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,for detecting some physical property for one particle,or the detection of for the second particle.

,11 O

12 O

We know that the projection of theWigner functiononto any Lagrangian plane produces a classical probability distribution.So, constructing our observables from regions on such a planeshould lead to measurements that satisfy the CHSH inequality.

In the case of a bipartite system, we can define such Lagrangian planes as , such that

with conjugate variables, and ,such that are symplectic transformations.

1'p 2'pjj xx '

Now define projection operators, and ,for the variable to be in the interval a or band the four observables,which take the value +1

jaP jbPjq'

IPO jajaˆˆ2ˆ

In terms of the Wigner function:

)','(''''ˆ21211

aa pqWdpdpdqdqP

0'ˆ,'ˆ 21 qq

Projection onto the a probability. ,',' 21 qqProjection onthe interval a

21 ',' qq

Since the product of commuting observables are also observables:

th

these can be measured and their expectation values satisfy

= .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.

,)','(''''ˆˆ212121 pqWdpdpdqdqPP

aaaa

with

This argument does not depend on whetherthe bipartite state is entangled,or not.

What if the Wigner function itself is positive definite?Then, even projectors over noncommuting variables,such that

)','(''''ˆ21211

aa pqWdpdpdqdqP

,)','(''''ˆ21211

bb pqWdqdqdpdpP

and

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 stateshas a positive Wigner function:The P-function with positive coefficients.

But this condition is only sufficient, not necessary.

WignerQ = Husimi P graincoarse

graincoarse

So, probabilities obtained over intervals forsqueezed 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:

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

)()()(and)()()( 22112211 xxx WWW

Then, after a rotation,4

The reduced density is just a section, so

,2

,22

,2

)(' 21212

21211

qqppqqpp

.2

,22

,2

)(' 112

1111

qpqp

To show how classical an entanglement can be,choose a simple Gaussian state, the product ofHO ground states:

is also a Gaussian, with elliptic level curves that are also rotated. After rotation and the partial trace:

,1

exp1

)( 22

j

jj

jjjW pqx

.2

1

2exp

2

1)(

22

pj

j

qjjjj

or

‘ ‘

.2

111

2exp

2

1)('

2

1

21

2

12111

pq

The narrowing of the Gaussian shows that the state is not pure.The Wigner function is more intuitive:Obtained by taking the Fourier transform,

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!

.)(

2

)(

2exp

)(

2)(' 2

121

21

21

21

21

2111

qp

x

W

.1

exp1

)( 22

j

jj

jjjW pqx

Go back to the product state of both

So, the original EPR state does not violate Bell inequalitiesfor 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 operatorsfor a given subspace?

The fact that the Wigner function

is symmetric with respect to the origin implies that .0ˆ,'ˆ 0 R .0'ˆ,'ˆso 011 R

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 originimplies 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 subsystemscannot distinguish this pure state from a classical distribution,but reflection eigenvalues are purely quantum.

Violation of CHSH:

Recall that the correlation for reflection measurementson either subsystem is given by the Wigner function:

We have already examined this at the origin.The decay of the Wigner function for large argumentsimplies that

sinks from 2, its maximal classical value, obtained at the origin,to its limiting value, 1.

Butbecause the expansion,

leads to

Thus, the smoothed EPR state can be measuredin ways that lead to violation of Bell’s inequalities.Banaszek and Wodkiewicz have proposed an experimentin quantum optics to achieve this.

A note on classicality versus hidden variables:

Bell’s inequalities set limits to the correlationsof 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 entanglementof the system under consideration with an uncontroled system, labledthe environment.

Generally we do not know the initial state of the environment:

Further averages, beyond the implicit averagein the reduced density operator.

A simple example: Weak scattering of many light particles.

If the duration of a single scattering processis short compared to the typical time scalesof the system evolving by itself:

scatt.|ˆ

ˆ,ˆˆ

tiH

ti

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 transportinformation about the system!

[Joos, inGiulini et. al.]

Consider a single scattering event from the system,if it is initially in the eigenstate .Then, if is the initial state of the environment,

n

0

,ˆ' 00 nn Snnn

where is the scattering operator (S-matrix)for this configuration of the system.

nS

For a general initial state,

,'0 nnn

nn

ncnc

so, the reduced density operator changes accordingly:

,''ˆ *

,

*

,

mnccmncc nmnmmn

nmmn

sys

because .''''tr nmmn

Thus, the matrix elements of the density operator evolve as

.'' 00 nmnmnmnmnm SS

If the overlap is close to unity,

,100 nmSS

then the effect of many collisions, with the ratewill be:

,

.exp1 tnmt

nmnm

nmt

scatt.|ˆ

with .1 00 nmSS

For the diagonal terms,Thus the trace of is not affected and only the offdiagonal terms decay with decoherence.

.0 mn

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,Then, for a single scattering event:

).(q

.)'(*)()'(*)()',( ' qq SSqqqqqq

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:

qkkikkSkkSq 'exp)',()',( 0

Then, for ,kkk

.''exp*'

'' qqkkiSS kk

kkqq

22*

''

''2

1''1 qqkkqqkkikk

kk

So, averaging over many scattering events:

.'exp)'(*)()',( 2qqtqqqq

)',('',| 2 qqqqqqt env

The Lindblad equation has the general form:

The Lindblad operators, account for the effectof the external environment on the reduced density operator,

Consider the case,Then the position representation for the environmental termbecomes

.

:ˆˆ qL

.)',('2

)',(''22

',| 22

222

qqqqqqqqqqqqt env

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 forare singular in the chord representation.

Therefore, these will be represented by their Weyl symbol in the following formulae.

LH ˆandˆ

In the case where the Lindblad operators are linear functions of :ˆandˆ qp

Note that the Hamiltonian is evaluated at the chord tips:

In terms of the double phase space variableand

This is now compactified through the definition ofthe 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.

x

Hx J

x

Hx Jand

A classical canonical transformation, C ,is evolved continuously by the trajectory pairs:if the evolution is generated by H(x).

)(:)(': 0 xKCKxxtCxxK tttt

Mecânica quântica:

tt KCKtC ˆˆˆ)(ˆ

,)(e)( txtx

tK '

Nonunitary double phase space Hamiltonian?

xyx

Hamilton’s equations: )( yyxx

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:

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).

Espaço simples

Espaço duplo

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,

Thus, the amplitude for long chords of the classically evolving chord functionis dampened by the decoherence functional:

taken over trajectory pairs, or a single trajectory in double phase space.The latter interpretation is mandatory, in the presence of dissipation.

),(0 tThe dissipative part of the double Hamiltonianexpands , while its Fourier transform,contracts.

),(0 txW

But this is coarse-grained by

.Τ.F

in the convolution for the Wigner function:

.Τ.F ),(),( 0 txWtxW

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 windowthat coarsegrains the Wigner function iswhich equals 0 at t=0 .

,)(det tM

When then this Gaussian could bethe Wigner function of a pure squeezed state,so that the evolved Wigner functioncould be identified with a Husimi function.

,1)(det tM

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 Hamiltonianand the Lindblad operators.

A longer time makes all P-functions positive.

If the Lindblad operators, are all self-ajoint,then decoherence and diffusion, but no dissipation.

Example:Evolution of the Wigner functionFor the “Schrödinger cat”

,ˆkL

)0ˆandˆˆ( HqL

DampedHarmonic Oscillator

7.Semiclasical Markovian Wigner and Chord functions

Insert the semiclassical chord functionthe semiclassical chord function into the the LindbladLindblad equation and perform the integrals by the method of stationary phasethe 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 functionand the chord function are just WKB wave functions:

y

)(xy

x

j

ji

j xSxaxW )(exp)()(

x

Sxy j

jj

J)(

k

ki

k )(exp)()( A

,)(

kkk y

yx Jor else:

Semiclassical evolution of the chord functionemploys 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:

So we can includein the double phase space Hamiltonian.

For general Lindblad operators, the open term can alsobe evaluated by stationary phase as:

Stationary phase evaluation of the commutator is

If we ignore the Hamiltonian motion,then we can consider the action,to be constant. Then, only the WKB amplitudesevolves as

and

This procedure is analogous to that leading to the Trotter formulafor path integrals.

If is the WKB-evolution of a branch of the initial chord function, in double phase space, then

)0,( j

),(0 tj

Here the decoherence functionaldecoherence functional is

and

Cancelation of long chords Cancelation of quantum correlations

recall that ))((),( 00 tt jj

),(0 tj The dissipative part of the double Hamiltonianexpands , while its Fourier transform,contracts.

),(0 txW j

But this is coarse-grained by

{.Τ.F }

In the convolution for the Wigner function:

{.Τ.F } ),(),( 0 txWtxW jj