PHYS 15C: Quantum Mechanics Spring 016 Problem Set 1 Solutions TA: Chaitanya Murthy April 7, 016 Problem 1. Correlations for two spin-1/ s The singlet state of the two spin-1/ particles can be written as ψ = 1 + aa ; ab aa ; + ab where + aa ; ab denotes the state where Alice s particle has spin up along â, and Bob s particle has spin down along â, and so on. a The probability that Alice and Bob will measure the spins of their respective particles up along â and ˆb is P ++ = + aa ; + bb ψ = 1 + bb ab 1 Take â in the z-direction, and ˆb in the y, z-plane. Let θ be the angle between the two, so that â ˆb = cos θ. Then + b = exp i θσ x +a = [ cosθ/1 sinθ/σ x +a = cosθ/ + a sinθ/ a, up to an overall phase. Therefore, + b a = sin θ/ = 1 1 cos θ. Using this in eq. 1, P ++ = 1 1 â ˆb b Alice measures her particle to have spin down along the â-axis; this projects the spin singlet to ψ aa ; + ab Thus, the conditional probability that Bob measures his particle to have spin up along ˆb, given that Alice measured spin down along â, is, using eq., + bb + ab = cos θ/ = 1 1 + â ˆb c There are four possible measurement outcomes, ++,, +, and +. The first two are assigned a value +1, and the second two 1. Thus Dâ, ˆb = P ++ + P P + + P + 1
PHYS 15C: Quantum Mechanics Spring 016 Problem Set 1 Solutions Page By symmetry, P = P ++ and P + = P +, so Following part a, Dâ, ˆb = P ++ P + 3 P + = + aa ; bb ψ = 1 bb ab = 1 cos θ/ = 1 1 + â ˆb otice that this joint probability is half the conditional probability of part b, as it should be. Using this with the result of part a in eq. 3, we get Dâ, ˆb = â ˆb Problem. Three photons: beyond Bell In the H, V basis, the three-photon state is ψ = 1 H 1 H H 3 + V 1 V V 3 a The 5-degree rotated basis is Inverting this, H = H + V, V H V = H = H + V, V = H V 5 Substituting into eq., and dropping the ket symbols to simplify notation, H 1 + V 1H + V H 3 + V 3 + H 1 V 1H V H 3 V 3 Terms with an odd number of V s cancel between the two parts; terms with an even number of V s add: H 1H H 3 + H 1V V 3 + V 1H V 3 + V 1V H 3 Thus, in an ααα experiment, the outcomes H H H, H V V, V H V and V V H each occur with probability 1/, while the remaining outcomes H H V, H V H, V H H and V V V never occur. b We have so L = H = H i V H + i V, R = R + L R L, V = 6 i Using eqs. 5 and 6 in eq., R 1 + L 1 R + L H 3 + V 3 R 1 L 1 R L H 3 V 3 Expanding as before, ψ = 1 [ R 1R V 3 + L 1L V 3 + R 1L H 3 + L 1R H 3 7
PHYS 15C: Quantum Mechanics Spring 016 Problem Set 1 Solutions Page 3 c From eq. 7, in a ββα experiment, the outcomes RRV, LLV, LRH and RLH each occur with probability 1/, while the remaining outcomes RRH, LLH, LRV and RLV never occur. Photons are indistinguishable, so βαβ and αββ experiments must give the same results with the labels appropriately permuted. That is, if β, β = R/L and α = H, V, then the probability of observing βαβ in a βαβ experiment is the same as that of observing ββ α in the ββα experiment, and so on. Mathematically, this is ensured by permutation symmetry of ψ. d The functions A i λ take the values ±1 for measurement outcomes H and V of the ith photon, and B i λ take the values ±1 for measurement outcomes L and R. The four outcomes of the ββα experiment that occur with nonzero probability are RRV, LLV, LRH and RLH, and B 1 B A 3 = 1 for each of these. Thus, B 1 λb λa 3 λ = 1 Similarly, B 1 A B 3 = A 1 B B 3 = 1 e Since [B i λ = 1, the above results imply that A 1 A A 3 = A 1 B B 3 B 1 A B 3 B 1 B A 3 = 1 3 = 1 The results of an ααα experiment that would be consistent with local hidden variables are therefore H H V, H V H, V H H and V V V. f According to part a, quantum mechanics predicts that in an ααα experiment, the outcomes H H H, H V V, V H V and V V H each occur with probability 1/. Each of these has A 1 A A 3 = +1 According to part e, any local hidden variables theory predicts that these four outcomes will never occur. g The conflict in Bell s inequality is for statistical predictions, while in this problem the conflict arises even in definite predictions.
PHYS 15C: Quantum Mechanics Spring 016 Problem Set 1 Solutions Page Problem 3. Searching a small quantum phonebook a Consider the general case where there are entires in the phonebook. The probability that you find your friend s name on the first lookup is p 1 = 1/. The probability of finding it on the second lookup is 1 1 p = = 1 1 since this involves picking any of the 1 incorrect entries the first time, and then picking the correct one among the 1 remaining entries the second time. Similarly, 1 1 p 3 = = 1 1 and in the same way, p = = p = 1 Of course, you never need more than 1 lookups. Thus, The average number of lookups needed is then n = Setting =, 1 n p n = 1 p 1 = 1 p n = 1 n + 1 = 1 = 1 n = 9 1 + + 1 = b When x x 0 that is, when x x 0 = 0, U F0 x 0 x 0 0 = x 0 and U F0 x 1 x 1 0 = x 1 When x = x 0, U F0 x 0 0 x 0 0 1 = x 0 1 and U F0 x 0 1 x 0 1 1 = x 0 0 Therefore, U F0 acts on the state x 0 1 / as 1 when x x 0, but as 1 when x = x 0. In symbols, U F0 x 0 1 / ŨF 0 x 0 1 /, where ŨF 0 = 1 x 0 x 0 acts only in the computational subspace. c In general, the initial state is s = 1 1 x x=0
PHYS 15C: Quantum Mechanics Spring 016 Problem Set 1 Solutions Page 5 and the Grover operation is The state after a single iteration is therefore For the special case =, we see that G F0 = U s Ũ F0 = s s 11 x 0 x 0 ψ 1 = G F0 s = s s 11 x 0 x 0 s = s s x 0 s + x 0 x 0 s = 1 s + x 0 ψ 1 = x 0 so in this case the Grover algorithm locates your friend s name with unit probability after a single iteration. d For general, G F0 x 0 = U s x 0 = s s 1 x 0 = x 0 s s x 0 = x 0 s Together with the result of part c and linearity, this yields the state after T = iterations: ψ = G F0 ψ 1 = 1 G F0 s + G F0 x 0 = 1 [ 1 s + x 0 + [ x 0 s For the special case =, the first term above vanishes, leaving ψ = x 0 s The probability of finding your friend s name after two iterations of Grover s algorithm is therefore which is as bad as guessing blindly. P = x 0 ψ = 1 1 = 1 If we keep going, we find that P 3 = 1/, P = 1, and so on, with two 1/ s following each 1. This type of cyclic behavior will always occur for any, because we are iterating a unitary transformation on a finite-dimensional Hilbert space.