Multi-Qubit Systems & Tensor Products
How you combine qubits into registers, and why the space grows exponentially
By the end of this topic you'll be able to
Once we understand how to combine two systems A and B into a composite system AB, Binney and Skinner observe, we are in a position to build systems of arbitrary complexity — AB can be combined with some further system C to make ABC, and so on indefinitely. Two qubits are the simplest non-trivial case of this construction.
If |A;i⟩ and |B;j⟩ are basis states of A and B, the symbolic product |A;i⟩|B;j⟩ denotes the state of AB in which A is in |A;i⟩ and B is in |B;j⟩. A general state of the composite system is then a sum Σᵢⱼ cᵢⱼ|A;i⟩|B;j⟩ over every such pairing, not just a single one.
Binney and Skinner make the scaling vivid with a pair of meshed gear wheels: if wheel A has NA teeth and wheel B has NB teeth, an uncorrelated pair needs only NA + NB amplitudes to describe, one set per wheel. But once the wheels are meshed and their orientations become correlated, describing the pair takes NA × NB numbers, one for every combination of tooth positions.
Qubits obey exactly this arithmetic. Two qubits do not live in a 4-dimensional space by simple addition of two 2-dimensional spaces; they live in the tensor product H_A ⊗ H_B, of dimension 2×2=4, with basis {|00⟩,|01⟩,|10⟩,|11⟩} — every combination of the individual basis states.
A general 2-qubit state is c₀₀|00⟩+c₀₁|01⟩+c₁₀|10⟩+c₁₁|11⟩, subject to Σ|cᵢⱼ|²=1. When the two qubits happen to be independent, this reduces to the special case (α|0⟩+β|1⟩)⊗(γ|0⟩+δ|1⟩) — but, exactly as with the gear wheels, most valid two-qubit states cannot be factored this way; the ones that cannot are entangled, the subject of the next topic.
In a product state, all four outcomes get some probability — set by multiplying A's and B's independently.
The scaling compounds relentlessly with size: three qubits span 8 dimensions, and n qubits span 2ⁿ. Ten classical bits need only 10 numbers to describe fully; ten qubits need 2¹⁰ = 1024 complex amplitudes, and by sixty-four qubits the count has already passed 10¹⁹.
This is precisely the gap that makes quantum computing interesting and classical simulation hard in equal measure. A modest number of qubits can hold a state that would overwhelm any classical computer's memory to write down exactly — but a measurement still returns only n classical bits, never the full amplitude vector, so extracting anything useful from that vastness is the entire difficulty a quantum algorithm has to solve.
Try It Yourself
Compute the explicit 2-qubit state formed by taking the tensor product of |+⟩ = (|0⟩+|1⟩)/√2 on qubit 1 and |0⟩ on qubit 2.
- 1Write out |+⟩⊗|0⟩ and distribute, term by term, exactly like multiplying out (a+b)·c: (1/√2)(|0⟩+|1⟩) ⊗ |0⟩ = (1/√2)(|0⟩⊗|0⟩) + (1/√2)(|1⟩⊗|0⟩).
- 2Using the |ab⟩ shorthand for |a⟩⊗|b⟩, this is (1/√2)|00⟩ + (1/√2)|10⟩.
- 3As a 4-entry vector in the {|00⟩,|01⟩,|10⟩,|11⟩} basis, the coefficients are (1/√2, 0, 1/√2, 0) — nonzero only on the two basis states where qubit 2 is |0⟩, since qubit 2 was prepared definitely in |0⟩.
- 4Check normalization: |1/√2|² + 0 + |1/√2|² + 0 = 1/2 + 1/2 = 1. ✓
|+⟩⊗|0⟩ = (1/√2)|00⟩ + (1/√2)|10⟩ — a product state (qubit 1 in |+⟩, qubit 2 in |0⟩, entirely uncorrelated), not an entangled one.
Reference
| n-qubit space | Tensor product of n single-qubit spaces | |
| 2-qubit basis | All combinations of individual basis states | |
| Product state | A multi-qubit state that factors into independent single-qubit states | |
| Entangled state | A joint state with no independent single-qubit description |
Quick Check
How many complex amplitudes are needed to describe a general 5-qubit state?
What distinguishes a product state from an entangled state?