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Multi-Qubit Quantum Systems

From tensor products to entangled EPR pairs — an interactive guide to composite quantum systems, the CNOT gate, and the mathematics of entanglement

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⊗ Why Tensor Products?

In classical computing, combining two bits is simple: you just list them side by side. A pair of bits has 4 possible values (00, 01, 10, 11). In quantum mechanics, the situation is richer because of superposition.

When we combine two quantum systems, the resulting state space is not a simple Cartesian product — it is the tensor product \(\otimes\) of the individual Hilbert spaces. This mathematical operation captures all the ways the two systems can be correlated, including entanglement.

Key idea: The tensor product of two vector spaces \(\mathcal{H}_A\) and \(\mathcal{H}_B\) produces a new space \(\mathcal{H}_A\otimes\mathcal{H}_B\) whose dimension is the product of the original dimensions: \(\dim(\mathcal{H}_A\otimes\mathcal{H}_B) = \dim(\mathcal{H}_A)\times\dim(\mathcal{H}_B)\).

⊗ Tensor Product of Two Qubits

A single qubit \(|\psi\rangle = \alpha|0\rangle+\beta|1\rangle\) lives in \(\mathbb{C}^2\). When we combine two qubits, the tensor product gives us \(\mathbb{C}^2\otimes\mathbb{C}^2 = \mathbb{C}^4\), with the computational basis:

$$|00\rangle,\quad|01\rangle,\quad|10\rangle,\quad|11\rangle$$

The tensor product of two individual states is computed by multiplying every component of the first vector with every component of the second:

$$|\psi\rangle\otimes|\phi\rangle = \begin{pmatrix}\alpha\\\beta\end{pmatrix}\!\otimes\!\begin{pmatrix}\gamma\\\delta\end{pmatrix} = \begin{pmatrix}\alpha\gamma\\\alpha\delta\\\beta\gamma\\\beta\delta\end{pmatrix}$$

which represents the state \(\alpha\gamma|00\rangle + \alpha\delta|01\rangle + \beta\gamma|10\rangle + \beta\delta|11\rangle\).

⚡ Interactive Tensor Product Calculator

Choose a state for each qubit and see the tensor product computed step by step.

Qubit A:
Qubit B:
✎ Worked Example — Compute \(|+\rangle\otimes|1\rangle\)
1
Write \(|+\rangle\) as a column vector: \(|+\rangle = \tfrac{1}{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}\).
2
Write \(|1\rangle\) as a column vector: \(|1\rangle = \begin{pmatrix}0\\1\end{pmatrix}\).
3
Multiply each entry of the first vector by the entire second vector: \(\tfrac{1}{\sqrt2}\begin{pmatrix}1\cdot\begin{pmatrix}0\\1\end{pmatrix}\\[4pt]1\cdot\begin{pmatrix}0\\1\end{pmatrix}\end{pmatrix} = \tfrac{1}{\sqrt2}\begin{pmatrix}0\\1\\0\\1\end{pmatrix}\)
4
Read off the state: \(\tfrac{1}{\sqrt2}(|01\rangle + |11\rangle)\). Try it in the calculator above — select |+⟩ for A and |1⟩ for B!

✦ Important Properties

  • Bilinearity: \((\alpha|\psi_1\rangle+\beta|\psi_2\rangle)\otimes|\phi\rangle = \alpha(|\psi_1\rangle\otimes|\phi\rangle)+\beta(|\psi_2\rangle\otimes|\phi\rangle)\)
  • Not commutative: In general \(|a\rangle\otimes|b\rangle \neq |b\rangle\otimes|a\rangle\) (the ordering of subsystems matters).
  • Not surjective: Not every vector in the tensor product space can be written as a single tensor product. Those that cannot are entangled (see Tab ⑤).
Common misconception: The tensor product is not the same as the Cartesian product or the outer product, although it is related to both. It produces a new vector space, not a set of pairs.
💡 Pause & Think
If a single qubit lives in \(\mathbb{C}^2\), what is the dimension of the state space for a system of 3 qubits?
6
8
9
3

⊕ From 1 Qubit to \(n\) Qubits

An \(n\)-qubit quantum register lives in the tensor product of \(n\) copies of \(\mathbb{C}^2\):

$$\mathcal{H}_n = \underbrace{\mathbb{C}^2\otimes\mathbb{C}^2\otimes\cdots\otimes\mathbb{C}^2}_{n\text{ times}} = (\mathbb{C}^2)^{\otimes n}$$

The dimension of this space is \(2^n\). The most general \(n\)-qubit state is a superposition over all \(2^n\) computational basis states:

$$|\psi\rangle = \sum_{x\in\{0,1\}^n} c_x\,|x\rangle, \qquad \sum_{x}|c_x|^2=1$$

Each \(c_x\) is a complex amplitude — to fully describe the state classically, you need all \(2^n\) of them.

📈 The Exponential Explosion

This exponential scaling is the fundamental reason quantum computers can be powerful. Click a number of qubits to see how the dimension grows.

🔍 Explore the Computational Basis

Select the number of qubits, then click a basis state to see its tensor-product decomposition.

Qubits:

⬡ Quantum Registers as Subsystems

A multi-qubit system can be divided into subsystems (also called registers). For example, a 4-qubit state can be viewed as:

  • Four individual qubits: \(\mathcal{H}_A\otimes\mathcal{H}_B\otimes\mathcal{H}_C\otimes\mathcal{H}_D\)
  • Two 2-qubit registers: \(\mathcal{H}_{AB}\otimes\mathcal{H}_{CD}\)
  • One 3-qubit register and one qubit: \(\mathcal{H}_{ABC}\otimes\mathcal{H}_D\)
Partial measurement: When you measure one register of a composite system, the other register collapses into a state that depends on the measurement outcome. This is the physical basis of entanglement.

≡ Reference: Dimensions by Qubit Count

Qubits \(n\)Basis states \(2^n\)Amplitudes needed
122
244
388
53232
101,0241,024
201,048,576≈ 10⁶
50≈ 10¹⁵≈ 10¹⁵
3002³⁰⁰More than atoms in the universe

⊕ What is the CNOT Gate?

The Controlled-NOT (CNOT) gate is the most important two-qubit gate in quantum computing. It operates on a pair of qubits: a control qubit and a target qubit.

Rule: The CNOT flips the target qubit if and only if the control qubit is \(|1\rangle\). Otherwise, it leaves both qubits unchanged.

In computational-basis notation:

$$\mathrm{CNOT}\,|a,b\rangle \;=\; |a,\; a\oplus b\rangle$$

where \(\oplus\) denotes addition modulo 2 (the XOR operation).

⊞ Truth Table

Click any row to see the CNOT action on the circuit below.

Control InTarget InControl OutTarget Out
|0⟩|0⟩|0⟩|0⟩
|0⟩|1⟩|0⟩|1⟩
|1⟩|0⟩|1⟩|1⟩
|1⟩|1⟩|1⟩|0⟩

⚡ Interactive Circuit

Select an input state and watch the CNOT gate in action.

Input:

▦ Matrix Representation

In the ordered basis \(\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\}\) the CNOT is:

$$\mathrm{CNOT}=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}$$

Hover over elements to see which basis transition they encode.

Key Properties

  • Unitary: \(\mathrm{CNOT}^\dagger\!\cdot\!\mathrm{CNOT}=I\)
  • Self-inverse: \(\mathrm{CNOT}^{2}=I\) — applying it twice returns the original state
  • Entangling: it can create entanglement from separable states (see Tab ④)

✦ CNOT on Superpositions

The real power of CNOT appears when the control qubit is in superposition. Click to see what happens:

Key insight: When the control is in superposition, CNOT creates entanglement — the output cannot be factored as a product of single-qubit states. This is exactly how Bell pairs are created!
💡 Pause & Think
If the control qubit is in state \(|0\rangle\) (not in superposition), can CNOT ever create entanglement?
Yes — CNOT always entangles
No — the output is always separable
It depends on the target qubit

⟁ The EPR Circuit

An EPR pair (Einstein–Podolsky–Rosen pair), also called a Bell pair, is the simplest entangled two-qubit state. It is created by just two gates:

  1. Apply a Hadamard gate \(H\) to the first qubit.
  2. Apply a CNOT with the first qubit as control and the second as target.
Starting from \(|00\rangle\), this circuit produces the Bell state \(|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}\!\left(|00\rangle+|11\rangle\right)\).

▶ Step-by-Step Construction

Click the steps below or use the buttons to watch the circuit build the EPR pair.

What's happening

✦ All Four Bell States

By choosing different input pairs, the same \(H\)+CNOT circuit produces all four maximally-entangled Bell states:

InputNameState
|00⟩|Φ⁺⟩\(\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)\)
|01⟩|Ψ⁺⟩\(\tfrac{1}{\sqrt2}(|01\rangle+|10\rangle)\)
|10⟩|Φ⁻⟩\(\tfrac{1}{\sqrt2}(|00\rangle-|11\rangle)\)
|11⟩|Ψ⁻⟩\(\tfrac{1}{\sqrt2}(|01\rangle-|10\rangle)\)
Orthonormal basis: The four Bell states form a complete orthonormal basis for the 4-dimensional two-qubit space. Any two-qubit state can be decomposed into a superposition of Bell states.

✦ How Entanglement Is Born

Watch the quantum state evolve as it passes through the Hadamard and CNOT gates. The entanglement link appears when the state can no longer be factored.

q1 q2 |0⟩ |0⟩ H |+⟩ |0⟩ Entangled! |Φ⁺⟩ Separable Superposed Entangled

⚡ Try Different Inputs

Select a starting two-qubit state to see which Bell state the circuit produces.

🎲 Measure an EPR Pair

Experience entanglement first-hand! Start with \(|\Phi^+\rangle=\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)\), then measure one qubit and watch the other instantly collapse.

q₁
Qubit 1
q₂
Qubit 2
|00⟩: 0 |11⟩: 0 Total: 0

Run many measurements to see the 50/50 statistics emerge — but always perfectly correlated!

🎯 Challenge: Predict the Measurement

Test your intuition! You'll be given a two-qubit state and told what Qubit 1 measured. Can you predict Qubit 2?

Click "New Round" to start!
Correct: 0 Total: 0 Streak: 0

⟐ What is Entanglement?

A two-qubit state \(|\psi\rangle\) is called separable (not entangled) if it can be written as a tensor product of two single-qubit states:

$$|\psi\rangle = |\alpha\rangle\otimes|\beta\rangle$$

for some single-qubit states \(|\alpha\rangle\) and \(|\beta\rangle\).

Definition: A state that cannot be written as such a product is called entangled.

Entanglement is a uniquely quantum phenomenon with no classical counterpart. When two qubits are entangled, the outcome of measuring one qubit is correlated with the outcome of measuring the other — no matter how far apart they are.

⦿ Non-local Correlations

Measurement outcomes are correlated across any distance. Measuring one qubit instantly constrains the other, yet no information travels faster than light.

✖ No Cloning

Entangled states cannot be perfectly copied. The no-cloning theorem ensures quantum information is fundamentally different from classical bits.

⌘ Monogamy

A qubit maximally entangled with one partner cannot be entangled with another. Entanglement is a limited resource that must be shared carefully.

⚡ Resource for Computation

Entanglement powers teleportation, superdense coding, and quantum error correction. It is the key resource behind quantum advantage.

⚡ Separable vs. Entangled — Visual Comparison

Click each example to see whether it is separable or entangled, and why.

⊡ The Determinant Criterion

Write a general two-qubit state as:

$$|\psi\rangle = \alpha|00\rangle+\beta|01\rangle+\gamma|10\rangle+\delta|11\rangle$$

Arrange the amplitudes in a \(2\times 2\) matrix and compute its determinant:

$$M=\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix},\qquad \det M = \alpha\delta-\beta\gamma$$
Criterion: The state is separable ⟺ \(\alpha\delta-\beta\gamma=0\). If this quantity is nonzero, the state is entangled.

Why? If \(|\psi\rangle=(a|0\rangle+b|1\rangle)\otimes(c|0\rangle+d|1\rangle)\), then \(\alpha=ac,\;\beta=ad,\;\gamma=bc,\;\delta=bd\), so \(\alpha\delta-\beta\gamma=acbd-adbc=0\). A nonzero value means no such factorisation exists.

✎ Worked Example — Is \(\tfrac{1}{2}(|00\rangle+|01\rangle+|10\rangle+|11\rangle)\) entangled?
1
Identify the amplitudes: \(\alpha = \tfrac12,\;\beta = \tfrac12,\;\gamma = \tfrac12,\;\delta = \tfrac12\).
2
Form the amplitude matrix: \(M = \begin{pmatrix}\tfrac12 & \tfrac12 \\[4pt] \tfrac12 & \tfrac12\end{pmatrix}\).
3
Compute the determinant: \(\det M = \tfrac12 \cdot \tfrac12 - \tfrac12 \cdot \tfrac12 = \tfrac14 - \tfrac14 = 0\).
4
Since \(\det M = 0\), the state is separable. It factors as \(\tfrac{|0\rangle+|1\rangle}{\sqrt2}\otimes\tfrac{|0\rangle+|1\rangle}{\sqrt2} = |+\rangle\otimes|+\rangle\). Try it in the detector below!

🔬 Interactive Entanglement Detector

Enter the amplitudes of any two-qubit state and check whether it is entangled.

💡 Pause & Think
The state \(\tfrac{1}{2}(|00\rangle+|01\rangle+|10\rangle+|11\rangle)\) has amplitudes \(\alpha=\beta=\gamma=\delta=\tfrac{1}{2}\). Is it entangled?
Yes — it has four nonzero amplitudes
No — \(\alpha\delta-\beta\gamma = \tfrac14 - \tfrac14 = 0\)
Cannot tell without measuring

Theorem: \(|\Phi^+\rangle\) is Entangled

We prove that \(|\Phi^+\rangle=\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)\) cannot be written as a tensor product of two single-qubit states. Click each step to reveal it.

Step 1 — Assume Separability

Assume for contradiction that \(|\Phi^+\rangle\) is separable. Then there exist single-qubit states

$$|\alpha\rangle=a|0\rangle+b|1\rangle,\qquad |\beta\rangle=c|0\rangle+d|1\rangle$$

with \(|\Phi^+\rangle=|\alpha\rangle\otimes|\beta\rangle\).

Step 2 — Expand the Tensor Product

Expanding:

$$|\alpha\rangle\otimes|\beta\rangle = ac\,|00\rangle + ad\,|01\rangle + bc\,|10\rangle + bd\,|11\rangle$$
Step 3 — Match Coefficients

Comparing with \(|\Phi^+\rangle=\tfrac{1}{\sqrt2}|00\rangle+0\cdot|01\rangle+0\cdot|10\rangle+\tfrac{1}{\sqrt2}|11\rangle\) gives the system:

$$ac=\tfrac{1}{\sqrt2},\quad ad=0,\quad bc=0,\quad bd=\tfrac{1}{\sqrt2}$$
Step 4 — Derive the Contradiction

From \(ad=0\): either \(a=0\) or \(d=0\).

  • Case 1 — \(a=0\): then \(ac=0\neq\tfrac{1}{\sqrt2}\). Contradiction.
  • Case 2 — \(d=0\): then \(bd=0\neq\tfrac{1}{\sqrt2}\). Contradiction.
Step 5 — Conclusion

Since the assumption of separability leads to a contradiction in every case, we conclude:

\(|\Phi^+\rangle=\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)\) is entangled. \(\blacksquare\)

Quick Check via the Determinant

We can verify the same result algebraically. For \(|\Phi^+\rangle\) the amplitude matrix is:

$$M=\begin{pmatrix}\tfrac{1}{\sqrt2}&0\\[4pt]0&\tfrac{1}{\sqrt2}\end{pmatrix}$$

Its determinant:

$$\det M=\tfrac{1}{\sqrt2}\cdot\tfrac{1}{\sqrt2}-0\cdot0=\tfrac{1}{2}\neq 0$$
Because \(\det M=\tfrac12\neq0\), the state is entangled. The Bell states achieve the maximum \(|\det M|=\frac12\) — they are maximally entangled.
Score: 0 / 7

Quiz 1 — Tensor Product Basics

What is \(|1\rangle\otimes|0\rangle\) as a column vector?

\(\begin{pmatrix}1\\0\\0\\0\end{pmatrix}\) — this is \(|00\rangle\)
\(\begin{pmatrix}0\\1\\0\\0\end{pmatrix}\) — this is \(|01\rangle\)
\(\begin{pmatrix}0\\0\\1\\0\end{pmatrix}\) — this is \(|10\rangle\)
\(\begin{pmatrix}0\\0\\0\\1\end{pmatrix}\) — this is \(|11\rangle\)

Quiz 2 — Hilbert Space Dimension

How many complex amplitudes are needed to describe a general 5-qubit state?

5
10
32
25

Quiz 3 — CNOT Action

What is \(\mathrm{CNOT}\,|1,0\rangle\)?

|00⟩
|10⟩
|11⟩
|01⟩

Quiz 4 — Superposition Through CNOT

Applying CNOT to \(\tfrac{1}{\sqrt2}(|00\rangle+|10\rangle)\) yields:

\(\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)\)
\(\tfrac{1}{\sqrt2}(|00\rangle+|10\rangle)\)
\(\tfrac{1}{\sqrt2}(|01\rangle+|11\rangle)\)
\(\tfrac{1}{\sqrt2}(|10\rangle+|01\rangle)\)

Quiz 5 — Entanglement Detection

Is \(\tfrac12(|00\rangle+|01\rangle+|10\rangle+|11\rangle)\) entangled?

No — it is separable \(\bigl(\tfrac{|0\rangle+|1\rangle}{\sqrt2}\otimes\tfrac{|0\rangle+|1\rangle}{\sqrt2}\bigr)\)
Yes — it is entangled
Cannot be determined without measurement

Quiz 6 — Bell State Identification

Which Bell state does \((H\otimes I)\cdot\mathrm{CNOT}\) produce from \(|11\rangle\)?

\(|\Phi^+\rangle=\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)\)
\(|\Psi^+\rangle=\tfrac{1}{\sqrt2}(|01\rangle+|10\rangle)\)
\(|\Phi^-\rangle=\tfrac{1}{\sqrt2}(|00\rangle-|11\rangle)\)
\(|\Psi^-\rangle=\tfrac{1}{\sqrt2}(|01\rangle-|10\rangle)\)

Quiz 7 — Tensor Product Properties

Is the tensor product commutative? That is, does \(|a\rangle\otimes|b\rangle = |b\rangle\otimes|a\rangle\) always hold?

Yes — order does not matter
No — \(|0\rangle\otimes|1\rangle = |01\rangle \neq |10\rangle = |1\rangle\otimes|0\rangle\)
Only for basis states, not superpositions

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