Fourier Quantum Transform
1. 2. 3. 4. 5. Quantum Algorithms: Deutsch-Jozsa Algorithm Quantum Fourier Transform Quantum Phase Estimation Order Finding and Factoring Other Applications & Hidden Subgroup Problem
Jungsang Kim Spring 2009
Quantum Information Science ECE @ Duke University
Overview of Quantum Algorithm
Quantum Algorithms can be powerful!!
With n-qubits, one can construct 2n-dimensional Hilbert space There is a possibility of applying a unitary operation to a large superposition of states, to evaluate the operation simultaneously This is referred to as Quantum Parallelism
BUT
Quantum measurements allow one to only measure one of these outcomes Effective quantum algorithms utilize quantum parallelism, but the answers eventually collapse to a global property of the operation and therefore could be read out with non-negligible probability O(1)
Some examples
Deutsch-Jozsa Algorithm Quantum Fourier Transform: Phase estimation, order-finding, factoring, discrete logarithm, etc (Hidden subgroup problem)
Jungsang Kim Spring 2009 Quantum Information Science ECE @ Duke University
Deutsch-Jozsa Algorithm I
Problem Statement
f(x) is a function from x {0 ,1, 2 , L , 2 n 1} to {0,1} such that f(x) is either balanced or constant Balanced means f(x) has half (2n-1) 1s and half 0s Constant means f(x) is either identically equal to 1 or equal to 0 How many evaluations of f(x) does it take to differentiate a balanced function from constant function with certainty?
Classical answer is 2n-1 +1, and Quantum answer is 1!! Key to algorithm is a unitary matrix Uf that evaluates f(x) U f x, y = x, y f ( x ) x x y f(x) y f(x)
Uf y
U f x, y = ( 1)
f (x)
y f (x )
x, y
f (x ) If y = ( 0 1 ) 2 then y f (x ) = ( 1) y and
0 1 0 1
0 0 1 1
0 1 1 0
Jungsang Kim Spring 2009
Quantum Information Science ECE @ Duke University
Deutsch-Jozsa Algorithm II
Full Circuit implementing Deutsch-Jozsa Algorithm
0 1
n
H n
x
Uf
x y f (x )
H n
0
n
H
1
y
2
H0 =
3
0 +1 ,H 1 = 0 1 2
Look at state evolution:
0 = 0 1
x 0 1 1 = x n 2 2 ( 1) f ( x ) x 0 1 2 = x 2 2n
2 xz H x = z ( 1) z
H n x = H n x1 , x2 ,L, xn
3 = x z
Observation:
( 1)
x z + f ( x )
2n
z 0 1 2
=
z1 , z 2 ,L, z n
( 1)x z +Lx z
11
nn
z1 , z 2 ,L, z n
2n
- If f(x) is constant, 0 3 = 1 otherwise
Jungsang Kim Spring 2009
0 3 = 0 !!!
Quantum Information Science ECE @ Duke University
Quantum Fourier Transform
1. 2. 3. 4. 5. Quantum Algorithms: Deutsch-Jozsa Algorithm Quantum Fourier Transform Quantum Phase Estimation Order Finding and Factoring Other Applications & Hidden Subgroup Problem
Jungsang Kim Spring 2009
Quantum Information Science ECE @ Duke University
Fourier Transform and FFT
Continuous Fourier Transform
Fourier Transform and Inverse Fourier Transform is defined by
F ( y ) = f (x )e 2iyx dx
f ( x ) = F ( y )e 2ixy dy
Discrete Fourier Transform
When the points are discrete, so are the Fourier transform
1 yk N
x je
j =0
N 1
2ijk N
1 xj N
yk e 2ikj
k =0
N 1
N
Discrete Fourier transform is multiplying a NxN matrix that takes N2 operations Fast Fourier transform exists, that can calculate all elements of the discrete Fourier transform in NlogN steps!!
Jungsang Kim Spring 2009 Quantum Information Science ECE @ Duke University
Quantum Fourier Transform
Quantum Fourier transform on an orthonormal basis to be a linear operator that transforms the basis
0 , 1 ,L, N 1
is defined
1 j N
1 N
N 1
e 2ijk
k =0
N 1
N
k
Equivalently, the action on an arbitrary state is given by
1 xj j N j =0
y
j =0
N 1
k
k
where the amplitudes yk are the discrete Fourier transform of the amplitudes xj. Quantum Fourier transform is a unitary operator!!
Reminder on Notations
N = 2n, the basis states are 0 , 1 ,L, 2 n 1 Binary notation for j = j1 j2 L jn = j1 2 n 1 + j2 2 n 2 + L + jn 20 Notation for binary fraction 0. j j L j = j 2 + j 2 2 + L + j 2 k 12 k 1 2 k
Jungsang Kim Spring 2009
Quantum Information Science ECE @ Duke University
Product Representation of QFT
Alternatively, the QFT is described as
1 2 1 2ijk 2n j1 , j2 ,L , jn n 2 e k 2 k =0 n 1 2ij kl 2 l l =1 111 = n 2 L e k1 L k n 2 k1 =0 k 2 =0 k n =0
n
1 = n2 2
k1 = 0 k 2 = 0
L e 2ijkl 2 kl
l
1
1
1
n
k n = 0 l =1
1 n 1 2ijkl 2l kl = n 2 e 2 l =1 kl =0
l 1n = n 2 0 + e 2ij 2 1 2 l =1 1 = n 2 0 + e 2i 0. jn 1 0 + e 2i 0. jn1 jn 1 L 0 + e 2i 0. j1 j2 L jn 1 2
[
]
(
)(
)(
)
Does it make sense??
Check the term corresponding to
Jungsang Kim Spring 2009
k = k1k 2 L k n
Quantum Information Science ECE @ Duke University
Quantum Circuit for QFT
Remember the product representation
j 1 0 + e 2i 0. jn 1 n2 2
(
)( 0
+ e 2i 0. jn1 jn 1 L 0 + e 2i 0. j1 j2 L jn 1
)(
)
Consider the gate Rk,
0 1 Rk k 0 e 2i 2
And the following circuit
j1 j2 jn 1 jn
H R2
M
M
L L
Rn-1
Rn H
0 + e 2i 0. j1L jn 1
L
Rn-2
Rn-1
L L L
H R2 H
0 + e 2i 0. j2 L jn 1 0 + e 2i 0. jn1 jn 1 0 + e 2i 0. jn 1
Swap the order of the bits to complete the QFT
Does it make sense??
Check what happens to each bit
Jungsang Kim Spring 2009 Quantum Information Science ECE @ Duke University
Efficiency of QFT
j1 j2 jn 1 jn
H R2
M
M
L L
Rn-1
Rn H
0 + e 2i 0. j1L jn 1
L
Rn-2
Rn-1
L L L
H R2 H
0 + e 2i 0. j2 L jn 1 0 + e 2i 0. jn1 jn 1 0 + e 2i 0. jn 1
Efficiency of QFT
Number of gates needed are n+(n-1)++1=n(n+1)/2 Hadamard and conditional rotations Each conditional rotation requires 2 CNOTs and few (4) single qubit rotations Total number of gates needed is still O(n2)
Comparison to Classical Fourier Transform
Fast Fourier transform needs O(NlogN) = O(n2n) operations QFT is exponentially fast compared to classical FT!!
Jungsang Kim Spring 2009 Quantum Information Science ECE @ Duke University
Quantum Fourier Transform
1. 2. 3. 4. 5. Quantum Algorithms: Deutsch-Jozsa Algorithm Quantum Fourier Transform Quantum Phase Estimation Order Finding and Factoring Other Applications & Hidden Subgroup Problem
Jungsang Kim Spring 2009
Quantum Information Science ECE @ Duke University
Phase Estimation Problem
Phase Estimation Problem
Unitary Operator U has an eigenvector u If is unknown, can we estimate ? with eigenvalue e2i
Available Resources (Assumptions)
A black box capable of preparing u j 2 operations Black boxes performing the controlled-U
Consider the circuit
0
First Register t qubits H H H H U2
0
0 0 0
M
L L L L L
(0 +e ( M (0 +e ( (0 +e ( (0 +e (
U2
t-1
2i 2t 1
2i 2 2 2i 21 2i 2 0
) )1 ) )1 ) )1 )
)1
2 2 2 2
Secoond Register
Jungsang Kim Spring 2009
u
U2
1
U2
2
u
Quantum Information Science ECE @ Duke University
Phase Estimation Algorithm
0
First Register t qubits H H H H U2
0
0 0 0
M
L L L L L
(0 +e ( M (0 +e ( (0 +e ( (0 +e (
U2
t-1
2i 2t 1
2i 2 2 2i 21 2i 2 0
) )1 ) )1 ) )1 )
)1
2 2 2 2
Secoond Register
u
U2
1
U2
2
u
Final State is given by 1 1 2i 2t 1 2i 2t 2 2i 2 0 0 +e 1 0 +e 1 L 0 +e 1 = t2 2t 2 2 Apply Inverse Fourier Transform to the top t qubits
(
)(
)(
)
2t 1 k =0
e
2ik
k
0 u
H
j
FT+ Uj
u
If is expressed to t bits, i.e. = 0.j1jt, t 1 2 1 2ik ~ e k u u 2t 2 k =0 Inverse FT gives exactly!!
Quantum Information Science ECE @ Duke University
Jungsang Kim Spring 2009
Performance of QPE
What if cannot be written exactly with t-bit binary expansion?
Let 0 b 2t 1 such that b/2t=0.b1b2bt is the best t-bit representation of i.e., 0 2 t where b 2t
After lots of algebra (see text book page 223-224), one can show that If we wish to approximate to an accuracy 2-n With probability of success at least 1- one needs
1 t = n + log 2 + 2
Overhead cost of succeeding with Probability better than 1- Number of bits needed to represent to accuracy 2-n
What if one cannot prepare the eigenstate u , but some state = cu u ? u ~ Phase estimation algorithm output state is c u u uu ~ Where u is a good approximation to u 2 Probability of measuring n accurate to n bits is at least cn (1 )
Quantum Information Science ECE @ Duke University
Jungsang Kim Spring 2009
Quantum Phase Estimation
Inputs:
A black box which performs a controlled-Uj operation for integer j An eigenstate u of U with eigenvalue e 2iu t = n+log(2+1/(2)) initialized qubits to 0
Outputs: Runtime
~ An n-bit approximation u to the phase u
O(t2) operations and one call to controlled-Uj black box Succeeds with probability at least 1- 1. 0 u Initial State t 1 2 1 t 2 j =0 j u 2. Create Superposition 2 1 1 2t 1 2t 1 2ij j u j u Apply black box 3. 2t 2 j =0 j U u = 2t 2 j =0 e ~ 4. u u Apply inverse Fourier transform ~ 5. u Measure first register
Jungsang Kim Spring 2009 Quantum Information Science ECE @ Duke University
Quantum Fourier Transform
1. 2. 3. 4. 5. Quantum Algorithms: Deutsch-Jozsa Algorithm Quantum Fourier Transform Quantum Phase Estimation Order Finding and Factoring Other Applications & Hidden Subgroup Problem
Jungsang Kim Spring 2009
Quantum Information Science ECE @ Duke University
Order Finding
Order in modular arithmatic
For positive integers x and N (x < N) with no common factors, The order of x modulo N is defined to be the least positive integer r such that xr = 1(mod N)
Order finding is believed to be a hard problem in a classical computer
L = [log(N)] is the number of bits needed to specify N No classical algorithm is known to solve the problem using resources polynomial in L
Examples of Order:
What is order of x=5 modulo N=21? 51=5 (mod 21) 52=25=21*1+4=4 (mod 21) 53=125=21*5+20=20 (mod 21) 54=625=29*21+16=16 (mod 21) 55=3125=148*21+17=17 (mod 21) 56=15625=744*21+1=1 (mod 21)!!
Quantum Information Science ECE @ Duke University
Jungsang Kim Spring 2009
Order-Finding Algorithm
Quantum algorithm for Order-Finding is phase estimation algorithm applied to U y xy (mod N )
L Note that when N y 2 1 , use convention that xy(mod N)=y
Consider the state
1 us r
2isk k exp r x mod N k =0
r 1
For integer 0 s r 1 these states are eigenstates of U, since 1 r 1 2isk k +1 U us = exp r x mod N r k =0
k=k+1
1 = r
2is (k '1) k ' exp 1 x mod N r k '=
r
2is 1 = exp rr
Jungsang Kim Spring 2009
2is 2isk ' k ' exp x mod N = exp 1 r us r k '=
r
Note : x r mod N = 1 mod N = x 0 mod N
Quantum Information Science ECE @ Duke University
Requirements for Phase Estimation
Inputs to Quantum Phase Estimation Algorithm were:
A black box which performs a controlled-Uj operation for integer j An eigenstate u of U with eigenvalue e 2iu t = 2L+1+log(2+1/(2)) qubits initialized to 0
Efficient procedures for implementing controlled-Uj
U is simply multiplication of x modulo N UJ is modular exponentiation process:
We need to compute
t 1
z y z U zt 2 U zt 1 2 LU z1 2 y
0
t 2
= z x zt 2 x zt 1 2 L x z1 2 y mod N
0
t 1
t 2
= z x z y mod N
2 4 2 Use modular multiplication to compute x (mod N ), x (mod N ),L , x (mod N ), j t 1 This computation takes O(L3) (O(L) numbers with O(L2) steps each) Perform modular multiplication
j
x z (mod N ) = x zt 2 (mod N ) x zt 1 2 (mod N ) L x z1 2 (mod N )
0
(
t 1
)(
t 2
)(
(
)
This computation takes O(L3) (O(L) numbers with O(L2) steps each) z Then apply uncomputation procedure to get ( z , y ) z , x y (mod N )
Jungsang Kim Spring 2009
)
Quantum Information Science ECE @ Duke University
Requirements for Phase Estimation
Preparation of the eigenstate
us
Preparation of the eigenstate u s Observe that 1 r 1 us = 1 r k =0
requires the knowledge of r!!
Now the order finding circuit:
Prepare t = 2L+1+[log(2+1/2)] qubits in the phase-estimation register (We will estimate the phase to 2L+1 bits) Prepare the state 1 in the second register Run the following circuit
Register 1 t qubits
0
H
j
FT+
Register 2 1 L qubits
xj modN
We obtain the estimate of the phase s r accurate to 2L+1 bits with prob. (1 ) r
Jungsang Kim Spring 2009 Quantum Information Science ECE @ Duke University
Obtaining Denominator from Ratio
The phase estimation algorithm gives s r accurate to 2L+1 bits
With this finite bit resolution, how do we determine the denominator r? Knowing a priori that is a rational number, can we efficiently find r? Answer is yes, using continued fractions algorithm!!
Continued fractions algorithm:
Describe real numbers in terms of integers alone, using the form
[a0 , a1 , L , aM ] a0 +
1 a1 + 1 a2 + 1 L+ 1 aM
s 1 2 r 2r
Example 31/13?
Theorem: Suppose s/r is a rational number such that
Then s/r is a convergent of the continued fraction for , and thus can be computed in O(L3) operations using the continued fraction algorithm. If s/r is determined to 2L+1 bits, this condition is satisfied, so the theorem holds
Jungsang Kim Spring 2009 Quantum Information Science ECE @ Duke University
Performance of the Algorithm
The order finding algorithm will output s/r for any s satisfying 0 s < r .
r cannot be correctly determined if s and r has common factor One can show that repetition should provide the correct answer very efficiently The number of repetition necessary is either logN, or even fixed!! (finite probability of success)
Jungsang Kim Spri...