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Quantifying Entanglement
Daniel Rohrlich
Atom Chip Group, Ben Gurion University, Beersheba, Israel
20 January 2010
Is the uncertainty principle a fundamental limit on what we can
measure? Or can we evade it? Einstein and Bohr debated this
question for years, and never agreed.
Is the uncertainty principle a fundamental limit on what we can
measure? Or can we evade it? Einstein and Bohr debated this
question for years, and never agreed.
Today we are certain that uncertainty will not go away.
Quantum uncertainty is even the basis for new technologies such
as quantum cryptology.
It may be that the universe is not only stranger than we imagine,
but also stranger than we can imagine.
Heisenberg’s uncertainty principle
1. If a lens with aperture θ focuses
light of wavelength λ, Rayleigh’s
criterion implies Δx ≥ λ/2sinθ.
2. A wave of wavelength λ has
momentum p = h/λ.
3. From geometry we see here that
Δp ≥ 2p sinθ.
Therefore (Δx)(Δp) ≥ h.
θ
p = h/λ.
EPR
In 1935, after failing for years to defeat the uncertainty principle,
Einstein argued that quantum mechanics is incomplete.
The famous “EPR” paper
EPR
In 1935, after failing for years to defeat the uncertainty principle,
Einstein argued that quantum mechanics is incomplete.
Note that [x, ˆp] ≠ 0, but [x2–x1, pˆ 2+pˆ 1] = [x2, pˆ 2] – [x1, pˆ1] = 0.
That means we can measure the distance between two particles
and their total momentum, to arbitrary precision.
EPR claimed that we can measure x2 or p2 without affecting
Particle 2 in any way, via Particle 1 (and vice versa).
If so, then x2 and p2 have simultaneous values.
Quantum mechanics has no place for simultaneous values of x2
and p2 , hence quantum mechanics is incomplete.
EPR
Some reactions to the EPR argument:
Bohr (1935) The uncertainty principle still applies.
Schrödinger (1935) “Entanglement”
Pauli (1954) “One should no more rack one's brain about the
problem of whether something one cannot know anything about
exists all the same, than about the ancient question of how many
angels are able to sit on the point of a needle. But it seems to me
that Einstein's questions are ultimately always of this kind.”
Bell (1964) Quantum mechanics contradicts EPR!
EPR
The state defined by EPR
EPR ( x1 , x2 )  δ( x2  x1  L) 
 ik ( x  x  L)
2
1

e
dk
is an example of an entangled state. It does not reduce to a
product of states ψ1(x1)ψ2(x2).
Let  AB be a state of two subsystems, A and B. If the Schmidt
decomposition,

AB

c j
j
A
j
B
,
j
of 
AB
contains more than one term, the state is entangled.
Entanglement of identical particles:
Consider an interferometer with four half-silvered mirrors.
ψ
φ
1
2
Two particles enter from opposite corners. Assume, at first, that
they are not identical. Each reflection induces a phase factor i.
Final state – measuring correlations
U
〰
φ
ψ
1
L
R
〰
2
D
〰
Initial state (if the particles are not identical):
ψ1φ 2
ψ
φ
2
1
Final state (if the particles are not identical):
1
L 1 i U 1 i D 1  R 1 i U 2  L 2  R 2  i D 2
4
U
ψ1


L
φ
R
2
D
(Assume that the path lengths are all equal.)

Final state (if the particles are not identical):
1
L 1 i U 1 i D 1  R 1 i U 2  L 2  R 2  i D 2
4
But if the particles are identical, we must either symmetrize this
state (e.g. for photons) or antisymmetrize it (e.g. for electrons).


Symmetrized state:
1
L 1 i U 1 i D 1  R
4 2


1
4 2

1
2 2
1
 i U
2
 L
2
 R
 i U 1  L 1  R 1  i D 1  L 2  i U
[ L 1 L
 L
1

R
2
2
U
 R
1
1
L
U
2
2
 D
U
1
1
D
D
2
2
 R
1
 D 1U
R
2
]
2
2
 iD
2
i D
2
2

 R
2

Final state (if the particles are not identical):
1
L 1 i U 1 i D 1  R 1 i U 2  L 2  R 2  i D 2
4
But if the particles are identical, we must either symmetrize this
state (e.g. for photons) or antisymmetrize it (e.g. for electrons).



Symmetrized state:
Pairs of photons exit the interferometer in the same state (same
port) at the same corner, or in opposite directions (L/R or U/D)
at ports on opposite corners.
Final state (if the particles are not identical):
1
L 1 i U 1 i D 1  R 1 i U 2  L 2  R 2  i D 2
4
But if the particles are identical, we must either symmetrize this
state (e.g. for photons) or antisymmetrize it (e.g. for electrons).


Antisymmetrized state:
1
L 1 i U 1 i D 1  R
4 2


1
4 2

1
2 2
1
 i U
2
 L
2
 R
 i U 1  L 1  R 1  i D 1  L 2  i U
[i L 1 U
i L
1

D
2
2
i U
i D
1
1
L
L
2
2
i D
i U
1
1
R
R
2
2
i R
1
D
i R 1 U
2
2
 iD
2
2
]
i D
2
2

 R
2

Final state (if the particles are not identical):
1
L 1 i U 1 i D 1  R 1 i U 2  L 2  R 2  i D 2
4
But if the particles are identical, we must either symmetrize this
state (e.g. for photons) or antisymmetrize it (e.g. for electrons).


Antisymmetrized state:
Pairs of electrons exit the interferometer in different states
(different ports) at the same corner, or in orthogonal directions
(L/D or U/R) at ports on opposite corners.
I. Neder, N. Ofek, Y. Chung, M. Heiblum, D. Mahalu, and
V. Umansky, Nature (London) 448, 333 (2007).

Bell’s theorem
A convenient system for formulating and testing Bell’s theorem:
two entangled photons with polarization state

AB


ε1 A ε 2 B  ε 2
2
1
ε
A 1 B

,
where ε1 A and ε 2 A are orthogonal polarization states of
photon A and ε1 B and ε 2 B are orthogonal polarization states
of photon B. The correlation between Photon A and Photon B in
this polarization state depends only on the angle θAB between
their respective polarization axes: it is C(A,B) = cos 2θAB , so the
photons are perfectly correlated when they arrive at parallel
polarizers.
But when the angle between the polarization axes is 120°, the
correlation is –1/2.
Now let Alice and Bob prepare many such pairs of
photons, and let Alice measure the polarization of
one in each pair (Photon A), and Bob measure the
polarization of the other one (Photon B), along one of
three polarization axes:
Alice
120°
120°
120°
Bob
drawings by Tom Oreb
© Walt Disney Co.
We know what correlations they will measure. But
can the photons carry local programs telling them
how to behave?
–120°
0°
120°
Photon A
yes
no
no
Photon B
yes
no
no
Alice
Example of a local program
with perfect correlation at 0°
and correlation –1/3 at 120°
Bob
drawings by Tom Oreb
© Walt Disney Co.
We know what correlations they will measure. But
can the photons carry local programs telling them
how to behave? NO!
–120°
0°
120°
Photon A
yes
no
no
Photon B
yes
no
no
Alice
Example of a local program
with perfect correlation at 0°
and correlation –1/3 at 120°
Bob
drawings by Tom Oreb
© Walt Disney Co.
Bell’s theorem and GHZ
We will consider a theorem similar to Bell’s, proved in 1988 by
D. Greenberger, M. Horne, and A. Zeilinger (GHZ).
It describes three spin-½ particles prepared in one laboratory and
sent to three different laboratories, operated by Alice, Bob, and
Claire. For simplicity, we assume the particles are not identical.
GHZ
Setting for Greenberger-Horne-Zeilinger (GHZ) paradox:
Alice
A
Bob
B
time
O
space
Claire
C
GHZ
It describes three spin-½ particles prepared in one laboratory and
sent to three different laboratories, operated by Alice, Bob, and
Claire. For simplicity, we assume the particles are not identical.
The spin state of the three particles is
GHZ 


2
1
A

B
1
0
where     and     .
0
1

C

A

B

C
,
GHZ
Alice can measure ˆ x(A) or ˆ (yA) on her system;
Bob can measure ˆ x(B) or ˆ (yB) on his system;
Claire can measure ˆ x(C) or ˆ (yC) on her system;
0 1
0  i
1 0 

ˆ
ˆ
ˆ
ˆ








,


,


where x 
and S i  ˆ i .
y 
z 



2
1 0
i 0 
 0  1
The result of each measurement can be 1 or –1.
Note ˆ x    , ˆ x    , ˆ y   i  , ˆ y   i  ,
1
0
where     and     .
0
1
GHZ
To prove:
( A ) ( B) ( C )
1. ˆ x ˆ x ˆ x | GHZ   | GHZ
( A ) ˆ ( B) ˆ ( C )
ˆ
|



2.
GHZ
x  y  y | GHZ
 ˆ (yA )ˆ x( B)ˆ (yC) | GHZ
 ˆ (yA )ˆ (yB)ˆ x(C) | GHZ
That is, Alice, Bob and Claire discover two rules:
( A) ( B) (C)
1.  x  x  x  1
(A) (B) (C)
(A) (B) (C)
(A) (B) (C)
2.  x  y  y  1   y  x  y   y  y  x
GHZ
But now



1   x( A ) (yB) (yC)  (yA ) x( B) (yC)  (yA ) (yB) x(C)
  x( A ) x( B) x(C)
 1 ,

 
( A) 2
because  x

( A) 2
y
 1, etc.
CONTRADICTION!

GHZ
But now



1   x( A ) (yB) (yC)  (yA ) x( B) (yC)  (yA ) (yB) x(C)

  x( A ) x( B) x(C)
 1 ,

 
( A) 2
because  x

( A) 2
y
 1, etc.
CONTRADICTION!
This contradiction arose because we tacitly assumed, with EPR,
that observables have values whether we measure them or not.
John Bell:
For me, it is so reasonable to assume that the
photons in those experiments carry with them
programs, which have been correlated in advance,
telling them how to behave. This is so rational that
I think that when Einstein saw that, and the others refused to
see it, he was the rational man. The other people, although
history has justified them, were burying their heads in the sand.
I feel that Einstein’s intellectual superiority over Bohr, in this
instance, was enormous; a vast gulf between the man who saw
clearly what was needed, and the obscurantist. So for me, it is a
pity that Einstein’s idea doesn’t work. The reasonable thing
just doesn’t work.
[Jeremy Bernstein, Quantum Profiles (Princeton: Princeton
University Press), 1991, p. 84.]
“Everybody talks about entanglement but nobody does
anything about it.”
But what can we do about it?
Lord Kelvin (Sir William Thomson ):
In physical science the first essential step in the
direction of learning any subject is to find principles
of numerical reckoning and practicable methods for
measuring some quality connected with it. I often
say that when you can measure what you are speaking about,
and express it in numbers, you know something about it; but
when you cannot measure it, when you cannot express it in
numbers, your knowledge is of a meagre and unsatisfactory
kind; it may be the beginning of knowledge, but you have
scarcely in your thoughts advanced to the state of Science,
whatever the matter may be.
[Popular Lectures and Addresses (1891-1894), Vol. 1,
"Electrical Units of Measurement", 1883-05-03]
Bell’s theorem and CHSH
J. Clauser, M. Horne, A. Shimony and R. Holt proved (1969) that
any bipartite correlation C(A,B), if it arises from local programs
carried by two subsystems, must obey the following inequality:
–2 ≤ C(A,B) + C(A′,B) + C(A,B′) – C(A′,B′) ≤ 2 ,
where A and A′ are observables with eigenvalues ±1 that Alice
can choose to measure, and B and B′ are observables with
eigenvalues ±1 that Bob can choose to measure.
Theorem (N. Gisin, A. Peres, S. Popescu, DR, 1991-2):
Correlations of any entangled state violate the CHSH inequality.
Bell’s theorem and CHSH
So is

AB

“more” entangled than

AB



2
1
A

10 
101

1
A
B

 
B
A
 

A
B


B

?
Bell’s theorem and CHSH
So is

AB

“more” entangled than

AB



2
1
A

10 
101

1
A
B

 
B
A
 

A

Theorem (A. Elitzur, S. Popescu and DR): 
by a mixture of pairs, some in the state 
local programs. Is it “less” entangled?
These are ad hoc rankings.
B
AB

B
AB

?
can be realized
and some carrying
But what is entanglement good for?
Two examples:
☻Teleportation
☻Quantum cryptography
Claim: Such applications of entanglement pave the way to a
measure of entanglement the way the applications of heat
differentials – heat engines – paved the way to the Carnot cycle.
Example of teleportation:
We approximate Captain Kirk
Example of teleportation:
We approximate Captain Kirk by a single “qubit” in an
unknown state in Alice’s laboratory. We have, initially,
a 
Captain
Kirk
K
b 
K
 
A


B
A

B

Shared
/ 2 .“ebit”
Alice measures and projects onto the state
1
   
,
K
A
K
A
2
thereby sending Captain Kirk to Bob’s laboratory:


a
B
b 
.
B
[What must Alice do if she projects onto


2
1
K

A
 
K

A
 or


2
1
K



2
1
A

K
K


A
 
A
 ?]
K

A
,
Example of quantum cryptography: “the one-time pad”.
100011110011110110
–10010101011010100
“10001001000100010”
Bob
“10001001000100010”
+10010101011010100
100011110011110110
Alice
Manipulating entanglement
Can Alice and Bob turn

AB

1

1
into

AB
or into

AB


2

1
A

B
 
A

B


 A  B   A  B
2


2
A

B
 
A

B

?
Manipulating entanglement
Can Alice and Bob turn

AB

1

1
into

AB
or into

AB


2

1
A

B
 

A
B


 A  B   A  B
2


2
A

B
 
A

B

?
Yes, these are local unitary transformations.
Any redefinition of Alice’s basis by Alice or of Bob’s basis by
Bob is a local unitary transformation.
Manipulating entanglement
Can Alice and Bob turn

AB


10 
101

1
into

AB
1


2
A

A
B

B
 
 
A

A
B

B

?

Yes, they have a chance , via local filtering. Let 0 and 1 be
states of a local filter for Alice; consider the following operation:
 0  cos  0  sin   1
A

A
A
0  
A
0
A
Manipulating entanglement
Can Alice and Bob turn

AB


10 
101

1
into

AB
1


2
A

A
B

B
 
 
A

A
B

B

?

Yes, they have a chance , via local filtering. Let 0 and 1 be
states of a local filter for Alice; consider the following operation:
 0  cos  0  sin   1
A



A
A
A
A
0  
1  
A
A
A
0
1
1   sin  
(for unitarity)
A
0  cos 
A
1
Manipulating entanglement
Unitary evolution takes the initial state

10  A 0
101
1
into

10 cos 
101
1
A

 
B
0  sin  
A

A
1 
0 
B
B
 

A
0 
B
,
now if Alice looks into the filter and doesn’t find the state 1 ,
the (unnormalized) bipartite state is reduced to

10 cos 
101
1
A

B
 
A

B
,
and by choosing cos α =1/10, Alice and Bob have recovered the
target state  AB. They succeed once in a hundred trials!
Manipulating entanglement
This is also called the “Procrustean” method of obtaining 
.
AB
Manipulating entanglement
This is also called the “Procrustean” method of obtaining 
.
AB
But collective methods are both kinder and also more efficient.
Suppose Alice and Bob share two  AB pairs; the overall state

1
10 
101
A
0 
B
 
A
0 
B
10  A 0

B
 
B
B

A
A
0 
contains the cross-terms

10

101
A

B

A


A


B

.
B

Manipulating entanglement
This is also called the “Procrustean” method of obtaining 
.
AB
But collective methods are both kinder and also more efficient.
Suppose Alice and Bob share two  AB pairs; the overall state

1
10 
101
A
0 
B
 
A
0 
B
10  A 0

B
 
B
A

B
A
0 
contains the cross-terms

10

101
A

A

B


A


B

.
B

Manipulating entanglement
This is also called the “Procrustean” method of obtaining 
.
AB
But collective methods are both kinder and also more efficient.
Suppose Alice and Bob share two  AB pairs; the overall state

1
10 
101
A
0 
 
B
A
0 
B
10  A 0

B
 
B
A

B
A
0 
contains the cross-terms

10

101
A


A
A

B


B

A


A


B
B

.
B

Manipulating entanglement
This is also called the “Procrustean” method of obtaining 
But collective methods are both kinder and also more efficient.
Suppose Alice and Bob share two  AB pairs; their chance of
getting the state

AB


 A  B   A  B
2
1
is also about one in a hundred, but the efficiency of collective
methods improves with the number of initial pairs!
.
AB
Manipulating entanglement
This is also called the “Procrustean” method of obtaining 
But collective methods are both kinder and also more efficient.
Suppose Alice and Bob share two  AB pairs; their chance of
getting the state

AB

.
AB

 A  B   A  B
2
1
is also about one in a hundred, but the efficiency of collective
methods improves with the number of initial pairs!
C. H. Bennett, H. J. Bernstein, S. Popescu, and B. Schumacher,
Phys. Rev. A 53, 2046 (1996): The efficiency approaches the
entropy of entanglement.
Entropy of entanglement
Definition: The entropy of entanglement of the state
p
AB

p  A  B  1 p  A  B
(and any state related to it by local unitary transformations) is
– p log2 p – (1– p) log2 (1– p) .
Theorem: k pairs in the state  p
AB
can be transformed into n
pairs in the state 1 / 2 AB (“generalized singlet” states or ebits)
with an efficiency that asymptotically approaches the entropy of
entanglement:
n
lim k   p log 2 p  1  p log 2 1  p  .
n , k 
Entropy of entanglement
Proof: Let us consider
p
k
AB


p
A

B
 1 p 
A

B

k
,
which we can expand via the binomial theorem:
p
k
AB
k


j 0
k!
p j / 2 1  p ( k  j ) / 2
j!(k  j )!
j
( k  j )




  A  B
 A  B
.
The combinatorial factor counts the number of terms with a
given spin component (net ↑ over ↓) and the probability of this
given spin component (which Alice and Bob independently can
measure) is
k!
p j 1  p k  j
j!(k  j )!
Entropy of entanglement
Let’s estimate which term in the binomial expansion is largest by
treating j as a continuous variable and using Sterling’s formula to
approximate the factorials:
k!
p j 1  p k  j  k k j  j (k  j ) k  j p j 1  p k  j
j!(k  j )!
 e k ln k  j ln j (k  j ) ln(k  j ) j ln p ( k  j ) ln(1 p )
Setting the derivative of the right side with respect to j to 0, we
obtain
0   ln j  ln(k  j )  ln p  ln(1  p) ,
which implies
ln j  ln( k  j )  ln p  ln(1  p )
j
p

,
k  j 1 p
and therefore j = pk.
Entropy of entanglement
From this approximate calculation we learn two important
lessons.
1. If we substitute j = pk back into
e k ln k  j ln j ( k  j ) ln(k  j ) j ln p  ( k  j ) ln(1 p )
we get 1, which means that the most probable term is also,
asymptotically, the only relevant term.
Entropy of entanglement
From this approximate calculation we learn two important
lessons.
2. The number n of ebits Alice and Bob can obtain from this
most relevant term is log2 of the number of terms with j = pk i.e.
k!
it is log2
with j = pk. We have
j!(k  j )!
k!
kk
n  log 2
 log 2
j!(k  j )!
( pk ) pk k (1  p )k (1 p )
 k log 2 k  pk log 2 ( pk )  k (1  p ) log 2 k (1  p)
 k  p log 2 p  (1  p ) log 2 (1  p ) .
Reversibility
The concentration of k “less-entangled” pairs to n ebits is
reversible, using quantum data compression followed by
teleportation. Asymptotically, Alice and Bob need n ebits to
recreate the original k pairs.
Thermodynamic analogue
The entropy of entanglement seems a plausible measure of
entanglement, but how do we know that it is not just another ad
hoc measure?
We can answer this question by remembering how Carnot
established the efficiency of an ideal heat engine operating
between two temperatures. He had before him a great principle,
the Second Law of thermodynamics: no perpetuum mobile.
Thermodynamic analogue
Two reversible heat engines:
Q2 out
Q1 in
Q2 out
T2
T2
Yields
work
W′ >W
per cycle
Yields
work W
per cycle
T1
T1
Q1 in
Thermodynamic analogue
Two reversible heat engines:
Q2 in
Q2 out
Q1 in
T2
T2
Yields
work
W′ >W
per cycle
Uses
work W
per cycle
T1
T1
Q1 out
Thermodynamic analogue
Two reversible heat engines:
Q2 in
Q2 out
Q1 in
T2
T2
Yields
work
W′ >W
per cycle
Uses
work W
per cycle
T1
T1
Q1 out
Thermodynamic analogue
The laws of physics do not allow Alice and Bob to create (or
increase) entanglement via local operations and classical
communication.
(Alice and Bob can destroy or decrease entanglement via local
operations.)
Thermodynamic analogue
Two reversible entanglement concentrations:
n′ ebits
out
n ebits
out
Yields
n′ > n
ebits per
k inputs
Yields n
ebits per
k inputs
k p
AB
pairs in
k p
AB
pairs in
Thermodynamic analogue
Two reversible entanglement concentrations:
n ebits
in
n′ ebits
out
Yields
n′ > n
ebits per
k inputs
Yields k
p
AB
pairs per
n inputs
k p
k p
AB
pairs out
AB
pairs in
Thermodynamic analogue
Two reversible entanglement concentrations:
n ebits
in
n′ ebits
out
Yields
n′ > n
ebits per
k inputs
Yields k
p
AB
pairs per
n inputs
k p
k p
AB
pairs out
AB
pairs in
Thermodynamic analogue
Since k pairs in the state  p
AB
and n ebits are asymptotically
equivalent, the measure of their entanglement must be the same.
entropy of entanglement is the most efficient possible. Thus, the
question of a measure of entanglement reduces to the question of
a measure of the entanglement of n ebits.
Could the measure be any function of n? No, because all our
manipulations are exact only in the asymptotic limit; in the limit
k,n → ∞ we can only define entanglement per system, not total
entanglement. Here, too, thermodynamics provides the formal
principle: in the thermodynamic limit, we can define only
intensive quantities. The entanglement of n ebits must be
proportional to n.
Summary
Entropy of entanglement is the unique asymptotic (intensive)
measure of entanglement for pure bipartite states.
BUT
This result has not been generalized to mixtures of bipartite
states.
Among multipartite states (even pure multipartite states) there
are inequivalent kinds of entanglement.