# Multi-Qubit States > One qubit lives in $\mathbb{C}^2$. Two qubits do **not** live in $\mathbb{C}^2\times\mathbb{C}^2$ — > they live in the *tensor product* $\mathbb{C}^2\otimes\mathbb{C}^2=\mathbb{C}^4$, and that single > fact is where the entire mystery of quantum computing begins. This lesson assembles the > $n$-qubit state space, pins down the big-endian computational basis you will see in every Braket > bit-string, and gives you a sharp, runnable test for the question that drives the rest of the > course: *is this state entangled?* ## Learning Objectives After this lesson you will be able to: 1. Construct the $n$-qubit state space $(\mathbb{C}^2)^{\otimes n}=\mathbb{C}^{2^n}$ and write its **computational basis** $\{|q_0\cdots q_{n-1}\rangle\}$ with correct **big-endian** indexing. 2. Distinguish **product (separable)** states from **entangled** states, and prove a given state is one or the other. 3. Express **local** operations as tensor products $A\otimes B$ and contrast them with **global** (entangling) operations. 4. Apply the **coefficient-matrix / Schmidt-rank** separability test to a bipartite pure state. 5. Build multi-qubit states in NumPy and decide separability numerically via matrix rank. --- ## Intuition A single qubit is a unit vector $\alpha|0\rangle+\beta|1\rangle$ in $\mathbb{C}^2$. The naïve guess for two qubits is "a pair of qubits," $(\mathbb{C}^2,\mathbb{C}^2)$ — four real parameters. That guess is **wrong**, and the reason is the whole story. Quantum mechanics says the joint system can be in any *superposition* of the four joint configurations $|00\rangle,|01\rangle,|10\rangle,|11\rangle$, so its state space is the $4$-dimensional span of those configurations: $\mathbb{C}^4$, not a pair of $\mathbb{C}^2$'s. The extra room — four complex amplitudes instead of two pairs — is exactly the room where **entanglement** hides. The headline consequence is exponential: $n$ qubits need $2^n$ complex amplitudes. Most of those states do **not** factor into "a state of qubit 0" times "a state of qubit 1" times … . A factorable state is *separable* (boring, classical-ish); a non-factorable one is *entangled* (the resource behind teleportation, dense coding, and quantum speedups — Term 2). Our job here is to make "factorable" precise and testable, because everything downstream is a refinement of that distinction. --- ## Theory ### The $n$-qubit state space By the composite-systems postulate ([1.1.4](../c01-postulates/04-composite-systems-postulate.md)), the state space of a system built from parts is the **tensor product** of the parts' spaces ([0.1.7](../../term-0-foundations/c01-linear-algebra/07-tensor-products.md)). For $n$ qubits, $$ \mathcal{H}_n = \underbrace{\mathbb{C}^2\otimes\mathbb{C}^2\otimes\cdots\otimes\mathbb{C}^2}_{n\text{ factors}} = (\mathbb{C}^2)^{\otimes n} = \mathbb{C}^{2^n}, \qquad \dim\mathcal{H}_n = 2^n . $$ The dimension **multiplies** ($2\cdot2\cdots = 2^n$), it does not add — this is the single most important difference between quantum and classical state spaces, and the origin of the simulation wall at $\sim 50$ qubits. ### The computational basis and big-endian indexing An orthonormal basis of $\mathcal{H}_n$ is obtained by tensoring the single-qubit basis kets in all $2^n$ ways: $$ |q_0 q_1\cdots q_{n-1}\rangle := |q_0\rangle\otimes|q_1\rangle\otimes\cdots\otimes|q_{n-1}\rangle, \qquad q_k\in\{0,1\}. $$ We use **big-endian** ordering ([Appendix C](../../appendix/C-notation-glossary.md)): **qubit 0 is leftmost / most significant**. The bit-string $q_0 q_1\cdots q_{n-1}$ is read as a binary integer that fixes the basis ket's position in the $2^n$-vector: $$ \operatorname{index}(|q_0\cdots q_{n-1}\rangle) = \sum_{k=0}^{n-1} q_k\, 2^{\,n-1-k}. $$ For two qubits this gives $|00\rangle\to 0$, $|01\rangle\to 1$, $|10\rangle\to 2$, $|11\rangle\to 3$, so $$ |00\rangle=(1,0,0,0)^T,\ \ |01\rangle=(0,1,0,0)^T,\ \ |10\rangle=(0,0,1,0)^T,\ \ |11\rangle=(0,0,0,1)^T, $$ exactly the Kronecker-product coordinates of [0.1.7](../../term-0-foundations/c01-linear-algebra/07-tensor-products.md). This is also precisely how Braket's `measurement_counts` keys read (qubit 0 first), so the convention costs us nothing later. A general $n$-qubit pure state is $$ |\Psi\rangle = \sum_{x\in\{0,1\}^n} c_x\,|x\rangle, \qquad \sum_x |c_x|^2 = 1, \quad c_x\in\mathbb{C}. $$ ### Local vs global operations An operation acting on each qubit **independently** is a tensor product of single-qubit operators. To apply $A$ to qubit 0 and $B$ to qubit 1 of a two-qubit register, $$ (A\otimes B)\big(|\psi\rangle\otimes|\phi\rangle\big) = (A|\psi\rangle)\otimes(B|\phi\rangle), $$ and "do nothing to qubit 1" is just $A\otimes I$. These are **local** operations. By the mixed-product identity ([Appendix E](../../appendix/E-math-identities.md)), $(A\otimes B)(C\otimes D)=(AC)\otimes(BD)$ — local operations compose factor-by-factor and therefore **can never produce entanglement** (we prove the relevant direction below). A **global** operation is any unitary on $\mathcal{H}_n$ that is *not* of the product form $A\otimes B$; the canonical example is $\mathrm{CNOT}$ (Term 2.1), the entangling two-qubit gate. ### Product (separable) vs entangled states A bipartite pure state $|\Psi\rangle\in\mathcal{H}_A\otimes\mathcal{H}_B$ is a **product (separable)** state if it factors, $$ |\Psi\rangle = |\psi\rangle_A\otimes|\phi\rangle_B , $$ and **entangled** otherwise. (For $n>2$ qubits "separable" usually means *fully* separable, $|\psi_0\rangle\otimes\cdots\otimes|\psi_{n-1}\rangle$; one may also speak of separability across a particular bipartition $A\,|\,B$. We treat the bipartite case as the fundamental one.) **Local operations preserve separability.** If $|\Psi\rangle=|\psi\rangle\otimes|\phi\rangle$ then $(A\otimes B)|\Psi\rangle=(A|\psi\rangle)\otimes(B|\phi\rangle)$ is again a product. So no amount of single-qubit gates, applied in parallel, can entangle initially unentangled qubits — you need a global gate. ∎ ### The separability / Schmidt-rank test We need a *decidable* criterion, not "stare at it and try to factor." Arrange the amplitudes of a bipartite state into a matrix. Writing the basis of $\mathcal{H}_A$ as $\{|i\rangle\}$ ($d_A$ of them) and of $\mathcal{H}_B$ as $\{|j\rangle\}$ ($d_B$ of them), $$ |\Psi\rangle = \sum_{i=0}^{d_A-1}\sum_{j=0}^{d_B-1} c_{ij}\,|i\rangle_A\otimes|j\rangle_B, \qquad C := (c_{ij}) \in \mathbb{C}^{d_A\times d_B}. $$ $C$ is the **coefficient matrix**. The criterion is: $$ \boxed{\ |\Psi\rangle \text{ is a product state} \iff \operatorname{rank}(C) = 1.\ } $$ **Proof.** ($\Rightarrow$) If $|\Psi\rangle=|\psi\rangle\otimes|\phi\rangle$ with $|\psi\rangle=\sum_i a_i|i\rangle$, $|\phi\rangle=\sum_j b_j|j\rangle$, then $c_{ij}=a_i b_j$, i.e. $C = a\,b^{\top}$ is an outer product of nonzero vectors, which has rank $1$. ($\Leftarrow$) If $\operatorname{rank}(C)=1$, then $C=a\,b^{\top}$ for some nonzero $a\in\mathbb{C}^{d_A}$, $b\in\mathbb{C}^{d_B}$ (every rank-$1$ matrix is an outer product). Then $c_{ij}=a_i b_j$ and $$ |\Psi\rangle=\sum_{ij}a_i b_j|i\rangle|j\rangle = \Big(\sum_i a_i|i\rangle\Big)\otimes\Big(\sum_j b_j|j\rangle\Big), $$ a product state. ∎ The integer $\operatorname{rank}(C)$ — equivalently the number of nonzero **singular values** of $C$ — is the **Schmidt rank**. Rank $1$ means separable; rank $\ge 2$ means entangled, and a *larger* rank (with balanced singular values) means *more* entangled. Computing rank via the singular value decomposition (SVD) is robust to floating-point noise, which is why the code below uses a tolerance. The full theory — the Schmidt decomposition itself and entanglement entropy — is [Lesson 1.4.4](04-schmidt-decomposition.md). > **Why rank, not "can I factor it"?** Factoring by hand is error-prone and does not scale. The rank > of $C$ is a single number computable in $O((d_Ad_B)\min(d_A,d_B))$ time, gives a yes/no answer, and > generalizes immediately to a *quantitative* measure of entanglement. It is the workhorse of this > whole course. --- ## Worked Examples ### Example 1 — Two ways to read $|10\rangle$, and a local flip Take the two-qubit basis ket $|10\rangle$. **Big-endian:** qubit 0 is $1$, qubit 1 is $0$, binary $10_2=2$, so it sits at index $2$: $|10\rangle=(0,0,1,0)^T$. As a tensor product, $|10\rangle=|1\rangle\otimes|0\rangle=\binom{0}{1}\otimes\binom{1}{0}=(0,0,1,0)^T$ — consistent. Now apply the **local** operation "flip qubit 1, leave qubit 0," i.e. $I\otimes X$: $$ (I\otimes X)\,|1\rangle\otimes|0\rangle = (I|1\rangle)\otimes(X|0\rangle) = |1\rangle\otimes|1\rangle = |11\rangle . $$ A product went to a product, as local operations must. ### Example 2 — Is $\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)$ entangled? Its coefficient matrix (rows indexed by qubit 0, columns by qubit 1) is $$ C = \frac{1}{\sqrt2}\begin{pmatrix} c_{00} & c_{01}\\ c_{10} & c_{11}\end{pmatrix} = \frac{1}{\sqrt2}\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}. $$ $\det C = \tfrac12\neq 0$, so $\operatorname{rank}(C)=2$. By the test, the state is **entangled** — this is the Bell state $|\Phi^+\rangle$ of [Lesson 1.4.2](02-bell-states.md). Contrast with $\tfrac12(|00\rangle+|01\rangle+|10\rangle+|11\rangle)$, whose coefficient matrix $C=\tfrac12\binom{1\ \,1}{1\ \,1}$ has $\det C = 0$ and rank $1$ — **separable**, in fact $|+\rangle\otimes|+\rangle$. ### Example 3 — A three-qubit state across a bipartition Consider $|\Psi\rangle=\tfrac{1}{\sqrt2}(|000\rangle+|111\rangle)$ (the GHZ state, Term 2.3). Group it as qubit $0$ ($A$) versus qubits $1,2$ ($B$). The coefficient matrix has shape $2\times 4$ with the single nonzero entries $c_{0,\,00}=c_{1,\,11}=\tfrac1{\sqrt2}$: $$ C=\frac1{\sqrt2}\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\end{pmatrix},\qquad \operatorname{rank}(C)=2. $$ So $|\Psi\rangle$ is entangled across the cut $A\,|\,B$. The lesson: separability is **relative to a bipartition**, and the coefficient matrix is reshaped accordingly. We reuse this reshape trick constantly. --- ## Hands-on (Python) ```python import numpy as np from functools import reduce ket0 = np.array([1, 0], dtype=complex) ket1 = np.array([0, 1], dtype=complex) def tensor(*ops): """Kronecker product of any number of kets/operators. Big-endian: the FIRST argument is qubit 0 (most significant).""" return reduce(np.kron, ops) def basis_ket(bits): """Build a computational-basis ket from a bit-string like '101'.""" return tensor(*[ket0 if b == '0' else ket1 for b in bits]) # Big-endian indexing: |10> sits at integer index 2 = binary 10. print(np.real(basis_ket('10'))) # [0. 0. 1. 0.] print(int('10', 2)) # 2 -> the position of the lone 1 ``` ```python # Local (product) operations are tensor products; e.g. I ⊗ X flips qubit 1 only. I = np.eye(2, dtype=complex) X = np.array([[0, 1], [1, 0]], dtype=complex) state = basis_ket('10') # |10> flipped = tensor(I, X) @ state # apply X to qubit 1 print(np.real(flipped)) # [0. 0. 0. 1.] = |11> ``` ```python def coefficient_matrix(state, dA, dB): """Reshape a bipartite state vector into its coefficient matrix C (dA x dB).""" return state.reshape(dA, dB) def schmidt_rank(state, dA, dB, tol=1e-9): """Number of nonzero singular values of C = the Schmidt rank. rank 1 -> product (separable); rank >= 2 -> entangled (across the A|B cut).""" C = coefficient_matrix(state, dA, dB) s = np.linalg.svd(C, compute_uv=False) # singular values, descending return int(np.sum(s > tol)) # Product state |+>|0>: expect rank 1. plus = (ket0 + ket1) / np.sqrt(2) print(schmidt_rank(tensor(plus, ket0), 2, 2)) # 1 -> separable # Bell state (|00>+|11>)/√2: expect rank 2. bell = (basis_ket('00') + basis_ket('11')) / np.sqrt(2) print(schmidt_rank(bell, 2, 2)) # 2 -> entangled # Equal superposition |+>|+> over 2 qubits: separable despite "looking" mixed. equal = 0.5 * (basis_ket('00') + basis_ket('01') + basis_ket('10') + basis_ket('11')) print(schmidt_rank(equal, 2, 2)) # 1 -> separable ``` ```python # Separability is RELATIVE to a bipartition. GHZ = (|000>+|111>)/√2, # cut qubit 0 (dim 2) vs qubits 1,2 (dim 4): ghz = (basis_ket('000') + basis_ket('111')) / np.sqrt(2) print(schmidt_rank(ghz, 2, 4)) # 2 -> entangled across 0|12 ``` > **Floating-point note.** Never test singular values for exact zero. Use a tolerance (here `1e-9`); > simulator amplitudes and any non-trivial preparation accumulate rounding error. This mirrors how the > real Schmidt-decomposition code in [Lesson 1.4.4](04-schmidt-decomposition.md) treats tiny singular > values as numerical zeros. --- ## Exercises **E1 (easy).** Write the four-qubit basis ket $|1011\rangle$ as a $16$-vector index, and as a tensor product of single-qubit kets.

Solution

**E2 (easy).** How many *real* parameters does a normalized $n$-qubit pure state have, after removing the global phase? Evaluate for $n=1,2,3$.

Solution

A state has $2^n$ complex amplitudes $= 2\cdot 2^n$ real numbers. Normalization removes $1$; the irrelevant global phase removes another $1$. So $2^{n+1}-2$ real parameters: $n=1\Rightarrow 2$ (the Bloch sphere), $n=2\Rightarrow 6$, $n=3\Rightarrow 14$.

**E3 (medium).** Show that $|\psi\rangle=\tfrac{1}{\sqrt2}(|00\rangle+|10\rangle)$ is separable by factoring it, then confirm $\operatorname{rank}(C)=1$.

Solution

$\tfrac1{\sqrt2}(|00\rangle+|10\rangle)=\tfrac1{\sqrt2}(|0\rangle+|1\rangle)\otimes|0\rangle =|+\rangle\otimes|0\rangle$. Coefficient matrix $C=\tfrac1{\sqrt2}\binom{1\ \,0}{1\ \,0}$ has identical nonzero structure only in its first column, so its columns are linearly dependent ⇒ rank $1$. Separable. ✓

**E4 (medium).** The state $\tfrac{1}{\sqrt3}(|00\rangle+|01\rangle+|10\rangle)$ — is it entangled? Justify with the rank test.

Solution

$C=\tfrac1{\sqrt3}\begin{psmallmatrix}1&1\\1&0\end{psmallmatrix}$. $\det C=\tfrac13(1\cdot0-1\cdot1) =-\tfrac13\neq0$, so $\operatorname{rank}(C)=2$ ⇒ **entangled**. (It cannot be written as a single product; there is no way to get $|00\rangle,|01\rangle,|10\rangle$ but not $|11\rangle$ from a product.)

**E5 (hard).** Prove that a single-qubit-on-each-side local unitary $U_A\otimes U_B$ leaves the **Schmidt rank** of any bipartite pure state unchanged. (Hence local operations cannot change *whether* a state is entangled.)

Solution

Let $|\Psi\rangle$ have coefficient matrix $C$. Acting with $U_A\otimes U_B$ sends $c_{ij}\mapsto \sum_{kl}(U_A)_{ik}(U_B)_{jl}c_{kl}$, i.e. $C\mapsto U_A\,C\,U_B^{\top}$. Multiplying by the invertible matrices $U_A$ (left) and $U_B^{\top}$ (right) is multiplication by full-rank matrices, which preserves matrix rank: $\operatorname{rank}(U_A C U_B^{\top})=\operatorname{rank}(C)$. So the Schmidt rank — hence separable-vs-entangled — is invariant under local unitaries. ∎

**E6 (hard).** Generalize the rank test to detect whether an $n$-qubit state is entangled across the bipartition "qubit $0$ vs the rest." Describe the reshape and the rank you'd compute, and what rank range is possible.

Solution

Reshape the $2^n$-vector into a $2\times 2^{n-1}$ matrix $C$ (rows = qubit 0's basis, columns = the other $n-1$ qubits' joint basis). Compute $\operatorname{rank}(C)$ (equivalently the number of nonzero singular values). Since $C$ has $2$ rows, the rank is $1$ or $2$: rank $1$ ⇒ qubit 0 is unentangled from the rest; rank $2$ ⇒ entangled across that cut. (For a balanced cut into $k$ and $n-k$ qubits, the matrix is $2^k\times 2^{n-k}$ and the rank can be up to $\min(2^k,2^{n-k})$.)

--- ## Checkpoint 1. Why is the two-qubit state space $\mathbb{C}^4$ and not "two copies of $\mathbb{C}^2$"? What is $\dim$ for $n$ qubits? 2. Give the big-endian index of $|110\rangle$. 3. What is the difference between a local and a global operation? Why can't a local operation create entanglement? 4. State the coefficient-matrix separability test and the meaning of Schmidt rank. 5. Is separability an absolute property of a multi-qubit state, or relative to something?

Answers

1. The composite postulate uses the *tensor product*, whose dimension multiplies: $2\cdot2=4$ for two qubits and $2^n$ for $n$. The joint system can be in any superposition of the $4$ (resp. $2^n$) joint configurations. 2. $110_2 = 6$, so $|110\rangle$ is at index $6$. 3. Local $=A\otimes B$ (acts factor-wise); global $=$ any unitary not of product form (e.g. CNOT). Local ops send products to products: $(A\otimes B)(|\psi\rangle\otimes|\phi\rangle) =(A|\psi\rangle)\otimes(B|\phi\rangle)$, so they can't entangle initially unentangled qubits. 4. Form $C=(c_{ij})$ from $|\Psi\rangle=\sum c_{ij}|i\rangle|j\rangle$; the state is a product iff $\operatorname{rank}(C)=1$. The Schmidt rank is $\operatorname{rank}(C)=$ number of nonzero singular values; $1$ = separable, $\ge 2$ = entangled. 5. Relative to a **bipartition** $A\,|\,B$ (and for $n>2$, "fully separable" means it factors across every qubit). The coefficient matrix is reshaped to match the chosen cut.

--- ## Further Reading - **[NC]** Nielsen & Chuang, §2.2.8 (composite systems) and §2.5 (the Schmidt decomposition). - **[Pre]** Preskill, *Ph219*, Ch. 2 — bipartite systems and the meaning of separability. - **[Wat]** Watrous, *Theory of Quantum Information*, Ch. 2 — rigorous tensor-product formalism. - [0.1.7 Tensor Products](../../term-0-foundations/c01-linear-algebra/07-tensor-products.md) — the construction this lesson builds on, including the Kronecker product in coordinates. --- ← Prev: [POVMs & Generalized Measurement](../c03-measurement/03-povm-generalized.md) · Up: [Term 1](../README.md) · Next: [Bell States](02-bell-states.md) →