MASSACHUSETTS INSTITUTE OF TECHNOLOGY MIT 2.111/8.411/6.898/18.435 Quantum Information Science I September 30, 2010 Problem Set #4 Solution (due in class, 07-Oct-10) 1. Measurement in the Bell basis Show that the circuit ________ L H _____ ___ ________ L _____ ___ performs a measurement in the basis of the Bell states. Specifically, show that this circuit results in a y measurement being performed with four operators fMkg such that Mk Mk are the four projectors onto the Bell states. Answer: The action of the circuit on any input state jφi is jφi ! CNOT jφi ! (H1 ⊗ I2)CNOT jφi ! Mk(H1 ⊗ I2)CNOT jφi (1) where M1 = j00ih00j,M2 = j01ih01j, M3 = j10ih10j, M4 = j11ih11j.(Normalization factor is ignored.) Combining (H1 ⊗ I2)CNOT into Mk, we see that the total action is equivalent to the following four operators 0 1 M = j00i(h00j(H1 ⊗ I2)CNOT ) = p j00i(h00j + h11j) (2) 1 2 0 1 M = j01i(h01j(H1 ⊗ I2)CNOT ) = p j01i(h01j + h10j) (3) 2 2 0 1 M = j10i(h10j(H1 ⊗ I2)CNOT ) = p j10i(h00j − h11j) (4) 3 2 0 1 M = j11i(h11j(H1 ⊗ I2)CNOT ) = p j11i(h01j − h10j) (5) 4 2 0 y 0 Therefore, it is easy to see that (Mk) Mk are projections onto the Bell states. 1 (M 0 )yM 0 = (j00i + j11i)(h00j + h11j) (6) 1 1 2 1 (M 0 )yM 0 = (j01i + j10i)(h01j + h10j) (7) 2 2 2 1 (M 0 )yM 0 = (j00i − j11i)(h00j − h11j) (8) 3 3 2 1 (M 0 )yM 0 = (j01i − j10i)(h01j − h10j) (9) 4 4 2 2. Schmidt numbers and LOCC. Recall that the Schmidt number of a bi-partite pure state is the number of non-zero Schmidt components. Prove that the Schmidt number of a pure quantum state 2 cannot be increased by local unitary transforms and classical communication. The Schmidt number is strictly nonincreasing under more general conditions, namely, for arbitrary local operations and classical communication (LOCC); you are welcome to prove this also, but that is not required for credit. The Schmidt number is one measure of how entangled a bi-partite quantum state is. Answer: Suppose that a bi-partite state jφiAB has Schmidt decomposition as n AB X A B jφ i = λij i ij i i (10) i A B , where n is the Schmidt number, λi > 0 for i = 1:::n, and j i i(j i i) form an orthonormal set. After local unitary transformation and classical communication, the state is changed to n n ~AB X A A B B X ~A ~B jφ i = λiU j i iU j i i = λij i ij i i (11) i i A B ~A ~B Because U and U are local unitary operations, j i i and j i i still form two set of orthonormal vectors. Therefore, the above equation gives the Schmidt decomposed form of jφ~ABi and the Schmidt number remains n. If arbitrary local operations and classical communication are allowed, the state is changed into n n ¯AB X A A B B X ¯A ¯B jφ i = λiM j i iM j i i = λij i ij i i (12) i i A B ¯A ¯B Here M , M are general local operations, not necessarily unitary. Therefore, j i i and j i i are no longer orthonormal set. ¯A However, the reduced denstiy matrix of A still lives in the space spanned by j i i A ¯AB ¯AB ρ¯ = trBjφ ihφ j (13) n n X X X ¯B ¯B ¯A ¯A = (λiλjhkj i ih j jki)j i ih j j (14) i j k ¯A 0 A The n vectors fj i ig span a space of dimension m ≤ n. The rank n of the reduced density matrixρ ¯ is less than the dimension of the space m. Therefore n0 ≤ n. The rank of the reduced density matrixρ ¯A is equal to the Schimdt number of jφ¯ABi. Therefore, the Schimdt number of a state after arbitrary LOCC cannot be increased. 3. Reversible circuits. (a) Construct a reversible circuit which, when two bits x and y are input, outputs (x; y; c; x ⊕ y), where c is the carry bit when x and y are added. Answer: 3 (b) Construct a reversible circuit using Fredkin gates to simulate a Toffoli gate. Answer: (c) Construct a quantum circuit to add two two-bit numbers x and y modulo 4. That is, the circuit should perform the transformation jx; yi ! jx; x + y mod 4i. Answer: x = x1x2 y = y1y2 (15) 4. Quantum circuit for the Hamming weight. Construct a quantum circuit that performs the following unitary transformation: jzij0i ! jzijw(z)i ; 4 where w(z) denotes the Hamming weight of z (the number of ones in its binary representation). Try to do this for the general case of z being represented by n qubits, for arbitrary n, but if you cannot think of a clever solution (which takes avantage of quantum gates, versus just classical ones), just give a circuit for n = 3. Answer: 9 8 > > ¯ ¯ ¡¡¡ ¯ ¡¡¡ ¡¡¡ jÞ i > > ½ > > > > = < º º jÞ i jÞ i º º º º > > > > > > > > ; : ¡¡¡ ¡¡¡ ¯ ¯ ¡¡¡ ¯ jÞ i Ò 9 8 ¡¡¡ ¡¡¡ ¡¡¡ j¼i > > Ê Ê ½ ½ > > > > > > > > > > > > > > = < ¡¡¡ ¡¡¡ ¡¡¡ j¼i Ê Ê ¾ ¾ Ý jÛ ´Þ µi j¼i FÌ FÌ > > º > > º º > > º º º > > º º º > > > > > > > > ; : ¡¡¡ ¡¡¡ ¡¡¡ j¼i Ê Ê Ò Ò where 1 0 R = k (16) k 0 e2πi=2 'FT' is the quantum fourier transform circuit(see page 219 of book 'Quantum computation and quantum information'). A classical way to do this with n = 3 would be.