Solution (Due in Class, 07-Oct-10)

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 † measurement being performed with four operators {Mk} such that Mk Mk are the four projectors onto the Bell states. Answer: The action of the circuit on any input state |φi is

|φi → CNOT |φi → (H1 ⊗ I2)CNOT |φi → Mk(H1 ⊗ I2)CNOT |φi (1)

where M1 = |00ih00|,M2 = |01ih01|, M3 = |10ih10|, M4 = |11ih11|.(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 = |00i(h00|(H1 ⊗ I2)CNOT ) = √ |00i(h00| + h11|) (2) 1 2 0 1 M = |01i(h01|(H1 ⊗ I2)CNOT ) = √ |01i(h01| + h10|) (3) 2 2 0 1 M = |10i(h10|(H1 ⊗ I2)CNOT ) = √ |10i(h00| − h11|) (4) 3 2 0 1 M = |11i(h11|(H1 ⊗ I2)CNOT ) = √ |11i(h01| − h10|) (5) 4 2

0 † 0 Therefore, it is easy to see that (Mk) Mk are projections onto the Bell states. 1 (M 0 )†M 0 = (|00i + |11i)(h00| + h11|) (6) 1 1 2 1 (M 0 )†M 0 = (|01i + |10i)(h01| + h10|) (7) 2 2 2 1 (M 0 )†M 0 = (|00i − |11i)(h00| − h11|) (8) 3 3 2 1 (M 0 )†M 0 = (|01i − |10i)(h01| − h10|) (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 |φiAB has Schmidt decomposition as

n AB X A B |φ i = λi|ψi i|ψi i (10) i

A B , where n is the Schmidt number, λi > 0 for i = 1...n, and |ψi i(|ψi i) form an orthonormal set. After local unitary transformation and classical communication, the state is changed to

A B ˜A ˜B Because U and U are local unitary operations, |ψi i and |ψi i still form two set of orthonormal vectors. Therefore, the above equation gives the Schmidt decomposed form of |φ˜ABi and the Schmidt number remains n. If arbitrary local operations and classical communication are allowed, the state is changed into

A B ¯A ¯B Here M , M are general local operations, not necessarily unitary. Therefore, |ψi i and |ψi i are no longer orthonormal set. ¯A However, the reduced denstiy matrix of A still lives in the space spanned by |ψi i

¯A 0 A The n vectors {|ψi i} 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 |φ¯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 |x, yi → |x, 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:

|zi|0i → |zi|w(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

> >

= <

jÞ i jÞ i

º º

> >

; :

¡¡¡ ¡¡¡ ¯ ¯ ¡¡¡ ¯

jÞ i

9 8

¡¡¡ ¡¡¡ ¡¡¡

j¼i

> > Ê Ê

½ ½

> >

= <

¡¡¡ ¡¡¡ ¡¡¡

j¼i

Ê Ê

¾ ¾

jÛ ´Þ µi j¼i

FÌ

> >

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