Tensor Products and Multi-Qubit State Space

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Tensor Products and Multi-Qubit State Space

How single-qubit spaces combine into the 2^n-dimensional space multi-qubit states live in, and why entanglement means "not a tensor product."

Suggested before this lesson

Unitary Matrices

Everything so far has lived in $\mathbb{C}^2$ , a single qubit's state space. This lesson covers the operation that builds up the state space for multiple qubits: the tensor product, written $\otimes$ .

Tensor Product of Vectors

Given $|\psi\rangle = \begin{pmatrix}\alpha_0\\\alpha_1\end{pmatrix}$ and $|\phi\rangle = \begin{pmatrix}\beta_0\\\beta_1\end{pmatrix}$ , their tensor product is a 4-entry vector:

$|\psi\rangle \otimes |\phi\rangle = \begin{pmatrix}\alpha_0\beta_0 \\ \alpha_0\beta_1 \\ \alpha_1\beta_0 \\ \alpha_1\beta_1\end{pmatrix}$

every pairwise product of one entry from each vector, in a fixed order. Apply this to the computational basis states and you get exactly the 2-qubit basis you already know from every circuit diagram in this course:

$|00\rangle$ is shorthand for the 4-entry vector $(1,0,0,0)^\mathsf{T}$ , $|01\rangle$ for $(0,1,0,0)^\mathsf{T}$ , and so on, each ket labeling one axis of $\mathbb{C}^4$ .

Why Dimension Doubles Per Qubit

One qubit: $\mathbb{C}^2$ . Two qubits: $\mathbb{C}^2\otimes\mathbb{C}^2 = \mathbb{C}^4$ . $n$ qubits: tensor $n$ copies of $\mathbb{C}^2$ together and you get $\mathbb{C}^{2^n}$ . This is the mathematical reason simulating $n$ qubits classically requires tracking $2^n$ complex amplitudes, and why every visualization in this course that shows exact amplitudes (the Exact State table) grows exponentially with qubit count: the state genuinely lives in a $2^n$ -dimensional space, there's no shortcut representation for a generic state.

Tensor Product of Gates

Applying a gate to one qubit while leaving another untouched is the tensor product of that gate with the identity. qc.h(0) on a 2-qubit circuit ( $H$ on qubit 0, nothing on qubit 1) is the $4\times4$ matrix $I \otimes H$ (Qiskit's own little-endian qubit ordering puts qubit 0's gate on the right of the tensor product, qubit 1's on the left). Two independent single-qubit gates applied together, qc.h(0) and qc.x(1) in the same moment, compose as $X\otimes H$ . This is also why a broadcast like qc.h([0, 1, 2]) amounts to a single $8\times8$ matrix $H\otimes H\otimes H$ acting on all three qubits' joint state at once, exactly the uniform-superposition construction the algorithms track uses constantly.

Entanglement Is "Not a Tensor Product"

Not every valid 2-qubit state is some $|\psi\rangle\otimes|\phi\rangle$ . Take the Bell state:

$\tfrac{1}{\sqrt2}(|00\rangle + |11\rangle) = \tfrac{1}{\sqrt2}\begin{pmatrix}1\\0\\0\\1\end{pmatrix}$

Suppose this were $|\psi\rangle\otimes|\phi\rangle$ for some $|\psi\rangle=(\alpha_0,\alpha_1)^\mathsf{T}$ , $|\phi\rangle=(\beta_0,\beta_1)^\mathsf{T}$ . Matching entries against the formula above: $\alpha_0\beta_0=\tfrac{1}{\sqrt2}$ , $\alpha_0\beta_1=0$ , $\alpha_1\beta_0=0$ , $\alpha_1\beta_1=\tfrac{1}{\sqrt2}$ . From $\alpha_0\beta_0\neq0$ , neither $\alpha_0$ nor $\beta_0$ is zero. But then $\alpha_0\beta_1=0$ forces $\beta_1=0$ , and $\alpha_1\beta_0=0$ forces $\alpha_1=0$ , which makes $\alpha_1\beta_1=0$ , contradicting $\alpha_1\beta_1=\tfrac{1}{\sqrt2}$ . No such $|\psi\rangle,|\phi\rangle$ exist.

This is the formal definition of entanglement: a multi-qubit state is entangled exactly when it can't be factored as a tensor product of individual single-qubit states, i.e. exactly when there's no way to describe "qubit 0's state" and "qubit 1's state" separately. The Bell state above is the canonical example, and it's exactly the state the entanglement lesson right after this track builds with $H$ then $CX$ .

Build a two-qubit state

Pick a state for each qubit and watch the four joint amplitudes of $|\psi\rangle \otimes |\phi\rangle$ compute live, then try the Bell state and see why no combination of buttons can reach it.

Qubit 0 (left)

Qubit 1 (right)

Joint amplitudes

$∣ + ⟩ \otimes ∣ + ⟩$

$∣00 ⟩$

0.500

$∣01 ⟩$

0.500

$∣10 ⟩$

0.500

$∣11 ⟩$

0.500

Pick a single-qubit state for each qubit and watch the four joint amplitudes update live, every entry is just one amplitude from the left state times one from the right. Then flip to the Bell state: it is a perfectly valid 2-qubit state, but no button combination above builds it, which is exactly what "entangled" means.

Check your understanding

What is

|1\rangle \otimes |0\rangle

|01\rangle

|10\rangle

|11\rangle

|00\rangle

Check your understanding

Is the state

\frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)

entangled?

Yes, since it has 4 nonzero terms.

No: it factors as

|+\rangle \otimes |+\rangle

, an equal superposition on each qubit independently, so it's a product state, not entangled.

Impossible to tell without measuring it.

Yes, any state with more than one nonzero amplitude is entangled.

Tensor Product of Vectors

Given

|\psi\rangle = \begin{pmatrix}\alpha_0\\\alpha_1\end{pmatrix}

and

|\phi\rangle = \begin{pmatrix}\beta_0\\\beta_1\end{pmatrix}

, their tensor product is a 4-entry vector:

|\psi\rangle \otimes |\phi\rangle = \begin{pmatrix}\alpha_0\beta_0 \\ \alpha_0\beta_1 \\ \alpha_1\beta_0 \\ \alpha_1\beta_1\end{pmatrix}

|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

|00\rangle

is shorthand for the 4-entry vector

(1,0,0,0)^\mathsf{T}

|01\rangle

for

(0,1,0,0)^\mathsf{T}

, and so on, each ket labeling one axis of

\mathbb{C}^4

Why Dimension Doubles Per Qubit

One qubit:

\mathbb{C}^2

. Two qubits:

\mathbb{C}^2\otimes\mathbb{C}^2 = \mathbb{C}^4

n

qubits: tensor

n

copies of

\mathbb{C}^2

together and you get

\mathbb{C}^{2^n}

. This is the mathematical reason simulating

n

qubits classically requires tracking

2^n

complex amplitudes, and why every visualization in this course that shows exact amplitudes (the Exact State table) grows exponentially with qubit count: the state genuinely lives in a

2^n

-dimensional space, there's no shortcut representation for a generic state.

Tensor Product of Gates

Applying a gate to one qubit while leaving another untouched is the tensor product of that gate with the identity. qc.h(0) on a 2-qubit circuit (

H

on qubit 0, nothing on qubit 1) is the

4\times4

matrix

I \otimes H

(Qiskit's own little-endian qubit ordering puts qubit 0's gate on the right of the tensor product, qubit 1's on the left). Two independent single-qubit gates applied together, qc.h(0) and qc.x(1) in the same moment, compose as

X\otimes H

. This is also why a broadcast like qc.h([0, 1, 2]) amounts to a single

8\times8

matrix

H\otimes H\otimes H

acting on all three qubits' joint state at once, exactly the uniform-superposition construction the algorithms track uses constantly.

Entanglement Is "Not a Tensor Product"

Not every valid 2-qubit state is some

|\psi\rangle\otimes|\phi\rangle

. Take the Bell state:

\tfrac{1}{\sqrt2}(|00\rangle + |11\rangle) = \tfrac{1}{\sqrt2}\begin{pmatrix}1\\0\\0\\1\end{pmatrix}

Suppose this were

|\psi\rangle\otimes|\phi\rangle

for some

|\psi\rangle=(\alpha_0,\alpha_1)^\mathsf{T}

|\phi\rangle=(\beta_0,\beta_1)^\mathsf{T}

. Matching entries against the formula above:

\alpha_0\beta_0=\tfrac{1}{\sqrt2}

\alpha_0\beta_1=0

\alpha_1\beta_0=0

\alpha_1\beta_1=\tfrac{1}{\sqrt2}

. From

\alpha_0\beta_0\neq0

, neither

\alpha_0

nor

\beta_0

is zero. But then

\alpha_0\beta_1=0

forces

\beta_1=0

, and

\alpha_1\beta_0=0

forces

\alpha_1=0

, which makes

\alpha_1\beta_1=0

, contradicting

\alpha_1\beta_1=\tfrac{1}{\sqrt2}

. No such

|\psi\rangle,|\phi\rangle

exist.

H

then

CX

Build a two-qubit state

Pick a state for each qubit and watch the four joint amplitudes of $|\psi\rangle \otimes |\phi\rangle$ compute live, then try the Bell state and see why no combination of buttons can reach it.

Qubit 0 (left)

Qubit 1 (right)

Joint amplitudes

$∣ + ⟩ \otimes ∣ + ⟩$

$∣00 ⟩$

0.500

$∣01 ⟩$

0.500

$∣10 ⟩$

0.500

$∣11 ⟩$

0.500

Check your understanding

What is

|1\rangle \otimes |0\rangle

|01\rangle

|10\rangle

|11\rangle

|00\rangle

Check your understanding

Is the state

\frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)

entangled?

Yes, since it has 4 nonzero terms.

No: it factors as

|+\rangle \otimes |+\rangle

, an equal superposition on each qubit independently, so it's a product state, not entangled.

Impossible to tell without measuring it.

Yes, any state with more than one nonzero amplitude is entangled.