quantum computing mas 725 hartmut klauck ntu 19.3.2012
TRANSCRIPT
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Quantum ComputingMAS 725Hartmut KlauckNTU19.3.2012
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Simon’s Problem
Given: Black box function f:{0,1}n!{0,1}n Promise: there is an s2{0,1}n with s0n
For all x: f(x)=f(x©s) For all x,y: x y©s ) f(x) f(y)
Find s !
Example: f(x)=2bx/2c Then for all k: f(2k)=f(2k+1); s=00 ... 01
Simon’s algorithm solve the problem in time/queries poly(n) Every classical randomized algorithm (even with errors) needs
(2n/2) queries to the black box
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The quantum algorithm
Start with|0ni|0ni Apply Hn to the first n qubits Apply Uf Measure qubits n+1,...,2n Result is some f(z) There are z, z©s with f(z)=f(z©s) Remaining state on the first n qubits is
(1/21/2 |zi + 1/21/2|z©si) |f(z)i Forget |f(z)i
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The quantum algorithm
State: 1/21/2 |zi + 1/21/2|z©si Every z2{0,1}n is equally likely to have been selected
by the measurement (f(z) fixes z and z©s) We really get a probability distribution on the above
states for a random z Measuring now would just give us a random z from
{0,1}n [f(z) is chosen randomly in measurement 1, then with prob. ½ we get: z, with prob. 1/2: z©s, resulting in a uniformly random z]
How can we get information about s? Apply Hn
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The quantum algorithm
State: 1/21/2 |zi + 1/21/2|z©si z uniformly random Apply Hn
Result: y y |yi with y=1/21/2 ¢1/2n/2 (-1)y¢z + 1/21/2¢1/2n/2(-1)y¢(z©s)
=1/2(n+1)/2 ¢(-1)y¢z [1+(-1)y¢s]
y¢z=i yizi
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The quantum algorithm
y y |yi with y=1/2(n+1)/2 ¢(-1)y¢z [1+(-1)y¢s] Case 1: y¢s odd ) y=0 Case 2: y¢s even ) y=§ 1/2(n-1)/2
Measure now[we are now independent of z]
Result: some y: yi si ´ 0 mod 2 Hence we get the following equation over Z2
yi si ´ 0 mod 2 For a random y
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Postprocessing
Resulting equation yi si ´ 0 mod 2 All y with yi si ´ 0 mod 2 have the same probability Knowing this equation reduces the number of
candidates for s by 1/2
We iterate this: Repeat n-1 times Solve the resulting linear system
If we get n-1 linearly independent equations, then s is determined
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Simon‘s Algorithmus
H|0ni
|0niU_f
n-1times,thensolvethesystemofequations
H
y(1),...,y(n-1)
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Analysis
s is determined as soon as we have n-1 independent equations. Coefficients of equations are randomly y(j)1,...,y(j)n under the condition y(j)¢s=0 mod 2[i.e. from a subspace U of dim. n-1 in (Z2)n]
Probability that y(j+1) is linear independent from y(1),...,y(j): Vj=span[y(1),...,y(j)] has dim. j Prob, that a random y(j+1) from U is in Vj:
2j/2n-1
Total probability of all being independent: j=1,...,n-1 (1-2j-1/2n-1)= j=1,...,n-1 (1-1/2j)
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Analysis
Total probability of all equations being independent:j=1,...,n-1 (1-1/2j)
This is at least1/2 ¢ (1-[j=2,...,n-1 1/2j])¸ 1/4[Use (1-a)(1-b)¸ 1-a-b für 0<a,b<1]
I.e. with probability at least 1/4 we find n-1 linear independent equations, and we can compute s
Use Gaussian elimination O(n3) or other methods O(n2.373)
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Variation
Decision problem: With probability 1/2: s=0n
With prob. 1/2: s uniform from {0,1}n-{0n} Decide between the two cases
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Lower bound
Consider any randomized algorithm that computes s, given oracle access to f
Fix some f=fs for every s If there is a randomized algorithm with T queries and success
probability p (both in the worst case), then there is a deterministic algorithm with T queries and success probability p for randomly chosen s
Let r2{0,1}m be the string of random bits used Es Er [Success for fs with random r]=p
) there is a fixed r with Es [Success fs with r]¸ p
Fix r ) determinististic algorithm
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Lower bound
s is uniformly random from {0,1}n-{0n} Fix any f=fs Given a deterministic query algorithm, success probability
1/2 for random s Consider the situation when k queries have been asked Fixes queries/answers (xi,f(xi)) If there are xi,xj with f(xi)=f(xj), then algo. stops, success Otherwise: all f(xi) are different,
never xi©xj=s, number of pairs is Hence there are at least 2n-1- possible s s is uniformly random from the remaining s
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Lower bound
There are still 2n-1- ¸ 2n-k2 posssible s s uniformly random among those Query xk+1 (may depend on previous queries and
answers) For every xk+1 there are k candidates s‘(1),...,s‘(k):
s‘(j)=xj©xk+1 for s Hence we find the real s with prob. · k/ (2n-k2)
[over choice of s]
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Lower bound
Probability to find the real s · k/(2n-k2) Total success probability:
If T<2n/2/2 the success probability is too small
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Variation
Decision problem: With probability 1/2: s=0n
with prob. 1/2: s uniform from {0,1}n-{0n} Algorithm decides between the two cases Analysis similar, with less than 2n/2/2 queries error
larger than 1/4
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Summary
Simon‘s problem can be solved by a quantum algorithm with time O(n2.373) and O(n) queries with success probability 0.99
Every classical randomized algorithm with success probability 1/2 needs (2n/2) queries
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Algorithms so far
Deutsch-Josza and Simon: DJ: f balanced or constant S: f has „Period“ s (over (Z2)n) First Hadamard, then Uf, then Hadamard and
measurement D-J: black box with output (-1)f(x)
S: standard black box
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Order finding over ZN
Given numbers x, N, x<N Order r(x) of x in ZN:
min. r: xr =1 mod N „Period“ of the powers of x We will use a black box Ux,N that computes
Ux,N |ji|ki= |ji|xjk mod Nix and N have no common divisors
Quantum algorithm to find r(x) ? Note: we will not really need a black box, since
Ux,N can be computed easily
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But first….
We need to say what it means to have an efficient quantum algorithm
Don’t want to count queries to a black box, but just computation time (or space)
Efficient classical computation is captured by complexity classes like P or BPP
P : problems solvable by poly-time Turing machines BPP : problems solvable by poly-time randomized
Turing machines with bounded error The classes are believed to be the same
(derandomization)
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Computing with circuits
Circuit: inputs x1,…,xn2{0,1}n
Gates g1,…,gm Gate: takes inputs or output of a previous gate, computes a
function {0,1}2{0,1} Gates form a directed acyclic graph Output gate gm Size: m (corresponds to sequential computation time) Circuit bases (allowed function for the gates):
AND, OR, NOT NAND All Boolean function with 2 inputs
Size changes only by a constant factor with the basis
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Probabilistic circuits
Additional inputs r1,…,rm
For all inputs x1,…,xn : if r1,…,rm are uniform random bits, the correct result will be computed with probability 2/3
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Circuit families
A circuit Cn computes on inputs with n bits Circuit family (Cn)n2 N
Uniformity condition: Cn can be computed in polynomial time from n (given in unary)
Without this condition circuit families can compute everything
Now we can define P,BPP in terms of circuits Polynomial size and uniform
Equivalent to Turing machine definitions
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Facts about circuits
Almost all function {0,1}n{0,1} need circuit size (2n/n) Established by a counting argument
This bound is tight for non-uniform circuits, i.e., every function f:{0,1}n{0,1} can be computed in size O(2n/n)
We don‘t know any explicit function that needs !(n) circuit size
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Quantum circuits
n qubits initialized with the input (classical state) s qubits workspace At all times there is a global state on n+s qubitts Unitary operations (on 1, 2, or 3 qubits)
U1,…,UT; given together with choice of the qubits Applying operation Ui : take the tensor product with
the identity on the other qubits, multiply with the current state in the order: 1,...,T
One or more fixed qubits are measured in the end (standard basis)
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Quantum circuits
Uniform families defined as before, but we need to restrict the set of allowed unitaries (since the set of all unitaries on a single qubit is already not even countable)
Class BQP: functions computable by uniform families of polynomial size quantum circuits with error < 1/3
EQP: same, but no error allowed It is possible to show that these classes coincide with
definitions for them based on quantum Turing machines
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Relationships between complexity classes Pµ BPPµ BQP µ PSPACE
Pµ EQPµ BQP
All inclusions except the first and the last need to be proved Conclusion: BQPdoes not contain uncomputable functions Widely believed that P=BPP On the other hand the factorization problem is BQP, not known to
be in BPP Generally considered (very) unlikely BQP=PSPACE, or NPµBQP,
i.e. not likely that we can solve NP-complete problems
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Simulating quantum circuits
Theorem:Every quantum circuit with m gates and n+s can be simulated by a deterministic circuit of size m¢2O(n+s)
This implies that at most exponential speedups are possible
Uniformity is preseved by the simulation Idea: store the global quantum state on n+s qubits
explicitly (with limited precision), apply the m unitary operations one after another by performing matrix multiplication (with limited precision)
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P vs. BQP
Simulation of classical circuits Problem:
Quantum circuits are reversible (up to the final measurement)
„Fan-Out“ is not implementable due to no-cloning (i.e., using a computation result several time is not directly possible)
Solution: Simulate a classical circuit first by a classical reversible circuit
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Simulation
Toffolli Gate:maps a,b,c to a,b, (aÆb)©c
The Toffolli gate is reversible [given a,b,d can compute c=(aÆb)©d ]
The gate is universal [AND: set c=0, NOT: set c=1, b=1]
Fan-out:To copy a set b=1, c=0 (copies classical bits)
Classical reversible circuits are also quantum circuits Simulation of probabilistic circuits is immediate, hence
BPP µ BQP Measure 1/21/2 ( |0ki+|1ki) to get k copies of a random bit
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Which classes of unitaries are universal? We can use one of the following
CNOT and every unitary gate on 1 Qubit CNOT, Hadamard, plus O(1) rotation gates (approximately
universal) Toffoli Gate and Hadamard gates (approximately universal)
We can approximate any circuit with 2 qubit gates to any precision using only gates from one of the above sets with limited overhead
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Approximate computation
What is the influence of error?
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Limited precision
Suppose a quantum circuit computes |Ti=UT UT-1 U1 |xi |0…0i
Ui unitary Instead of UT apply VT
errors due to implementation while simulating the circuit with limited precision
Result is VT|T-1i=|Ti+|ETi, where |ETi=(VT-UT) |T-1i (not normalized)
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Limited precision
Result VT|T-1i=|Ti+|ETi, where |ETi=(VT-UT) |T-1i
Use Vi instead of Ui for all i: |1i=V1|0i=|1i+|E1i |2i=V2|1i=|2i+|E2i+V2|E1i |Ti=VT|T-1i
=|Ti+|ETi+VT|ET-1i++ VTV2 |E1i Hence k|Ti-|Ti k· i=1…T k|Eii k=i=1…T k(Vi-Ui) |i-1i k
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Approximating unitaries
Let U be an arbitrary unitary on n qubits And U‘ be any operator What is the approximation error? Spectral norm kUk=maxx:kxk=1 kU x k Approximation error: kU – U‘ k
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Total approximation error
i=1…T k(Vi-Ui) |i-1i k· i=1…T k(Vi-Ui) k If kVi-Uik· /T, then the total distance is at most k|Ti-|Ti k· implies what error? Measure all n+s qubits in the standard basis Measurement result a appears with probability
P(a)=|h a|Ti|2 resp. Q(a)=|h a|Ti|2
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Total approximation error
Measurement result a appears with probabilityP(a)=|h a|Ti|2 resp. Q(a)=|h a|Ti|2
Hence the total error is at most a|P(a)-Q(a)|· 2 k|Ti-|Ti k·2
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Conclusion
For polynomial time computations we need to apply unitaries with precision 1/poly(n) in the spectral norm
In particular: quantum computing is not an analogue model of computation requiring infinite precision
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Efficiency of approximating
Number of gates from a finite set we need to simulate any 1-qubit gate? Depends on the required precision
[Solovay, Kitaev] show -approximation with log2(1/) gates
If poly(n) gates have to be approximated with error 1/poly(n) we only need an overhead factor log2(n)