Bell State Probability

Classical Mechanics Level 2

In the product of spin states 1 2 ( ∣ ↑ ⟩ ⊗ ∣ ↑ ⟩ + ∣ ↑ ⟩ ⊗ ∣ ↓ ⟩ ) \frac{1}{\sqrt{2}} (|\uparrow\rangle\otimes|\uparrow\rangle + |\uparrow\rangle\otimes|\downarrow\rangle ) , what is the probability that a Bell measurement will find the Bell state ∣ Φ 1 ⟩ |\Phi_1\rangle ?

1 1 1 4 \frac14 1 2 \frac12 0 0

Matt DeCross by Matt DeCross , 27, USA

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1 solution

Matt DeCross
May 10, 2016

To solve this problem, rewrite the given state as a superposition of Bell states. Recall that the Bell states are:

∣ Φ 0 ⟩ = I ⊗ σ 0 ∣ Φ 0 ⟩ = 1 2 ( ∣ ↑ ⟩ ⊗ ∣ ↑ ⟩ + ∣ ↓ ⟩ ⊗ ∣ ↓ ⟩ ) ∣ Φ 1 ⟩ = I ⊗ σ 1 ∣ Φ 0 ⟩ = 1 2 ( ∣ ↑ ⟩ ⊗ ∣ ↓ ⟩ + ∣ ↓ ⟩ ⊗ ∣ ↑ ⟩ ) ∣ Φ 2 ⟩ = I ⊗ σ 2 ∣ Φ 0 ⟩ = i 2 ( ∣ ↑ ⟩ ⊗ ∣ ↓ ⟩ − ∣ ↓ ⟩ ⊗ ∣ ↑ ⟩ ) ∣ Φ 3 ⟩ = I ⊗ σ 3 ∣ Φ 0 ⟩ = 1 2 ( ∣ ↑ ⟩ ⊗ ∣ ↑ ⟩ − ∣ ↓ ⟩ ⊗ ∣ ↓ ⟩ ) \begin{aligned} |\Phi_0\rangle &= I \otimes \sigma_0 |\Phi_0\rangle = \frac{1}{\sqrt{2}} (|\uparrow\rangle \otimes |\uparrow\rangle + |\downarrow\rangle \otimes |\downarrow\rangle) \\ |\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) \\ |\Phi_3\rangle &=I \otimes \sigma_3 |\Phi_0\rangle = \frac{1}{\sqrt{2}} (|\uparrow\rangle \otimes |\uparrow\rangle - |\downarrow\rangle \otimes |\downarrow\rangle) \end{aligned}

The given state can be seen to be equivalent to 1 2 ∣ Φ 0 ⟩ + 1 2 ∣ Φ 1 ⟩ − i 2 ∣ Φ 2 ⟩ + 1 2 ∣ Φ 3 ⟩ \frac12|\Phi_0\rangle + \frac12 |\Phi_1\rangle - \frac{i}{2} |\Phi_2\rangle + \frac12|\Phi_3\rangle , and so the probability of finding it in any one Bell state is 1 4 \frac14 . One could also see the answer immediately because the given state is a product state and therefore not entangled, so it should not have a high likelihood of being in any particular entangled state.

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