# Introduction to Quantum Information Processing C

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"Introduction to Quantum Information Processing C"

```					       Introduction to
Quantum Information Processing
CS 467 / CS 667
Phys 667 / Phys 767
C&O 481 / C&O 681

Lecture 10 (2008)
Richard Cleve
DC 2117
cleve@cs.uwaterloo.ca
1
Order-finding via
eigenvalue estimation

2
Order-finding problem
Let M be an m-bit integer
Def: ZM* = {x  {1,2,…, M  1} : gcd(x,M ) = 1} (a group)
Def: ordM (a) is the minimum r > 0 such that ar = 1 (mod M )

Order-finding problem: given a and M, find ordM (a)

Example: Z21* = {1,2,4,5,8,10,11,13,16,17,19,20}
The powers of 10 are: 1, 10, 16, 13, 4, 19, 1, 10, 16, …
Therefore, ord21 (10) = 6

Note: no classical polynomial-time algorithm is known
for this problem                                            3
Order-finding algorithm (1)
Define: U (an operation on m qubits) as: Uy = ay mod M 
r 1
Define:  1     e 2i 1/ r  j a j mod M
j 0

r 1
Then U  1      e 2i 1/ r  j a j 1 mod M
j 0
r 1
  e 2i 1/ r e 2i 1/ r  j 1 a j 1 mod M
j 0

 e 2i 1/ r   1

4
Order-finding algorithm (2)
2n qubits   corresponds to the mapping:
xy  xax y mod M
Moreover, this mapping can
U         n qubits
be implemented with roughly
O(n2) gates

The phase estimation algorithm yields a 2n-bit estimate of 1/r
From this, a good estimate of r can be calculated by taking
the reciprocal, and rounding off to the nearest integer
Exercise: why are 2n bits necessary and sufficient for this?

Problem: how do we construct state 1 to begin with?
5
Bypassing the need for 1 (1)
r 1
Let     1   e 2i 1/ r  j a j mod M    Can still uniquely
j 0
r 1                          determine k and r,
 2   e 2i 2 / r  j a j mod M   provided they have
j 0
                                    no common factors
r 1
(and O(log n) trials
 k   e 2i k / r  j a j mod M
j 0
suffice for this)
      r 1
 r   e 2i r / r  j a j mod M
j 0

Any one of these could be used in the previous procedure,
to yield an estimate of k/r, from which r can be extracted
What if k is chosen randomly and kept secret?                       6
Bypassing the need for 1 (2)
Returning to the phase estimation problem, suppose that
1 and 2 have respective eigenvalues e2i1 and e2i2,
and that 11 + 22 is used in place of an eigenvalue:
0 H               †
0 H             FM
0 H
11 + 22                 U
What will the outcome be?

It will be an estimate of   1 with probability |1 |2
2 with probability |2 |2     7
Bypassing the need for 1 (3)
Using the state

yields results equivalent to choosing a k at random
1 r
Is it hard to construct the state       k ?
r k 1

In fact, it’s easy, since
1 r          1 r r 1 2i k / r  j j
  k  r  e
r k 1         k 1 j 0
a mod M  1

This is how the previous requirement for 1 is bypassed

8
Quantum algorithm for order-finding
inverse QFT
0   H                                     H   measure these qubits and
0   H                            H   4       apply continued fractions*
0   H              H   4   8                algorithm to determine a
0                                             quotient, whose
0
1
Ua,M                                denominator divides r

Ua,M y = ay mod M
Number of gates for a constant success probability is:
O(n2 log n loglog n)
* For a discussion of the continued fractions algorithm, please see
Appendix A4.4 in [Nielsen & Chuang]
9
Reduction from factoring
to order-finding

10
The integer factorization problem
Input: M (n-bit integer; we can assume it is composite)

Output: p, q (each greater than 1) such that pq = N

Note 1: no efficient (polynomial-time) classical algorithm
is known for this problem

Note 2: given any efficient algorithm for the above, we can
recursively apply it to fully factor M into primes* efficiently

* A polynomial-time classical algorithm for primality testing exists
11
Factoring prime-powers
There is a straightforward classical algorithm for factoring
numbers of the form M = pk, for some prime p

What is this algorithm?

Therefore, the interesting remaining case is where M has
at least two distinct prime factors

12
Numbers other than prime-powers
Proposed quantum algorithm (repeatedly do):
1. randomly choose a  {2, 3, …, M1}
2. compute g = gcd(a,M)
3. if g > 1 then
output g, M/g                             Analysis:
else
compute r = ordM(a) (quantum part)        we have M | a r1
if r is even then
compute x = a r/2 1 mod M      so M | (a r/2+1)(a r/21)
compute h = gcd(x,M)
if h > 1 then output h, M/h     thus, either M | a r/2 +1
or gcd(a r/2 +1,M)
is a nontrivial factor of M
latter event occurs with probability  ½   13
Introduction to
Quantum Information Processing
CS 467 / CS 667
Phys 667 / Phys 767
C&O 481 / C&O 681

Lecture 11 (2008)
Richard Cleve
DC 2117
cleve@cs.uwaterloo.ca
14
Universal sets of gates

15
A universal set of gates (1)
Main Theorem: any unitary operation U acting on k qubits
can be decomposed into O(4k) CNOT and one-qubit gates

Proof sketch (for a slightly worse bound of O(k24k)) :

We first show how to simulate a controlled-U, for any one-
qubit unitary U

Straightforward to show: every one-qubit unitary matrix
can be expressed as a product of the form

eiδ    0  eiα / 2      0   cosθ / 2 sin θ / 2 eiβ / 2       0 
                                                                       
0     eiδ   0       e iα / 2   sin θ / 2 cosθ / 2  0   e iβ / 2 
16
A universal set of gates (2)
This can be used to show that, for every one-qubit unitary U,
there exist A, B, C, and , such that:
• ABC=I
i A X B X C = U, where X  
0 1     Exercise: show
• e                               1 0
     
         how this follows

The fact implies that

P                                  1 0 
                              where P  
 0 e iλ 

U             C        B        A                        

17
A universal set of gates (3)
Controlled-U gates can also simulate controlled-controlled-V
gates, for an arbitrary unitary one-qubit unitary V:

                                 where V = U 2

V               U        U†       U

18
A universal set of gates (4)
Example: Toffoli gate         a              a
―controlled-controlled-NOT‖
b              b
c              c  (ab)

In this case, the one-qubit gates can be:

1 1 1                      1   0 
H   1  1                 T 
2                         0 eiπ / 4 


19
A universal set of gates (5)
From the Toffoli gate, generalized Toffoli gates (which are
controlled-controlled- ... -NOT gates) can be constructed:

a1
a2
a3


:
ak
b1
:
bk2
c
c  (a1  a2  ...  ak)
20
A universal set of gates (6)
From generalized Toffoli gates, generalized controlled-U
gates (controlled-controlled- ... -U) can be constructed:
1   0 0 0 0 0   0     0 
                          
0   1 0 0 0 0   0     0 


0   0 1 0 0 0   0     0 
                          
0   0 0 1 0 0   0     0 
0                     0 
0           0                      0 0 0 1 0   0         
U               U            0   0 0 0 0 1 0       0 
                          
0   0 0 0 0 0 U 00   U 01 
0                    U11 
    0 0 0 0 0 U10         

21
A universal set of gates (7)
The approach essentially enables any k-qubit operation of the
simple form     1 0 0 0 0 0 0 0
                           
 0 U 00   0 0 U 01 0 0    0
0 0       1 0   0   0 0   0
                           
0 0       0 1   0   0 0   0
0 U       0 0 U11   0 0   0
     10                    
0 0       0 0   0   1 0   0
                           
0 0       0 0   0   0 1   0
0 0                       1
          0 0   0   0 0    

to be computed with O(k2) CNOT and one-qubit gates

In a spirit similar to Gaussian elimination, any 2k2k unitary
matrix can be decomposed into a product of O(4k) of these
22
A universal set of gates (8)
This completes the proof sketch*
Thus, the set of all one-qubit gates and the CNOT gate are
universal in that they can simulate any other gate set

Question: is there a finite set of gates that is universal?

Answer 1: strictly speaking, no, because this results in only
countably many quantum circuits, whereas there are
uncountably many unitary operations on k qubits (for any k)

 Actually we proved a slightly worse bound of O(k24k)
23
Approximately universal gate sets
Answer 2: yes, for universality in an approximate sense ...

To be continued ...

24
Introduction to
Quantum Information Processing
CS 467 / CS 667
Phys 667 / Phys 767
C&O 481 / C&O 681

Lecture 12 (2008)
Richard Cleve
DC 2117
cleve@cs.uwaterloo.ca
25
Approximately universal
sets of gates

26
Universal gate sets
The set of all one-qubit gates and the CNOT gate are
universal in that they can simulate any other gate set

Quantitatively, any unitary operation U acting on k qubits
can be decomposed into O(4k) CNOT and one-qubit gates

Question: is there a finite set of gates that is universal?

Answer 1: strictly speaking, no, because this results in only
countably many quantum circuits, whereas there are
uncountably many unitary operations on k qubits (for any k)

27
Approximately universal gate sets
Answer 2: yes, for universality in an approximate sense
As an illustrative example, any rotation can be approximated
within any precision by repeatedly applying
cos( 2π)  sin( 2π)
R                   
 sin( 2π) cos( 2π) 
some number of times

In this sense, R is approximately universal for the set of
all one-qubit rotations: any rotation S can be approximated
within precision  by applying R a suitable number of times

It turns out that O((1/)c) times suffices (for a constant c)
28
Approximately universal gate sets
In three or more dimensions, the rate of convergence with
respect to  can be exponentially faster

1     0 
Theorem 2: the gates CNOT, H, and T  
 0 e iπ / 4 

are approximately universal, in that any
unitary operation on k qubits can be
simulated within precision  by applying
O(4k log c(1/)) of them (c is a constant)

[Solovay, 1996][Kitaev, 1997]
29
Complexity classes

30
Complexity classes
Recall:
• P (polynomial time): problems solved by O(nc)-size
classical circuits (decision problems and uniform circuit
families)
• BPP (bounded error probabilistic polynomial time):
problems solved by O(nc)-size probabilistic circuits that
err with probability  ¼
• BQP (bounded error quantum polynomial time):
problems solved by O(nc)-size quantum circuits that err
with probability  ¼
• PSPACE (polynomial space): problems solved by
algorithms that use O(nc) memory.                           31
Summary of previous containments
P  BPP  BQP  PSPACE  EXP
EXP
We now consider further
structure between P and                      PSPACE
PSPACE

Technically, we will restrict                  BQP
our attention to languages
(i.e. {0,1}-valued problems)                   BPP

Many problems of interest can                  P
be cast in terms of languages
For example, we could define
FACTORING = {(x,y) :  2  z  y, such that z divides x}   32
NP
Define NP (non-deterministic polynomial time) as
the class of languages whose positive instances have
―witnesses‖ that can be verified in polynomial time
Example: Let 3-CNF-SAT be the language consisting of all
3-CNF formulas that are satisfiable
3-CNF formula:
f x1 ,..., xn   x1  x3  x4   x2  x3  x5     x1  x5  xn 
f x1 ,..., xn  is satisfiable iff there exists b1 ,...,bn  0,1
such that f b1 ,...,bn   1

No sub-exponential-time algorithm is known for 3-CNF-SAT
But poly-time verifiable witnesses exist (namely, b1, ..., bn)         33
Other “logic” problems in NP
• k-DNF-SAT:
f x1 ,..., xn   x1  x3  x4   x2  x3  x5     x1  x5  xn 
 But, unlike with k-CNF-SAT, this one is known to be in P

• CIRCUIT-SAT:
1      Λ           Λ                              output
bit
0            Λ                  Λ      Λ

Λ      Λ           Λ            Λ
 All known
1
algorithms
1                 Λ             Λ                  exponential-
0             Λ            Λ            Λ           time
34
“Graph theory” problems in NP

• k-COLOR: does G have a k-coloring ?
• k-CLIQUE: does G have a clique of size k ?
• HAM-PATH: does G have a Hamiltonian path?
• EUL-PATH: does G have an Eulerian path?

35
“Arithmetic” problems in NP
• FACTORING = {(x, y) :  2  z  y, such that z divides x}

• SUBSET-SUM: given integers x1, x2 , ..., xn, y, do there exist
i1, i2 , ..., ik {1, 2,... , n} such that xi1+ xi2 + ... + xik = y?

• INTEGER-LINEAR-PROGRAMMING: linear programming
where one seeks an integer-valued solution (its existence)

36
P vs. NP
All of the aforementioned problems have the property that
they reduce to 3-CNF-SAT, in the sense that a polynomial-
time algorithm for 3-CNF-SAT can be converted into a poly-
time algorithm for the problem

Example:         algorithm for 3-COLOR

algorithm for
3-CNF-SAT

If a polynomial-time algorithm is discovered for 3-CNF-SAT
then a polynomial-time algorithm for 3-COLOR easily follows
In fact, this holds for any problem X  NP, hence 3-CNF-SAT
is NP-hard ...                                           37
P vs. NP
All of the aforementioned problems have the property that
they reduce to 3-CNF-SAT, in the sense that a polynomial-
time algorithm for 3-CNF-SAT can be converted into a poly-
time algorithm for the problem

Example:              algorithm for X

algorithm for
3-CNF-SAT

If a polynomial-time algorithm is discovered for 3-CNF-SAT
then a polynomial-time algorithm for 3-COLOR easily follows
In fact, this holds for any problem X  NP, hence 3-CNF-SAT
is NP-hard ... Also NP-hard: CIRCUIT-SAT, k-COLOR, ... 38
FACTORING vs. NP
Is FACTORING NP-hard too?
PSPACE
If so, then every problem in
NP is solvable by a poly-time
quantum algorithm!          3-CNF-SAT
NP            co-NP

But FACTORING has
not been shown to be
NP-hard                   FACTORING
P
Moreover, there is ―evidence‖
that it is not NP-hard:
FACTORING  NPco-NP

If FACTORING is NP-hard then NP = co-NP                       39
FACTORING vs. co-NP
FACTORING = {(x, y) :  2  z  y, s.t. z divides x}

co-NP: languages whose negative
PSPACE
instances have ―witnesses‖ that can
be verified in poly-time                     NP        co-NP

Question: what is a
good witness for the
negative instances?              FACTORING
P
p1, p2 , ..., pm of x will work

Can verify primality and compare
p1, p2 , ..., pm with y, all in poly-time
40
Introduction to
Quantum Information Processing
CS 467 / CS 667
Phys 667 / Phys 767
C&O 481 / C&O 681

Lecture 13 (2008)
Richard Cleve
DC 2117
cleve@cs.uwaterloo.ca
41
More state distinguishing
problems

42
More state distinguishing problems
Which of these states are distinguishable? Divide them into
equivalence classes:
1. 0 + 1                     5. 0       with prob. ½
0 + 1 with prob. ½
2. −0 − 1
6. 0         with prob. ¼
3. 0 with prob. ½
1 with prob. ½                 1         with prob. ¼
0 + 1   with prob. ¼
0 − 1   with prob. ¼
4. 0 + 1 with prob. ½
0 − 1 with prob. ½
7. The first qubit of 01 − 10

43
This is a probabilistic mixed state
Density matrix formalism

44
Density matrices (1)
Until now, we’ve represented quantum states as vectors
(e.g. ψ, and all such states are called pure states)

An alternative way of representing quantum states is in terms
of density matrices (a.k.a. density operators)

The density matrix of a pure state ψ is the matrix  = ψψ

Example: the density matrix of 0 + 1 is
 α 2 αβ  


α 

ρ    α β        2 
β        α β β 
          

45
Density matrices (2)
How do quantum operations work using density matrices?

Effect of a unitary operation on a density matrix:
applying U to  yields U U
†

(this is because the modified state is UψψU† )
Effect of a measurement on a density matrix:
measuring state  with respect to the basis 1, 2,..., d,
yields the k th outcome with probability k k

(this is because k k = kψψk = kψ2 )
—and the state collapses to kk                           46
Density matrices (3)
A probability distribution on pure states is called a mixed state:
( (ψ1, p1), (ψ2, p2), …, (ψd, pd))
The density matrix associated with such a mixed state is:
d
ρ   pk ψ k ψ k
k 1

Example: the density matrix for ((0, ½ ), (1, ½ )) is:
1 1 0 1 0 0 1 1 0
20 0   2 0 1   2 0 1 
                  

Question: what is the density matrix of
((0 + 1, ½ ), (0 − 1, ½ )) ?                          47
Density matrices (4)
How do quantum operations work for these mixed states?
Effect of a unitary operation on a density matrix:
applying U to  still yields U U
†

This is because the modified state is:
d
 d             t
pU ψ
k 1
k   k   ψ k U  U   pk ψ k ψ k U  UρU t
t

 k 1         
Effect of a measurement on a density matrix:
measuring state  with respect to the basis 1, 2,..., d,
still yields the k th outcome with probability k k
Why?                                                           48
Recap: density matrices
Quantum operations in terms of density matrices:
• Applying U to    yields U U†
• Measuring state  with respect to the basis 1, 2,..., d,
yields: k th outcome with probability k k
—and causes the state to collapse to kk

Since these are expressible in terms of density matrices alone
(independent of any specific probabilistic mixtures), states with
identical density matrices are operationally indistinguishable

49
problems …

50
State distinguishing problems (1)
The density matrix of the mixed state          d
((ψ1, p1), (ψ2, p2), …,(ψd, pd)) is: ρ   pk ψ k ψ k
k 1
Examples (from beginning of lecture):
1 1 1
1. & 2. 0 + 1 and −0 − 1 both have ρ  
2 1 1

3. 0 with prob. ½
1 with prob. ½
4. 0 + 1 with prob. ½
0 − 1 with prob. ½        1 1 0
ρ 
2 0 1 

6. 0         with prob. ¼
1         with prob. ¼
0 + 1   with prob. ¼
0 − 1   with prob. ¼                                   51
State distinguishing problems (2)
Examples (continued):

5. 0       with prob. ½
0 + 1 with prob. ½

1 1 0 1 1 / 2 1 / 2 3 / 4 1 / 2
has:          2 1 / 2 1 / 2  1 / 2 1 / 4
2 0 0                              

7. The first qubit of 01 − 10 ...? (later)

52
Characterizing density matrices
Three properties of  :
d
ρ   pk ψ k ψ k
• Tr = 1 (Tr M = M11 + M22 + ... + Mdd )
k 1
•  =† (i.e.  is Hermitian)
•    0, for all states 

Moreover, for any matrix  satisfying the above properties,
there exists a probabilistic mixture whose density matrix is 

Exercise: show this

53
Taxonomy of various
normal matrices

54
Normal matrices
Definition: A matrix M is normal if M†M = MM†
Theorem: M is normal iff there exists a unitary U such that
M = U†DU, where D is diagonal (i.e. unitarily diagonalizable)
 λ1       0    0
0     λ2       0
D                   
             
                   
0         0    λd 
Examples of abnormal matrices:
eigenvectors:
1   1 is not even          1 1 is diagonalizable,
0   1 diagonalizable       0 2 but not unitarily
                                                           55
Unitary and Hermitian matrices
Normal:       λ1   0     0     with respect to some
0     λ2    0     orthonormal basis
M                 
          
                
0     0     λd 

Unitary: M†M = I which implies |k |2 = 1, for all k

Hermitian: M = M† which implies k  R, for all k

Question: which matrices are both unitary and Hermitian?

Answer: reflections (k  {+1,1}, for all k)
56
Positive semidefinite
Positive semidefinite: Hermitian and k  0, for all k

Theorem: M is positive semidefinite iff M is Hermitian and,
for all , M  0

(Positive definite: k > 0, for all k)

57
Projectors and density matrices
Projector: Hermitian and M 2 = M, which implies that M is
positive semidefinite and k  {0,1}, for all k
d
Density matrix: positive semidefinite and Tr M =1, so     λ
k 1
k   1

Question: which matrices are both projectors and density
matrices?

Answer: rank-1 projectors (k = 1 if k = j; otherwise k = 0)

58
Taxonomy of normal matrices
normal

unitary            Hermitian

positive
reflection           semidefinite

density
projector
matrix

rank one
projector             59
Introduction to
Quantum Information Processing
CS 467 / CS 667
Phys 667 / Phys 767
C&O 481 / C&O 681

Lecture 14 (2008)
Richard Cleve
DC 2117
cleve@cs.uwaterloo.ca
60
Bloch sphere for qubits

61
Bloch sphere for qubits (1)
Consider the set of all 2x2 density matrices 

They have a nice representation in terms of the Pauli matrices:

0 1                 1 0                  0  i 
σx  X              σz  Z               σy  Y  
1 0
                 0  1
                 i 0 
Note that these matrices—combined with I—form a basis for
the vector space of all 2x2 matrices

We will express density matrices    in this basis

Note that the coefficient of I is ½, since X, Y, Y are traceless
62
Bloch sphere for qubits (2)
I  c x X  c yY  c z Z
We will express ρ 
2
First consider the case of pure states , where, without
loss of generality,  = cos()0 + e2isin()1 (,   R)

 cos 2θ     e i 2φcosθsin θ  1 1  cos2θ  e i 2φsin 2θ 
ρ   i 2φ                           i 2φ                        
e cosθsin θ        sin θ      2 e sin 2θ  1  cos2θ  
2

Therefore cz = cos(2), cx = cos(2)sin(2), cy = sin(2)sin(2)
These are polar coordinates of a unit vector (cx , cy , cz)  R3
63
Bloch sphere for qubits (3)
0
+ = 0 +1
–i         – = 0 – 1
+i = 0 + i1
+                       –   –i = 0 – i1
+i

1
Note that orthogonal corresponds to antipodal here

Pure states are on the surface, and mixed states are inside
(being weighted averages of pure states)
64
Basic properties of the trace

65
Basic properties of the trace
d
The trace of a square matrix is defined as Tr M    M
k 1
k ,k

It is easy to check that
Tr M  N   Tr M  Tr N and Tr M N   Tr N M 

         λ
d
The second property implies Tr M   Tr U MU 1
k
k 1

Calculation maneuvers worth remembering are:
Tr a b M   b M a                           
and Tr a b c d  b c d a

Also, keep in mind that, in general, Tr M N   Tr M Tr N

66
Partial trace (1)
Two quantum registers (e.g. two qubits) in states  and 
(respectively) are independent if then the combined system
is in state  =  
In such circumstances, if the second register (say) is discarded
then the state of the first register remains 
In general, the state of a two-register system may not be of the
form   (it may contain entanglement or correlations)
We can define the partial trace, Tr2  , as the unique linear
operator satisfying the identity Tr2( ) =          index means
2nd system
For example, it turns out that                                           traced out
Tr2(                                                 )    1 1 0
00         11          00                   = 
2 0 1 
1           1            1           1
11
2           2            2           2
          67
Partial trace (2)
Example: discarding the second of two qubits
 1 0 0 0                  0 1 0 0
Let A0 = I0           and A1 = I1  0 0 0 1
0 0 1 0                          
For the resulting quantum operation, state   becomes 

For d-dimensional registers, the operators are Ak = Ik ,
where 0, 1, …, d1 are an orthonormal basis

68
Partial trace (3)
For 2-qubit systems, the partial trace is explicitly
 ρ00,00   ρ00,01   ρ00,10   ρ00,11 
ρ         ρ01,01   ρ01,10   ρ01,11   ρ00,00  ρ01,01   ρ00,10  ρ01,11 
Tr2  01,00                             
 ρ10,00   ρ10,01   ρ10,10   ρ10,11   ρ10,00  ρ11,01
                  ρ10,10  ρ11,11 

                                   
 ρ11,00   ρ11,01   ρ11,10   ρ11,11 
and
 ρ00,00   ρ00,01   ρ00,10   ρ00,11 
ρ         ρ01,01   ρ01,10   ρ01,11   ρ00,00  ρ10,10   ρ00,01  ρ10,11 
Tr1  01,00                             
 ρ10,00   ρ10,01   ρ10,10   ρ10,11   ρ01,00  ρ11,10
                  ρ01,01  ρ11,11 

                                   
 ρ11,00   ρ11,01   ρ11,10   ρ11,11 
69
POVMs
(Positive Operator Valued Measurements)

70
POVMs (1)
Positive operator valued measurement (POVM):
m
Let A1, A2 , …, Am be matrices satisfying  j
At Aj  I
j 1
Then the corresponding POVM is a stochastic operation on 
       
that, with probability Tr Aj ρ At produces the outcome:
j

j (classical information)
A j ρ At
j        (the collapsed quantum state)

Tr A j ρ At
j   
Example 1: Aj = jj (orthogonal projectors)

This reduces to our previously defined measurements …   71
POVMs (2)
When Aj = jj are orthogonal projectors and  = ,

       
Tr Aj ρ At = Trjjjj
j
= jjjj
= j2

A j ρ At               φj φj ψ ψ φj φj
Moreover,                                                 φj φj
j


Tr A j ρ A   t
j               φj ψ
2

72
POVMs (3)
Example 3 (trine state “measurent”):

Let 0 = 0, 1 = 1/20 + 3/21, 2 = 1/20  3/21
2 1 0
Define A0 =2/300    
3 0 0 
    
1  2 3  2                        1  2 3  2
A1=2/311                  A2=2/322           
4  2    6                        4  2    6
t        t        t
Then   A0 A0  A1 A1  A2 A2  I

If the input itself is an unknown trine state, kk, then the
probability that classical outcome is k is 2/3 = 0.6666…
73
General quantum
operations

74
General quantum operations (1)
General quantum operations (a.k.a. “completely positive
trace preserving maps”, “admissible operations” ):
m
Let A1, A2 , …, Am be matrices satisfying    Atj A j  I
j 1

m

Then the mapping    Aj  Atj is a general quantum op
j 1

Example 1 (unitary op): applying U to        yields U U   †

75
General quantum operations (2)
Example 2 (decoherence): let A0 = 00 and A1 = 11

This quantum op maps  to 0000 + 1111

 α 2 αβ    α 2   0 
For ψ =   0 + 1,         2          2
α β β   0
                  β 

Corresponds to measuring  ―without looking at the outcome‖

After looking at the outcome,  becomes 00 with prob. ||2
11 with prob. ||2
76
General quantum operations (3)
Example 4 (discarding the second of two qubits):
 1 0 0 0                  0 1 0 0
Let A0 = I0           and A1 = I1  0 0 0 1
0 0 1 0                          

State     becomes 

State      1
2
00    1
2
 
11     1
2
00    1
2
11             1 1 0
becomes 
2 0 1 


Note 1: it’s the same density matrix as for ((½ , 0), (½ , 1))
Note 2: the operation is the partial trace Tr2 
77
Distinguishing mixed states

78
Distinguishing mixed states (1)
What’s the best distinguishing strategy between these two
mixed states?
0       with prob. ½                           0 with prob. ½
0 + 1 with prob. ½                           1 with prob. ½
3 / 4 1 / 2                                    1 1 0
ρ1                                              ρ2  
1 / 2 1 / 4
                                    2 0 1 

1
+
1 also arises from this                0
orthogonal mixture:                            … as does 2 from:
0
0 with prob. cos2(/8)                        0 with prob. ½
1 with prob. sin2(/8)                        1 with prob. ½
79
Distinguishing mixed states (2)
We’ve effectively found an orthonormal basis 0, 1 in
which both density matrices are diagonal:
1   1
cos 2 π / 8       0             1 1 0                    +

ρ2                                  
ρ1  
      0

sin 2 π / 8       2 0 1 
                         0

Rotating 0, 1 to 0, 1 the scenario can now
0
be examined using classical probability theory:
Distinguish between two classical coins, whose probabilities
of ―heads‖ are cos2(/8) and ½ respectively (details: exercise)

Question: what do we do if we aren’t so lucky to get two
density matrices that are simultaneously diagonalizable?
80
Simulations among operations

81
Simulations among operations (1)
Fact 1: any general quantum operation can be simulated
by applying a unitary operation on a larger quantum system:

input                    output

0
U
0
0

Example: decoherence
α 2   0 
0 + 1              ρ       2
0
      β 

0
82
Simulations among operations (2)
Fact 2: any POVM can also be simulated by applying a unitary
operation on a larger quantum system and then measuring:

input                        quantum output

0
U        j   classical output
0
0

83
Separable states

84
Separable states
A bipartite (i.e. two register) state  is a:
• product state if    = 
m
• separable state if        p
j 1
j   j    j   ( p1 ,…, pm  0)
(i.e. a probabilistic mixture
of product states)
Question: which of the following states are separable?
ρ1  1  00  11  00  11 
2

ρ2  1  00  11  00  11   1  00  11  00  11 
2                          2

85

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