Download 4.Classical entanglement

4.Classical entanglement
CHSH version of the Bell inequalities:
These deal with correlations between
a set of four classical probabilities.
In particular, the either-or observable, O1  1,
for detecting some physical property for one particle,
or the detection of O2  1 for the second particle.
We know that the projection of theWigner function
onto any Lagrangian plane
produces a classical probability distribution.
So, constructing our observables from regions on such a plane
should lead to measurements that satisfy the CHSH inequality.
In the case of a bipartite system, we can define
such Lagrangian planes as q'1 , q'2  , such that
with conjugate variables, p'1 and p'2 ,
such that x j  x ' j are symplectic transformations.
Now define projection operators, P̂ja and P̂jb ,
for the variable q' j to be in the interval a or b
and the four observables,
which take the value +1
qˆ'1 , qˆ'2   0
 Oˆ ja  2Pˆja  Iˆ
In terms of the Wigner function:
Pˆ1a   dq'1  dq'2  dp'1 dp'2 W (q' , p')
a
Projection on
the interval a
Projection onto the q'1 , q'2 , a probability.
Since the product of commuting observables are also observables:
with
Pˆ1a Pˆ2 a   dq'1  dq'2  dp'1 dp'2 W (q' , p' ),
a
a
these can be measured and their expectation values satisfy
t
h
 2.
=
Because these are purely classical probabilities,
no term with a (-4)-coefficient can be positive.
This is the classic argument for the CHSH form
of Bell’s inequality.
Measurements of commuting variables cannot violate CHSH.
This argument does not depend on whether
the bipartite state is entangled,or not.
What if the Wigner function itself is positive definite?
Then, even projectors over noncommuting variables,
such that
Pˆ1a   dq'1  dq'2  dp'1 dp'2 W (q' , p')
a
and
Pˆ1b   dp'1  dp'2  dq'1 dq'2 W (q' , p') ,
b
will lead to classical correlations, within the CHSH inequalities.
Does this mean that there is no entanglement?
What are the conditions for a positive Wigner function?
The condition for the Wigner function of a pure state
to be positive is that it is a Gaussian
(a generalized coherent state, or squeezed state)
(Hudson: Rep. Math. Phys. 6 (1974) 249)
What about a mixed state?
Evidently, any probability distribution over coherent states
has a positive Wigner function:
The P-function with positive coefficients.
 Wigner 
 P
Q = Husimi 
coarsegrain
coarsegrain
But this condition is only sufficient, not necessary.
So, probabilities obtained over intervals for
squeezed states will always be classically correlated:
they must satisfy the CHSH inequality.
See the paper by Bell in
Speakable and unspeakable in quantum mechanics
Bell also produces the example of a Fock state
for which such probabilities violate CHSH.
Is a multidimensional squeezed state
too ‘classical’ to be entangled?
Consider any simple, L=2, product state:
W (x)  W1 (x1 )W2 (x2 ) and  ( )  1 (1 ) 2 (2 )
Then evolve this state with the Hamiltonian,
(classical, or Weyl representation)
Being quadratic, this merely rotates both the p and the q coordinates
(classically) in the argument of
Then, after a  4 rotation,
  p1   p 2  q1   q 2    p1   p 2  q1   q 2 
  2 
 ,
 ' ( )  1 
,
,
2
2  
2
2 

The reduced density is just a section, so
  p1  q1    p1  q1 
  2 
 .
1 ' ( )  1 
,
,
 2 2  2 2
To show how classical an entanglement can be,
choose a simple Gaussian state, the product of
HO ground states:
 j 2
1
1 2
W j (x j ) 
exp 
qj 
pj,

 j 
 
or
 j
1
 j ( j ) 
exp 
2
 
‘
2

1  pj  
  
   .
 2   j  2  
  qj 
2
‘
is also a Gaussian, with elliptic level curves that are also rotated.
After rotation and the partial trace:
2
2



1
1  2  q1  1  1 1  p1  
     
  .
 '1 (1 ) 
exp 
2
  2    1 2  2  


The narrowing of the Gaussian shows that the state is not pure.
The Wigner function is more intuitive:
Obtained by taking the Fourier transform,

2 12
 2 12  2 
2 p12
 q1  .
W '1 (x1 ) 
exp 
 
(1  2 )
  (1  2 )  (1  2 )  
This broader Gaussian still integrates to one.
(It could be obtained as an average over
pure Gaussian Wigner functions.)
Is this a freak?
Nothing could be more classical for a start (positive Wigner function)
and then a classical rotation produces entanglement!
Go back to the product state of both
 j 2
1
1 2
W j (x j ) 
exp 
qj 
pj .

 j 
 
So, the original EPR state
does not violate Bell inequalities
for the measurement of any four observables
defined by intervals of position or momentum.
But it is technically entangled.
Bell leaves open the possibility that
other unusual observables may lead
to inequality violations for such a state.
What about generalized parities:
the eigenvalues of reflection operators
for a given subspace?
The fact that the Wigner function


is symmetric with respect to the origin implies that ˆ ' , Rˆ0  0 .


so ˆ '1 , Rˆ '01  0.
Hence, there is a finite probability
of obtaining the -1 eigenvalue,
if a parity measurement
is performed on subsystem-1.
The same also holds for subsystem-2.
The fact that the Wigner function is symmetric about the origin
implies that all the pure states, into which ̂ ' can be decomposed,
must have pure parity, but they are not all even.
Thus, we need a common basis for all these operators:
the product of an even-odd basis for both subsystems,
leading to the table:
Conclusion
The original EPR state is truly quantum,
i.e. correctly described as entangled,
just as the Bohm version of EPR.
The secret lies in the property that is measured:
Generalized position measurements on the subsystems
cannot distinguish this pure state from a classical distribution,
but reflection eigenvalues are purely quantum.
Violation of CHSH:
Recall that the correlation for reflection measurements
on either subsystem is given by the Wigner function:
We have already examined this at the origin.
The decay of the Wigner function for large arguments
implies that
sinks from 2, its maximal classical value,
obtained at the origin,
to its limiting value, 1.
But
because the expansion,
leads to
Thus, the smoothed EPR state can be measured
in ways that lead to violation of Bell’s inequalities.
Banaszek and Wodkiewicz have proposed an experiment
in quantum optics to achieve this.
A note on classicality versus hidden variables:
Bell’s inequalities set limits to the correlations
of any possible classical-like theory.
This is much stronger than my presentation,
but the inequalities must include
the classical system that corresponds directly
to the quantum system under consideration.
5.Decoherence:
the Lindblad Equation
Decoherence results through entanglement
of the system under consideration
with an uncontroled system, labled
the environment.
Generally we do not know the initial state
of the environment:
Further averages, beyond the implicit average
in the reduced density operator.
A simple example: Weak scattering of many light particles.
If the duration of a single scattering process
is short compared to the typical time scales
of the system evolving by itself:

[Joos, in
Giulini
et. al.]

ˆ
ˆ
ˆ
i
 H , ˆ  i
|scatt.
t
t
The last term accounts for the total effect
of many scattering events,
in which the system is dynamically insensitive,
i.e. no recoil.
Nonetheless, the scatterers transport
information about the system!
Consider a single scattering event from the system,
if it is initially in the eigenstate n .
Then, if 0 is the initial state of the environment,
n  0  n 'n  n Sˆn  0 ,
where Ŝ n is the scattering operator (S-matrix)
for this configuration of the system.
For a general initial state,


  cn n   0   cn n  ' n ,
n
n

so, the reduced density operator changes accordingly:
ˆ sys   cm* cn n m   cm* cn 'm 'n n m ,
n,m
because tr 'n 'm  'm 'n .
n,m
Thus, the matrix elements of the density operator evolve as
nm  nm 'm 'n  nm 0 Sm Sn 0 .
If the overlap is close to unity,
0 Sm Sn 0  1   ,
then the effect of many collisions, with the rate  ,
will be:
nm  nm 1    t  nm exp   t  .
ˆ

|scatt.     nm
t
with
  1  0 Sm Sn 0
 .
For the diagonal terms, n  m    0.
Thus the trace of ̂ is not affected
and only the offdiagonal terms decay with decoherence.
Generally, the greater the difference between n and m,
the faster is the decay.
For the scattering off a particle,
the scattering depends on its position,
determined by its wave function,  (q).
Then, for a single scattering event:
 (q, q' )   (q) * (q' )   (q) * (q' )  Sq' Sq  .
If the scattering interaction is translationally invariant,
the S-matrix in the momentum representation
depends on the position of the scaterrer by a phase factor:
S q (k , k ' )  S0 (k , k ' ) exp  ik  k 'q 
Then, for     k k ,
k
 Sq' Sq     k *k ' exp  ik  k 'q  q'.
kk '
1

2
2
   k  1  i k  k 'q  q'  k  k ' q  q ' 
2


kk '
*
k'
So, averaging over many scattering events:


 (q, q' )   (q) * (q' ) exp  t q  q'2 .

2

|env  q, q'   q  q'  (q, q' )
t
The Lindblad equation has the general form:
The Lindblad operators,
account for the effect
of the external environment on the reduced density operator, ̂ .
Consider the case, Lˆ   qˆ :
Then the position representation for the environmental term
becomes
2

2

2
2
2
q  q'  (q, q' ) .
|env  q, q' 
2q ' q  q '  q  ( q, q ' )  
t
2
2


The same form as obtained for a weakly scattering environment.
Hermitian Lindblad operators lead to decoherence,
but no dissipation.
Not so with the master equation for quantum optics:
(A single cavity field mode interacting with 2-level atoms)
in terms of the field mode operators,
and
The Hamiltonian is just the harmonic oscillator,
so this is a quantum damped harmonic oscilator,
allowing for emission and absorption of photons
(depending on the temperature, through A).
6.Linblad Equation
for the Chord Function
This depends on product formulae for the chord representation,
where the delta-function eliminates the free side of the polygon
of the original cocycle.
In the unitary part of the Lindblad equation
there are products of two operators
and of three in the open part.
The problem is that common forms for Hˆ and Lˆ
are singular in the chord representation.
Therefore, these will be represented
by their Weyl symbol in the following formulae.
In the case where the Lindblad operators
are linear functions of pˆ and qˆ :
Note that the Hamiltonian is evaluated at the chord tips:
In terms of the double phase space variable
and
This is now compactified through the definition of
the double phase space Hamiltonian:
In the absence of dissipation,
will be constants of the full classical motion,
generated by
in double phase space.
Each of these reduced Hamiltonians
generate independent motions for each chord tip.
H
x  J 
x

and
H
x  J 
x

A classical canonical transformation, C ,
is evolved continuously
by the trajectory pairs: x (t ) e x (t ) ,
if the evolution K't is generated by H(x).
Kt : x0  xt 
 C ' (t ) : x  x  Kt  C  K  t ( x )
Mecânica quântica:
Cˆ (t )  Kˆ t Cˆ Kˆ t
Nonunitary double phase space Hamiltonian?
 x y     x
Hamilton’s equations:
x    x
y   y (    )
Hamiltonian motion in double phase space is compatible
with contraction of the centre-Wigner plane,
together with expansion of the chord plane.
The master equation for the chord function is thus
or, alternatively,
that is,
Exemplo: Hamiltonianas quadráticas:
Espaço simples

Espaço duplo
Em geral, os movimentos de x e y
estão acoplados, mas y=0
é sempre um plano invariante,
onde a evolução é gerada por H’(x).
Solution for a quadratic Hamiltonian:
The unitary evolution of the system is simply:
given in terms of the classical Poisson brackets,
just as for the Wigner function.
The open term, for each Lindblad operator, is
Then the exact general solution is simply obtained
from the classical evolution,
as
Thus,
the amplitude for long chords
t
of
h the classically evolving chord function
is
u dampened by the decoherence functional:
s
,
taken over trajectory pairs,
or a single trajectory in double phase space.
The latter interpretation is mandatory, in the presence of dissipation.
The dissipative part of the double Hamiltonian
0
0
expands  ( , t ) , while its Fourier transform, W ( x, t )
contracts.
But this is coarse-grained by
F.Τ .
in the convolution for the Wigner function:
W ( x, t ) W ( x, t )  F . Τ .
0
Since, the classical evolution is linear,
the decoherence functional is a quadratic function of the chords.
Therefore,

is a Gaussian in chord space, which narrows in time.
Its Fourier transform is:
where,
The width of the Gaussian decoherence window
that coarsegrains the Wigner function is  det M (t ) ,
which equals 0 at t=0 .
When det M (t )  1, then this Gaussian could be
the Wigner function of a pure squeezed state,
so that the evolved Wigner function
could be identified with a Husimi function.
Then, the evolved Wigner function must be positive!
The time for positivity
is independent of the initial pure state.
This time does depend on both the Hamiltonian
and the Lindblad operators.
A longer time makes all P-functions positive.
If the Lindblad operators, Lˆk ,
are all self-ajoint,
then decoherence
and diffusion,
but no dissipation.

Example:
Evolution of the Wigner function
For the “Schrödinger cat”
( Lˆ  qˆ and
Hˆ  0)
Damped
Harmonic
Oscillator
7.Semiclasical Markovian
Wigner and Chord functions
Insert the semiclassical chord function into
the Lindblad equation and
perform the integrals
by the method of stationary phase:
Semiclassical pure state:
Chords and centres are conjugate coordinates for double phase space.
Pictured in double phase space, both the Wigner function
and the chord function are just WKB wave functions:
y
y j ( x)  J j 
S j
x
y (x)
j
x
or else:

W ( x)   a j ( x) exp i S j ( x)
 k
 k
xk ( y ) 
J
,
y

 ( )   A k ( ) exp i  k ( )
k

Semiclassical evolution of the chord function
employs the solution of the Hamilton-Jacobi equation.
In terms of the original Hamiltonian, this is
in the nondissipative case, or
This is an ordinary H-J equation in double phase space:
Stationary phase evaluation of the commutator is
For general Lindblad operators, the open term can also
be evaluated by stationary phase as:
So we can include
in the double phase space Hamiltonian.
If we ignore the Hamiltonian motion,
then we can consider the action,
to be constant. Then, only the WKB amplitudes
evolves as
and
This procedure is analogous to that leading to the Trotter formula
for path integrals.
0

If j ( , t ) is the WKB-evolution
of a branch of the initial chord function,
in double phase space, then
recall that
 j ( ,0)
 0j ( , t )   0j ( (t ))
Here the decoherence functional is
and
Cancelation of long chords

Cancelation of quantum correlations
The dissipative part of the double Hamiltonian
0
0
expands  j ( , t ) , while its Fourier transform, W j ( x, t )
contracts.
But this is coarse-grained by
F.Τ .{
}
In the convolution for the Wigner function:
W j ( x, t )  W ( x, t )  F . Τ . {
0
j
}