A two-qubit system was originally in the state 34|00⟩−5√4|01⟩+14|10⟩−14|11⟩, and then we measured the first qubit to be 0. Now, if we measure the second qubit in the standard basis, what is the probability that the outcome is 0?
To find the probability that the outcome of measuring the second qubit is 0, we need to first calculate the conditional probability given that the first qubit was measured to be 0.
The given state of the two-qubit system is |Ψ⟩ = 34|00⟩ - 5√4|01⟩ + 14|10⟩ - 14|11⟩.
We want to calculate the conditional probability P(II=0|I=0) which is the probability of observing the second qubit in the state 0 given that the first qubit was measured to be 0.
To calculate this probability, we need to normalize the state after measuring the first qubit to be 0.
Step 1: Normalize the state after the first qubit measurement:
Since the first qubit was measured to be 0, we need to find the normalized projection of the state |Ψ⟩ onto the subspace spanned by the states |00⟩ and |10⟩.
First, let's find the projection coefficient α_00:
α_00 = ⟨00|Ψ⟩ = ⟨00|(34|00⟩ - 5√4|01⟩ + 14|10⟩ - 14|11⟩)
= 34⟨00|00⟩ - 5√4⟨00|01⟩ + 14⟨00|10⟩ - 14⟨00|11⟩
= 34 * 1 - 5√4 * 0 + 14 * 0 - 14 * 0 (since ⟨ij|kl⟩ = 1 for i=k, j=l and 0 otherwise)
= 34
Now, let's find the projection coefficient α_10:
α_10 = ⟨10|Ψ⟩ = ⟨10|(34|00⟩ - 5√4|01⟩ + 14|10⟩ - 14|11⟩)
= 34⟨10|00⟩ - 5√4⟨10|01⟩ + 14⟨10|10⟩ - 14⟨10|11⟩
= 34 * 0 - 5√4 * 0 + 14 * 1 - 14 * 0 (since ⟨ij|kl⟩ = 1 for i=k, j=l and 0 otherwise)
= 14
Next, we need to normalize the state in the subspace spanned by the states |00⟩ and |10⟩.
|Ψ_normalized⟩ = (α_00|00⟩ + α_10|10⟩) / sqrt(|α_00|^2 + |α_10|^2)
= (34|00⟩ + 14|10⟩) / sqrt(34^2 + 14^2)
= (34|00⟩ + 14|10⟩) / sqrt(1172)
Step 2: Calculate the probability of the second qubit being 0:
Now that we have the normalized state of the system after measuring the first qubit, we can calculate the probability of measuring the second qubit to be 0. In general, the probability of measuring a quantum state |ψ⟩ in the standard basis state |0⟩ is given by the square of the projection coefficient:
P(II=0|I=0) = |⟨0|Ψ_normalized⟩|^2
= |⟨0| (34|00⟩ + 14|10⟩) / sqrt(1172) |^2
= | (34/ sqrt(1172)) ⟨0|00⟩ + (14/ sqrt(1172)) ⟨0|10⟩ |^2
= | (34/ sqrt(1172)) * 1 + (14/ sqrt(1172)) * 0 |^2 (since ⟨0|ij⟩ = 1 for j=0 and 0 otherwise)
= | (34/ sqrt(1172)) |^2
= (34^2 / 1172)
= 1156 / 1172
≈ 0.9871
Therefore, the probability that the outcome of measuring the second qubit in the standard basis is 0, given that the first qubit was measured to be 0, is approximately 0.9871.