Id: hw.tex,v 1.4 009/0/09 04:31:40 ike Exp 1 MIT.111/8.411/6.898/18.435 Quantum Information Science I Fall, 010 Sam Ocko November 15, 010 Problem Set # 8 Solutions 1. (a) The eigenvectors of S 1 S are 11, 10, 10 + 01 having eigenvalues 1, 1, 1, 3. We note that these are the same as the eigenvectors of the SWAP operator, where they have eigenvalues 1, 1, 1, 1. Therefore,, 10 01 S 1 S = SWAP 1. Setting J(t) = 1 for a time of π we apply the unitary: e i π S 1 S 4 = e i( SWAP 1) π 4 e i πswap SWAP. (b) We can implement the SWAPgate by applying the hamiltonian for half the time. To within a phase SWAP e iπswap 4 = 1 iswap. To apply a ControlledT gate, we implement the following circuit: FIG. 1. Quantum circuit which gives Controlled-Not gate To see that this circuit works, we see that 1 SWAPZ1 SWAP (1 iswap) Z 1 (1 iswap) 1 (Z 1 Z ) i (Z 1 + Z ) SWAP = 1 ((Z 1 Z ) i (Z 1 + Z )) = i 0 0 0 0 1 0 0 0 0 1 0 0 0 0 i
Id: hw.tex,v 1.4 009/0/09 04:31:40 ike Exp Therefore, 1 0 0 0 1 0 0 0 i 0 0 0 S 1 S 0 1 0 0 0 i 0 0 0 1 0 0 SWAPZ1 SWAP 0 0 i 0 0 0 1 0 0 0 1 0 = 0 0 0 i 0 0 0 i 0 0 0 i i 0 0 0 0 i 0 0 0 0 i 0 Cont-Z 0 0 0 i And by putting Hadamards on both sides, we turn a Controlled-Z gate into a Controlled- Not gate.. When a = b = 1 4, c = 1 4, H = (1 Z1)(1 Z) 4 1 4 = 11 11 1 4. Turning on the Hamiltonian for a time of π we apply the unitary: 1 0 0 0 e iπh e iπ 11 11 0 1 0 0 = 0 0 1 0 = Cont-Z 0 0 0 e iπ By putting Hadamards on both sides of a Controlled-Z gate, we get a Controlled-Not gate.
Id: hw.tex,v 1.4 009/0/09 04:31:40 ike Exp 3 3. (a) [H, a] = ω ( a a a aa a ) = ω (( aa 1 ) a aa a ) = ωa Therefore, when H ψ = E ψ, we use the commutation relations [a, a ] = 1 H (a ψ ) = (ah ωa) ψ = (ae ωa) ψ = (E ω) (a ψ ) so a ψ is an eigenstate with energy E ω. (b) We can similarly show a ψ is an eigenstate with energy E + ω using [H, a ] = ωa : H ( a ψ ) = ( a H + ωa ) ψ = ( a E + ωa ) ψ = (E + ω) ( a ψ ) Therefore, using induction, n = (a ) n is an eigenstate with energy n ω. We must now prove the normalization n n = 1. To do so, we use the identity a ( a ) k ( ) = k a k 1 + ( ) a k a, and by induction: ( a ) n 0 = 0 a n ( a ) n 0 = ( 0 a n 1 n ( a ) n 1 ( + a ) ) n a 0 = n 0 a n 1 ( a ) n 1 0 =... = 0 0 Therefore, n n = =1 and is a normalized eigenstate with energy n ω.
Id: hw.tex,v 1.4 009/0/09 04:31:40 ike Exp 4 4. (a) We use the identity a n = n n 1 : (b) a α = e α (c) We note that α = e α a n = e α e α likewise for β. We also note that n=1 n 1 = e α (n 1)! (α ) n n n = e α n =0 e α α α = n! n = ( ) α n = 1 ( ) αa n 0 = e α e a α 0, Ba B = ( BaB ) = (acos(θ) + bsin(θ)) = a cos(θ) + b sin(θ), likewise for Bb B. Therefore, using the fact that a, b commute: +1 n = α α B α β = Be α e αa e β e βb 00 = e α + β e α(a cos(θ)+b sin(θ)) e β( a sin(θ)+b cos(θ)) B 00 = e αcos(θ) βsin(θ) + αsin(θ)+βcos(θ) e a (αcos(θ) βsin(θ)) e b (αsin(θ)+βcos(θ)) 00 = αcos(θ) βsin(θ) αsin(θ) + βcos(θ)
MASSACHUSETTS INSTITUTE OF TECHNOLOGY MIT.111/8.411/6.898/18.435 Quantum Information Science I November 4, 010 Problem Set #8 (due in class, TUESDAY 16-Nov-10) P1: (Universality of Heisenberg Hamiltonian) Consider the two-qubit Heisenberg Hamiltonian H(t) = J(t) S 1 S = J(t) ] [X 1 X + Y 1 Y + Z 1 Z. (1) 4 (a) Show that a swap operation U can be implemented by turning on J(t) for an appropriate amount of time τ, to obtain U = exp( iπs 1 S ). (b) Compute the swap gate, obtained by turning on J(t) for time τ/. Together with arbitrary single qubit gates, the swap gate is universal for quantum computation. P: (NMR controlled-not) Give an explicit sequence of single qubit rotations which realize a controlled-not between two spins evolving under the Hamiltonian H = az 1 + bz + cz 1 Z. Simplify the result as much as you can, to reduce the number of single qubit rotations. P3: (Simple harmonic oscillators) Consider the Hamiltonian H = ωa a, and let n be an energy eigenstate with energy n ω. (a) Compute [H, a] = Ha ah and use the result to show that if ψ is an eigenstate of H with energy E n ω, then a n ψ is an eigenstate with energy E n ω. (b) Show that n = (a ) n 0. P4: (Coherent states) The coherent state α of a simple harmonic oscillator is defined as α = e α / n, () where n is an n-photon energy eigenstate. (a) Prove that a coherent state is an eigenstate of the photon annihilation operator, that is, show a α = λ α for some constant λ. (b) Compute α α. (c) Let B = exp [ θ ( a b ab )] be the beamsplitter operator coupling two simple harmonic oscillators. Using the fact that BaB = a cos θ + b sin θ and BbB = a sin θ + b cos θ, compute B α β where α and β are coherent states of the two systems. Express the result in a simple form. P5: (Recent implementations of quantum algorithms and protocols). Find a recent paper in the literature about an implementation of a quantum algorithm (preferably) or a quantum protocol, involving more than one qubit. Write a short (< 500 word) summary of it, on the QIS wiki. See instructions on the course homepage, http://web.mit.edu/.111/