Download Lecture 12

Introduction to
Quantum Information Processing
CS 467 / CS 667
Phys 467 / Phys 767
C&O 481 / C&O 681
Lecture 12 (2005)
Richard Cleve
DC 3524
[email protected]
Course web site at:
http://www.cs.uwaterloo.ca/~cleve/courses/cs467
1
Contents
• Correction to Lecture 11: 2/3  2/3
• Distinguishing mixed states
• Basic properties of the trace
• Recap: general quantum operations
• Partial trace
• POVMs
• Simulations among operations
• Separable states
• Continuous-time evolution
2
• Correction to Lecture 11: 2/3  2/3
• Distinguishing mixed states
• Basic properties of the trace
• Recap: general quantum operations
• Partial trace
• POVMs
• Simulations among operations
• Separable states
• Continuous-time evolution
3
Correction to Lecture 11: 2/3 instead of 2/3
General quantum operations (III)
Example 3 (trine state “measurent”):
Let 0 = 0, 1 = 1/20 + 3/21, 2 = 1/20  3/21
Define A0 =2/300
2 1 0

3 0 0
1  2 3  2
A1=2/311  

4  2
6
Then A0
t
t
1  2 3  2
A2=2/322  

4  2
6
t
A0  A1 A1  A2 A2  I
The probability that state k results in “outcome” Ak is 2/3,
and this can be adapted to actually yield the value of k with
this success probability
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• Correction to Lecture 11: 2/3  2/3
• Distinguishing mixed states
• Basic properties of the trace
• Recap: general quantum operations
• Partial trace
• POVMs
• Simulations among operations
• Separable states
• Continuous-time evolution
5
Distinguishing mixed states (I)
What’s the best distinguishing strategy between these two
mixed states?
0
with prob. ½
0 + 1 with prob. ½
3 / 4 1 / 2
ρ1  

1
/
2
1
/
4


0 with prob. ½
1 with prob. ½
1
1 also arises from this
orthogonal mixture:
1 1 0
ρ2  
2 0 1
+
0
… as does 2 from:
0
0 with prob. cos2(/8)
1 with prob. sin2(/8)
0 with prob. ½
1 with prob. ½
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Distinguishing mixed states (II)
We’ve effectively found an orthonormal basis 0, 1 in
which both density matrices are diagonal:
cos 2 π / 8

0
ρ2  

2


0
sin
π
/
8


1 1 0
ρ1  
2 0 1
1
1
Rotating 0, 1 to 0, 1 the scenario can now
be examined using classical probability theory:
+
0
0
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?
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• Correction to Lecture 11: 2/3  2/3
• Distinguishing mixed states
• Basic properties of the trace
• Recap: general quantum operations
• Partial trace
• POVMs
• Simulations among operations
• Separable states
• Continuous-time evolution
8
Basic properties of the trace
The trace of a square matrix is defined as Tr M 
d
M
k 1
It is easy to check that
k ,k
Tr M  N   Tr M  Tr N and Tr M N   Tr N M 

 λ
The second property implies Tr M   Tr U 1MU 
d
k 1
k
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
9
• Correction to Lecture 11: 2/3  2/3
• Distinguishing mixed states
• Basic properties of the trace
• Recap: general quantum operations
• Partial trace
• POVMs
• Simulations among operations
• Separable states
• Continuous-time evolution
10
Recap: general quantum ops
General quantum operations (a.k.a. “completely positive
trace preserving maps” ):
m
Let A1, A2 , …, Am be matrices satisfying
t
A
 j Aj  I
j 1
m
t
ρ

A
ρ
A
Then the mapping
 j j is a general quantum op
j 1
Example: applying U to
 yields U U
†
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• Correction to Lecture 11: 2/3  2/3
• Distinguishing mixed states
• Basic properties of the trace
• Recap: general quantum operations
• Partial trace
• POVMs
• Simulations among operations
• Separable states
• Continuous-time evolution
12
Partial trace (I)
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
For example, it turns out that
Tr2( 
1
2
00 
1
2
 
11 
1
2
00 
1
2
11
)
1 1 0
= 
2 0 1
2nd system
traced out
13
Partial trace (II)
We’ve already seen this defined in the case of 2-qubit systems:
discarding the second of two qubits
 1 0 0 0
0 1 0 0
Let A0 = I0  
and
A
=
I

1

1


0
0
1
0
0
0
0
1




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
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Partial trace (III)
For 2-qubit systems, the partial trace is explicitly
 ρ00,00
ρ
Tr2  01,00
 ρ10,00

 ρ11,00
ρ00,01
ρ01,01
ρ10,01
ρ11,01
ρ00,10
ρ01,10
ρ10,10
ρ11,10
ρ00,11 
ρ01,11   ρ00,00  ρ01,01

ρ10,11   ρ10,00  ρ11,01

ρ11,11 
ρ00,10  ρ01,11 
ρ10,10  ρ11,11 
ρ00,01
ρ01,01
ρ10,01
ρ11,01
ρ00,10
ρ01,10
ρ10,10
ρ11,10
ρ00,11 
ρ01,11   ρ00,00  ρ10,10

ρ10,11   ρ01,00  ρ11,10

ρ11,11 
ρ00,01  ρ10,11 
ρ01,01  ρ11,11 
and
 ρ00,00
ρ
Tr1  01,00
 ρ10,00

 ρ11,00
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• Correction to Lecture 11: 2/3  2/3
• Distinguishing mixed states
• Basic properties of the trace
• Recap: general quantum operations
• Partial trace
• POVMs
• Simulations among operations
• Separable states
• Continuous-time evolution
16
POVMs (I)
Positive operator valued measurement (POVM):
m
t
A
Let A1, A2 , …, Am be matrices satisfying  j A j  I
j 1
Then the corresponding POVM is a stochastic operation on 
that, with probability Tr Aj ρ Atj produces the outcome:


j (classical information)
A j ρ Atj

Tr A j ρ Atj

(the collapsed quantum state)
Example 1: Aj = jj (orthogonal projectors)
This reduces to our previously defined measurements …
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POVMs (II)
When Aj = jj are orthogonal projectors and  = ,


Tr Aj ρ Atj = Trjjjj
= jjjj
= j2
Moreover,
A j ρ Atj

Tr A j ρ A
t
j


φj φj ψ ψ φj φj
φj ψ
2
 φj φj
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POVMs (III)
Example 3 (trine state “measurent”):
Let 0 = 0, 1 = 1/20 + 3/21, 2 = 1/20  3/21
Define A0 =2/300
2 1 0

3 0 0
1  2 3  2
A1=2/311  

4  2
6
Then
t
t
1  2 3  2
A2=2/322  

4  2
6
t
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…
19
• Correction to Lecture 11: 2/3  2/3
• Distinguishing mixed states
• Basic properties of the trace
• Recap: general quantum operations
• Partial trace
• POVMs
• Simulations among operations
• Separable states
• Continuous-time evolution
20
Simulations among operations (I)
Fact 1: any general quantum operation can be simulated
by applying a unitary operation on a larger quantum system:
input


0
0
0
Example: decoherence
0 + 1
0
output
U
discard
α 2
ρ
 0
0 
2
β 
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Simulations among operations (II)
Fact 2: any POVM can also be simulated by applying a unitary
operation on a larger quantum system and then measuring:
input

0
0
0
U

quantum output
j
classical output
22
• Correction to Lecture 11: 2/3  2/3
• Distinguishing mixed states
• Basic properties of the trace
• Recap: general quantum operations
• Partial trace
• POVMs
• Simulations among operations
• Separable states
• Continuous-time evolution
23
Separable states
A bipartite (i.e. two register) state  is a:
• product state if
 = 
• separable state if  
m
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  12  00  11  00  11 
ρ2  12  00  11  00  11   12  00  11  00  11 
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• Correction to Lecture 11: 2/3  2/3
• Distinguishing mixed states
• Basic properties of the trace
• Recap: general quantum operations
• Partial trace
• POVMs
• Simulations among operations
• Separable states
• Continuous-time evolution
25
Continuous-time evolution
Although we’ve expressed quantum operations in discrete
terms, in real physical systems, the evolution is continuous
Let H be any Hermitian matrix and t  R
1
Then eiHt is unitary – why?
 λ1



†
D


H = U DU, where



λd 
eiλ1t


t
† iDt
iHt

Therefore e = U e U = U 
U
iλd t 

e


0
(unitary)
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