Tensor Products (Multi-Qubit States)

Overview

A single qubit has states like $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$. But real quantum computation becomes interesting only when we combine qubits into joint systems.

This page introduces the mathematical rule for combining quantum systems: the tensor product. You will learn why it is required (not optional), how it acts on basis states, and how the compact notation $|00\rangle$, $|01\rangle$, $|10\rangle$, $|11\rangle$ arises.

Intuition

For two classical bits, the joint state is a pair like $(0,1)$. There are four possibilities: $(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$.

For two qubits, the joint system also has four computational-basis outcomes: $00$, $01$, $10$, $11$. But the state is not just “a pair of qubit states.”

Why not?

• A qubit state is a vector (a rule for probabilities across outcomes).
• A two-qubit state must be a single object that predicts probabilities for joint outcomes.
• Those probabilities must allow superposition and interference across the four joint outcomes.

So the state space of a joint system must be a space where you can:

1. represent the four joint basis outcomes, and
2. take linear combinations of them with complex amplitudes.

That is exactly what the tensor product construction provides.

A useful way to think about it: “two qubits” means the number of basis outcomes multiplies (2 outcomes per qubit → $2\times2=4$ outcomes total), and therefore the number of amplitude coordinates multiplies too.

Formal Description

Single-qubit basis

Recall the computational basis for one qubit:

$$|0\rangle,\ |1\rangle.$$

A general single-qubit state is

$$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle.$$

Tensor product of basis states

For a two-qubit system, the computational basis is built from tensor products of the one-qubit basis states:

$$|0\rangle\otimes|0\rangle,\quad |0\rangle\otimes|1\rangle,\quad |1\rangle\otimes|0\rangle,\quad |1\rangle\otimes|1\rangle.$$

We use the shorthand:

$$|0\rangle\otimes|0\rangle = |00\rangle,\quad |0\rangle\otimes|1\rangle = |01\rangle,\quad |1\rangle\otimes|0\rangle = |10\rangle,\quad |1\rangle\otimes|1\rangle = |11\rangle.$$

Here is what the notation means in words:

• $|01\rangle$ means: “first qubit is in the basis state $|0\rangle$ and second qubit is in the basis state $|1\rangle$.”
• The left digit refers to the first subsystem; the right digit refers to the second.

Tensor product distributes over addition

The tensor product is linear in each input. The most important expansion rule is:

$$(\alpha|0\rangle+\beta|1\rangle)\otimes|0\rangle = \alpha(|0\rangle\otimes|0\rangle) + \beta(|1\rangle\otimes|0\rangle).$$

In shorthand:

$$(\alpha|0\rangle+\beta|1\rangle)\otimes|0\rangle = \alpha|00\rangle + \beta|10\rangle.$$

This is how “a state of qubit A and a state of qubit B” becomes a single joint state.

General two-qubit state

A general pure state of two qubits can be written as a linear combination of the four basis states:

$$|\Psi\rangle = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle + \alpha_{10}|10\rangle + \alpha_{11}|11\rangle.$$

Each coefficient $\alpha_{ij}$ is a complex amplitude.

If you measure both qubits in the computational basis, then

$$P(ij) = |\alpha_{ij}|^2,$$

and normalization requires

$$|\alpha_{00}|^2+|\alpha_{01}|^2+|\alpha_{10}|^2+|\alpha_{11}|^2 = 1.$$

Cartesian product vs tensor product

A Cartesian product of sets lists pairs, like $(0,1)$. It describes classical joint possibilities.

A tensor product of vector spaces creates a space where you can form superpositions of joint basis states. That is why quantum joint states use tensor products rather than simple pairs.

Worked Example

Let qubit A be in

$$|\psi\rangle = \tfrac{3}{5}|0\rangle + \tfrac{4}{5}|1\rangle$$

and qubit B be in $|0\rangle$.

The joint state is the tensor product:

$$|\Psi\rangle = |\psi\rangle\otimes|0\rangle.$$

Expand using linearity:

$$|\Psi\rangle = \left(\tfrac{3}{5}|0\rangle + \tfrac{4}{5}|1\rangle\right)\otimes|0\rangle = \tfrac{3}{5}|00\rangle + \tfrac{4}{5}|10\rangle.$$

Now measure both qubits in the computational basis.

• Outcome $00$ occurs with probability $|3/5|^2=9/25$.
• Outcome $10$ occurs with probability $|4/5|^2=16/25$.
• Outcomes $01$ and $11$ have probability 0.

This matches the intuition: the second qubit is definitely 0, while the first is probabilistic.

Turtle Tip

Common Pitfalls

Quick Check

What’s Next

Tensor products tell us how to build joint state spaces. Next we’ll classify two-qubit states into two fundamentally different types: product states (which factor into single-qubit states) and entangled states (which do not).