|q⟩ Bad Qubits

beginner · Programming · Two Qubits & Tensor Products

Product States

When you add a second qubit to a system, the combined state lives in a space that is the tensor product of the two single-qubit spaces. The most structured two-qubit states are product states — those where the joint state can be written as a tensor product of independent single-qubit states.

What makes a state a product state

A two-qubit state ∣ψ⟩|\psi\rangle is a product state if there exist single-qubit states ∣ϕA⟩|\phi_A\rangle and ∣ϕB⟩|\phi_B\rangle such that

∣ψ⟩=∣ϕA⟩⊗∣ϕB⟩.|\psi\rangle = |\phi_A\rangle \otimes |\phi_B\rangle.

The symbol ⊗\otimes is the tensor product, but when the context is clear you can just write ∣ϕA⟩∣ϕB⟩|\phi_A\rangle|\phi_B\rangle or ∣ϕAϕB⟩|\phi_A \phi_B\rangle.

To see how the amplitudes are computed, expand using the computational basis. If ∣ϕA⟩=α∣0⟩+β∣1⟩|\phi_A\rangle = \alpha|0\rangle + \beta|1\rangle and ∣ϕB⟩=γ∣0⟩+δ∣1⟩|\phi_B\rangle = \gamma|0\rangle + \delta|1\rangle, then

= \alpha\gamma\,|00\rangle + \alpha\delta\,|01\rangle + \beta\gamma\,|10\rangle + \beta\delta\,|11\rangle.$$ Every amplitude in the four-dimensional state is a **product** of an amplitude from qubit $A$ and an amplitude from qubit $B$. That multiplicative structure is the signature of a product state. <Callout type="tip"> A quick test: write out the four amplitudes $a_{00}, a_{01}, a_{10}, a_{11}$ of a two-qubit state and check whether $a_{00}\,a_{11} = a_{01}\,a_{10}$. If so, the state factors into a product; if not, the qubits are **entangled** and no factoring is possible. </Callout> ## A concrete example Take $|\phi_A\rangle = |+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $|\phi_B\rangle = |1\rangle$. The product is $$|+\rangle \otimes |1\rangle = \tfrac{1}{\sqrt{2}}|0\rangle \otimes |1\rangle + \tfrac{1}{\sqrt{2}}|1\rangle \otimes |1\rangle = \tfrac{1}{\sqrt{2}}\,|01\rangle + \tfrac{1}{\sqrt{2}}\,|11\rangle.$$ The amplitudes are $a_{00} = 0$, $a_{01} = \tfrac{1}{\sqrt{2}}$, $a_{10} = 0$, $a_{11} = \tfrac{1}{\sqrt{2}}$. Checking: $a_{00}\,a_{11} = 0 = a_{01}\,a_{10}$, confirming this is a product state. Crucially, preparing $|+\rangle \otimes |1\rangle$ requires **only single-qubit gates** — one Hadamard on qubit 0 and one $X$ on qubit 1. The qubits never interact. That independence is exactly what "product" means: the subsystems carry no correlation. Contrast this with a Bell state such as $\tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$, where $a_{00} = \tfrac{1}{\sqrt{2}}$, $a_{11} = \tfrac{1}{\sqrt{2}}$, $a_{01} = 0$, $a_{10} = 0$, so $a_{00}\,a_{11} = \tfrac{1}{2}$ but $a_{01}\,a_{10} = 0$. Those are not equal, so the Bell state is **not** a product state — it is entangled, and no combination of single-qubit operations can produce it from $|00\rangle$. ## Try it Prepare the 2-qubit product state $|+\rangle|1\rangle$ — qubit 0 in $|+\rangle$ and qubit 1 in $|1\rangle$. Because this is a product state, a single gate on each qubit is enough. <RunnableExample /> After running, the state vector should show $|01\rangle$ and $|11\rangle$ each with amplitude $\tfrac{1}{\sqrt{2}} \approx 0.707$, confirming the factored structure.

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