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χ ψ 2.
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17
This means first of all that every quantum computation is inherently probabilis-
tic. Observing (a part of) the quantum memory isnot exactly thesame as “printing
the output”. We must plan a series of runs of the same quantum program and the
subsequent classical processing of the observed results, and we can hope only to get
the desired answer with probability close to one.
Furthermore, thismeansthatbyimplementing quantumparallelismsimplemind-
edly as in (14), and then observing the memory as if it were the classical n–bit
register, we will simply get some value F(x) with probability 1/N. This does not
use the potential of the quantum parallelism. Therefore we formulate a corrected
version of this notion, leaving more flexibility and stressing the additional tasks of
the designer, each of which eventually contributes to the complexity estimate.
2.3. Quantum parallel processing: version II. To solve efficiently a prob-
lem involving properties of the graph of a function F, we must design:
(i) An auxiliary unitary operator U carrying the relevant information about the
graph of F.
(ii) A computationally feasible realization of U with the help of standard quantum
gates.
(iii) A computationally feasible realization of the input subroutine.
(iv) A computationally feasible classical algorithm processing the results of many
runs of quantum computation.
All of this must be supplemented by quantum error–correcting encoding, which
wewillnot address here. Inthenext sectionwe willdiscuss somestandard quantum
subroutines.
3. Selected quantum subroutines
3.1. Initialization. Using the same conventions as in (14) and the subsequent
H⊗n,
comments, in particular, the identification H = we have
n 1
N−1 ⊗n
1 1 1
x = ǫ ...ǫ = ( 0 + 1 ) . (15)
n−1 0
√N | i √N | i (cid:18)√2 | i | i (cid:19)
xX=0 ǫiX=0,1
In other words,
N−1
1
(n−1) (0)
x = U ...U 0...0 (16)
√N | i 1 1 | i
xX=0
where U : H H is the unitary operator
1 1 1
→
1 1
0 ( 0 + 1 ), 1 ( 0 1 ),
| i 7→ √2 | i | i | i 7→ √2 | i−| i
18
(i)
and U = id U id acts only on the i–th qubit.
1 ⊗···⊗ 1 ⊗···⊗
Thus making the quantum gate U act on each memory bit, one can in n steps
1
initialize our register in the state which is the superposition of all 2n classical states
with equal weights.
3.2. Quantum computations of classical functions. Let be a finite basis
B
ofclassicalgatescontainingone–bitidentityandgeneratingallBooleancircuits, and
F : Fm Fn a function. We will describe how to turn a Boolean circuit of length
2 → 2
L calculating F into another Boolean circuit of comparable length consisting only
of reversible gates, and calculating a modified function, which however contains all
information about the graph of F. Reversibility means that each step is a bijection
(actually, an involution) and hence can be extended to a unitary operator, that is,
a quantum gate. For a gate f, define f( x,y ) = x,f(x)+y as in 2.2(D) above.
| i | i
3.2.1. Claim. A Boolean circuite of length L in the basis can be pro-
S B
cessed into the reversible Boolean circuit of length O((L+m+n)2) calculating a
S
permutation H : Fm+n+L Fm+n+L with the following property:
2 → 2 e
H(x,y,0)= (x,F(x)+y,0) = (F(x,y),0).
e
Here x,y,z have sizes m,n,L respectively.
Proof. We will understand L here as the sum of sizes of the outputs of all
gates involved in the description of . We first replace in each gate f by its
S S
reversible counterpart f. This involves inserting extra bits which we put side by
side into a new register of total length L. The resulting subcircuit will calculate
e
a permutation K : Fm+L Fm+L such that K(x,0) = (F(x),G(x)) for some
2 → 2
function G (garbage).
Nowaddtothememoryonemoreregisterofsizenkeepingthevariabley.Extend
K to the permutation K : Fm+L+n Fm+L+n keeping y intact: K : (x,0,y)
2 → 2 7→
(F(x),G(x),y).Clearly, K is calculated by the same boolean circuit as K, but with
extended register.
Extend this circuit by the one adding the contents of the first and the third
register: (F(x),G(x),y) (F(x),G(x),F(x)+y). Finally, build the last extension
7→
which calculates K¯−1 and consists of reversed gates calculating K in reverse order.
This clears the middle register (scratchpad) and produces (x,0,F(x) + y). The