Quantum Computing 3.1 -- Entangled or not?

Computer Science Level 4

Which of the possible answers is an entangled state of a qubit pair?

Details: A qubit pair state ∣ ψ ⟩ |\psi\rangle is separable if there is a representation as a tensor product ∣ ψ 1 ⟩ ⊗ ∣ ψ 2 ⟩ |\psi_1 \rangle \otimes |\psi_2\rangle , so that ∣ ψ ⟩ = ∣ ψ 1 ⟩ ⊗ ∣ ψ 2 ⟩ = ( α 1 ∣ 0 ⟩ + β 1 ∣ 1 ⟩ ) ⊗ ( α 2 ∣ 0 ⟩ + β 2 ∣ 1 ⟩ ) = α 1 α 2 ∣ 00 ⟩ + α 1 β 2 ∣ 01 ⟩ + β 1 α 2 ∣ 10 ⟩ + β 1 β 2 ∣ 11 ⟩ α 1 , α 2 , β 1 , β 2 ∈ C \begin{aligned} |\psi\rangle &= |\psi_1 \rangle \otimes |\psi_2\rangle\\ &= (\alpha_1 |0\rangle + \beta_1 |1\rangle) \otimes (\alpha_2 |0\rangle + \beta_2|1\rangle) \\ &= \alpha_1 \alpha_2 |00\rangle + \alpha_1 \beta_2 |01\rangle + \beta_1 \alpha_2 |10\rangle + \beta_1 \beta_2 |11\rangle\\ & \alpha_1, \alpha_2, \beta_1,\beta_2 \in \mathbb{C} \end{aligned} ∣ ψ ⟩ |\psi\rangle is entangled , if such a representation is not possible. ( ∣ ψ ⟩ ≠ ∣ ψ 1 ⟩ ⊗ ∣ ψ 2 ⟩ ) (|\psi\rangle \not= |\psi_1 \rangle \otimes |\psi_2\rangle) .

1 2 ( ∣ ⁣ 00 ⟩ + i ⁣ ∣ ⁣ 01 ⟩ + i ⁣ ∣ ⁣ 10 ⟩ − ⁣ ∣ ⁣ 11 ⟩ ) \frac{1}{2} (\mid\! 00\rangle+i\!\mid\! 01\rangle+i\!\mid\! 10\rangle-\!\mid\! 11\rangle) 1 2 ( ∣ ⁣ 00 ⟩ − ⁣ ∣ ⁣ 10 ⟩ ) \frac{1}{\sqrt{2}} (\mid \! 00 \rangle - \!\mid \! 10\rangle) 1 2 ( ∣ ⁣ 00 ⟩ + ⁣ ∣ ⁣ 01 ⟩ + ⁣ ∣ ⁣ 10 ⟩ + ⁣ ∣ ⁣ 11 ⟩ ) \frac{1}{2} (\mid \! 00 \rangle+\!\mid\! 01 \rangle+\!\mid \! 10 \rangle +\!\mid\! 11\rangle) 1 2 ( ∣ ⁣ 01 ⟩ + ⁣ ∣ ⁣ 10 ⟩ ) \frac{1}{\sqrt{2}} (\mid \! 01 \rangle +\!\mid \! 10\rangle)

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

Markus Michelmann
Jul 11, 2018

For three of the four possible answers, a decomposition in single states can be found: 1 2 ( ∣ 00 ⟩ − ∣ 10 ⟩ ) = 1 2 ( ∣ 0 ⟩ − ∣ 1 ⟩ ) ⊗ ∣ 0 ⟩ 1 2 ( ∣ 00 ⟩ + ∣ 01 ⟩ + ∣ 10 ⟩ + ∣ 11 ⟩ ) = 1 2 ( ∣ 0 ⟩ + ∣ 1 ⟩ ) ⊗ 1 2 ( ∣ 0 ⟩ + ∣ 1 ⟩ ) 1 2 ( ∣ 00 ⟩ + i ∣ 01 ⟩ + i ∣ 10 ⟩ − ∣ 11 ⟩ ) = 1 2 ( ∣ 0 ⟩ + i ∣ 1 ⟩ ) ⊗ 1 2 ( ∣ 0 ⟩ + i ∣ 1 ⟩ ) \begin{aligned} \frac{1}{\sqrt{2}} (|00\rangle - |10\rangle) &= \frac{1}{\sqrt{2}} (|0\rangle - |1\rangle) \otimes |0\rangle \\ \frac{1}{2} (|00\rangle + |01\rangle + |10\rangle + |11\rangle) &= \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \otimes \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \\ \frac{1}{2} (|00\rangle +i |01\rangle + i|10\rangle - |11\rangle) &= \frac{1}{\sqrt{2}}(|0\rangle +i |1\rangle) \otimes \frac{1}{\sqrt{2}}(|0\rangle +i |1\rangle) \end{aligned} We are now trying to represent the fourth state as a tensor product 1 2 ( ∣ 01 ⟩ + ∣ 10 ⟩ ) = ! α 1 α 2 ∣ 00 ⟩ + α 1 β 2 ∣ 01 ⟩ + β 1 α 2 ∣ 10 ⟩ + β 1 β 2 ∣ 11 ⟩ ⇒ α 1 α 2 = 0 and β 1 β 2 = 0 ⇒ α 1 or α 2 = 0 and β 1 or β 2 = 0 ⇒ α 1 β 2 = 0 or β 1 α 2 = 0 ⇒ ∣ ψ ⟩ = ∣ 01 ⟩ or ∣ ψ ⟩ = ∣ 10 ⟩ \begin{aligned} & & \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle) &\stackrel{!}{=} \alpha_1 \alpha_2 |00\rangle + \alpha_1 \beta_2 |01\rangle + \beta_1 \alpha_2 |10\rangle + \beta_1 \beta_2 |11\rangle \\ \Rightarrow & & \alpha_1 \alpha_2 = 0 \quad &\text{and} \quad \beta_1 \beta_2 = 0 \\ \Rightarrow & & \alpha_1 \text{ or } \alpha_2 = 0 \quad &\text{and} \quad \beta_1 \text{ or } \beta_2 = 0\\ \Rightarrow & & \alpha_1 \beta_2 = 0 \quad &\text{or} \quad \beta_1 \alpha_2 = 0 \\ \Rightarrow & & |\psi\rangle = |01\rangle \quad &\text{or} \quad |\psi\rangle = |10\rangle \end{aligned} Thus we receive a contradiction. Therefore, this state must be entangled. In particular, this state has the special name ∣ Ψ + ⟩ |\Psi +\rangle and belongs to the four Bell states ∣ Φ ± ⟩ = 1 2 ( ∣ 00 ⟩ ± ∣ 11 ⟩ ) ∣ Ψ ± ⟩ = 1 2 ( ∣ 01 ⟩ ± ∣ 10 ⟩ ) \begin{aligned} |\Phi_\pm\rangle &= \frac{1}{\sqrt{2}}(|00\rangle \pm |11\rangle) \\ |\Psi_\pm\rangle &= \frac{1}{\sqrt{2}}(|01\rangle \pm |10\rangle) \end{aligned} which are all maximally entangled.

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