Suppose a two-qubit system is in the state |Φ⟩ = (|00⟩ + |11⟩ )/ √2. Consider projective measurements with outcome ±1. Calculate the maximal violation of the CHSH inequality and find the measurement setting that can result the maximal violation.
I am new to quantum computing.I don't know how to solve this problem.
I's a tough one and one I struggled with too. Calculating the maximal violation of the CHSH inequality for the given two-qubit state |Φ⟩ = (|00⟩ + |11⟩ )/ √2, we need to consider four different measurement settings and compute the corresponding expectation values.
The CHSH inequality is expressed as follows:
CHSH = E(A, B) + E(A, B') + E(A', B) - E(A', B') ≤ 2,
where A, A' are the measurement settings for the first qubit there, and B, B' are the measurement settings for the second qubit, and E(A, B) is the expectation value for the measurement settings A and B. If that makes sense?
Now considering the four measurement settings:
Setting 1: A = σz (measurement in the computational basis for the first qubit) B = σz (measurement in the computational basis for the second qubit)
In this case, the expectation value E(A, B) can be calculated as follows:
E(A, B) = ⟨Φ|A ⊗ B|Φ⟩ = ⟨Φ|(σz ⊗ σz)|Φ⟩.
Expanding the tensor product and substituting |Φ⟩ = (|00⟩ + |11⟩ )/ √2:
E(A, B) = ⟨Φ|(σz ⊗ σz)|Φ⟩ = (1/√2)(⟨00|(σz ⊗ σz)|00⟩ + ⟨00|(σz ⊗ σz)|11⟩ + ⟨11|(σz ⊗ σz)|00⟩ + ⟨11|(σz ⊗ σz)|11⟩) = (1/√2)(1 + 1 + 1 - 1) = 2/√2 = √2.
Setting 2: A = σx (measurement in the Hadamard basis for the first qubit) B = σz (measurement in the computational basis for the second qubit)
In this case, the expectation value E(A, B) can be calculated as follows:
E(A, B) = ⟨Φ|A ⊗ B|Φ⟩ = ⟨Φ|(σx ⊗ σz)|Φ⟩.
Using the same steps as above, we find:
E(A, B) = -1/√2.
Setting 3: A = σz (measurement in the computational basis for the first qubit) B = σx (measurement in the Hadamard basis for the second qubit)
In this case, the expectation value E(A, B) can be calculated as follows:
E(A, B) = ⟨Φ|A ⊗ B|Φ⟩ = ⟨Φ|(σz ⊗ σx)|Φ⟩.
Using the same steps as above, we find:
E(A, B) = -1/√2.
Setting 4: A = σx (measurement in the Hadamard basis for the first qubit) B = σx (measurement in the Hadamard basis for the second qubit)
In this case, the expectation value E(A, B) can be calculated as follows:
E(A, B) = ⟨Φ|A ⊗ B|Φ⟩ = ⟨Φ|(σx ⊗ σx)|Φ⟩.
Using the same steps as above, we find:
E(A, B) = -√2.
Now, let's calculate the maximal violation of the CHSH inequality:
CHSH = E(A, B) + E(A, B') + E(A', B) - E(A', B') = √2 + (-1/
Hope that helps you.