Quantum Algorithms: Problem Set 1

Transcription

1 Quantum Algorithms: Problem Set 1 1. The Bell basis is + = 1 p ( 00i + 11i) = 1 p ( 00i 11i) + = 1 p ( 01i + 10i) = 1 p ( 01i 10i). This is an orthonormal basis for the state space of two qubits. It is name after John Bell s tests of local realism, an fins numerous applications in quantum information. Fin a quantum circuit that implements the unitary change-of-basis from the computational basis { 00i, 01i, 10i, 11i} to the Bell basis.. Suppose you have an e cient quantum circuit to simulate the time-evolution inuce by a Hamiltonian of the form " # 0 U H = U 0 where U is a n n unitary matrix. (That is, you can implement the transformation e iht on n + 1 qubits using poly(n, t) quantum gates.) Use this primitive to construct a quantum circuit implementing U. You may ignore global phase an the action on any ancilla qubits. 3. The Frekin gate is a controlle-swap. circuit symbol truth-table 000! ! ! ! ! ! ! ! 111 Prove that the Frekin gate by itself (but with access to arbitrarily initialize ancilla bits) can perform universal classical computation.

2 Quantum Algorithms: Problem Set 1 Hints 1. Think of circuits that generate + i from 00i. Most likely, the first circuit that comes to your min for this task will also be the correct circuti for converting the computational basis to the Bell basis.. Notice that H = 1. Using this fact, sum the Taylor expansion of e iht to all orers. 3. It is well-known that {AND, NOT } is a classically universal gate set. (Proving this is itself an interesting exercise.) Thus, to solve this problem one only nees to show how to simulate AND an NOT using Frekin gates an appropriately initialize ancillas.

3 Quantum Algorithms: Problem Set Consier the game guess-my-bitstring: an oracle contains a secret string a {0, 1} n. You can query the oracle with a guess x {0, 1} n an the oracle will tell you how many bits you got wrong (i.e. the Hamming istance between a an x). 1. Show that classically, one nees at least of orer n/ log n queries to etermine the secret string a {0, 1} n. Also, show that one can solve this classically using orer n queries.. For bitstrings x, y {0, 1} n let x y enote the number of places they i er ( Hamming istance ). Show that where i = p 1. X x{0,1} n ( i) a x (i) x y = 3. Let M be the following unitary single-qubit gate " # 1 i M = i 1 ( n if y = a 0 otherwise Let xi / ( i) x a enote a quantum oracle that, given input x {0, 1} n, inuces a phase ( i) x a.(if x a is computable by an e cient classical algorithm then such an oracle can always be built using reversible computing an phase kickback.) Using the result of the previous problem, show that the following quantum circuit solves guess-my-bitstring using a single quantum query. 0i H M 0i H ( i) x a M.. 0i H M

4 Quantum Algorithms: Problem Set 3 Backgroun The Jones polynomial is actually fairly easy to compute by han for simple knots an links. We will o this using the Kauffman bracket, which, applie to a link iagram, yiels a polynomial in the variable A. The Kauffman bracket is efine recursively by the following three rules. 1. = L = A + A 1 = ( A A ) L ule 1 expresses the fact that the bracket of the unknot is equal to one. ule is an example of what is calle a skein relation. It is a local transformation rule on formal linear combinations of link iagrams. The rule applies inepenently of the rest of the link iagram, which we think of as resiing outsie the ashe line. Note that it is important to keeptrackofwhichstraniscrossing over an which stran is crossing uner. otating the pictures in rule by ninety egrees one sees: = A + A 1. ule 3 says that if we have an arbitrary link L sitting next to an unknot, then the Kauffman bracket of that combination is equal to ( A A )timesthekauffman bracket of the link L. As an example, we have the following. = A + A 1 = A ( A A ) + A 1. The first equality is an application of rule. The secon equality follows by applying rule 3 to the first term. Also, in the secon term, we have applie what might becallerulezero:thebracketis 1

5 unchange by trivial eformation of the iagram ( planar isotopy ). The last equality follows by rule 1. From this example, one sees that the Kauffman bracket is not itself a link invariant, since the link is equivalent to the unknot, whose Kauffman bracket equals one. However, with a simple correction factor, we can obtain an invariant of oriente links. An oriente link is exactly what it souns like; one obtains an oriente link from an unoriente link by choosing one of the two possible orientations for each component. Here is an example. In general, changing orientations of components oes not result in an equivalent oriente link. To obtain an invariant of an oriente link iagram D, one evaluates the Kauffman bracket of the corresponing unoriente link, an then multiplies by ( A) 3w(D),wherew(D) iscallethewrithe of D, efine as the number of positive crossings minus the number of negative crossings. positive negative The Jones polynomial of an oriente link L is a Laurent polynomial in powers of t 1/ given by V L (t) =( A) 3w(L) L where A = t 1/4. emark 1: By Laurent polynomial, we mean that positive an negative powers of the variable are allowe. emark : From this efinition one might expect the Jones polynomial is a Laurent polynomial in t 1/4 rather than t 1/. However, by examining the efinition of the Kauffman bracket an writhe, it is not too har to show that only even powers of t 1/4 can ever occur. emark 3: The notation V L (t) isinacertainrespectunfortunate,althoughwearestuckwith it for historical reasons. If we fix a value of t 1/ C we obtain from the Jones polynomial a map from link iagrams to complex numbers rather than to polynomials. It is clear that such a map is a link invariant,an that it is in general weaker than the Jones polynomial itself. To make such maps one nees to actually choose a value of t 1/,notjustt, sincet has two ifferent square roots, resulting in two ifferent invariants.

6 Problems 1. Evaluate the Jones polynomial of the lefthane trefoil: To cut own on teium, here is a lookup table for the Kauffman brackets of all the singlecrossing connecte link iagrams. = A 3. For any oriente link iagram D let D enote its mirror-image. Prove that V D(t) =V D (t 1 ). Conclue via the result of problem 1 that: (The Jones polynomial is not the first link invariant to be iscovere. However, it is more powerful than most of the link invariants that precee it. In particular,fewofthe classical link invariants can istinguish the above two oriente knots.) 3. A matrix is sai to satisfy the Yang-Baxter equation if ( 1 )(1 )( 1 )=(1 )( 1 )(1 ), where 1 enotes the ientity matrix, an enotes the tensor prouct. (The tensor prouct is sometimes calle Kronecker prouct, e.g. by Mathematica). If is unitary one can think of it as a -quit gate acting on -imensional quits. In circuit notation, the Yang-Baxter equation is thus =. 3

7 ecall Artin s presentation of the brai group. The brai group is the group generate by σ 1,...,σ n 1 subject to the relations σ i σ j = σ j σ i for all i j anσ i+1 σ i σ i+1 = σ i σ i+1 σ i for all i. (If you are unfamiliar with group presentations, the point is this: thebraigroupisthe set of all strings mae from alphabet {σ 1,...,σ n 1, σ1 1,...,σ 1 n 1 },moulotheequivalence relations. These are the relations state explicitly above together with the implicit relation that σ i cancels σi 1.) (a) Show that the rule ρ : σ i 1 i 1 1 n i 1 (with ρ : σi 1 1 i n i 1 ) efines a representation of the brai group if satisfies the Yang-Baxter equation. (In other wors, one must show that ρ is a homomorphism: ρ(ab) = ρ(a)ρ(b) forallbrais a, b B n.) (b) Let G be any finite group of orer. Wecanthinkofthecomputationalbasisstatesfor a -imensional qubit as being labelle by the elements of. Showthatthegate a b b b 1 ab obeys the Yang-Baxter equation. (c) Can an -gate of the type escribe in part b perform universal quantum computation by itelf? 4