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1. You are given one of two quantum states of a single qubit: either φ = 0 or ψ = cos θ 0 + sin θ 1 . You want to make a single measurement that best distinguishes between these two states, i.e., you want to find the best basis for a measurement to distinguish the two states. So let’s making measure the qubit in basis |vi = α 0 + β 1 , |v⊥ i, where α, β are to be determined for optimal success. For outcome |vi we guess that the qubit was in state |0i; for outcome |v⊥ i we guess that the qubit was in state |ψi. Determine the optimal measurement basis given this procedure. You can take α and β to be real numbers, in which case the normalization |α|2 + |β |2 = 1 implies that you can write α and β as e.g. α = sin γ and β = cos γ. Hint: you will need to first construct the probability of a correct measurement in this situation. You should convince yourselves that this is given by Pr[qubit was|0i] Pr[measure|vi|qubit was|0i] + Pr[qubit was|ψi] Pr[measure|v⊥ i|qubit was|ψi]. where, e.g., Pr[measure|vi|qubit was|0i] = | v 0 |2 . If the state you are presented is either φ or ψ with 50% probability each, what is the probability that your measurement correctly identifies the state? Solution: Answer: We measure the qubit in basis |vi = α 0 + β 1 , |v⊥ i. For outcome |vi we guess that the qubit was in state |0i; for outcome |v⊥ i we guess that the qubit was in state |ψi. The optimal measurement basis is v =cos γ 0 + sin γ 1 , where γ = 21 (θ − π2 ). Graphically, this means that the angle bisector between v and v⊥ is the same as that between 0 and ψ . Let’s prove that this choice of measurement basis v , v⊥ indeed maximizes the probability that our guess is correct: Pr[qubit was|0i] Pr[measure|vi|qubit was|0i] + Pr[qubit was|ψi] Pr[measure|v⊥ i|qubit was|ψi] = 12 | v 0 |2 + | v⊥ ψ |2 = 12 | v 0 |2 + 1 − | v ψ |2 . Let v = α 0 + β 1 . We want to choose α, β to maximize | v 0 |2 − | v ψ |2 = |α|2 − |α cos θ + β sin θ |2 = |α|2 (1 − cos2 θ ) − |β |2 sin2 θ − cos θ sin θ (ᾱβ + α β̄ ) , Using 1 − cos2 θ = sin2 θ and assuming without loss of generality sin θ ≥ 0, this is equivalent to maximizing sin θ (|α|2 − |β |2 ) − cos θ (ᾱβ + α β̄ ). We may assume α, β ∈ ℜ, since for fixed magnitudes |α|, |β |, we want to choose arguments such as to either maximize (if cos θ < 0) or minimize (if cos θ > 0) the real part of ᾱβ . Therefore let α = cos γ, β = sin γ for some angle γ. We wish to maximize sin θ cos 2γ − cos θ sin 2γ = sin(θ − 2γ) . Therefore let γ = 21 (θ − π2 ). Geometrically, the vector v + v⊥ is a multiple of 0 + ψ . The success probability with this choice of γ is cos2 ( π4 − θ2 ) = 12 (1 + sin θ ). Without the assumption sin θ > 0, the success probability is 21 (1 + |sin θ |). C/CS/Phys 191, Fall 2008, 1 Any two quantum states, in an arbitrary dimension, lie in a plane. By rotational symmetry the above q 1 argument applies: the optimal measurement gives a success probability of 2 (1 + 1 − | φ ψ |2 ). φ and ψ are orthogonal, hence perfectly distinguishable. This Note that this probability is one if probability is 1/2 if | φ ψ | = 1. In this case there is no optimal basis, you can do no better than just guessing randomly. Note that allowing multiple measurements does not increase the optimum success probability – only the first measurement gives information about the original state. Also, a unitary transformation, followed by a measurement, is the same a measurement in a rotated basis – so allowing unitary transformations does not increase the optimum success probability either. 2. Prove that the Bell state |ψ − i is rotationally invariant: i.e. |ψ − i = √1 (|vv⊥ i − |v⊥ vi). 2 2 |α| + |β |2 = 1), |v⊥ i = −β̄ |0i + ᾱ|1i, Answer: For |vi = α|0i + β |1i (with normalization condition up to choice of phase. (Indeed, hv⊥ |vi = β α − αβ = 0.) Then |vv⊥ i − |v⊥ vi = (α|0i + β |1i) ⊗ (−β̄ |0i + ᾱ|1i) − (−β̄ |0i + ᾱ|1i) ⊗ (α|0i + β |1i) = (−α β̄ + β̄ α)|00i + (|α|2 + |β |2 )|01i − (|β |2 + |α|2 )|10i + (β ᾱ − ᾱβ )|11i = |01i − |10i , where in the last equality we used the normalization condition. Hence, |Ψ− i is rotationally invariant; for any unitary operator U on C 2 , (U ⊗U)|Ψ− i = |Ψ− i. 3. Show that if U and V are unitary, then so is U ⊗V . If U and V are both unitary, then each can be written in some basis as a diagonal matrix with only complex phases eiφn along the diagonal. Since operators in a tensor product act on different Hilbert spaces, we can diagonalize both of them independently. Let eiφn be the diagonal elements of U, and let eiφm be the diagonal elements of V . Then we use the rules for tensor products of matrices to see that U ⊗V is a diagonal matrix with diagonal elements eiφn eiφm . Therefore it is also a diagonal matrix with only complex phases along the diagonal. So it is unitary. 4. Write the 4 × 4 matrix of the unitary operation on two qubits resulting from performing a Hadamard transform on the first qubit and a phase flip on the second qubit 1 √ 2 1 1 1 −1 ⊗ 1 0 0 −1 1 0 1 0 1 1 1 0 −1 0 −1 =√ 1 0 1 0 2 1 −1 0 −1 0 −1 = 1 0 1 0 1 0 −1 0 −1 =√ 2 1 0 −1 0 0 −1 0 1 C/CS/Phys 191, Fall 2008, 2