Download Atomic entangled states

Quantum information Theory:
Separability and distillability
J. Ignacio Cirac
Institute for Theoretical Physics
University of Innsbruck
KIAS, November 2001
SFB Coherent Control
€U TMR
Entangled states
Superposition principle in Quantum Mechanics:
If the systems can be in
j0i
or
j1i
then they can also be in
Two or more systems:
A
B
c
j
0
i
+
c
j
1
i
0
1
entangled states
If the systems can be in
j
0
iA
j
0
iB
or
j
1
iA
j
1
iB
j
0
i
j
0
i
+
c
j
1
i
j
1
i
then they can also be in c
0
A
B
1
A
B
Entangled states possess non-local (quantum) correlations:
A
B
The outcomes of measurements in A and B are correlated.
In order to explain these correlations classically (with a
realistic theory), we must have non-locality.
Fundamental implications: Bell´s theorem.
Applications
Secret communication.
Alice
Bob
j
Á
i
=
j
0
;
1
i
+
j
1
;
0
i
1. Check that particles are indeed entangled.
Correlations in all directions.
No eavesdropper present
2. Measure in A and B (z direction):
Alice
Bob
0
1
1
1
0
0
1
1
1
0
Send secret messages
Given an entangled pair, secure secret communication is possible
Computation.
ouput
quantum
processor
input
jªouti
j
ª
i
=
U
j
ª
i
o
u
t
i
n
jªini
A quantum computer can perform
ceratin tasks more efficiently
A quantum computer can do the same as a classical computer ... and more
- Factorization (Shor).
- Database search (Grover).
- Quantum simulations.
Precission measurements:
We can measure more precisely
Efficient communication:
1
Alice
101
1
Alice
101
+
0 0
1
Bob
0 0
1 Bob
Entangled state
We can use less resources
Problem: Decoherence
environment
A
B
The systems get entangled with the environment.
j
©
i
j
E
i
!
j
ª
i
A
B
E
A
B
E
Reduced density operator:
½
=
t
r
(
j
ª
i
h
ª
j
)
=
6
j
©
i
h
©
j
E
A
B
Solution: Entanglement distillation
½
Idea:
½
½
...
local operation
½
environment
(classical communication)
Distillation:
j
0
;
1
i
+
j
1
;
0
i
local operation
...
Fundamental problems in Quantum Infomation:
Separability and distillability
SEPARABILITY
A
½
B
DISTILLABILITY
½
½
...
½
j
©
i
=
j
0
;
1
i
+
j
1
;
0
i
½
Are these systems entangled?
Can we distill these systems?
Additional motivations: Experiments
Separability:
Ion traps
Cavity QED
Optical lattices
Magnetic traps
Atomic ensembles
Distillability: quantum communication.
jÁi
Long distance Q. communication?
NMR
Quantum dots
Josephson junctions
Basic properties:
Th. Physics
Mathematics
Separability
This talk
Distillability
Quantum Information
Algorithms, etc:
Computer Science
Physical implementations:
Th. Physics
Exp. Physics
Q. Optics
Condensed Matter
NMR
Outline
Separability.
Distillability.
Gaussian states.
Separability.
Distillability.
Multipartite case:
1. Separability
½
Are these systems entangled?
1.1 Pure states
Product states are those that can be written as:
j
©
i
=
j
a
i
j
b
i
Otherwise, they are entangled.
Examples:
Product state:
j
0
iA
j
0
iB
j
0
i
j
0
i
+
c
j
1
i
j
1
i
Entangled state: c
0
A
B
1
A
B
Entangled states cannot be created by local operations.
j
a
i
j
b
i
!
j
a
(
t
)
i
j
b
(
t
)
i
1.2 Mixed states
½
In order to create an entangled state, one needs interactions.
Separable states are those that can be prepared by LOCC out of a product state.
Otherwise, they are entangled.
X
¸0
=
p
j
a
i
h
a
j
j
b
i
h
b
jwhere p
k
A state is separable iff ½
k
k
k
k
k
(Werner 89)
k
Problem: given ½, there are infinitely many decompositions
X
½=
¸ kjª k i hª k j
ª
j
ª
i
=
±
spectral decomposition h
k
j
k
;
j
qk j©k i h©kj
©
j
©
i
=
6
±
k
j
k
;
j
need not be orthogonal h
k
X
=
k
= :::
X
½=
pkjak i hak j - jbk i hbkj
k
22
=
C
C
Example: two qubits ( H
)
00
01
10
11
0
1
7122
1B
1322C
B
C
½
= @
2
0 2231A
2217
1
2
½= (j0; 0i h0; 0j + j1; 1i h1; 1j) + j+ ; + i h+ ; + j
5
5
1
+ (j0; ¡ i h0; ¡ j + j1; ¡ i h1; ¡ j)
10
where
1
p
j
§
i=
(
j
0
i§
j
1
i
)
2
1.3 Separability: positive maps
A linear map L
:A
(
H
) A
(
H
)is called positive
½
¸0
!L
(
½
)
¸0
½
½
A
B
state
Extensions
state
A
A
½
B
B
state
?
: need not be positive, in general
0
!
(
L
1
)
(
½
)
¸
0
A postive map is completely positive if: ½
A
B̧
A
B
½ is separable iff (
for all positive maps
L
1
)
(
½
)
¸
0
(Horodecki 96)
However, we do not know how to construct all positive maps.
Example: Any physical action.
½
¸0
L
(
½
)
¸0
A
A
½
B
B
state
state
Any physical action can be described in terms of a completely positive map.
Example: transposition (in a given basis)
X
½
= ½
iih
jj
i
;
jj
i
;
j
1
5
X
T
(
½
)
= ½
j
j
i
h
i
j
i
;
j
µ
2
1¡ i
1+ i
3
;
¶
1
5
!
µ
2
1+ i
1¡ i
3
¶
i
;
j
Is positive
A
A
B
B
Extension: partial transposition.
X
X
Is called
½
=
½
j
i
;
j
i
h
k
;
l
j
(
T
1
)
(
½
)
=
½
j
k
;
j
i
h
i
;
l
j
,
then
i
j
k
l
i
j
k
l
0
1
B
B
@
C
C
A
transposes
the blocks
partial transposition
Example:
0
1
B
1 B0
0
2@
1
0
0
0
0
0
0
0
0
1
0
1
1 0
C
B
0 C 1 B0 0
!
0 A 2@
0 1
1
0 0
Partial transposition is positive, but not completely positive.
0
1
0
0
1
0
0C
C
0A
1
What is known?
PPT
In general
NPT
- Low rank
- Necessary or sufficient
conditions
?
2x2 and 2x3
Gaussian states
(Horodecki and Peres 96)
PPT
(Horodecki 97)
NPT
(Giedke, Kraus, Lewenstein, Cirac, 2001)
PPT
NPT
2. Distillability
½
½
...
½
j
©
i
=
j
0
;
1
i
+
j
1
;
0
i
Can we distill MES using LOCC?
PPT states cannot be distilled. Thus, there are bound entangled states.
(Horodecki 97)
There seems to be NPT states that cannot be distilled.
(DiVincezo et al, Dur et al, 2000)
2.1 NPT states
(IBM, Innsbruck 99)
We just have to concentrate on states with non-positive partial transposition.
T
A
¸ 0then there exists A and B, such that
Idea: If ½
·
Z
Ţ
A
y
y
½
=
(
A
B
)
½
(
A
B
)
with ~
d
¹
(
U
U
)
½
~
(
U
U
)¸
0
U
Physically, this means that
½
A
B
U
U
random
the same random
and still has non-positive partial transposition.
Thus, we can concentrate on states of the form:
Z
d¹ U(U - U)~
½(U - U)y = aP+ + bP¡
1
d
¡
where b
= t
r
(
P
~
)
¡½
(
x
)
=
P
+
x
P
We consider the (unnormalized) family of states: ½
+
¡
2
2
=
C
C
Qubits: H
T
A
½
¸0
T
A
½
¸0
3
distillable
x
N
y
A
B
)
½
(
A
B
)
!
j
©
i
h
©
j
one can easily find A, B such that (
3
3
=
C
C
Higher dimensions: H
T
A
½
¸0
T
A
½
¸0
2
x
3
?
NPT
distillable
Idea: find A, B such that they project
T
A
2 2
¸0
onto C
with ½
-C
there is a strong evidence that they are not distillable: for any finite N, all
T
A
2 2
¸0
projections onto C
have ½
-C
What is known?
PPT
In general
NPT
?
2xN
Gaussian states
(Horodecki 97, Dur et al 2000)
PPT
NPT
(Giedke, Duan, Zoller, Cirac, 2001)
PPT
NPT
3. Gaussian states
squeezed states:
(2-mode approximation)
Light source:
1
X
n
j
v
a
c
i
=
¸
j
n
;
n
i
y
y
¸
(
a
b
¡
a
b
)
j
ª
i
=
e
n
=
0
Gaussian state:
Decoherence: photon absorption, phase shifts
¡H
½
=
e
=
H
(
X
;
P
;
X
;
P
)
where H
a
a
b
b
is at most quadratic in
y
y
a
+
a a
¡
a
p
X
=
;
P
=
ip
a
a
2
2
Internal levels can be approximated
Atomic ensembles:
by continuous variables in Gaussian
states
Optical elements:
- Beam splitters:
- Lambda plates:
- Polarizers, etc.
Measurements: X, P
0
¡
H
0
¡
H
½
=
e
!
½
=
e
Gaussian
- Homodyne detection:
Gaussian
local
oscillator
We consider:
B
A
m modes
n modes
2
n
2
m
H
=
[
L
(
R
)
]
[
L
(
R
)
]
½ Gaussian
Is ½ separable and/or distillable?
3.1 What is known?
1 mode + 1 mode:
T
A
½ is separable iff ½
=
(
T
1
)
(
½
)
¸
0
(Duan, Giedke, Cirac and Zoller, 2000; Simon 2000)
2 modes + 2 modes:
There exist PPT entangled states.
(Werner and Wolf 2000)
3.2 Separability
CORRELATION MATRIX
R
=
(
X
;
P
;
X
;
:
:
:
;
X
;
P
;
:
:
:
)
a
a
a
b
b
1
1
2
1
1
All the information about ½ is contained in:
°
=
2
R
e
h
(
R
¡
d
)
(
R̄
¡
d̄
)
ithe „correlation matrix“.
®
®
®
;
¯
=
h
R
i
!
0
where d
®
®
µ
°=
2nX2n
A
CT
C
B
¶
2mX2m
For valid density operators: °¸iJ
is the „symplectic matrix“
=
J
©
J
©
:
:
:
©
J
where J
2
2
2
µ ¶
0¡1
=
and J
2
10
Given a CM, ° : does it correspond to a separable state (separable)?
(G. Giedke, B. Kraus, M. Lewenstein, and Cirac, 2001)
Idea: define a map
°
´°
0
°1
°2
°N
...
T
AN+ 1 = BN+ 1 = AN ¡ Re[CN (BN ¡ i J) ¡ 1CN ]
CN+ 1 = ¡ Im[CN (BN ¡ i J) ¡ 1CNT ]
Facts:
°Nis a CM of a separable state iff °
is too.
N
+
1
°
N
+
1is no CM
If °N is a CM of an entangled state, then either
or
°
N
+
1is a CM of an entangled state
!°
=
½
½
If ° is separable, then °
. This last corresponds to ½
N
1
1
A
B
(for which one can readily see that is separable)
CONNECTION WITH POSITIVE MAPS?
!
°
N
N
+
1
Map for CM‘s: °
¡
H ¡
H
N
N
+
1
=
e
!
½
=
e
Map for density operators: ½
N
N
+
1
(½N+ 1)AB = trBB~f [(½N )AB - (½N )A~B~]XB B~g
½
A
A~
½
B
B~
Non-linear
density operators
Gaussian
separable
A
A~
3.3 Distillability
(Giedke, Duan, Zoller, and Cirac, 2001)
T
¸0
Idea: take ½ such that ½
Two modes: N=M=1:
=
°
Symmetric states:°
A
B
B
A
T
½
s¸ 0
A
A
T
½
~
s¸ 0
Non-symmetric states:
B
A
B
A
distillable state.
T
½
¸0
T
½
s¸ 0
B
B
symmetric state.
General case: N,M
A
T
½
N
;
M̧0
B
A
T
½
¸0
1
;
1
two modes
T
¸0
½ is distillable if and only if ½
There are no NPT Gaussian states.
B
4. Multipartite case.
C
A
B
½
Are these systems entangled?
Fully separable states are those that can be prepared by LOCC out of a product state.
½=
N
X
pk jak i A hakj - jbki B hbkj - jck i Chckj
k= 1
We can also consider partitions:
Separable A-(BC)
Separable B-(AC)
C
A
Separable C-(AB)
C
B
A
C
B
A
B
N
X
p
j
a
i
h
a
j
j
'k
i
h
'k
j
k
k
A
k
B
C
N
X
p
j
b
i
h
b
j
j
'k
i
h
'k
j
k
k
B
k
A
C
N
X
p
j
c
i
h
c
j
j
'k
i
h
'k
j
k
k
C
k
A
B
k
=
1
k
=
1
k
=
1
4.1 Bound entangled states.
Consider
C
A
C
N
X
½=
A
B
pk jak i Ahakj - j' k i BCh' k j =
k= 1
N
X
B
pk jbki B hbkj - j' k i ACh' k j
k= 1
but such that it is not separable C-(AB).
QUESTIONS:
Is B entangled with A or C?
Is A entangled with B or C?
Is C entangled with A or B?
Consequence: Nothing can be distilled out of it. It is a bound entangled state.
4.2 Activation of BES.
(Dür and Cirac, 1999)
Consider
C
C
A
A
B
but A and B can act jointly
C
A
B
singlets
Then they may be able to distill GHZ states.
B
ACTIVATION OF BOUND ENTANGLED STATES
Distillable iff two groups
3 and 5 particles
Distillable iff two groups
35-45% and 65-55%
Distillable iff two groups
have more than 2 particles.
6
3 1
4
7
2
Not distillable
8
5
Not-distillable
6
4
6
3 1
7
4
2
3 1
2
Distillable
7
8
8
5
Distillable
4
2
5
6
3 1
Not-distillable
7
10
9
11
12
8
5
Not-distillable
6
3 1
4
2
Not-distillable
7
8
5
If two particles remain
separated not distillable.
Two parties can distill iff the
other join
3 1
2
3
6
5
10
4
7
9
8
11
(Shor and Smolin, 2000)
1
2
6
5
10
4
7
1
3
12
3
C
1
9
8
4
2
12
Superactivation
4
2
A
B
11
Two copies
4.3 Family of states
Define:
where
N
¡1
There are 2
parameters.
Any state can be depolarized to this form.
5. Conclusions
The separability problem is one of the most challanging problems in quantum
Information theory. It is relevant from the theoretical and experimental point of view.
Gaussian states:
Solved the separability and distillability problem for two systems.
Solved the separability problem for three (1-mode) systems
Maybe we can use the methods developed here to attack the general problem.
Multipartite systems:
New behavior regarding separability and bound entanglement.
Family of states which display new activation properties.
Innsbruck:
Geza Giedke
Hannover
M. Lewenstein
Barbara Kraus
Wolfgang Dür
Guifré Vidal
J.I.C.
Collaborations:
R. Tarrach (Barcelona)
P. Horodecki (Gdansk)
L.M. Duan (Innsbruck)
P. Zoller (Innsbruck)
SFB Coherent Control
€U TMR
EQUIP
KIAS, November 2001
Institute for Theoretical Physics
Postdocs:
- L.M. Duan (*)
- P. Fedichev
- D. Jaksch
- C. Menotti (*)
- B. Paredes
- G. Vidal
- T. Calarco
Ph D:
- W. Dur (*)
- G. Giedke (*)
- B. Kraus
- K. Schulze
P. Zoller
J. I. Cirac
FWF SFB F015:
„Control and Measurement of Coherent Quantum Systems“
€
EU networks:
„Coherent Matter Waves“, „Quantum Information“
EU (IST):
„EQUIP“
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