Entanglement for Pedestrians

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Transcript Entanglement for Pedestrians

Introduction to
Entanglement
Allan Solomon, Paris VI
Mathematics Prelude

Paradigm: Quantum mechanics & Maths
Quantum Field
Theory
For example,
Feynman Diagrams
involve
Analysis
Algebra
Riemann Zeta fns Hopf Algebra
and extensions
Braid Groups
Combinatorics
Topology
Graph Theory
Knot Theory
Counting
Mathematics Prelude


Entanglement
(Fr.) Intrication
(Eng.) Intricate = Complexity
involves Analysis, Algebra ,
Topology,Combinatorics
Borromean Rings – example of an entangled system
Physics Prelude: EPR paradox
Source emits spin singlet
Measurement by Alice on |y> determines Bob’s measurement.
The electrons are entangled.
“Entanglement is the
characteristic trait of
quantum mechanics”
Erwin Schroedinger, 1935.
“Interference (in Classical and
Quantum Mechanics) is just
the fact that the sum of
squares is not the square of
the sum.”
Richard Feynman
1 Vectors, Vector Spaces
A basic operation for vectors is addition.
For mathematicians therefore, vector
addition presents no surprises. For
physicists, vector addition is such a
remarkable property that in quantum
mechanics the phenomena it gives rise to
it go by many names, superposition rule,
interference, entanglement,…
Vector notation
Maths Notation
Physics Notation
1
0 
  | 0 , | 1 ,  | 0    | 1 
e1    e2    e1  e2   
0 
1
 
e1*  1 0 e2*  0 1
(e1 , e2 )  e1*e2
0 
 1 0 
1 
1
0 1 
e e   0 1  

0
0
0
 


*
1 2
 0 |  1|
 0 |1 
|0><1|
Dirac
2 Bipartite Spaces
V  V1  V2
Maths Notation
e1  e2
0 
1 
1
0
     
 0   1    0 
   
0 
 
Physics Notation
| 0 |1 
| 0,1 
If V1 has basis {ei} and V2 has basis {fj} then
{ei fj, i=1..m,j=1..n} is a basis for V1V2
Every vector in V1V2 is a sum of products; but not
every vector is a product. If it is a product, then it is
said to be non-entangled.
Bipartite Spaces: Entanglement

Example:
1
e1   
0 
Not
entangled
Entangled
0 
e2   
1
e1  e2  e2  e2  (e1  e2 )  e2
e1  e2  e2  e1
“entangled” means not factorizable
0 
1
 
1
 
0 
0 
1
 
0 
 
1
3 States, Pure and Mixed
(a) Pure States
Vectors correspond to Pure states:
example
 
1
0 
| y  i | i 
     0   1
 
 
 
We may equally represent a Pure State |y by the Operator
(Projector) |yy| which projects onto that state:

  *
P   
 

 *  * 
  *
*
 
 
*
P2  P
NOTE: trace P=1 (Normalization) and
P is Hermitian with (semi-)positive eigenvalues (0 and 1 ).
States
(b) Mixed States
We define a (mixed) state r as a positive matrix of trace 1
Note: A mixed state is a mixture of pure states.
1

2

UrU  



.




n 
i i

1


2

U
r U
.



n 

 1
0

 
 1

0
 
  2 
 U  1
.



 

0

 
 i i Pi
 1 i  0
.
0

 0

  ...  n 



0

which is a (convex) sum of pure states.


 U
.


1 
Mixed state is not a unique sum of pure states

Example:
0 
1 
| a   
0 
0 
 
1 
0 
1  
| b 
2 0 
1 
 
r   | a  a |   | b  b |
| a1   | a    | b 
| b1   | a    | b 
1
2
1
2
r  | a1  a1 |  | b1  b1 |
(    1)
4 Entropy of a State (Von Neumann Entropy)
State r has eigenvalues i
( r )  tr ( r log r )
  i (i log i )
usually log2
Claude Shannon
(E is entropy here!)
Example(a): Pure state
Every Pure State has Entropy Zero.
Example(b): Mixed state
1 0
r

0 0 
( r )  1 log 1  0 log 0  0
1 0 

 0 2 
( r )  1 log 1  2 log 2
r
( r )   log   (1   ) log(1   ) 0  ( r )  1
 12 0
(  1  )   12 log 12  (1  12 ) log(1  12 )  1
0 2 
Maximum entropy 1 for maximally random state.
John Von
Neumann
5 Measures of Entanglement
Intuitively we expect
Entangled?
E Measure
(1) (Pure) state
(1/2)(|0,0> + |0,1>)
No
0
(2) Bell state
(1/2)(|0,0> + |1,1>)
Yes
1
(3)  |0,0> +  (1- ) |1,1>)
Yes
0 .?. 1
It turns out that the (VN) Entropy gives a measure of
entanglement for pure states;
but not directly, as all pure states have entropy zero.
We must first take the Partial Trace
over one subsystem of the bipartite system.
6 Partial Trace
If V = VA  VB then trB(QA  QB)=QA tr(QB)
Extend to sums by linearity.
Pure States: QA =|u1><u2|
QB =|v1><v2|
then trB (QA  QB)= |u1><u2| <v2| v1>
Example (non-entangled state):
| a 
1
(| 0,0   | 0,1 )
2
1 / 2 1 / 2
1 / 2 1 / 2
1
r a  (| 0,0   | 0,1 )( 0,0 |   0,1 |)  
 0
0
2

0
 0
TrB ( r a ) 
0
0
0
0
0
0
0

0
1 0
1
(| 0  0 |  | 0  0 |)  

2
0 0 
Entanglement (Entropy of partially traced state) is 0.
Example: Bell state
|b>=(1/||
rb=(1/|| ||
TrBrb=(1/||  ||

( r b ) 
1 / 2 0 
 0 1 / 2


1
1
1
1
 log( )  log( )
2
2
2
2
John Stewart Bell
1
Example (Entangled state):
1
| b 
(| 0,0   | 0,1   | 1,0 )
3
1
rb  (| 0,0   | 0,1   | 1,0 )( 0,0 |   0,1 |   1,0 |
3
1
TrB ( rb )  (| 0  0 |  | 0  1 |  | 0  0 |  | 1  0 |  | 1  1 |)
3
2 / 3 1 / 3


1 / 3 1 / 3
Entropy of partially traced state is non-zero (=.55)
Entangled state (intermediate)
Pure state
2

 cos(  )


0



0




sin(  ) cos(  )
cos()|0,0>+sin()|1,1> =

0 0 sin(  ) cos(  )


0 0
0



0 0
0



2 
0 0
sin(  )

So this measure of entanglement
gives an intuitively correct variation
from 0 (non-entangled)
to maximum of 1 (Bell state) for
PURE States.
cos(  )




 0 




 0 




 sin(  ) 

Recap: Definitions





A pure state may be represented by a vector or a
positive matrix (Projection matrix) with eigenvalues 1,
0, … 0, 0, 0
A mixed state is a (convex) sum of pure states and may
be represented by a positive matrix of trace 1.
A pure state is separable (non-entangled) if it can be
written as a product of vectors (factorizable).
A mixed state is separable if it can be written as a
(convex) sum of separable (factorizable) pure states.
The expression of a mixed state as a convex sum of pure
states is not unique.
Measure of Entanglement


The entanglement y of a pure bipartite
state yVAVB is given by the Entropy of
the Partial Trace of y
The entanglement r of a mixed
bipartite state rVAVB is given by
 (r)=min{Siyi | r Siyi}

This definition of entanglement measure for Mixed States is very
difficult to apply, requiring infinite tests.

Example (revisited)
| a | 0,1 
| b 
1
(| 0,0 
2
r   | a  a |   | b  b |
  3 / 4,   1 / 4
 | 1,1 
(    1)
| a1   | a    | b 
| b1   | a    | b 
r1  2 | a1  a1 |  2 | b1  b1 |
1
E(r3/4 /4.5
1
E(r/.8 /.8.8
(this IS the min and therefore the entanglement )
7 Concurrence (Physics viewpoint)

Spin Flip Operation
| y   y | y *   | y~ 
qubit
 |   * |
2-qubit
 |   * | |    y   y | *   |  
~
~
Concurrence C = |  |  |
Factorizable state
~
flip
| 
 |  |  
Bell state
|  | 
 |  |  |  
flip
C=0
C=1
So Concurrence gives a measure of entanglement
for PURE states
Concurrence (Maths viewpoint)
Partial trace of Pure state r gives 2X2 matrix so
Entanglement determined by eigenvalue equation
D
±4D/
Recall
( r )   log   (1   ) log(1   )
Concurrence C2=4D
C varies from 0 to 1 so Concurrence gives a measure
of entanglement for PURE states
~
~
Equivalently (for pure states) C2=tr rr~    |    |  
Wootters’ Concurrence

Wootters[1,2] has shown that the form for C
C ( r )  max{0, 1  2  3  4 }
where the ’s are the square roots of the eigenvalues of
in descending order
rr~
Gives the entanglement for mixed states;
i.e. it gives the correct minimum over Pure States.
(Note the formula coincides with the previous for PURE states.)
[1] Hill, S and Wootters, WK, PhysRevLett 78,26,5022(1997)
[2] Wootters, WK, PhysRevLett 80,10,2245(1998)
Feynman Nobel Lecture


The Development of the Space-Time
View of Quantum Electrodynamics
“We have a habit in writing articles
published in scientific journals to make the
work as finished as possible, to cover all
the tracks, to not worry about the blind
alleys or to describe how you had the
wrong idea first, and so on.”
Tripartite entanglement



“Naïve Solution”
Extend Concurrence to 3-subspace Pure
states by summing over (3) Partial Traces
Example
1 / 3 (| 001   | 010   | 100 ) 
(1 / 3) (| 001   | 010   | 100 ) ( 001 |   010 |   100 |)
Three Partial Traces are equal,
each with Concurrence 2/3 ,
leading to a 3-concurrence of
(1/3) (2/3+2/3+2/3)=2/3

















1
3
0
0
0
0
0
1
1
3
3
1
1
3
3
0
0

0




0




0



0
Tripartite states
However ……….
The entangled state (/|000>+|111>)
 1



0
0
0
 2

has 3 equal partial traces



which is separable (concurrence =0).

 0


 0



 0


0
0
0
0
0
0

0 

0 

1 

2 
Borromean Rings analogy –
every cut leaves a separable system
A funny “Resource”
Eugene Wigner



Physicists do experiments on the principle
that they can be replicated in other
laboratories – invariance under
transformations.
In Quantum mechanics, we expect our
measurable quantities to be invariant
under Unitary Transformations (or antiunitary – Wigner)
This is NOT the case for Entanglement!
Open Problems



What is the significance of Entanglement
for Quantum Computing?
Find a measure of Entanglement for 3 (or
more) qubits (tripartite spaces,..).
Interaction with the Environment
(Dissipation of Entanglement)