Lecture 3.1 — Two Qubits and the Tensor Product

The Turning Point¶

Everything changes when we have more than one qubit.

In Chapter 2, we mastered the single qubit: a wave over two configurations, complex amplitudes, rotations on the Bloch sphere, interference, measurement. Beautiful physics, and the foundation of quantum sensing.

But a single qubit is not enough for quantum computing. With one qubit, you have 2 configurations. With two qubits, you have 4. With three, you have 8. With $N$ qubits, you have $2^N$ configurations.

This exponential scaling is where quantum computing gets its power — and where things become truly strange. A system of just 50 qubits has more configurations than there are atoms in the Earth. No classical computer can even write down the full quantum state.

Today we learn how to describe multi-qubit systems, and we’ll meet entanglement — correlations that have no classical explanation.

Part 1: Classical Correlations¶

Before quantum mechanics, let’s build our intuition with something tangible.

Two Marbles¶

I have two marbles: marble A (yours) and marble B (mine). Each marble is either blue (0) or red (1), chosen at random.

Individual probability distributions:

Since there are two marbles, each with two possible colors, the joint system has four possible outcomes — all possible pairs:

That’s $2 \times 2 = 4$ outcomes. For $N$ marbles: $2^N$ outcomes. This combinatorial explosion isn’t quantum mechanics yet — it’s just counting. But quantum mechanics inherits this structure.

Joint Probability Distributions¶

A joint probability distribution $P(i, j)$ assigns a probability to each of the four outcomes. It must satisfy:

The marginal distributions are obtained by summing over the other variable:

The marginals tell you about each marble individually. But they don’t fully determine the joint distribution — different joint distributions can have the same marginals. The extra information is in the correlations between the marbles.

The key question is: what joint probability distribution describes a given pair?

Correlated Outcomes¶

The answer is (C): outcomes (0,0) and (1,1) each with probability 1/2. We never see different colors.

Are these marbles independent?

Two systems are independent (uncorrelated) if the joint probability factors into a product of the marginals:

Let’s check. The marginals of distribution (C) are $P_A(0) = 0.5$, $P_A(1) = 0.5$, and $P_B(0) = 0.5$, $P_B(1) = 0.5$. The product gives $P_A(0) \cdot P_B(0) = 0.25 \neq 0.5 = P(0,0)$.

The joint distribution cannot be written as a product of marginals. These marbles are classically correlated.

Anti-Correlated Outcomes¶

The answer is (A): outcomes (0,1) and (1,0) each with probability 1/2.

Are they correlated? Try factoring: $P(i,j) = P_A(i) \cdot P_B(j)$. With $P_A(0) = 0.5$ and $P_B(0) = 0.5$, the product gives $P(0,0) = 0.25 \neq 0$. Doesn’t work. We cannot write this as a product. The marbles are classically correlated — this time, anti-correlated.

Classical Correlations Are Not Unusual¶

There’s nothing strange about classical correlations — they happen all the time. If I put one red marble and one blue marble into two boxes and shuffle them, the boxes are anti-correlated. Opening one box tells you the color of the other. This is perfectly described by a hidden variable: the shuffling procedure determined which marble went where, and opening a box just reveals a pre-existing fact.

Any classical correlation, no matter how strong, can be explained by a hidden variable. The correlation was established in the past, encoded in some shared information, and measurement just reads it out. This is the picture Einstein wanted for quantum mechanics too.

What we’ll discover in the next lecture — through Bell’s theorem — is that quantum mechanics produces correlations that are too strong to be explained this way. It’s not that quantum systems are correlated (classical systems are too). It’s that the amount and structure of quantum correlations exceed what any hidden variable model can produce.

Keep this in your back pocket: we’re about to see quantum states with exactly the same measurement statistics as these marbles — outcomes 00 and 11, or 01 and 10, each 50/50. The question will be: is the explanation the same?

Part 2: Now Qubits¶

One Qubit (Review)¶

A single qubit lives in a 2-dimensional Hilbert space:

It’s a wave over two configurations, fully described by a point on the Bloch sphere — 2 real degrees of freedom (after normalization and global phase). We can represent it as a column vector:

Quantum Mechanics Is a Wave over Configurations¶

Here is the key conceptual leap.

For a single qubit, the state is a superposition over two possibilities:

Each possibility gets a complex amplitude. The squared magnitudes give probabilities.

For two qubits, the state is a superposition over all four configurations of the pair:

Each configuration gets its own complex amplitude. This is the essential structure of quantum mechanics: it assigns a complex number — an amplitude with magnitude and phase — to every possible configuration of the system. The probabilities are the squared magnitudes: $P(01) = |c_{01}|^2$, etc.

For $N$ qubits, there are $2^N$ configurations, each with its own amplitude. This exponential growth is the fundamental resource of quantum computing — and the reason quantum systems are hard to simulate classically.

Quantum dynamics is a wave over configurations. Everything that can happen does happen, just with some complex amplitude (magnitude and phase).

Two Independent Qubits: The Tensor Product¶

Now suppose we have two qubits: qubit A in state $|\psi\rangle = a|0\rangle + b|1\rangle$ and qubit B in state $|\phi\rangle = c|0\rangle + d|1\rangle$.

How do we describe the joint system? We need a mathematical operation that combines configuration spaces. This is the tensor product, written $\otimes$:

The tensor product distributes just like ordinary multiplication — $(a + b)(c + d) = ac + ad + bc + bd$:

where $|0\rangle \otimes |0\rangle \equiv |00\rangle$, and so on.

This describes two independent particles that have never interacted. Each qubit has its own state, the amplitudes are just products of individual amplitudes ($\Psi_{ij} = u_i v_j$), and knowledge about one qubit tells you nothing about the other. This is the quantum analog of independent marbles.

You will see several equivalent notations:

We’ll mostly use the compact form $|01\rangle$.

The General Two-Qubit State¶

The tensor product describes two independent, non-interacting particles. But the most general two-qubit state is richer:

with four complex amplitudes satisfying $\sum|C_{ij}|^2 = 1$.

For a tensor product state, the amplitudes factor: $C_{ij} = u_i v_j$. But in general, the four amplitudes can be anything (subject to normalization). When they don’t factor, the state describes particles whose correlations go beyond anything a product description can capture. These are entangled states, and they arise from interactions between particles or from correlated creation processes (like photon pair generation).

Counting Degrees of Freedom¶

How much information does a two-qubit state carry?

Now compare: two independent qubits, each on its own Bloch sphere, have $2 + 2 = 4$ degrees of freedom. Where do the extra 2 come from?

Those extra 2 degrees of freedom encode correlations between the qubits. A product state like $|+\rangle \otimes |0\rangle$ uses only the 4 Bloch sphere DOFs — the correlations are zero, and the qubits are independent. But some two-qubit states use all 6 DOFs. These are the entangled states, and they’re what we’ll explore next.

The Tensor Product as a Matrix and a Vector¶

For two qubits with states $\vec{u} = \begin{pmatrix} u_0 \\ u_1 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} v_0 \\ v_1 \end{pmatrix}$, the tensor product defines a set of amplitudes:

We can arrange these amplitudes in two equivalent ways.

As a $2 \times 2$ matrix (the amplitude table):

Rows correspond to qubit A’s state ($|0\rangle_A$ or $|1\rangle_A$), columns to qubit B’s. For a product state, this matrix is an outer product of two vectors: $C = \vec{u}\,\vec{v}^T$.

As a 4-component column vector:

This is simply the matrix entries read off row by row into a single column. Both representations contain the same information. The vector form generalizes to $N$ qubits (a column of $2^N$ entries), while the matrix form is specific to two qubits — but it will be very useful when we introduce the determinant test for entanglement in the next lecture.

For $N$ qubits, the state vector has $2^N$ components. A classical computer simulating 50 qubits must store $2^{50} \approx 10^{15}$ complex numbers — about a petabyte of memory. A quantum computer with 50 qubits just... has 50 qubits. This is the source of quantum computational power: the state space grows exponentially, but the physical resources grow linearly.

Summary¶

1. Classical correlations arise when two systems share a common cause. The joint probability $P(i,j)$ cannot be written as $P_A(i) \cdot P_B(j)$, but there’s nothing mysterious — a hidden variable explains it. Classical correlations are perfectly normal; what Bell’s theorem will show is that quantum correlations are stronger than any hidden variable can produce.

2. Quantum mechanics is a wave over configurations. A two-qubit state assigns a complex amplitude to each of the four joint configurations $|00\rangle, |01\rangle, |10\rangle, |11\rangle$. For $N$ qubits: $2^N$ amplitudes.

3. The tensor product $\otimes$ describes two independent qubits: amplitudes multiply ($\Psi_{ij} = u_i v_j$), and the result can be written as a $2 \times 2$ matrix (outer product $\vec{u}\,\vec{v}^T$) or a 4-component vector.

4. A general two-qubit state has 6 real DOF = 2 (Bloch sphere A) + 2 (Bloch sphere B) + 2 (correlations). The extra DOFs encode correlations — and when the amplitudes don’t factor, we call the state entangled.

Homework¶

Problem 1: Tensor Product Practice¶

Compute the following tensor products and write the result as a 4-component column vector.

(a) $|0\rangle \otimes |1\rangle$

(b) $|1\rangle \otimes |+\rangle$ where $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$

(c) $|+\rangle \otimes |-\rangle$ where $|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$

(d) $|+\rangle \otimes |+\rangle$

Problem 2: Three Qubits¶

(a) How many computational basis states are there for three qubits? List them all.

(b) Write $|+\rangle \otimes |0\rangle \otimes |1\rangle$ as a sum of computational basis states.

(c) The GHZ state is $|GHZ\rangle = \frac{1}{\sqrt{2}}(|000\rangle + |111\rangle)$. Is it a product state? Try to factor it.

(d) If you measure all three qubits of $|GHZ\rangle$ in the computational basis, what outcomes are possible?

Looking Ahead¶

We’ve built the mathematical framework for two-qubit states: tensor products, the amplitude table, and the $6 = 4 + 2$ DOF counting that reveals where correlations live.

But we haven’t yet explored what makes quantum correlations different from classical ones. Next lecture, we’ll introduce the Bell states — the four maximally entangled two-qubit states — and discover that their correlations persist across every measurement basis simultaneously. This is something no classical system can do, and it leads directly to Einstein’s challenge, Bell’s theorem, and the foundations of quantum information science. -->