, leaving $(|++\rangle+|--\rangle)/\sqrt{2}$. This predicts matched X-basis outcomes too. The exercises below separate this basis-dependent structure from any claim of controllable communication.

Chapter assessment

Before moving on, you should be able to derive the Bell state from $|00\rangle$, name the basis order, compute its Z-basis probabilities, and explain why the result is not a separable product state. The exercises below test exactly those skills.

Summary

Next step

You can now describe the shared state used by quantum information protocols. Chapter 6 turns that resource into precise circuits for no-cloning reasoning, teleportation, and superdense coding.

References and further study

• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information.
• John Preskill, Lecture Notes for Physics 229: Quantum Information and Computation.
• IBM Quantum Learning, Quantum information for multiple systems.