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Tensor products

Combining qubits: |a⟩⊗|b⟩ as a 4-entry vector

about 5 min

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Two qubits, four amplitudes

The state of two qubits lives in a 4-dimensional space with basis ∣00⟩,∣01⟩,∣10⟩,∣11⟩\ket{00}, \ket{01}, \ket{10}, \ket{11}, in that order. The first label is the top qubit: ∣01⟩\ket{01} means the top qubit is in ∣0⟩\ket0 and the bottom one in ∣1⟩\ket1 (kets are column vectors: refresher). If the top qubit is in a0∣0⟩+a1∣1⟩{a_0\ket0 + a_1\ket1} and the bottom one in b0∣0⟩+b1∣1⟩{b_0\ket0 + b_1\ket1}, the pair is in the tensor product:

[a0a1]⊗[b0b1]=[a0b0a0b1a1b0a1b1]\begin{bmatrix}a_0\\a_1\end{bmatrix}\otimes\begin{bmatrix}b_0\\b_1\end{bmatrix} = \begin{bmatrix}a_0 b_0\\ a_0 b_1\\ a_1 b_0\\ a_1 b_1\end{bmatrix}

Each entry is “top amplitude times bottom amplitude” for that pair of labels. It’s just the distributive law: (a0∣0⟩+a1∣1⟩)(b0∣0⟩+b1∣1⟩)(a_0\ket0 + a_1\ket1)(b_0\ket0 + b_1\ket1) expanded, writing ∣x⟩∣y⟩\ket{x}\ket{y} as ∣xy⟩\ket{xy}. The questions below use ∣±⟩=12(∣0⟩±∣1⟩)\ket\pm = \tfrac{1}{\sqrt2}(\ket0 \pm \ket1).

Check yourself

Write ∣+⟩⊗∣0⟩\ket{+}\otimes\ket{0} (top qubit ∣+⟩\ket+, bottom ∣0⟩\ket0).

Hint
  1. Only basis states whose bottom label is 0 survive.
Show the answer

Answer: 12∣00⟩+12∣10⟩\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|10\rangle

12(∣0⟩+∣1⟩)∣0⟩=12(∣00⟩+∣10⟩)\tfrac{1}{\sqrt2}(\ket0+\ket1)\ket0 = \tfrac{1}{\sqrt2}(\ket{00} + \ket{10}).

Check yourself

Write ∣+⟩⊗∣−⟩\ket{+}\otimes\ket{-}.

Hint
  1. Multiply each top amplitude by each bottom amplitude: a0b0,a0b1,a1b0,a1b1a_0 b_0, a_0 b_1, a_1 b_0, a_1 b_1.
Show the answer

Answer: 12∣00⟩−12∣01⟩+12∣10⟩−12∣11⟩\frac{1}{2}|00\rangle - \frac{1}{2}|01\rangle + \frac{1}{2}|10\rangle - \frac{1}{2}|11\rangle

12(∣00⟩−∣01⟩+∣10⟩−∣11⟩)\tfrac12(\ket{00} - \ket{01} + \ket{10} - \ket{11}).

The factor test. A product state has α00α11=(a0b0)(a1b1)=(a0b1)(a1b0)=α01α10\alpha_{00}\alpha_{11} = (a_0b_0)(a_1b_1) = (a_0b_1)(a_1b_0) = \alpha_{01}\alpha_{10}. So if α00α11−α01α10≠0\alpha_{00}\alpha_{11} - \alpha_{01}\alpha_{10} \neq 0, the state cannot be a product: it is entangled. (The converse also holds: when the difference is 0, the state factors.)

Check yourself

Is 12(∣00⟩+∣01⟩+∣10⟩−∣11⟩)\tfrac12(\ket{00} + \ket{01} + \ket{10} - \ket{11}) a product state?

  1. ANo: it is a superposition of all four basis states, so it must be entangled.
  2. BNo: α00α11−α01α10=−12\alpha_{00}\alpha_{11} - \alpha_{01}\alpha_{10} = -\tfrac12, so it is entangled.
  3. CYes: every amplitude has size 12\tfrac12, like ∣+⟩∣+⟩\ket+\ket+.
Hints
  1. Read off α00,α01,α10,α11\alpha_{00}, \alpha_{01}, \alpha_{10}, \alpha_{11} in that order.
  2. Compute α00α11−α01α10\alpha_{00}\alpha_{11} - \alpha_{01}\alpha_{10}.
Show the answer

Answer: B. No: α00α11−α01α10=−12\alpha_{00}\alpha_{11} - \alpha_{01}\alpha_{10} = -\tfrac12, so it is entangled.

α00α11−α01α10=(12)(−12)−(12)(12)=−12≠0\alpha_{00}\alpha_{11} - \alpha_{01}\alpha_{10} = (\tfrac12)(-\tfrac12) - (\tfrac12)(\tfrac12) = -\tfrac12 \neq 0, so the state is entangled.

Why not the others

  • A. Right verdict, wrong reason. ∣+⟩∣+⟩\ket+\ket+ is also a superposition of all four basis states, yet it is a product. The test is α00α11−α01α10\alpha_{00}\alpha_{11} - \alpha_{01}\alpha_{10}, which here is −12-\tfrac12.
  • C. Equal sizes are not enough; the signs matter. ∣+⟩∣+⟩\ket+\ket+ has all four amplitudes +12+\tfrac12. Here α00α11=−14\alpha_{00}\alpha_{11} = -\tfrac14 but α01α10=+14\alpha_{01}\alpha_{10} = +\tfrac14, so no choice of factors works.

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