C/CS/Phys C191 Deutsch and Deutsch-Josza algorithms 10/20/07 Fall 2007 Lecture 17

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1 C/CS/Phs C9 Deutsch ad Deutsch-Josza algorithms 0/0/07 Fall 007 Lecture 7 Readigs Beeti et al., Ch Stolze ad Suter, Quatum Computig, Ch ) Nielse ad Chuag, Quatum Computatio ad Quatum Iformatio, Ch..4.3,.4.4 Deutsch s algorithm Deutsch s algorithm is a perfect illustratio of all that is miraculous, subtle, ad disappoitig about quatum computers. It calculates a solutio to a problem faster tha a classical computer ever ca. It illustrates the subtle iteractio of superpositio, phase-kick back, ad iterferece. Fiall, ufortuatel, is solves a completel poitless problem. Deutsch s algorithm aswers the followig questio: suppose f () is either costat or balaced, which oe is it? If f () were costat the for all the result is either 0 or. owever, if f () were balaced the for oe half of the iputs f () is 0, ad for the other half it is (which s correspod to 0 or is completel arbitrar). To aswer this questio classicall, we clearl eed to quer the fuctio for both = 0 ad =, hece two queries are required. Usig a quatum algorithm it turs out that we ca solve the problem with just oe quer of the fuctio. 0> U f > f() Figure : Quatum circuit for Deutsch s algorithm, testig whether a Boolea fuctio o oe qubit is costat or balaced. C/CS/Phs C9, Fall 007, Lecture 7

2 Aside: Deutsch s algorithm, as all kow quatum algorithms that provide epoetial speedup over classical sstems do, aswers a questio about a global propert of a solutio space. These are ofte called promise problems, whereb the structure of the solutio space is promised to be of some form ad b carefull usig superpositio, etaglemet ad iterferece we ca etract iformatio about that structure. The reaso these problems obtai epoetial improvemet over all kow classical algorithms is that classicall oe has to calculate ever poit i the solutio space i order to obtai full kowledge about this structure. Quatum mechaicall we calculate ever poit usig quatum parallelism. Ufortuatel this is ofte ot how most algorithms are phrased. Usuall we work with problems that are phrased of the form what gives a value of f () with the desired propert? Thus far, quatum computers ca ol provide square-root improvemet to such quer-based problems. Let ψ 0 be the iitial state vector ad ψ be the state of the sstem prior to applig Uf. Let ψ be the state of the sstem after applig U f ad ψ 3 be the state of the sstem prior to measuremet. Iput: ψ 0 = 0, It ma seem strage to start out with a result register of istead of 0, but igore this for ow, we will retur to it shortl. Appl the gate to the quer ad result registers to obtai: ( ) ψ = 0 + ( ) 0 The first qubit is referred to as the quer qubit, the secod as the result qubit. (Note that the term quer is used i two differet respects here - querig the fuctio f (), i.e., evaluatig it, ad querig the qubit, i.e., measurig it. Now, lets eamie f (): Suppose f ()=0 the f ()= 0 = ( ) ( ) = 0 Suppose f ()= the f ()= = ( ) ( ) 0 = 0 + Sice ± =() f (), we ca compactl describe this behavior for both istaces with the followig sigle formula: f ()=() f () ( 0 ) Thus, U f trasforms ( 0 ) ito: () f () ( ) 0 Or we ca write it all out i detail: [ U ( ) f 0 + ( ) ] 0 = [() f (0) ( ) 0 0 +() f () ( ) ] 0 Thus is a eample of backward sig propagatio or kick-back of phase from qubit to qubit - actuall a ver simple oe sice the phase is ol a global phase here at this poit. But below we shall see how it gets moved ito the state of qubit, resultig i a real overall kick-back from qubit to qubit. Now we look at what this state is for the two istaces of f beig costat or balaced. First, suppose f is costat, that is f (0)= f (). The: [() f (0) ( ) 0 0 +() f () ( ) ] 0 = () f (0) [ 0 ( 0 ) + ( 0 )] C/CS/Phs C9, Fall 007, Lecture 7

3 [ ( = ± ) ( )] = ± ( ) 0 + ( ) 0 Sice the first qubit is i + we ca aticipate that performig a adamard gate o qubit will the trasform this to 0. Suppose ow istead that f is balaced, that is f (0) f (), the: [() f (0) 0 ( 0 ) +() f () ( 0 ) ] = [() f (0) 0 ( 0 ) +() () f (0) ( 0 )] = () f (0) [ ( ) ( )] = ± [ ( ) ( )] ( = ± 0 ) ( 0 ) Sice the first qubit is i here, we ca aticipate that performig a adamard gate o qubit i this case will trasform this to. So it seems we ca get orthogoal states for qubit for the two differet istaces. So lets ow ru the qubit through a gate to get ψ 3 : ± ψ 3 = 0 ( 0 ) if f(0)= f () ± ( 0 ) if f(0) f () Sice i our case f (0) f ()=0 f (0)= f () we ca write this as [ 0 ψ 3 = ± f (0) f () ] ece it is possible to measure (the first, quer, qubit) to fid f (0) f (). Aside: Note that f (0) f () is a global propert of f (). Classicall it would require two evaluatios of f () to fid this aswer. Usig a quatum computer we are able to evaluate both aswers simultaeousl ad the iterfere these aswers to combie them together. Aother more subtle poit is that the phase of the result qubit trasfers to the quer qubit. This is a special case of phase kick back. I effect, the quer qubit acts as a cotrol of whether or ot to flip the result qubit. While the result qubit is potetiall flipped b the state of the quer qubit, the phase of the quer qubit is altered b the phase of the result (or target) qubit! This propert is also critical to Shor s algorithm. 3 Deutsch-Josza algorithm The Deutsch-Jozsa algorithm is a geeralizatio of Deutsch s algorithm to Boolea fuctios o qubits. Suppose f () : { } {0,} ad that f is either costat or balaced. The goal is determie which oe it is. Classicall it is eas to see that this would require (i worst case) querig just over half the solutio space, or / + queries. The Deutsch-Jozsa algorithm aswers this questio with just oe quer! We eed qubits ad oe additioal qubit: the former are the aalog of ad costitute a quer register, while the latter correspods to the result qubit i the Deutsch algorithm. The startig state of the sstem ψ 0 is fairl straightforward C/CS/Phs C9, Fall 007, Lecture 7 3

4 ψ 0 = 0 The smbolic otatio 0 simpl meas cosecutive 0 qubits. We the appl the trasform. This smbol meas to appl the gate to each of the qubits (i parallel, although this does ot matter. The ke is ol that the gate is applied oce to each qubit). I omework 3 ou showed that this trasform is: i () = i j j j ere is aother proof of this importat relatio: Cosider first a gate applied to a sigle qubit. We eed to multipl the compoet b if = 0 ad b if = : these two procedures ca be combied as i the Deutsch algorithm above b simpl multiplig b the factor () f (). The = ( 0 +() ) = () z z z where here z spas ol two values, 0 ad. This seems like otatioal overkill to represet a simple adamard gate,. owever, whe we geeralize the latter to the otatio pas off sice the above form ca immediatel be geeralized b summig over all possible combiatios of qubit basis states, i.e., over all -qubit states z: = () z z. z Each qubit cotributes a idepedet phase term, leadig to the dot product z. Comig back to Deutsch-Jozsa, we ow trasform ψ 0 with the -qubit ad -qubit adamard gates as: ψ = 0 = [ ] 0 {0,} Note that the otatio {0,} meas all possible bit strigs of size. For eample, for =, we have 00, 0, 0, ad. We the appl the trasform U f that implemets f () to obtai the state ψ : > Uf > f() Figure : Quatum circuit for the Deutsch-Jozsa algorithm, a -qubit geeralizatio of Deutsch s algorithm C/CS/Phs C9, Fall 007, Lecture 7 4

5 ψ = () {0,} f () [ ] 0 Fiall we appl aother trasform to obtai ψ 3 : [ z+ f () ψ 3 = z 0 ] () z {0,} {0,} The ke to the Deutsch-Jozsa algorithm is the followig rather subtle poit. We measure the probabilit amplitude of z = 0, the all zero state. (If the qubits are spis i a magetic field, this would be the groud state - correct?) Cosider the istace whe f () is costat. Sice z = 0, z must also be equal to zero ad hece () z+ f () is either or+for all values of, where holds for f ()= ad holds for f ()=0. I this case the amplitude for z = 0 is ± {0,} = ± which ehausts the probabilit amplitude for ψ 3. I other words, sice ψ 3 is ormalized to ad the amplitude of z = 0 alread gives probabilit, there ca be o other compoet i ψ 3 - all other amplitudes must be zero. ece whe ou measure the first qubits i the quer register, ou will obtai a zero (or more correctl, 0 ). Coversel, if f () is balaced the () z+ f () will be + for some values of ad for other values of. This is where the balaced requiremet comes ito pla. Sice all possible values are cosidered ad the fuctio is perfectl balaced, oe must have equal umbers of + ad. The amplitude of the all zero state z = 0 is the: + + = 0 where is the set of s such that f () is equal to 0 ad is the set of s where f () is equal to. ece ou will ot measure the all zero eigevalue 0 whe f () is balaced sice the probabilit amplitudes iterfere destructivel to produce a et probabilit amplitude of zero for the all zero state. What will be measured if the fuctio is balaced? Athig ecept the all zero eigevalue. At least oe qubit will result i a measuremet value of. Note that this algorithm does require ol oe quer of f (), i.e., of U f, but it requires the abilit to make a -qubit measuremet. C/CS/Phs C9, Fall 007, Lecture 7 5