Entanglement Projection

Classical Mechanics Level 2

Which of the following correctly rewrites the state ∣ ↑ ⟩ ⊗ ∣ ↓ ⟩ |\uparrow\rangle \otimes |\downarrow\rangle in the Bell basis?

1 2 ( ∣ Φ 1 ⟩ + ∣ Φ 2 ⟩ ) \frac{1}{\sqrt{2}} (\vert\Phi_1\rangle + \vert\Phi_2\rangle) 1 2 ( ∣ Φ 1 ⟩ − ∣ Φ 2 ⟩ ) \frac{1}{\sqrt{2}} (\vert\Phi_1\rangle - \vert\Phi_2\rangle) 1 2 ( ∣ Φ 1 ⟩ − i ∣ Φ 2 ⟩ ) \frac{1}{\sqrt{2}} (\vert\Phi_1\rangle - i\vert\Phi_2\rangle) 1 2 ( ∣ Φ 1 ⟩ + i ∣ Φ 2 ⟩ ) \frac{1}{\sqrt{2}} (\vert\Phi_1\rangle + i\vert\Phi_2\rangle)

Matt DeCross by Matt DeCross , 27, USA

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1 solution

Matt DeCross
May 10, 2016

Relevant wiki: Quantum Teleportation

The two relevant Bell states are:

∣ Φ 1 ⟩ = I ⊗ σ 1 ∣ Φ 0 ⟩ = 1 2 ( ∣ ↑ ⟩ ⊗ ∣ ↓ ⟩ + ∣ ↓ ⟩ ⊗ ∣ ↑ ⟩ ) ∣ Φ 2 ⟩ = I ⊗ σ 2 ∣ Φ 0 ⟩ = i 2 ( ∣ ↑ ⟩ ⊗ ∣ ↓ ⟩ − ∣ ↓ ⟩ ⊗ ∣ ↑ ⟩ ) \begin{aligned} |\Phi_1\rangle &= I \otimes \sigma_1 |\Phi_0\rangle =\frac{1}{\sqrt{2}} (|\uparrow\rangle \otimes |\downarrow\rangle + |\downarrow\rangle \otimes |\uparrow\rangle) \\ |\Phi_2\rangle &=I \otimes \sigma_2 |\Phi_0\rangle = \frac{i}{\sqrt{2}} (|\uparrow\rangle \otimes |\downarrow\rangle - |\downarrow\rangle \otimes |\uparrow\rangle) \end{aligned}

Verifying the answer:

1 2 ( ∣ Φ 1 ⟩ − i ∣ Φ 2 ⟩ ) = 1 2 ( ∣ ↑ ⟩ ⊗ ∣ ↓ ⟩ + ∣ ↓ ⟩ ⊗ ∣ ↑ ⟩ ) + 1 2 ( ∣ ↑ ⟩ ⊗ ∣ ↓ ⟩ − ∣ ↓ ⟩ ⊗ ∣ ↑ ⟩ ) = ∣ ↑ ⟩ ⊗ ∣ ↓ ⟩ \begin{aligned} \frac{1}{\sqrt{2}} (|\Phi_1\rangle - i |\Phi_2\rangle) &= \frac{1}{2} (|\uparrow\rangle \otimes |\downarrow\rangle + |\downarrow\rangle \otimes |\uparrow\rangle) + \frac{1}{2} (|\uparrow\rangle \otimes |\downarrow\rangle - |\downarrow\rangle \otimes |\uparrow\rangle) \\ &= |\uparrow\rangle \otimes |\downarrow\rangle \end{aligned}

2 Helpful 2 Interesting 2 Brilliant 4 Confused

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