Quantum entanglement Masatsugu Sei Suzuki Department of...

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Quantum entanglement Masatsugu Sei Suzuki

Department of Physics, SUNY at Binghamton (Date: November 30, 2015)

Here we discuss the physics of quantum entanglement. At first, undergraduate students who

just want to know the essential points of the quantum entanglement, may encounter some difficulty in understanding the definition of technical words, such as spooky action at a distance, non-locality, locality, hidden variable theory, separability, qubit, and so on. The definition of these words are given in the APPENDIX (source: Wikipedia).

The derivation of the Bell inequality is mathematically not so complicated. It is essential for ones to verify that the Bell inequality is not satisfied for the quantum entanglement phenomena from the experimental sides with the use of entangled spins or photons. So far so many books on the quantum entanglement, quantum information, and quantum computer have been published. Even after I read these books including textbooks of quantum mechanics, I have not understood sufficiently what is going on the spooky action at a distance. In order to teach the quantum entanglement undergraduate students, I felt it necessary to understand such a weirdness of the quantum entanglement in much more detail. While I struggled to understand the spooky action at a distance (named by Einstein), I had a good opportunity to read a book entitled, Einstein: His Life and Universe (by W. Issacson). Although this book is not written by a physicist doing the research on this topics (I am not sure whether Issacson is a physicist or not), I realize that the weirdness of the behavior of the quantum entanglement can be well described in this book. Of course, physicists who want to know the essence of the weird behavior based on the mathematics, may not be satisfied with the simple and clear explanations given by Issacson.

Here the content of the book is summarized as follows. (a) Quantum mechanics asserts that particles do not have a definite state except when

observed, and two particles can be in an entangled state so that the observation of one determines a property of the other instantly. As soon as any observation is made, the system goes into a fixed state.

(b) This may be conceivable for the microscopic quantum realm, but it is baffling when one

imagines the intersection between the quantum realm and observable everyday world. (c) The EPR paper would not succeed in showing that quantum mechanics was wrong. But

it did eventually become clear that quantum mechanics was incompatible with our common sense understanding of locality- our aversion to spooky action at a distance. The odd thing is that Einstein, apparently, was far more right than he hoped to be.

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http://en.wikipedia.org/wiki/John_Stewart_Bell

The Bell test experiments serve to investigate the validity of the entanglement effect in quantum mechanics by using some kind of Bell inequality. John Bell published the first inequality of this kind in his paper "On the Einstein-Podolsky-Rosen Paradox". Bell's Theorem states that a Bell inequality must be obeyed under any local hidden variable theory but can in certain circumstances be violated under quantum mechanics. The term "Bell inequality" can mean any one of a number of inequalities — in practice, in real experiments, the CHSH or CH74 inequality, not the original one derived by John Bell. It places restrictions on the statistical results of experiments on sets of particles that have taken part in an interaction and then separated. A Bell test experiment is one designed to test whether or not the real world obeys a Bell inequality. _____________________________________________________________________________ Alain Aspect (born 15 June 1947) is a French physicist noted for his experimental work on quantum entanglement. Aspect is a graduate of the École Normale Supérieure de Cachan (ENS Cachan). He passed the 'agrégation' in physics in 1969 and received his master's degree from Université d’Orsay. He then did his national service, teaching for three years in Cameroon.

http://en.wikipedia.org/wiki/Alain_Aspect

In the early 1980s, while working on his PhD thesis from the lesser academic rank of lecturer, he performed the elusive "Bell test experiments" that showed that Albert Einstein, Boris Podolsky and Nathan Rosen's reductio ad absurdum of quantum mechanics, namely that it implied 'ghostly action at a distance', did in fact appear to be realized when two particles were separated by an arbitrarily large distance (EPR paradox). A correlation between their wave functions remained, as they were once part of the same wave-function that was not disturbed before one of the child particles was measured.

If quantum theory is correct, the determination of an axis direction for the polarization measurement of one photon, forcing the wave function to 'collapse' onto that axis, will influence the measurement of its twin. This influence occurs despite any experimenters not knowing

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which axes have been chosen by their distant colleagues, and at distances that disallow any communication between the two photons, even at the speed of light.

Aspect's experiments were considered to provide overwhelming support to the thesis that Bell's inequalities are violated in its CHSH version. However, his results were not completely conclusive, since there were so-called loopholes that allowed for alternative explanations that comply with local realism.

Stated more simply, the experiment provides strong evidence that a quantum event at one location can affect an event at another location without any obvious mechanism for communication between the two locations. This has been called "spooky action at a distance" by Einstein (who doubted the physical reality of this effect). However, these experiments do not allow faster-than-light communication, as the events themselves appear to be inherently random. _____________________________________________________________________________ 1. What is quantum entanglement? (summary of this chapter)

Entanglement is one of the strangest predictions of quantum mechanics. Two objects are entangled if their physical properties are undefined but correlated, even when the two objects are separated by a large distance. No mechanism for entanglement is known, but so far experiments universally show that nonlocal entanglement is real. Something that happens to one particle does affect, instantaneously, what happens to the second particle, no matter how far it may be from the first one.

Fig. Two objects are entangled if their physical properties are undefined but correlated, even

when the two objects are separated by a large distance. The EPR paper would not succeed in showing the quantum mechanics was wrong. But it did eventually become clear that quantum mechanics was, as Einstein argued, incompatible with

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our common-sense understanding of locality - our aversion to spooky action at a distance. The odd thing is that Einstein, apparently, was far more right than he hoped to be. 2. EPR (Einstein, Podolsky, and Rosen) (1935)

In 1935, Einstein, Podolsky, and Rosen (EPR) pointed out the incompleteness of quantum mechanics. This argument is one of the most remarkable attacks on quantum theory ever launched. Recall that, according to quantum theory, measurements can be made of only one of any pair of complementary variables: position or momentum, energy or time. But the simultaneous measurement of both members of such a pair is impossible. The Einstein, Podolsky and Rosen (EPR) paper argued that such measurements were quite possible, and it gave a simple description of how to carry them out. Thus, according to their analysis, it was possible to obtain a more complete description of physical reality. Their conclusion was that quantum theory was incomplete. 3. Bohm’s version of EPR (1951) and Bell’s inequality

Their original gedanken experiment was revised by Bohm (1951) as the form of Stern-Gerlach measurements of two spin 1/2 particles; Bohm's version of EPR. Suppose that two particles with spin 1/2 are generated from the system with spin 0. Before the measurement, we do not know the spin directions of these two particles with spin 1/2. Suppose that the spin direction (the quantized axis +z) of the particle-1 is determined by Alice (the observer of the SG-1). Then the spin direction of the particle -2 can be uniquely determined by Bob (the second observer) to be the -z axis. This result is consistent with that derived from the spin angular momentum conservation. This result is rather different from that in quantum mechanics. After the first measurement (by Alice), the original state is changed into the new state. The second measurement is influenced by the first measurement. There is some probability of finding the partice-2 having the +z spin direction, as well as the probability of finding the particle having the -z direction.

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6

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7

of reproducing the expectations of quantum mechanics with a model in which the two particles can act as classical-like independent objects that cannot communicate after they have become separated.

http://www.youtube.com/watch?v=qXvZpn_dnMs&list=TLwELoPwvGH1pgMhPqpa6ojOCe2M6yhUU4

Suppose that each of the detectors is capable of measuring the spin of the approaching

particle in some direction that is only decided upon when the two particles are well separated from each other. The problem is to see whether it is possible to reproduce the expectations of quantum mechanics using a model in which the particles are regarded as unconnected independent classical-like particles, each one being unable to communicate with the other after they have separated. It turns out, because of a remarkable theorem due to the Northern Irish physicist John S. Bell, that it is not possible to reproduce the predictions of quantum theory in this way. Bell derived inequalities relating the joint probabilities of the results of two physically separated measurements that are violated by the expectations of quantum mechanics, yet which are necessarily satisfied by any model in which the two particles behave as independent entities after they have become physically separated. Thus, Bell-inequality violation demonstrates the presence of essentially quantum-theoretic effects—these being effects of quantum entanglements between physically separated particles—which cannot be explained by any model according to which the particles are treated as unconnected and independent actual things. 4. Locality (by Einstein)

Suppose that Alice is very far from Bob such that the propagation time for informing the result of Alice to Bob is long enough. Bob may measure the spin direction of the partricle-2, just before the result of the measurement by Alice is informed to Bob. In this case, if the spin direction of the particle 1 is measured by Alice as +z direction, before the second experiment (by Bob), we can conclude the spin direction of the particle-2 (measured by Bob) as -z direction according to the spin angular momentum conservation law. How can we explain such gedanken experiments in terms of quantum mechanics? Is there some possibility of hidden variables to determine the result of spin directions of the two particles during the production of two particles? (a)

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(b)

(c)

(d)

(e)

9

(f)

(g)

http://www.youtube.com/watch?v=qXvZpn_dnMs&list=TLwELoPwvGH1pgMhPqpa6ojOCe2M6yhUU4

Fig. Suppose that the distance between Alice and source is much shorter than that between Bob and source. The distance between Alice and Bob is long so that it takes long times for them to communicate with the speed of light. The particle 1 with up-state spin from the source moves to Alice, while the particle 2 with down-state spin from the source simultaneously moves to Bob. Because of the short distance, Alice first measures the

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spin direction of the particle 1 as the up-spin. At this time, the particle 2 does not reach Bob. Immediately after the measurement by Alice. Alice lets Bob know about her result at the speed of light. Before the light reaches Bob, the particle 2 reaches Bob. So Bob measures the spin direction of the particle 2 as the spin-down, before the information from Alice reaches Bob. At this moment, Bob comes to know the result of Alice without information from Alice.

5. Significance of the Bell's inequality (Aczel)

Bell’s theorem concerns a very general class of local theories with hidden, or supplementary, parameters. The assumption is as follows: suppose that the quantum theory is incomplete but that Einstein’s ideas about locality are preserved. We thus assume that there must be a way to complete the quantum description of the world, while preserving Einstein’s requirement that what holds true here cannot affect what holds true there, unless a signal can be sent from here to there (and such a signal, by Einstein’s own special theory of relativity, could not travel faster than light). In such a situation, making the theory complete means discovering the hidden variables, and describing these variables that make the particles or photons behave in a certain way.

Einstein had conjectured that correlations between distant particles are due to the fact that their common preparation endowed them with hidden variables that act locally. These hidden variables are like instruction sheets; and the particles’ following the instructions, with no direct correlations between the particles, ensures that their behavior is correlated. If the universe is local in its nature (that is, there is no possibility for super-luminal communication or effect, i.e., the world is as Einstein viewed it) then the information that is needed to complete the quantum theory must be conveyed through some pre-programmed hidden variables.

John Bell had demonstrated that any such hidden-variable theory would not be able to reproduce all of the predictions of quantum mechanics, in particular the ones related to the entanglement in Bohm’s version of EPR. The conflict between a complete quantum theory and a local hidden variables universe is brought to a clash through Bell’s inequality. 6. EPR argument on the simultaneous measurement (Bellac Quantum Physics) (a) Perfect anti-correlation

Let us suppose that we are capable of making a state

][2

1zzzz

of two identical spin-1/2 particles, with the two particles traveling with equal momenta in opposite directions. For example, they could originate in the decay of an unstable particle of

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zero spin and zero momentum, in which case momentum conservation implies that the particles move in opposite directions. An example which is simple theoretically (but not experimentally) is the decay of a 0 meson into an electron and a positron:

ee0 . Two experimentalists, conventionally named Alice and Bob, measure the spin component of each particle on a certain axis when the particles are very far apart compared with the range of the force and have not interacted with each other for a long time. For clarity, in this figure the axes used for spin measurement are taken to be perpendicular to the direction of propagation, though this is not essential.

Fig. Configuration of an EPR type of experiment. This figure is applicable to the SG system for the spin system as well as the polarization filter system for the photon.

Using a Stern–Gerlach device in which the magnetic field points in the direction a, Alice

measures the spin component on this axis for the particle traveling to the left, particle 1, while Bob measures the component along the b axis of the particle traveling to the right, particle 2.

Let us first study the case where Alice and Bob both use the z axis, zeba . We assume that

the decays are well separated in time, and that each experimentalist can know if he or she is measuring the spins of particles emitted in the same decay. In other words, each pair (e+, e−) is perfectly well identified in the experiment.

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Using her Stern–Gerlach device, Alice measures the z component of the spin of particle 1, azS , with the result +ħ/2 or -ħ/2, and Bob measures b

zS of particle 2. Alice and Bob observe a

random series of results +ħ/2 or −ħ/2. After the series of measurements has been completed, Alice and Bob meet and compare their results. They conclude that the results for each pair exhibit a perfect (anti-) correlation. When Alice has measured +ħ/2 for particle 1, Bob has measured −ħ/2 for particle 2 and vice versa. To explain this anticorrelation, let us calculate the result of a measurement in the state

][2

1zzzz

of the physical property ]ˆˆ[ bz

az SS , Hermitian operator acting in the tensor product space of the

two spins. It is found that is an eigenket of ]ˆˆ[ bz

az SS with eigenvalue -ħ2/4

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]ˆˆ[2b

zaz SS .

Measurement of ]ˆˆ[ bz

az SS must then give the result −ħ2/4, which implies that Bob must

measure the value −ħ/2 if Alice has measured the value +ħ/2 and vice versa. Within the limit of accuracy of the experimental apparatus, it is impossible that Alice and Bob both measure the value +ħ/2 or −ħ/2.

Upon reflection, this result is not very surprising. It is a variation of the game of the two customs inspectors. Two travelers 1 and 2, each carrying a suitcase, depart in opposite directions from the origin and eventually are checked by two customs inspectors Alice and Bob. One of the suitcases contains a red ball and the other a green ball, but the travelers have picked up their closed suitcases at random and do not know what color the ball inside is. If Alice checks the suitcase of traveler 1, she has a 50% chance of finding a green ball. But if in fact she finds a green ball, clearly Bob will find a red ball with 100% probability. Correlations between the two suitcases were introduced at the time of departure, and these correlations reappear as a correlation between the results of Alice and Bob.

However, as first noted by Einstein, Podolsky, and Rosen (EPR) in a celebrated paper (which used a different example, ours being due to Bohm), the situation becomes much less commonplace if Alice and Bob decide to use the x axis instead of the z axis for another series of

measurements. Since is invariant under the rotation, if Alice and Bob orient their Stern–

Gerlach devices in the x direction, they will again find that their measurements are perfectly anticorrelated, because

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4

]ˆˆ[2b

xax SS

For any direction n in the z-x plane, we also have the same relation

4

]ˆˆ[2ba SS nn

Measurement of ]ˆˆ[ ba SS nn must then give the result −ħ2/4, which implies that Bob must

measure the value −ħ/2 if Alice has measured the value +ħ/2 and vice versa, when the magnetic field is applied along the direction n.

Now we assume that

)cos,0,(sin 111 n )cos,0,(sin 222 n

Then we have

])cos((2

1)sin([

4]ˆˆ[ 2121

2

11 ba SS nn

where

)(2

1zzzz

When 21 , it is clear that is the eigenket.

((Mathematica)) Proof for 4

]ˆˆ[2ba SS nn

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(b) Possibility of simultaneous measurement: locality principle

The viewpoint underlying the EPR analysis of these results is that of “realism”: EPR assume that microscopic systems possess intrinsic properties which must have a counterpart in the physical theory. More precisely, according to EPR, if the value of a physical property can be predicted with certainty without disturbing the system in any way, there is an “element of

reality” associated with this property. For a particle of spin 1/2 in the state z . Sz is a

property of this type because it can be predicted with certainty that 2/zS . However, the

value of Sx in this same state cannot be predicted with certainty (it can be 2/ or 2/ with

50% probability of each); xS and zS cannot simultaneously have a physical reality. Since the

operators xS and zS do not commute, in quantum physics it is impossible to attribute

simultaneous values to them. In performing their analysis, EPR used a second hypothesis, the locality principle, which

stipulates that if Alice and Bob make their measurements in local regions of space-time which cannot be causally connected, then it is not possible that an experimental parameter chosen by Alice, for example the orientation of her Stern–Gerlach device, can affect the properties of particle 2.

According to the preceding discussion, this implies that without disturbing particle 2 in any

way, a measurement of azS by Alice permits knowledge of b

zS with certainty, and a

measurement of axS permits knowledge of b

xS with certainty. If the “local realism” of EPR is

accepted, the result of Alice’s measurement serves only to reveal a piece of information which was already stored in the local region of space-time associated with particle 2. A theory that is

Clear "Global` " ;

exp : exp . Complex re , im Complex re, im ;

110

;

201

;

121

2KroneckerProduct 1, 2 KroneckerProduct 2, 1 ;

x PauliMatrix 1 ; y PauliMatrix 2 ; z PauliMatrix 3 ;

n1 Sin Cos , Sin Sin , Cos . 0;

X1 n1 1 x n1 2 y n1 3 z FullSimplify;

X12 KroneckerProduct X1, X1 ;

X12. 12 Simplify

0 ,1

2,

1

2, 0

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more complete than quantum mechanics should contain simultaneous information on the values

of bxS and b

zS and be capable of predicting with certainty all the results of measurements of

these two physical properties in the local region of space-time attached to particle 2. The

physical properties bxS and b

zS then simultaneously have a physical reality, in contrast to the

quantum description of the spin of a particle by a state vector. EPR do not dispute the fact that quantum mechanics gives predictions that are statistically

correct, but quantum mechanics is not sufficient for describing the physical reality of an individual pair. Within the framework of local realism such as that defined above, the EPR argument is unassailable and the verdict incontestable: quantum mechanics is incomplete! Nevertheless, EPR do not suggest any way of “completing” it, and we shall see in what follows that local realism is in conflict with experiment.

According to local realism, even if an experiment does not permit the simultaneous

measurement of bxS and b

zS , these two quantities still have a simultaneous physical reality in the

local region of space-time attached to particle 2, and owing to symmetry the same is true for axS

and azS , of particle a. This ineluctable consequence of local realism makes it possible to prove

the Bell inequalities, which fix the maximum possible correlations given this hypothesis. (c) Rotational invariance of the Bell’s state and perfect correlation

The Bell’s state is an eigenvector of all components of the total spin with an eigenvalue 0. In other words, the Bell’s state is rotationally invariant,

][2

1][

2

121212111

zzzz nnnn

for individual measurement of nS 1ˆ of particle 1 and the same spin component nS 2

ˆ of

particle 2, where n is a arbitrary unit vector specified in the spherical polar coordinate. It

follows that the measurements of nS 1ˆ and nS 2

ˆ , always give results that sum to zero (spin

angular momentum conservation), i.e., if the measurement of nS 1ˆ gives the result of

2

,

then the measurement of nS 2ˆ always gives

2

. We say that these observables exhibit perfect

correlation. ((Note))

There are two significant factors in the quantum entanglements. The first is the invariance of the Bell’s state under the rotation. The second is that the spin operators at the same site are not commutable, which is the essential nature in the quantum mechanics.

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6. Hidden variable theory Suppose that Alice observes the spin—up state (denoted as }{ a using the SG with the

direction a (arbitrary). Since the spin angular momentum is conserved, she also knows that Bob will observe the spin-down without any knowledge of the result from Bob.

{a+, a-} Similarly, Alice observes the spin—down state (denoted as }{ a using the SG with the

direction a (arbitrary). Since the spin angular momentum is conserved, she also knows that Bob will observe the spin-up }{ a , without any knowledge of the result from Bob.

{a-, a+}.

7. Prediction using quantum mechanics

We consider the above argument using our knowledge of quantum mechanics. The original state before Alice measures,

],,[2

10 zzzz

Fig. Stern-Gerlach experiment. Note that z and z , for simplicity. zea .

After Alice measures using the SG with the z direction, she finds that the particle 1 is in the spin-up state. The state is collapsed into

211 , zzzz

y0>

Alice particle-1

SG-z

+>1

->1

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where the probability of finding the system in the state 1 is 1/2. Then Bob measures using

the SG with the direction b.

222 2sin

2cos zez i

b

222 2cos

2sin zez i

b

where a is along the z direction. The angles and are the polar angles that give the orientation of b. Note that

cosba

with 1 ba .

y1>

Bob particle-2

SG-b

b+>2

b->2

18

Fig. The direction of a for Alice and b for Bob. a is along the z direction. The angles and are the polar angles that give the orientation of b. Note that cosba

The probability that Bob measures +1/2 is

2sin)( 22

22

zPB bb

The probability that Bob measures -1/2 is

x

y

z a; Alice

q

dq

f df

r

drr cosq

r sinq

r sinq df

rdq

er b; Bob

ef

eq

19

)1(2

1)cos1(

2

1

2cos)( 22

22 babb zPB

In other words, these probabilities depend on the angle . When ba ,

1)( bBP

as Alice predicted. So the quantum mechanics is consistent with common sense. ((Note))

We note that the probability for finding the state b,z is evaluated as

2cos

2

1

2

1, 222

0

zzP bb

with

],,[2

10 zzzz .

When = 0, the probability becomes P = 1/2 (50%), which is consistent with the result described later. 8. One type of SG system for the measurement

We consider a two-electron system in a spin singlet state, with a total spin zero.

],,[2

1zzzz

20

Fig. A schematic of the EPR experiment in which A measures the spin of particle 1 and B

measures the spin of particle 2 We note that for any unit vector n,

][2

1

],,[2

1

2121nnnn

ie

zzzz

This implies the invariance of the Bell’s state under the rotation [ )(ˆ)(ˆˆ yz RRR ]

0

1

1

0

2cos

2sin

2cos

2sin

2cos

2sin

2cos

2sin

2sin

2cos

2cos

2sin

2sin

2cos

2cos

2sin

2cos

2sin

2sin

2cos

]

2

2

2

2

2

2

2121

i

i

i

i

i

i

i

iiii

e

e

e

e

e

e

e

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Supposespins is

this case

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Spin zero s

1

21

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e first term

certain that

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he SG with z-z}; {-z, z}.res the state

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the state ke

particles) or

and right, to of the appros are in full f

Alice measurn state usingz direction), B

Note that th–z.

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pin-up s that

of S2z,

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{a+, a-}

Fig. L

( _______{a-, a+}

Fig. L

( ((Quant

}

Left: config(Bob). {a+}

__________}

Left: config(Bob). {a-} (

tum mechan

[2

10,0

0,0, zz

a

a

guration for (particle-1).

___________

guration for (particle-1).

nics, non-lo

, zz

2

1

,2

1

z

the particle. {a-} (partic

__________

the particle{a+} (partic

cality))

], zz

, zzz

22

e-1 (Alice). cle-2).

___________

e-1 (Alice). cle-2).

,2

1 zz

Right: con

__________

Right: con

], zz

a

a

nfiguration f

___________

nfiguration f

for the parti

__________

for the parti

icle-2

__

icle-2

23

or

2

10,0,

2 xzP

The probability is 50 %. Then both models give the same value for the probability. ((Note))

Suppose that Alice first does her measurement and find the state z . After that, Bob does his

measurement. Before his measurement, the state of the system already collapse into the z .

212

10,0 zz .

Suppose that Bob first does his measurement and find the state z . After that, Alice does her

measurement. Before her measurement, the state of the system already collapse into the state

z .

122

10,0 zz .

9. Two types of SG systems for the measurement

For a particular pair, there must be a perfect matching between particle 1 and particle 2 to ensure zero total angular momentum. If the particle 1 is of type {+z, -x}, then the particle 2 must belong to type {-z, +x}, and so on. The results of correlation measurements can be reproduced if the particle 1 and particle 2 are matched, as follows.

Particle 1 Particle 2 Population Event

A BSpin zero source

particle-1 particle-2

a

b

a

b

24

{a+, b+} {a-, b-} N1 = 25% E1 {a+, b-} {a-, b+} N2 = 25% E2 {a-, b+} {a+, b-} N3 = 25% E3 {a-, b-} {a+, b+} N4 = 25% E4

such as configurations

Fig. Left: configuration for the particle-1 (Alice). Right: configuration for the particle-2

(Bob). {a+, b+} (particle-1). {a-, b-} (particle-2). The population N1.


(Bob). {a+, b-} (particle-1). {a-, b+} (particle-2). The population N2.

a

b

a

b

a

b

a

b

25


(Bob). {a-, b+} (particle-1). {a+, b-} (particle-2). The population N3.


(Bob). {a-, b-} (particle-1). {a+, b+} (particle-2). The population N4. We note that the population for each event is the same:

NNNNN 4321

Probability:

},{ 21 ba : N2,

},{ 21 ba : N1,

},{ 21 ba : N4,

},{ 21 ba : N3,

},{ 21 aa : N1, N2,

},{ 21 aa : N3, N4

},{ 21 ab : N3,

},{ 21 ab : N1,

a

b

a

b

a

b

a

b

_______ When a

{{{{

and that{..} nota ((Hidde(i)

Io

Io

},{ 21 ab :

},{ 21 ab :

__________

= ez and b =

{+z, +x} {+z, -x} {-z, +x} {-z, -x}

t each of theation provide

en variable t

If A makes mof S2z, 50 %

If A makes mof S2z, 50 %

N4

` N2

___________

= ex. we have

{- {- {+ {+

ese distinct ges a non-qua

theory; loca

measuremenof B's measu

Pa{+{+{-{-

measuremenof B's measu

N4,

N2,

__________

e

-z, -x} -z, +x}+z, -x}+z, +x}

groups of parantum descri

ality))

nts of S1z andurements wi

article-1+z, +x}+z, -x}-z, +x}-z, -x}

nts of S1z andurements wi

26

___________

N1

N2

N3

N4

rticles is proiption of the

d obtain the ill yield /

Par{-z{-z{+z{+z

d obtain the vill yield 2/

__________

= 25%= 25%= 25%= 25%

oduced in eq state of the

value 2/ ,2/ . The even

rticle-2 z, -x} z, +x} z, -x} z, +x}

value - 2/ ,2 . The event

___________

E1 E2 E3 E4

qual numberparticle.

, and B maknts E1 and E

, and B makts E3 and E4

__________

. Note that a

kes measuremE2 are allowe

kes measuremare allowed.

____

a new

ments ed.

ments .

(ii)

Io

Io

If A makes mof S2x, 50 %

If A makes mof S2x, 50 %

Pa{+{+{-{-

measuremenof B's meas

Pa{+{+{-{-


Pa


nts of S1x andurements wi


nts of S1x andurements wi

article-1

27

Par{-z{-z{+z{+z

d obtain the ill yield /

Par{-z{-z{+z{+z

d obtain the vill yield 2/

Par

rticle-2 z, -x} z, +x} z, -x} z, +x}

value 2/ ,2/ . The even

rticle-2 z, -x} z, +x} z, -x} z, +x}

value - 2/

2 . The event

rticle-2

, and B maknts E1 and E

, and B makts E2 and E4

kes measuremE3 are allowe

kes measuremare allowed.

ments ed.

ments .

_______(iii)

Iot

Iot

__________

If A makes mof S2x, 25 % to measure o

If A makes mof S2x, 25 % to measure o

{+{+{-{-

___________


of S2x ( 2/ )

Pa{+{+{-{-


of S2x ( 2/ )

Pa{+{+{-{-

+z, +x}+z, -x}-z, +x}-z, -x}

__________

nts of S1z andsurements w and E2 for B


nts of S1z andsurements w and E3 for B


28

{-z{-z{+z{+z

___________

d obtain the will yield 2/

Bob to meas

Par{-z{-z{+z{+z

d obtain the vwill yield 2/

Bob to meas

Par{-z{-z{+z{+z

z, -x} z, +x} z, -x} z, +x}

__________

value 2/ ,2 and 25 %sure of S2x (-

rticle-2 z, -x} z, +x} z, -x} z, +x}

value - 2/ ,2 and 25 %sure of S2x (-

rticle-2 z, -x} z, +x} z, -x} z, +x}

___________

, and B makwill yield

2/ ) and


2/ ).

__________

kes measurem2/ . E1 for


_____

ments r Bob

ments r Bob

_______(iv)

Iot

Iot

_______10. Q

In qu

__________

If A makes mof S2z, 25 % to measure o

If A makes mof S2z, 25 % to measure o

__________Quantum m

uantum mec

___________


of S2z ( 2/ )

Pa{+{+{-{-


of S2z ( 2/ )

Pa{+{+{-{-

___________mechanics (n

chanics, we e

__________

nts of S1x andsurements w

and E1 for B


nts of S1x andsurements w

and E2 for B


__________non-locality)

express the s

29

___________

d obtain the will yield 2/

Bob to meas

Par{-z{-z{+z{+z

d obtain the vwill yield 2/

Bob to meas

Par{-z{-z{+z{+z

___________)

state 0,0 as

__________

value 2/ ,2 and 25 %ure of S2z (-

rticle-2 z, -x} z, +x} z, -x} z, +x}

value - 2/

2 and 25 %ure of S2z (-

rticle-2 z, -x} z, +x} z, -x} z, +x}

__________

s

___________


2/ ) and


2/ ).

___________

__________



__________

_____

ments r Bob

ments r Bob

_____

The ket

(a) P

or

(b) P

[2

10,0

x and th

(2

1x

(2

1x

Probability:

0,0, zz

, xzP

Probability:

, zz

he ket x a

)zz

)zz

particle 1 (

2

1

,2

1

z

2

10,0

2

particle 1 (

], zz

are defined b

1z ) and p

, zzz

1z ) and p

30

by

article 2 (

,2

1 zz

article 2 (

2z ),

], zz

2z ),

or

_______

(c) P

or

(d) P

0,0, zz

, zzP

__________

Probability:

0,0, xz

, xzP

Probability:

2

1

,2

1

z

2

10,0

2

___________

particle 1 (

2

1

,2

1

x

z

4

10,0

2

particle 1 (

, zzz

__________

1z ) and p

2

1

,

z

zzx

1z ) and p

31

,2

1 zz

___________

article 2 (

,2

1 xz

article 2 (

], zz

__________

2x ),

], zzx

2x ),

_________________________

32

2

1

2

1

],,2

1,,

2

10,0,

zx

zzxzzzxzxz

or

4

10,0,

2 xzP

In conclusion, the non-quantum model in which each of the particles in the two-particle system has definite attributes, is able to reproduce the results of quantum mechanics. _____________________________________________________________________ 12. Three types of SG systems: Bell's inequality (Townsend, Sakurai)

We now consider more complicated situation where the model leads to predictions different from the usual quantum-mechanical predictions. We start with three unit vectors, a, b, and c, which are, in general, not mutually orthogonal. We define {a-, b+, c+} as follows. The Stern-Gerlach experiment along the a direction yields - 2/ . The Stern-Gerlach experiment along the b direction yields 2/ . The Stern-Gerlach experiment along the c direction yields 2/ . There must be a perfect matching in the sense that the other particles necessarily belongs to type {a-, b+, c+} to ensure zero angular momentum.

((Hidden variable theory))

Note that the angular momentum is conserved. If Alice measures the spin up state using the SG with a direction), Bob will measure the spin-down state if he uses the SG with a direction for the measurement. (i)

Particle 1 Particle 2 Population Events

{a+, b+, c+} {a-, b-, c-} N1 E1

A BSpin zero source


a

b c

a

b c

33

{a+, b+, c-} {a-, b-, c+} N2 E2 {a+, b-, c+} {a-, b+, c-} N3 E3 {a+, b-, c-} {a-, b+, c+} N4 E4 {a-, b+, c+} {a+, b-, c-} N5 E5 {a-, b+, c-} {a+, b-, c+} N6 E6 {a-, b-, c+} {a+, b+, c-} N7 E7 {a-, b-, c-} {a+, b+, c+} N8 E8

such as configurations _____________________________________________________________________________


(Bob). {a+, b+, c+} (particle-1). {a-, b-, c-} (particle-2) __________________________________________________________________________


(Bob). {a+, b+, c-} (particle-1). {a-, b-, c+} (particle-2) __________________________________________________________________________

a

b c

a

b c

a

b c

a

b c

34


(Bob). {a-, b-, c+} (particle-1). {a+, b+, c-} (particle-2) ________________________________________________________________________

Probability: particle 1 (1

a ) and particle 2 (2

b ),

8

1

43),(

iiN

NNP ba


a ) and particle 2 (2

c ),

8

1

42),(

iiN

NNP ca


c ) and particle 2 (2

b ),

8

1

73),(

iiN

NNP bc

We note that

)()( 734243 NNNNNN

since Ni is positive. Then we have

0),(),(),( bcbaca PPP .

a

b c

a

b c

35

This is the Bell's inequality. (ii)

Similarly we get the Bell’s inequality as

0),(),(),( cbcaba PPP .

This inequality can be derived as follows. The probability: particle 1 (1

b ) and particle 2

(2

c ) is given by

8

1

62),(

iiN

NNP cb

We note that

)()( 624342 NNNNNN

since Ni is positive. This leads to the above inequality. ((Note)) Possible configurations

a

b c

a

b c

N1

36

a

b c

a

b c

N2

a

b c

a

b c

N3

a

b c

a

b c

N4

a

b c

a

b c

N5

37

_____________________________________________________________________________

Before we discuss this case, we note that

][2

1],,[

2

10,0

2121nnnn iezzzz

Here we use the notations

a

b c

a

b c

N6

a

b c

a

b c

N7

a

b c

a

b c

N8

38

2sin

2cos

2sin

2cos

i

i

e

zezn

,

2cos

2sin

2cos

2sin

i

i

e

zezn

and

nn 2

sin2

cos

z

nn

2cos

2sin

ii eez

Then we have

][2

1

])[2

sin2

(cos2

1

}2

cos2

sin2

cos2

sin

2cos

2sin{[

2

1}

2cos

2sin

2sin

2cos

2cos

2sin{[

2

1

]2

sin2

][cos2

cos2

sin{[2

1

]}2

cos2

sin][2

sin2

{[cos2

1

],,[2

10,0

2121

2121

22

2121

2

21

2

2121

21

2

21

2

21

2211

2211

nnnn

nnnn

nnnnnn

nnnn

nnnnnn

nnnn

nnnn

i

i

iii

ii

iii

ii

ii

e

e

eee

ee

eee

ee

ee

zzzz

((Mathematica))

_______

In qu

except f (1)

Fig. A

2

__________

uantum mec

[2

10,0

for the phase

Alice [the sp

2] measure t

ba 0,0,

___________

chanics, we e

, aa

e factor.

pin state a

the SG exper

b

a

2

1

,2

1

__________

express the s

], aa ,

a of the pa

riments.

a

aab

,

39

___________

state 0,0 as

2

10,0

article 1] and

ba ,2

1

__________

s

,[ bb

d Bob [the sp

aab ],

___________

], bb

pin state

__________

b of the pa

_____

article

where

(2)

Fig. A

m

(3)

Fig. A

m

),(P ba

b 2

cos

b 2

sin a

Alice (the sp

measure the

ba 0,0,

),(P ba

Alice [the sp

measure the

2

1 ab

a sin2ab

a cos2ab

pin state

SG experim

b

a

2

1

,2

1

2

1 ab

pin state

SG experim

2

12 a

a2

n ab

a2

s ab

a of the pa

ments.

a

aab

,

2

12 a

a of the pa

ments.

40

sin2

12 b

article 1) an

ba ,2

1

cos2

12 b

article 1] an

2n 2 ab

nd Bob (the

aa ],

2s2 ab

nd Bob [the

spin b o

spin b o

of the partic

of the partic

cle 2)

cle 2]

(4)

Fig. A

m

_______Similarl

Fig. A

m

ba 0,0,

),(P ba

Alice [the sp

measure the

ba 0,0,

),(P ba

__________ly, we have

Alice [the sp

measure the

),(P ca

b

a

2

1

,2

1

2

1 ab

pin state

SG experim

b

a

2

1

,2

1

2

1 2 ab

___________

pin state

SG experim

2

1 2 ca

a

aab

,

2

12 a

a of the pa

ments.

ab

aab

,

2

12 a

__________

a of the pa

ments.

2sin

2

1 22 ac

41

ba ,2

1

cos2

12 b

article 1] an

ba ,2

1

sin2

1 22b

___________

article 1] an

c

aab ],

2s2 ab

nd Bob [the

aab ],

2ab

.

__________

nd Bob [the

spin b o

___________

spin c o

of the partic

__________

of the partic

cle 2]

_

cle 2]

_______

Fig. A

m

Then we

or

or

Suppose

__________

Alice [the sp

measure the

),(P bc

e have

sin2

sin2 ab

abcos1

coscos ac

e that 2ab

co1)( f

___________

pin state

SG experim

2

1 2 bc

sin2

n 22 ac

accos1

coss abcb

2 , acbc

co2)2os(

__________

c of the pa

ments.

2sin

2

1 22 cb

2cb

cbcos1

1b

c

os

42

___________

article 1] an

b

(Bell's ineq

__________

nd Bob [the

quality)

___________

spin b o

__________

of the partic

____

cle 2]

43

Fig. f()<0 for 0<</2, which means that the Bell's inequality is violated. So the quantum mechanical predictions are not compatible with the Bell's inequality.

13. Probability for finding two particles in the specified state

We assume the population Ni (i = 1, 2, 3, …., 8). For example, N1 is the population for the case when the particle 1 (spin 1/2) is in the states {a+, b+, c+} and the particle 2 (spin 1/2) is in the state {a-, b-, c-}.

Particle-1 Particle-2 Population

{a+, b+, c+} {a-, b-, c-} N1 {a+, b+, c-} {a-, b-, c+} N2 {a+, b-, c+} {a-, b+, c-} N3 {a+, b-, c-} {a-, b+, c+} N4 {a-, b+, c+} {a+, b-, c-} N5 {a-, b+, c-} {a+, b-, c+} N6 {a-, b-, c+} {a+, b+, c-} N7 {a-, b-, c-} {a+, b+, c+} N8

Then we get the probability for each case,

8

1

43),(

iiN

NNP ba ,

8

1

21),(

iiN

NNP ba ,

8

1

42),(

iiN

NNP ca

p

4

p

2

3 p4

pq

1

2

3

4

f q

44

8

1

31),(

iiN

NNP ca ,

8

1

87),(

iiN

NNP ba ,

8

1

65),(

iiN

NNP ba ,

8

1

86),(

iiN

NNP ca ,

8

1

75),(

iiN

NNP ca ,

8

1

65),(

iiN

NNP ab ,

8

1

21),(

iiN

NNP ab ,

8

1

62),(

iiN

NNP cb ,

8

1

51),(

iiN

NNP cb ,

8

1

87),(

iiN

NNP ab ,

8

1

43),(

iiN

NNP ab ,

8

1

84),(

iiN

NNP cb ,

8

1

73),(

iiN

NNP cb ,

8

1

75),(

iiN

NNP ac ,

8

1

31),(

iiN

NNP ac ,

8

1

73),(

iiN

NNP bc ,

8

1

51),(

iiN

NNP bc ,

8

1

86),(

iiN

NNP ac ,

8

1

42),(

iiN

NNP ac ,

8

1

84),(

iiN

NNP bc ,

8

1

62),(

iiN

NNP bc .

Note that for example, ),( ba P is the probability of finding the particle 1 in the state a

and the particle 2 in the state b . Then we have the inequalities such as

0

)(),(),(),(

8

1

54

8

1

514243

ii

ii

N

NN

N

NNNNNNPPP cbcaba

45

0

)(),(),(),(

8

1

72

8

1

734342

ii

ii

N

NN

N

NNNNNNPPP bcbaca

0

)(),(),(),(

8

1

81

8

1

844221

ii

ii

N

NN

N

NNNNNNPPP cbcaba

,

0

)(),(),(),(

8

1

81

8

1

844331

ii

ii

N

NN

N

NNNNNNPPP bcbaca

0

)(),(),(),(

8

1

81

8

1

844331

ii

ii

N

NN

N

NNNNNNPPP cbbaac

It is noted that the sign of the following equation cannot be determined,

8

1

4543

8

1

544243

2

)(),(),(),(

ii

ii

N

NNNN

N

NNNNNNPPP cbcaba

14. Bell’s inequality (I): McIntyre, Quantum Mechanics

46

Bell's argument relies on observers A (Alice) and B (Bob) making measurements along a set of different directions. We consider three directions a, b, and c. Each observer makes measurements of the spin projection along one of these three directions, chosen randomly. Any single observer's result can be only spin up or spin down along that direction, but we record the results independent of the direction of the SG analyzers, so we denote one observer's result simply as + or -, without noting the axis of measurement. The results of the pair of measurements from one correlated pair of particles are denoted + -, for example, which means observer A recorded a+ and observer B recorded a-. There are only four possible system results: ++, +-, -+, or --. Even more simply, we classify the results as either the same, ++, --, or opposite, +-, or -+.

((Hidden variable theory)) Note that the angular momentum is conserved. If Alice measures the spin up state using the SG with a direction), Bob will measure the spin-down state using the SG with a direction. There are eight types measurements.

Particle 1 Particle 2 Population

{a+, b+, c+} {a-, b-, c-} N1 Type-1 {a+, b+, c-} {a-, b-, c+} N2 Type-2 {a+, b-, c+} {a-, b+, c-} N3 Type-3 {a+, b-, c-} {a-, b+, c+} N4 Type-4 {a-, b+, c+} {a+, b-, c-} N5 Type-5 {a-, b+, c-} {a+, b-, c+} N6 Type-6 {a-, b-, c+} {a+, b+, c-} N7 Type-7 {a-, b-, c-} {a+, b+, c+} N8 Type-8

A BSpin zero source


a

b c

a

b c

47

Fig. Type-1. Left: configuration for the particle-1 (Alice). Right: configuration for the

particle-2 (Bob). {a+, b+, c+} (particle-1). {a-, b-, c-} (particle-2) Type-1: N1

{a+, b+, c+} (particle-1)

{a-, b-, c-} (particle-2) The probability of observing the same spin directions: Psame

The probability of observing the opposite spin directions: Popp

Popp=9/9=1, Psame = 0

with the combinations

{a+, a-}, {a+, b-}, {a+, c-}, {b+, a-}, {b+, b-}, {b+, c-}, {c+, a-} {c+, b-}, {c+, c-},

Type-8: N8

a

b c

a

b c

48

Fig. Type-8 Left: configuration for the particle-1 (Alice). Right: configuration for the particle-2 (Bob). {a-, b-, c-} (particle-1). {a+, b+, c+} (particle-2)

Popp=9/9=1, Psame = 0

with the combinations

{a-, a+}, {a-, b+}, {a-, c+}, {b-, a+}, {b-, b+}, {b-, c+}, {c-, a+} {c-, b+}, {c-, c+},

Type-2: N2

Fig. Type-2 Left: configuration for the particle-1 (Alice). Right: configuration for the

particle-2 (Bob). {a+, b+, c-} (particle-1). {a-, b-, c+} (particle-2)

{a+, b+, c-} (particle-1)

{a-, b-, c+} (particle-2)

Popp=5/9, {a+, a-},{a+, b-}{b+, a-},{b+, b-}, {c-, c+}

Psame = 4/9 {a+, c+},{b+, c+}{c-, a-},{c-, b-}, Type-3 – 7 (similar to Type-2), N3, N4, N5, N6, N7

{a+, b-, c+} {a-, b+, c-} Type-3 {a+, b-, c-} {a-, b+, c+} Type-4 {a-, b+, c+} {a+, b-, c-} Type-5 {a-, b+, c-} {a+, b-, c+} Type-6 {a-, b-, c+} {a+, b+, c-} Type-7

49

Popp=5/9, Psame = 4/9

Then we have

9

4)(

9

4

9

4

)(9

4)(

9

4

)(9

4)(0

8

1

81

8

1

8187654321

8

1

76543281

ii

ii

ii

same

N

NN

N

NNNNNNNNNN

N

NNNNNNNNP

or

9

4sameP

and

9

5)(

9

4

9

5

)(95

)(94

)(95

)(1

8

1

81

8

1

8765432181

8

1

76543281

ii

ii

ii

opp

N

NN

N

NNNNNNNNNN

N

NNNNNNNNP

or

9

5oppP

50

_________________________________________________________________________ ((Prediction by Quantum mechanics)) evaluation of Psame and Popp

The values of Psame and Popp can be predicted from the quantum mechanics as follows. The probability:

2sin

2

1

2

1

2

1),( 222 abP

baabba

2cos

2

1

2

1

2

1),( 222 abP

baabba

2cos

2

1

2

1

2

1),( 222 abP

baabba

2sin

2

1

2

1

2

1),( 222 abP

baabba

Then we get

2sin

2sin

2

12),(),( 22 abab

same PPP

baba

2cos

2cos

2

12),(),( 22 abab

opp PPP

baba

for the configuration {a, b}.

The angle ab between the measurement directions of the observers A and B is = 0 in 1/3

of the measurements, bc = 2/3 in 1/3 and ca = 4/3 in 1/3. So the average probabilities are

9

4

2

13

sin3

2

2

0sin

3

13

sin3

1

3sin

3

1

2sin

3

1

22

222

cabcabsameP

51

9

5

2

13

cos3

2

2

0cos

3

12

cos3

1

2cos

3

1

2cos

3

1

22

222

cabcaboppP

These predictions of quantum mechanics are inconsistent with the range of possibilities that we derived for local hidden variable theories. 15. Bell's inequality (II)

We have shown the inequality based on the same argument which is derived above

0),(),(),( cbcaba PPP .

This can be rewritten as

02

cos2

1

2sin

2

1

2cos

2

1 222 bcacab .

When

acbcab , abac 2 , ab ,

then we have

0cos2

3sin

2cos

2

1)( 222

f .

We make a plot of f() as a function of . It is clearly shown that f() becomes negative between

/4 and /2.

52

Fig. Plot of )(f as a function of . It becomes negative between /4 and /2. This means

that the quantum mechanics is inconsistent with hidden variable theory (locality). 16. Spin correlation function (I)

Measurements of the spin components along two arbitrary directions a and b are performed

on two spin 1/2 particles in the singlet 0,0 . The results of each measurement finds the parallel

spin-up or spin-down along that particular axis. Denoting ),( baP the probabilities of

obtaining ±1 along a for particle 1 and ±1 along b for particle 2, the average value for the product of spins is given by

),(),(),(),(

),(

babababa

ba

PPPP

PPE opposame

Since

2sin

2

1),( 2 abP

ba ,

2sin

2

1),( 2 abP

ba (parallel spin alignment)

2cos

2

1),( 2 abP

ba ,

2cos

2

1),( 2 abP

ba (antiparallel spin alignment)

we get

p

4

p

2

3 p

4p

q

0.2

0.4

0.6

0.8

1.0

f q

53

)(

cos2

cos2

sin

),(),(),(),(),(

22

ba

bababababa

ab

abab

PPPPE

if a and b are unit vectors. E(a, b) is the spin correlation function and defined by

).()(ˆ)(ˆ),( 21 bababa E

17. Derivation of the spin correlation function (II): )(ˆ)(ˆ),( 21 baba E

Here we show that

abcos4

0,0)ˆ)(ˆ(0,02

21

bSaS ,

where ab is the angle between the unit vectors a and b. Therefore we have the spin correlation function by

)(ˆ)(ˆ

0,0)(ˆ)(ˆ0,0

0,0)ˆ)(ˆ(0,04

),(

21

21

212

ba

ba

bSaSba

E

where

aSa 11ˆ2

)(ˆ

, bSb 22ˆ2

)(ˆ

((Proof))

cossin

sincos

2ˆ

2)ˆ(

i

i

e

enσnS .

54

cos2

)ˆ

cos2

ˆ

zz

zz

nS

nS

Therefore

ab

ab

cos2

ˆ

cos2

ˆ

abSa

abSa

where ab is the angle between a and b (see the APPENDIX for the derivation in detail). Here

we write the state 0,0 as

],,[2

10,0 aaaa ,

We see that

],,[2

1

2

],)ˆ(,)ˆ[(2

10,0)ˆ( 111

aaaa

aaaSaaaSaS

or

],,[2

1

2)ˆ(0,0 1 aaaaaS

,

where

aaaaaS ,2

,)ˆ( 1

, aaaaaS ,

2,)ˆ( 1

.

Note that

55

],,

,,[22

][22

][22

)ˆ(2

1)ˆ(

2

1

],,)[ˆ(2

10,0)ˆ(

221

221

221221

22

baabbaab

baabbaab

babbaba

babbaba

abSaabSa

aaaabSbS

with

22

22222222

22

])[ˆ()ˆ(

babbab

abbbbbSabS

22

22222222

22

])[ˆ()ˆ(

babbab

abbbbbSabS

Then we have

ab

abab

cos4

)2

sin2

(cos4

][8

],,

,,][,,[8

]0,0ˆ][,,[22

0,0)ˆ)(ˆ(0,0

2

222

22222

2

221

abababab

baabbaab

baabbaabaaaa

bSaaaabSaS

Alternatively, we have the same result as follows.

56

ab

abab

cos4

)coscos(8

]ˆˆ[4

]ˆˆ[4

],ˆ,,ˆ,[4

],ˆ,,ˆ,

,ˆ,,ˆ,[4

],ˆ,ˆ][,,[8

0,0)ˆ)(ˆ(0,0

2

2

221

22

22

22

22

2

21

abSaabSa

abSaaaabSaaa

aabSaaaabSaa

aabSaaaabSaa

aabSaaaabSaa

aabSaabSaaaabSaS

since 1 aa , and 1 aa . Thus we have the spin correlation function,

ba

bSaS

baba

ab

E

cos

0,0)ˆ)(ˆ(0,04

)(ˆ)(ˆ),(

212

21

_____________________________________________________________________________ ((Mathematica-1))

57

((Mathematica-2))

Clear"Global`";

exp_ : exp . Complexre_, im_ Complexre, im;

1z 10; 2z 0

1; Sx

—

2 0 1

1 0; Sy

—

2 0 0

;

Sz —

2 1 0

0 1; nx Sin Cos; ny Sin Sin;

nz Cos;

1 1

2KroneckerProduct1z, 2z KroneckerProduct2z, 1z;

1 MatrixForm

012

12

0

58

18. Proof of Bell's theorem in terms of the hidden variables

Sn nx Sx ny Sy nz Sz FullSimplify

12— Cos,

12 — Sin, 1

2 — Sin,

12— Cos

K1 KroneckerProductSz, Sn Simplify; K1 MatrixForm

14—2 Cos 1

4 —2 Sin 0 0

14 —2 Sin 1

4—2 Cos 0 0

0 0 14—2 Cos 1

4 —2 Sin

0 0 14 —2 Sin 1

4—2 Cos

Transpose1.K1.1 Simplify

14—2 Cos

012

12

0

Sn nx Sx ny Sy nz Sz FullSimplify

12— Cos,

12 — Sin, 1

2 — Sin,

12— Cos

K1 KroneckerProductSz, Sn Simplify; K1 MatrixForm

14—2 Cos 1

4 —2 Sin 0 0

14 —2 Sin 1

4—2 Cos 0 0

0 0 14—2 Cos 1

4 —2 Sin

0 0 14 —2 Sin 1

4—2 Cos


14—2 Cos


14—2 Cos

59

In obtaining the Bell’s inequality, we have never used quantum mechanics. We have only assumed the Einstein’s locality principle and an underlying hidden variable model. Consequently, a Bell inequality is a constraint that any physical theory that is both, local and realistic, has to satisfy. This inequality can be violated by a quantum state. Hence the quantum mechanics is compatible with an underlying local realistic model.

The spin-up and down states are described by

1),(1 av , 1),(2 bv

where v is n-vector of hidden. Since the angular momentum is conserved,

),(),( 21 bvbv

The spin correlation is defined by

),(),()()()(),( 2121 bvavvvbaba ndE

where )(v is the probability density and

1)( νnd

Using the angular momentum conservation, we have

),(),()()()( 1121 bvavvvba nd

Then we get

)],(),()[,()(

)],(),(),(),()[()()()()(

111

11112121

cvbvavvv

cvavbvavvvcaba

n

n

d

d

Since 1),(21 bv

60

)],(),(1)[,(),()(

)],(),(),(),()[,(),()(

)],(),()[,(),()()()()()(

1111

111111

112

112121

cvbvbvavvv

cvbvbvbvbvavvv

cvbvbvavvvcaba

n

n

n

d

d

d

Then

)()(1

)],(),(1)[(

)],(),(1)[(

|)],(),(1[),(),()(

)],(),(1)[,(),()()()()()(

21

21

11

1111

11112121

cb

cvbvvv

cvbvvv

cvbvbvavvv

cvbvbvavvvcaba

n

n

n

n

d

d

d

d

where

),(),( 21 cvcv

or

),(1),(),( cbcaba EEE (Bell's theorem)

since

),(),( 12 cvcv ,

where the spin correlation function,

)()(),( 21 baba E , )()(),( 21 caca E

and

)()(),( 21 cbcb E

((Note))

In quantum mechanics, we know that

61

baba ),(E

Then we have

)()(),(),( cbacbacaba EELHS

cb 1RHS

We now choose that

0ba , cossin bac

with

1bb

Fig. The geometry of three vectors a, b and c. The vector a is perpendicular to the vector b.

The angle between c and b is . Thus

sin caLHS , cos1RHS .

LHS

a

b

c

62

Fig. LHS and RHS as a function of the angle between b and c. LHS>RHS.

LHS>RHS except for = 0 and /2. This means that the quantum mechanics is inconsistent with hidden variable theory (locality). 19. Classical model (the Bell inequality)

It is instructive to consider an example of correlation in classical physics for which the Bell inequality is of course satisfied. Consider two wheels which spin with angular momentum J and –J about their common axle, so that the total angular momentum of the system is zero, just as in the quantum mechanics spin example. Now let the two wheels fly apart, and measure the sign of the component of each wheel’s angular momentum along arbitrary directions a for the first wheel and b for the second. Now consider an ensemble of such wheel and axle systems with the axles distributed isotropically in space, and let an and bn be the signs of the angular momentum components for the n-th axle. So, just as in the quantum mechanics case, an and bn are always ±1.

If we draw great circles on the unit sphere whose planes are perpendicular to a and b, the surface of the sphere is divided into four lunes of aperture ab ab , ab , and ab .

LHS

RHS

p

6p

3p

2

y

0.2

0.4

0.6

0.8

1.0

63

Fig. Classical example of the Bell inequality for two wheels which fly apart with angular

momenta J and –J. The angle between the vectors a and b is ab . (from M. Redhead,

Incompleteness, Nonlocality, and Realism). If the axes cuts the sphere in the region of the two shaded lunes of aperture ab , then clearly

nnba will be +1, while if it cuts the sphere in the region of the two unshaded lunes of aperture

ab , nnba will be -1. With an isotropic distribution of axle directions we have then the

simple result

ab

abab

nnbaE

21

2

)1)((2)1(2

),(

ba

(Classical)

64

The reason why the Bell inequality is violated in the quantum mechanics is due to the angular dependence of the correlation coefficients With the choice of the four directions a, a, b, and b’ as shown in this figure.

Fig. Special choice of directions for illustrating the violation of the Bell’s inequality. It is easily checked that LHS of the Bell inequality comes out equal to 2; so in this particular example the Bell inequality is saturated but not, of course, violated. We have

2

)20(2

2

)(2

2

)2

1()2

1(

)2

1()2

1()','(),'()',(),(

''''

'''

'

babaabab

baba

ababEEEE babababa

The reason why the Bell inequality is violated in QM is due to the angular dependence of

the correlation coefficients specified in

abE cos),( ba , (QM)

For small angles, ab , the QM prediction hangs on to perfect anti-correlation more tightly than

in the classical example. Thus we get

a'

a b

b'

65

21),(

2abE

ba

So the correlation is proportional to 2ab rather than ab .

Fig. ),( baS vs the angle between the unit vectors a and b. ab . The QM prediction

(denoted by red; abS cos),( ba ). The classical prediction (denoted by blue).

20. Bell's states for the spin 1/2 system

The Bell's states are defined by

E a,b

ab3

2 2 2

3

2

1.

0.5

0.5

1.

66

)(2

1

)(2

1

)(2

1

)(2

1

)(2

1

)(2

1

)(2

1

)(2

1

2121

212112

2121

212112

2121

212112

2121

212112

zzzz

zzzz

zzzz

zzzz

zzzz

zzzz

zzzz

zzzz

Using the total spin angular momentum along the z axis, which is given by

zzz SSS 21ˆˆˆ . or zzz SSS 221

ˆ11ˆˆ

we have

12

2121

21212112

)(2

)(2

1)ˆˆ(ˆ

zzzz

zzzzSSS zzz

12

2121

21212112

)(2

)(2

1)ˆˆ(ˆ

zzzz

zzzzSSS zzz

,

0)(2

1)ˆˆ(ˆ

21212112 zzzzSSS zzz ,

67

0)(2

1)ˆˆ(ˆ

21212112 zzzzSSS zzz .

So that 12

and 12

are the eigenkets of zS with the eigenvalue 0. On the other hands,

12

and 12

are not the eigenkets of zS .

((Mathematica))

Clear"Global`";

exp_ :

exp . Complexre_, im_ Complexre, im;

1 10;

2 01;

B1

1

2KroneckerProduct1, 1

KroneckerProduct2, 2 MatrixForm

120012

B2

1



68

21. Mathematics for spin 1/2 system (1) Inner product between eigenkets of spin 1/2 system.

The eigenkets n and n for the spin 1/2 system are obtained as

2sin

2cos

ien , and

2cos

2sin

ien

012

120

B3

1



1200

12

B4

1



012

12

0

69

Here we choose a phase factor for n which is a little different from the conventional

notation

2cos

2sin

ien

We now calculate the inner product

2sin

2cos

ai

a

ae

a ,

2sin

2cos

bi

b

be

b ,

2cos

2sin

ai

a

ae

a ,

2cos

2sin

bi

b

be

b

Then we have the scalar product

)2

cos(

2sin

2sin

2cos

2cos

2sin

2cos

2sin

2cos

ba

baba

ai

a

bib

a

b

ee

ab

,

)2

sin(

2cos

2sin

2sin

2cos

2sin

2cos

2cos

2sin

ba

baba

ai

a

bib

a

b

ee

ab

70

)2

sin(

2sin

2cos

2cos

2sin

2cos

2sin

2sin

2cos

ba

baba

ai

a

bib

a

b

ee

ab

,

)2

cos(

2cos

2cos

2sin

2sin

2cos

2sin

2cos

2sin

ba

baba

ai

a

bib

a

b

ee

ab

(2) Matrix elements of spin operator

)cos(2

)]2

(sin)2

([cos2

][2

][2

ˆˆ[2

)(ˆ2

ˆ2

ˆ

22

22

ba

baba

abab

abbaabba

abbbσaabbbσa

abbbbbσa

abσaabSa

where σS ˆ2

ˆ , a and b are unit vectors. bbbσ ˆ , and bbbσ ˆ . Similarly we

have

71

)sin(2

)]2

sin()2

cos()2

cos()2

[sin(2

][2

][2

ˆˆ[2

)(ˆ2

ˆ2

ˆ

**

ba

babababa

abababab

abbaabba

abbbσaabbbσa

abbbbbσa

abσaabSa

)sin(2

)]2

sin()2

cos()2

cos()2

[sin(2

][2

][2

ˆˆ[2

)(ˆ2

ˆ2

ˆ

**

ba

babababa

abababab

abbaabba

abbbσaabbbσa

abbbbbσa

abσaabSa

72

)cos(2

)]2

(cos)2

([sin2

][2

][2

ˆˆ[2

)(ˆ2

ˆ2

ˆ

22

22

ba

baba

abab

abbaabba

abbbσaabbbσa

abbbbbσa

abσaabSa

(3) Probability

(i) 2

0,0,),( baba P

ab

aabaaaba

ba

2

1

],,2

1,,

2

1

0,0,

Then we have the probability

2

sin2

1

2

1),( 22 baP

abba

(ii) 2

0,0,),( baba P

ab

aabaaababa

2

1

],,2

1,,

2

10,0,

Then we have the probability

73

2

cos2

1

2

1),( 22 baP

abba

(iii) 2

0,0,),( baba P

ab

aabaaababa

2

1

],,2

1,,

2

10,0,

2

cos2

1

2

1),( 22 baP

abba

(iv) 2

0,0,),( baba P

ab

aabaaababa

2

1

],,2

1,,

2

10,0,

2

sin2

1

2

1),( 22 baP

abba

22. CHSH inequality

The CHSH inequality can be used in the proof of Bell's theorem, which states that certain consequences of entanglement in quantum mechanics cannot be reproduced by local hidden theories. Experimental verification of violation of the inequalities is seen as experimental confirmation that nature cannot be described by local hidden variables theories. CHSH stands for John Clauser, Michael Horne, Abner Shimony, and Richard Holt, who described it in a much-cited paper published in 1969. They derived the CHSH inequality, which, as with John Bell's original inequality (Bell, 1964), is a constraint on the statistics of "coincidences" in a Bell test experiment which is necessarily true if there exist underlying local hidden variables (local realism). This constraint can, on the other hand, be infringed by quantum mechanics. http://en.wikipedia.org/wiki/CHSH_inequality

The usual form of the CHSH inequality is:

74

The usual form of the CHSH inequality is given by

})]',(),()[',()]'(),()[,(){(

)'()'()()'()'()()()(

)','(),'()',(),(

221221

21212121

bvbvavbbvavvv

babababa

babababa

nd

EEEES

where

),(),()()()(),( 2121 bvavvvbaba ndE

We note that

),(2 bv )',(2 bv )',(),( 22 bvbv )',(),( 22 bvbv

1 1 0 2 1 -1 2 0 -1 1 -2 0 -1 -1 0 2

)',(),( 22 bvbv and )',(),( 22 bvbv can only be ±2 and 0, and 0 and ±2, respectively.

Then we get the CHSH inequality,

2S

((Note))

In quantum mechanics, we have

baba ),(E

S can be rewritten as

'''' babababa S

The CHSH inequality is predicted as

− 2 ≤ S ≤ 2,

75

from the local hidden theories. On the other hand, the mathematical formalism of quantum

mechanics predicts a maximum value for S of 22 , which is greater than 2, and CHSH violations are therefore predicted by the theory of quantum mechanics. ((Maximal violation)) quantum mechanics

Here we consider the operator defined by

)]'(ˆ)(ˆ)['(ˆ)]'(ˆ)(ˆ)[(ˆˆ221221 bbabba Q .

We calculate the square of this operator based on the quantum mechanics.

)]'(ˆ)(ˆ)[(ˆ)]'(ˆ)(ˆ)['(ˆ

)]'(ˆ)(ˆ)['(ˆ)]'(ˆ)(ˆ)[(ˆ

)]'(ˆ)(ˆ)['(ˆ)]'(ˆ)(ˆ)[(ˆ

)]}'(ˆ)(ˆ)['(ˆ)]'(ˆ)(ˆ)[(ˆ{

)]}'(ˆ)(ˆ)['(ˆ)]'(ˆ)(ˆ)[(ˆ{ˆ

221221

221221

222

21

222

21

221221

2212212

bbabba

bbabba

bbabba

bbabba

bbabba

Q

Here we use the formula

)(ˆ1)(ˆˆ baσbabσaσ i

So we have

11ˆˆ aaaσaσ

Then we get

)](ˆ)'(ˆ)(ˆ)'(ˆ)'(ˆ)(ˆ)(ˆ)'(ˆ

)](ˆ)'(ˆ)'(ˆ)(ˆ)'(ˆ)(ˆ)'(ˆ)(ˆ14

)](ˆ)'(ˆ)'(ˆ)(ˆ)[(ˆ)'(ˆ

)](ˆ)'(ˆ)'(ˆ)(ˆ)['(ˆ)(ˆ

)](ˆ)'(ˆ)'(ˆ)(ˆ12[)](ˆ)'(ˆ)'(ˆ)(ˆ12[ˆ

22112211

22112211

222211

222211

222222222

bbaabbaa

bbaabbaa

bbbbaa

bbbbaa

bbbbbbbb

Q

or

)]'(ˆ),(ˆ)]['(ˆ),(ˆ[14ˆ2211

2 bbaa Q

76

Since

)'(ˆ2

)]'(ˆ1)'[()'(ˆ1)'(

ˆ'ˆ'ˆˆ]'ˆ,ˆ[

1

11

111111

aaσ

aaσaaaaσaa

aσaσaσaσaσaσ

i

ii

we get

)'ˆ)(ˆ(''414

)]'(ˆ)]['(ˆ[414ˆ

21

212

nσnσbbaa

bbσaaσ

Q

We note that

'0,0)'ˆ)(ˆ(0,0 21 nnnσnσ

with

'

'

aa

aa

n , '

''

bb

bbn

.

Then we have

)]')('()'')([(44

)'()'(442

babababa

bbaa

Q

8sinsin44)'()'(44 '.,'.,2 bbaabbaa Q

When 2'.,

aa , 2'.,

bb , 2Q has a maximum (=8). Thus the average value of the operator

2Q over the state vector is given by

82 Q or 222 Q .

((Note)) The significance of the commutation relation in quantum mechanics

77

The fact that S>2 in the quantum mechanics is closely related to the operators which are not commutable. In the local hidden theory where the commutation relations hold,

0)]'(ˆ),(ˆ[ 11 aa , 0)]'(ˆ),(ˆ[ 22 bb .

which can be regarded as the classical case, we have

42 Q , or 22 Q .

The value of 2 is called the Tsirelson bound for the CHSH inequality. The Tsirelson bound is named after B.S. Tsirelson (or Boris Cirel’son). 23. Derivation of the Bell’s original inequality from CHSH inequality. Here we derive the original form in which Bell presented his inequality, from the CHSH inequality

2)','(),'()',(),( babababa EEEE .

By interchanging a and a’, we have

2)','(),'()',(),( babababa EEEE (1)

By interchanging b and b’, we have also

2),'()','()',(),( babababa EEEE (1)

Here we define

)',(),( baba EEX , )','(),'( baba EEY

Then we get

2YX , 2YX

The region for satisfying the inequality is shaded by green in the {X,.Y} plane.

Fig. T

Mathem

or

We now

Then we

The region w

matically, this

2 YX ,

),'( ba EE

w consider an

1

),(

aba

n

d

d

E

e have

where YX

s region can

, or

2)','( ba

n experiment

()(

,()(

()(

21

1

21

vvv

avvv

ba

n

n

d

2Y , YX

be also deno

r Y 2

),( ba EE

t for which a

),

),()

)

2

av

ava

ab

.

78

2Y , is the

oted by the i

X2

)',( baE

a = b,

e same as the

inequality

e region wheere YX 2

79

)',(1

)',(12

)',(12

)',(),(2)','(),'(

bb

bb

bb

bbabbababa

E

E

E

EEEE

So we get

)',(1)','(),'( bbbaba EEE

When aa' and cb' , we get the original Bell’s inequality

),(1),(),( cbcaba EEE )

24. Evaluation of S

Here we evaluate the value of S. We choose the directions of the unit vectors a, b, a', and, b' as shown in Fig.

80

Fig. 4'''

babaab for the electron spins.

a = z, a' = x, The angle between a and b is .

sin)2

cos()',(

sin)2

cos(),'(

cos)','(),(

ba

ba

baba

E

E

EE

leading to

x

y

a

b

a'

b'

2

2

81

)4

sin(22

)cos(sin2

cossinsincos

)','(),'()',(),(

babababa EEEES

We make a plot of S as a function of . The value of |S| becomes larger than 2 for 2/0

and 2/3 .

When = /4, we have

2

1)','(),'(),( bababa EEE

2

1)

4

3cos()','(

baE

Then we have

22

)','(),'()',(),(

babababa EEEES

Fig. Bell’s inequalities are violated by certain quantum prediction for 222 S .

_____________________________________________________________________________ We now discuss the quantum entanglement for the photon system.

S

4 2

3

4

5

4

3

2

7

42

3

2

1

1

2

3

82

Part II Entangled Photon 25. Entangled state of photon 25.1 The one-photon kets:

0

1x , for the linear polarization along the x axis

1

0y for the linear polarization along the y axis

iR

1

2

1 for the right-hand circularly polarization,

i

L1

2

1, for the left-hand circularly polarization,

Note that

RRJ z ˆ.

LLJ z ˆ

with

][2

1yixR

][2

1yixL

The angular momentum: zzz JJJ 21ˆˆˆ , zzz JJJ 2121

ˆ11ˆˆ

For the Bell's state,

),,(2

1RLLR

we have

83

0

],)(,)[(2

1

),,(2

1)ˆˆ(ˆ

21

RLLR

RLLRJJJ zzz

Thus is the eigenstate of zJ with the eigenvalue 0.

((Note)) There are still two more Bell’s states;

)(2

1

)(2

1

212112

212112

LLRR

LLRR

We apply the operator zJ to these two states,

12

2112

2

],,[2

12

),,(2

1)ˆˆ(ˆ

LLRR

LLRRJJJ zzz

and

12

2112

2

],,[2

12

),,(2

1)ˆˆ(ˆ

LLRR

LLRRJJJ zzz

So that 12 and 12 are not the eigenkets of zJ .

25.2 The two-photon kets (Bell's states)

84

Test of the Bell inequalities are typically carried out on entangles states of photons. The Bell's inequality is violated, in such a way that the quantum mechanical predictions are fulfilled within error limits.

The Bell’s state ],,[2

1RLLR can be rewritten under the basis of { xx, , yx, ,

xy, , and yy, } as follows.

1

0

0

1

2

1

),,(2

1),,(

2

1

)],,,,(

),,,,[(22

1

)](())([(22

1

][2

1

],,[2

1

2121

21212121

21212121

11222211

yyxxyyxx

yyxyiyxixx

yyxyiyxixx

yixyixyixyix

RLLR

RLLR

85

0

1

1

0

2

0

0

2

1

),,(2

),,(2

)],,,,(

),,,,[(22

1

)](())([(22

1

][2

1

],,[2

1

2121

21212121

21212121

11222211

i

i

i

xyyxi

xyyxi

yyxyiyxixx

yyxyiyxixx

yixyixyixyix

RLLR

RLLR

under the basis of { xx, , yx, , xy, , and yy, }.

((Note))

The other two Bell states can be also rewritten under the basis of { xx, , yx, , xy, , and

yy, } as follows.

1

1

2

111

2

1,

i

i

iiRRRR

1

1

2

111

2

1,

i

i

iiLRLR

86

1

1

2

111

2

1,

i

i

iiRLRL

1

1

2

111

2

1,

i

i

iiLLLL

Then we have

1

0

0

1

2

1,,(

2

112

RLLR

0

1

1

0

2,,(

2

112

iRLLR

1

0

0

1

2

1,,(

2

112

LLRR

0

1

1

0

2,,(

2

112

iLLRR

((Mathematica))

87

Clear"Global`";


X 10; Y 0

1; R

1

2 1;

L 1

2 1

;

1 1

2KroneckerProductR, L KroneckerProductL, R;

1 MatrixForm

12

0012

11 1

2KroneckerProductX, X KroneckerProductY, Y;

11 MatrixForm

12

0012

88

25.3 Invariance of the Bell’s state under the rotation for photon

(from Bellac, quantum physics)

(a) The Bell’s state under the rotation about the z axis

),,(2

),,(2

1RLLR

ixyyx

We start with the Bell’s state defined by

2 1


2 MatrixForm

0

2

2

0

21

1

2KroneckerProductX, Y KroneckerProductY, X;

21 MatrixForm

0

2

2

0

89

0

1

1

0

2

1

0

1

0

0

2

1

0

0

1

0

2

1

0

1

1

0

2

1

1

0

0

1

2

1

),,(2

1xyyx

The rotated states are given by

sin

cossincos yx ,

cos

sincossin

2yx .

Then the rotated state ' under the rotation about the z axis is given by

90

0

1

1

0

2

1

cossin

cos

sin

cossin

2

1

cossin

sin

cos

cossin

2

1

sin

cos

cos

sin

2

1

cos

sin

sin

cos

2

1

),,(2

1

'

2

2

2

2

.

Thus the state is invariant under the rotation.

Here we note that can be rewritten as

),,(2

RLLRi

Since R and L are expressed by

iyixR

1

2

1][

2

1,

i

yixL1

2

1][

2

1

we get

91

ii

i

i

i

i

iiii

RLLRRLLR

0

1

1

0

2

1

1

22

1

1

1

22

1

11

22

111

22

1

2

1

2

1),,(

2

1

or

),,(2

RLLRi

Under the rotation around the z axis,

Re

ie

i

i

i

RR

i

i

1

2

1

cossin

sincos

2

1

[2

1

'

,

92

Le

ie

i

i

i

LL

i

i

1

2

1

cossin

sincos

2

1

[2

1

'

.

Under the rotation around the z axis, then we have

],,[2

)','','(2

'

RLLRi

RLLRi

It is found that the state is invariant under the rotation about the z axis.

(b) Invariance of the Bell’s state under the rotation about the z axis

),,(2

1),,(

2

1RLLRyyxx

93

1

0

0

1

2

1

1

0

0

0

2

1

0

0

0

1

2

1

1

0

1

0

2

1

0

1

0

1

2

1

),,(2

1yyxx

1

0

0

1

2

1

cos

cossin

cossin

sin

2

1

sin

cossin

cossin

cos

2

1

cos

sin

cos

sin

2

1

sin

cos

sin

cos

2

1

),,(2

1

'

2

2

2

2

Here we note that can be rewritten as

),,(2

1RLLR

since

94

1

0

0

1

2

1

1

1

22

1

1

1

22

1

11

22

111

22

1

2

1

2

1),,(

2

1

i

i

i

i

iiii

RLLRRLLR

or

),,(2

1RLLR

Under the rotation around the z axis, then we have

],,[2

1

)','','(2

1

'

RLLR

RLLR

which means that the state is invariant under the rotation about the z axis.

),,(2

1)','','(

2

1RLLRRLLR

26. Correlation function E(a, b) for the Bell’s state (I) for the photon

Tests of the Bell inequalities are typically carried out on entangles states of photons. We can label the linear polarization states of the photons as follows. We assign the value +1 if a measurement by Alice (Bob) finds the photon to be horizontally polarized along an axis labeled

by the upolarize

Here we

Fig. R

p

We call

unit vector aed, that is pol

(),( ba PE

e calculate E

Red arrows pperpendicula

the angle th

cos' xx

sin'y

a (b) and a larized along

(), ba P

),( baE for th

parallel to thar to a or b)

hat a makes w

sin yx

cos yx

value -1 if g an axis per

(), ba P

he two-photo

he vector a o(vertical,-a-

with the x ax

sin

cos,

cos

siny

95

f the measurrpendicular t

(), ba P

on entangles

or b (horizon, -b).

xis with the

rement findsto a (b). We

), ba

states.

ntal, +a, +b).

e linear pola

s the photon define

The blue ar

arization stat

n to be verti

rrows (which

tes

ically

h is

96

Fig. Definition of a'x and a'y .

We call the angle that b makes with the x axis with the linear polarization states

sin

cossincos" yxx ,

cos

sincossin" yxy ;

Fig. Definition of b"x and b"y .

x' a

y' a

x'' by'' b

97

sinsin

cossin

sincos

coscos

"'",'21

xxxx

coscos

sincos

cossin

sinsin

"'",'21

yyyy

cossin

sinsin

coscos

sincos

"'",'21

yxyx

coscos

sincos

cossin

sinsin

"'",'21

yyyy

For the Bell's state of photon given by

],,[2

1],,[

2

1yyxxRLLR

(Aspect et al. used this Bell's state) we calculate the expectation value E(a, b) as follows. Note that

'x a 'y a

"x b "y b

(i) Calculation of ba ,",' xx

98

)cos(2

1

]sinsincos[cos2

1

],",',",'[2

1",'

yyxxxxxxxx

and the joint probability

)(cos2

1)(cos

2

1",'),( 222

abxxP ba

with

ab

where

",', xx ba

(ii) Calculation of ba ,",' yy

)cos(2

1

]coscossin[sin2

1

],",',",'[2

1",'

yyyyxxyyyy


)(cos2

1)(cos

2

1",'),( 222

abyyP ba

where

",', yy ba

99

(iii) Calculation of ba ,",' yx

)sin(2

1

]cossinsincos[2

1

],",',",'[2

1",'

yyyxxxyxyx


)(sin2

1)(sin

2

1",'),( 222

abyxP ba

where

",', yx ba

(iv) Calculation of ba ,",' xy

)sin(2

1

]sincoscossin[2

1

],",',",'[2

1",'

yyxyxxxyxy


)(sin2

1)(sin

2

1",'),( 222

abxyP ba

where

",', xy ba

From the above calculations, we get the correlation function

100

)2cos(

)(sin)(cos

),(),(),(),(),(22

ab

abab

PPPPE

bababababa

for the Bell's state,

],,[2

1],,[

2

1yyxxRLLR

Note that the result for the spin -1 photons is different from that of spin 1/2 particles. When 0ba

1),( baE

0),(),( baba PP , 2

1),(),( baba PP

((Mathematica))

Clear"Global`";


X 10; Y 0

1; X1 Cos X Sin Y;

Y1 Sin X Cos Y; X2 Cos X Sin Y;

Y2 Sin X Cos Y;

R 1

2 1; L

1

2 1

;

1 1


1 MatrixForm

12

0012

101

_____________________________________________________________________ 27. Correlation function E(a, b) for the Bell’s state (II) of photon

We consider another Bell's state defined by

],,[2

1),,(

2xyyxRLLR

i

(i) Calculation of ba,",' xx

1 KroneckerProductX1, X2 Simplify;

Transpose1.1 Simplify

Cos 2

2 KroneckerProductY1, Y2 Simplify;


Cos 2

3 KroneckerProductX1, Y2 Simplify;


Sin 2

4 KroneckerProductY1, X2 Simplify;


Sin 2

102

ab

xyxxyxxxxx

sin2

1

)sin(2

1

]cossinsin[cos2

1

],",',",'[2

1",'

with

ab

Then we get

abP 2sin2

1),( ba

(ii) Calculation of ",' yy

ab

xyyyyxyyyy

sin2

1

)sin(2

1

]sincoscos[sin2

1

],",',",'[2

1",'

abP 2sin2

1),( ba

(iii) Calculation of ba,",' yx

103

ab

xyyxyxyxyx

cos2

1

)cos(2

1

]sinsincos[cos2

1

],",',",'[2

1",'

abP 2cos2

1),( ba

(iv) Calculation of ba,",' xy

ab

xyxyyxxyxy

cos2

1

)cos(2

1

)coscossin(sin2

1

],",',",'[2

1",'

abP 2cos2

1),( ba

From the above calculations, we get the correlation function

)2cos(

)(sin)(cos

),(),(),(),(),(22

ab

abab

PPPPE

bababababa

for the Bell's state given by

],,[2

1),,(

2xyyxRLLR

i

When 0ba

104

1),( baE

2

1),(),( baba PP , 0),(),( baba PP

((Mathematica))

Clear"Global`";

exp_ :

exp . Complexre_, im_ Complexre, im;

X 10; Y 0

1; X1 Cos X Sin Y;

Y1 Sin X Cos Y; X2 Cos X Sin Y;

Y2 Sin X Cos Y;

R 1

2 1; L

1

2 1

;

1

2

KroneckerProductR, L KroneckerProductL, R;

1 MatrixForm

012

12

0

105

_____________________________________________________________________________28. Experimental result by Aspect et al

Using the correlation function E(a, b), we define S as

)','(),'()',(),( babababa EEEES

where S involves four measurements in four different orientations of the polarizers. Evaluate S

for the set of orientations in which ''' babaab = 22.5° and 'ab = 67.5°. It can be shown

that

2S

in a local hidden variable theory (CHSH inequality, Clauser, Horne, Shimony, Holt, 1969). The experiment of Aspect et al. yields

1 KroneckerProductX1, X2 Simplify;


Sin 2

2 KroneckerProductY1, Y2 Simplify;


Sin 2

3 KroneckerProductX1, Y2 Simplify;


Cos 2

4 KroneckerProductY1, X2 Simplify;


Cos 2

106

tSexp = 2.697 ± 0.015.

This particular choice of angles yields the greatest conflict between quantum mechanical

calculation of S and the Bell’s inequality 2S .

Fig. 5.22''' babaab for the photon polarizations.

We know that

)(2cos)2cos(),( baba abE

with

],,[2

1],,[

2

1yyxxRLLR

O

a

b

a'

b'a

aa

107

Then we have

)2cos()2cos()2cos()2cos(

)','(),'()',(),(

'''' babaabab

EEEES

babababa

When 5.22''' babaab . 5.673' ab , we get

)6cos()2cos(3 S =2.82843 (= 22 ) > 2. (1)

((Mathematica))

There is no classical correlation which can reproduce the quantum correlations; the quantum correlations are too strong to be explained. ((Comment-1)) S. Weinberg [Lectures Notes on Quantum Mechanics (Cambridge University Press, 2013)]. "Because the polarizers in this experiment were not perfectly efficient, the expected value was only 2.70 ± 0.05. The experimental result for the left-hand side of Eq. (1) was 2.697 ± 0.0515, in good agreement with quantum mechanics, and in clear disagreement with the inequality (1) satisfied by all local hidden variable theories." ((Comment-2))

p

8p

4

3 p8

p

2

5 p8

3 p4

7 p8

pq

-2.5

-2.

-1.5

-1.

-0.5

0.5

1.

1.5

2.

2.5

f q

108

Bell’s theorem puts forward an inequality. The inequality compares the sum, denoted by S, of possible results of the experiment—outcomes on the detector held by Alice, and the one held by Bob:

2S .

According to Bell’s theorem, if the inequality above is violated, that is, the sum of the particular responses for Alice and Bob is greater than two or less than minus two, as a result of some actual experiment with entangled particles or photons, that result constitutes evidence of non-locality, meaning that something that happens to one particle does affect, instantaneously, what happens to the second particle, no matter how far it may be from the first one. The conclusion from Bell’s theorem is that hidden variables and a locality assumption had no place within the quantum theory, which was incompatible with such assumptions. Bell’s theorem was thus a very powerful theoretical result in physics. ____________________________________________________________________ 29. Bell's inequality for the entangled state of photon

We take a to mean horizontal polarization parallel to Alice (or Bob) polarizer, while a

means vertical polarization to Alice (or Bob).

Fig. Left: configuration for the photon-1 (Alice). Right: configuration for the photon-2 (Bob).

{a+, b+, c+} (photon-1). {a-, b-, c-} (photon-2)

A BEntangled pair source

photon-1 photon-2

a

b c

a

b c

a

b c

a

b c

109


{a+, b+, c-} (photon-1). {a-, b-, c+} (photon-2)


{a-, b-, c+} (photon-1). {a+, b+, c-} (photon-2)

Photon-1 Photon-2 Population

{a+, b+, c+} {a-, b-, c-} N1 {a+, b+, c-} {a-, b-, c+} N2 {a+, b-, c+} {a-, b+, c-} N3 {a+, b-, c-} {a-, b+, c+} N4 {a-, b+, c+} {a+, b-, c-} N5 {a-, b+, c-} {a+, b-, c+} N6 {a-, b-, c+} {a+, b+, c-} N7 {a-, b-, c-} {a+, b+, c+} N8

Using the same procedures previously, we get the inequality such that

a

b c

a

b c

a

b c

a

b c

110

0

)(),(),(),(

8

1

81

8

1

844221

ii

ii

N

NN

N

NNNNNNPPP cbcaba

or

0),(),(),( cbcaba PPP

which is called the Bell inequality. (a)

We assume that the Bell entangled state is given by

],,[2

1],,[

2

1yyxxRLLR ,

In this case, we have

abP 2sin2

1),( ba , acP 2cos

2

1),( ca , abP 2sin

2

1),( cb

Then the Bell's inequality can be expressed by

0)sincos(sin2

1 222 bcacab

Suppose that

acbcab , ab , 2bc 3ac

)2sin3cos(sin2

1)( 222 f

We make

111

From the above figure, f() becomes negative near 0 and . So there is a clear contradiction between the HV prediction and the quantum mechanics. (b)

We assume that the Bell entangled state is given by

],,[2

1),,(

2xyyxRLLR

i ,

In this case, we have

abP 2cos2

1),( ba , acP 2sin

2

1),( ca , abP 2cos

2

1),( cb

Then the Bell's inequality can be expressed by

0)cossin(cos2

1 222 bcacab

Suppose that

acbcab , ab , 2bc 3ac

)2cos3sin(cos2

1)( 222 f

We make a plot of )(f as a function of .

p4

p2

3 p4

pq

-0.4

-0.2

0.2

0.4

0.6f q

112

From the above figure, f() becomes negative near /4 and 3/4. So there is a clear contradiction between the HV prediction and the quantum mechanics 30. Experiment of Aspect et al. 30.1 Apparatus http://roxanne.roxanne.org/epr/experiment.html

The source consists of excited calcium atoms which have a valence electron in the (4p)2 1S0 state. Notice that the calcium atom's valence electron has been placed into an excited state where it has no net angular momentum (J = 0). As the atom de-excites, the electron cascades into the 4s4p 1P1 state and releases a green photon at 551.3 nm. This state then decays back to the (4s)2 1S0 state releasing a blue photon at 422.7 nm. Because the total angular momentum at the beginning and end of the cascade is zero, the two photons emitted must each have opposite

p4

p2

3 p

4p

q

0.2

0.4

0.6

0.8

1.0

f q

113

angular momentum and thus the two photons must be circularly polarized in opposite directions. The filters in figure 1 only allow the transmission of one of the two colors. Thus, these filters guarantee that emitted photon pairs travel to opposite detectors such that the green photons go along path A and the blue photons go along path B.

The squares in figure 1 labeled PA1 and PA2 are polarization filters set at orientations a and b respectively. These filters only transmit light whose polarization is aligned along their axis (parallel to a for PA1) and will reflect light whose polarization is perpendicular to their axis. Since circularly polarized light can be seen as a composition of two linear polarization states, these filters will split apart the circularly polarized light into two linear states and send them in different directions. If a single photon passes through one of these filters, it will be split and detected at one of the other photo-multiplier tubes. If the photon is detected as having been transmitted through the filter, it is assumed that it's polarization was parallel to the filter and a +1 is registered at the coincidence counter. If, on the other hand, the single photon is detected at the PMT tube along the reflected path, a -1 is sent to the coincidence counter.

Before remarking on how the results are recorded at the coincidence counter, what do we expect to see? Recall that the two emitted photons are circularly polarized in opposite directions (i.e. one is rotating clockwise and the other is rotating counter-clockwise). Notice that the detectors on the "A" side and "B" side of the experiment are facing each other. So, if A sees a photons polarization as spinning clockwise, then B will see the polarization of that same photon as spinning in the opposite direction! Since we are creating photons with exactly opposite polarizations and since we are looking at these polarizations using oppositely facing filters, the two photons in this experiment should register the same polarization states and thus whichever linear component is measured on side A must also have the same value on side B. After the PMT tubes send their result to the coincidence counter, there is one more requirement which must be met before an event is recorded. One only wants to consider events which are space-like separated. Since the polarization filters are separated by 13 meters, it would take a signal 40 ns traveling at the speed of light to go from one of the filters to the other one. To ensure that there is no communication between events which happen at the two filters, the coincidence counter only accepts results if the time delay between receiving the signals from the PMT tubes on side A and B is less than 20 ns. If the signals arrive with a time separation greater than this, the event is thrown out. If an event does occur within 20 ns, then the result from side A (either +1 or -1) is multiplied by the result from side B and an average value is found from repeated measurements.

Now, given all of these measurements, how do we extract a meaningful result? In the following two sections, we will work out what we expect to see in this experiment as a function of the filter orientations a and b. Using the data, how do we derive an average value for the experiment? The definition of the expectation value (or average value) is the sum of all the resulting values times the probability for that value. So, the expectation value, as a function of the filter orientations is:

114

),(),(),(),(),( bababababa PPPPE ,

where P++ is the probability that both detectors registered a "+" result, P-- is the probability that both detectors registered a "-" result, etc. These probabilities can be measured by looking at how many counts of a particular type were measured and dividing this by the total number of counts recorded. For example, P_+ would be the number of times the left detector registered a -1 at the same time as the right detector registered a +1 (by at the same time I mean within the 20ns limit mentioned above). 30.2 The electron configuration of Ca is given by 1s2 2s2p6 3s2p6 4s2. (a) 4s4s

D0 x D0 = D0 l = 0

D1/2 x D1/2 = D1 + D0 s = 1, s = 0

l = 0 and s = 1 D0 x D1 = D1 j = 1 4 3S1

l = 0 and s = 0

D0 x D0 = D0 j = 0 4 1S0

(b) 4s4p

D1 x D0 = D1 l = 1 D1/2 x D1/2 = D1 + D0 s = 1, s = 0

l = 1 and s = 1

D1 x D1 = D2 + D1 + D0 j = 2 4 3P2 (g = 3/2) j=1 4 3P1 (g = 3/2) j = 0 4 3P0

l = 1 and s = 0

D1 x D0 = D1 j = 1 4 1P1

115

(c) (4p)2

D1 x D1 = D2 + D1 + D0 l = 2, 1, 0 D1/2 x D1/2 = D1 + D0 s = 1, s = 0

l = 2 and s = 1

D2 x D1 = D3 + D2 + D1 j = 3 4 3D3 j=2 4 3D2 j = 1 4 3D1

l = 2 and s = 0

D2 x D0 = D2 j = 2 4 1D2 j= 1 4 1D1 j = 0 4 1D0

l = 1 and s = 1

D1 x D1 = D2 + D1 + D0 j = 2 4 3P2 j= 1 4 3P1 j = 0 4 3P0

l = 1 and s = 0

D1 x D0 = D1 j = 1 4 1P2

l = 0 and s = 1

D0 x D1 = D1 j = 1 4 3S1

l = 0 and s = 0

D0 x D0 = D0

j = 0 4 1S0

116

((White)) H.E. White, Introduction to atomic spectra (McGraw-Hill, NY 1934).

= 551.3 nm between 4 1S0 (l = 0, s = 0, j = 0, (4p)2) and 4 1P1 (l = 1, s = 0, j = 1, 4s4p)

= 422.6 nm between 4 1P1 (l = 1, s = 0, j = 1, 4s4p) and 4 1S0 (l = 0, s = 0, j = 0, (4s)2)

30.3 Emission of RHC and LHC photons

551.3 nm

422.7 nm

4 1S0 j=0

4 1 P1 j=1

4 1S0 j=0

Laser excitation

Ca

118

(a) The emission of RA (RHC photon)

119

(i) The state 0,0 mj (ii) The state 1,1 mj .

Transition from the state 0,0 mj to the state 1,1 mj . RA: RHC photon emerges

because of the angular momentum conservation. (b) The emission of RB (RHC photon)

120

(i) The state 1,1 mj . (ii) The state 0,0 mj .

Transition from the state 1,1 mj to the state 0,0 mj . RA: RHC photon emerges due

to the angular momentum conservation. (c) The emission of LA (LHC photon)

121

(i) The state 0,0 mj . (ii) The state 1,1 mj .

Transition from the state 0,0 mj to the state 1,1 mj . LA: RHC photon emerges

due to the angular momentum conservation. (d) The emission of LB (LHC photon)

122

____________________________________________________________________________ 30.4 Polarizations of the cascade photons.

The physics of the atomic transitions dictates an additional condition: the two=photon state must have even parity. This means that when the coordinate system is changed from right-handed to left-handed, the state remains invariant. We have the Bell’s state,

124

Fig. Configuration of polarizations of entangled photons. The sense of the angular

momentum J for R and L is defined by the right-hand rule. The thumb of the right

hand gives the sense of J. The states are given by

][2

1,,[

2

1yyxxLLRR BABA

][2

,,[2

1yxxy

iLLRR BABA

][2

1,,[

2

1yyxxRLLR BABA

][2

,,[2

1yxxy

iRLLR BABA

Note that Aspect et al. use the Bell's state given

][2

1,,[

2

1yyxxRLLR BABA

125

or

1 2

I II

S

ba1

1

1

1

x

z

y

126

Measurement of the polarization of 1 along the orientation a and of polarization of 2 along the orientation b;

'x a , 'y a ,

"x b , "y b ,

The probability to find +1 or -1 for 1 (measured along a) and +1 or -1 for 2 along b).

)(cos2

1),(),( 2 bababa PP

)(sin2

1),(),( 2 bababa PP

31. Conclusion

Quantum mechanics asserts that particles do not have a definite state except when observed, and two particles can be in an entangled state so that the observation of one determines a property of the other instantly. As soon as any observation is made, the system goes into a fixed state.

The EPR paper would not succeed in showing that quantum mechanics was wrong. But it did eventually become clear that quantum mechanics was incompatible with our common sense understanding of locality - our aversion to spooky action at a distance. The odd thing is that Einstein, apparently, was far more right than he hoped to be.

_______________________________________________________________________ REFERENCES 1. A. Einstein, B. Podolsky, and N. Rosen, Phys. Rev. 47, 777-780 (1935). 2. David. Bohm, Quantum Theory (Dover Publication, Inc, New York, 1979). 3. John S. Townsend , A Modern Approach to Quantum Mechanics, second edition

(University Science Books, 2012). 4. J.J. Sakurai and J. Napolitano, Modern Quantum Mechanics, second edition (Addison-

Wesley, New York, 2011). 5. David H. McIntyre, Quantum Mechanics A Paradigms Approach (Pearson Education,

Inc., 2012). 6. S. Weinberg; Lectures on Quantum Mechanics (Cambridge University Press, 2013). 7. M. Bell, L. Gottfried, and M. Veltman, John S. Bell on the foundation of quantum

mechanics (World Scientific, Singapore, 2001). 8. J.S. Bell, Speakable and unspeakable quantum mechanics (Cambridge University Press,

Cambridge, 1987).

127

9. A. Aspect, P. Grangier, and G. Roger, Phys. Rev. Lett. 49, 91–94 (1982)," Experimental Realization of Einstein-Podolsky-Rosen-Bohm Gedankenexperiment: A New Violation of Bell's Inequalities

10. R. Penrose, The Road to Reality (Jonathan Cape, London, 2004). 11. A.D. Aczel, Entanlement The Greatest Mystery in Physics (Four Walls Eight Window,

New York, 2001). 12. P. Ghose, Testing quantum mechanics on new ground (Cambridge University Press,

Cambridge, 1999). 13. G. Greenstein and A.G. Zajonc, The Quantum Challenge (John and Bartlett Publishers,

Boston, 1997). 13 J. Audretsch, ed. Entangled World (Wiley-VCH, 2006). 14. J. Audretsch, ed. Entangled Systems, New Directions in Quantum Physics (Wiley-VCH,

2005). 15. W. Issacson, Einstein: His Life and Universe (Simon & Schuster, New York, 2007).. 16. J. Audretsch, Entangled Systems New Directions in Quantum Physics. 17. A. Aspect, PITP lectures on Quantum Phenomena, University of British Columbia (May

25, 2012). 18. R.P. Feynman, R.,B. Leighton, and M. Sands, The Feynman Lectures in Physics, 6th

edition (Addison Wesley, Reading Massachusetts, 1977). Vol.III. Chapter 18. 19. M.L. Bellac, Quantum Physics (Cambridge University, 2006). 20 F. Laloe, Do we really understand Quantum Mechanics (Cambridge University Press,

2012). 21. V. Scarani, Quantum Physics A First Encounter (Oxford, 2003). 22. M. Beck, Quantum Mechanics Theory and Experiment (Oxford, 2012). 23. M. Readhead, Incompleteness, Nonlocality, and Realism (Oxford, 1987). 24. D.L. Hemmick and A.M. Shakur, Bell’s Theorem and Quantum Realism: Reassessment

in Light of the Schrödinger Paradox (Springer, 2012). 25. D. Bruß and G. Leuchs edited, Lectures on Quantum Information (Wiley-VCH, Verlag,

2007). (1) Quantum Entanglement Documentary- Atomic Physics and Reality

http://www.youtube.com/watch?v=BFvJOZ51tmc

where J. Bell, D. Bohm, A. Aspect, J. Wheeler talked about physics. (2) Fabric of the Cosmos-Quantum Leap (Brian Greene)

http://www.youtube.com/watch?v=NbIcg0XsbFQ _____________________________________________________________________________

128

APPENDIX-A Glossary Definitions of key words in quantum entanglement from the Wikipedia Action at a distance

In physics, action at a distance is the nonlocal interaction of objects that are separated in space. This term was used most often in the context of early theories of gravity and electromagnetism to describe how an object responds to the influence of distant objects. More generally "action at a distance" describes the failure of early atomistic and mechanistic theories which sought to reduce all physical interaction to collision. The exploration and resolution of this problematic phenomenon led to significant developments in physics, from the concept of a field, to descriptions of quantum entanglement and the mediator particles of the standard model. Spooky action at a distance [spukhafte Fernwirkung (in german)]

Entanglement arises naturally when two particles are created at the same point and instant in space, for example. Entangled particles can become widely separated in space. But even so, the mathematics implies that a measurement on one immediately influences the other, regardless of the distance between them. Einstein and co-authors pointed out that according to special relativity, this was impossible and therefore, quantum mechanics must be wrong, or at least incomplete. Einstein famously called it spooky action at a distance. The basic idea here is to think about the transfer of information. Entanglement allows one particle to instantaneously influence another but not in a way that allows classical information to travel faster than light. This resolved the paradox with special relativity but left much of the mystery intact. Separability Different particles or systems that occupy different regions in space have an independent reality. Locality An action involving one of these particles or systems cannot influence a particle or system in another part of space unless something travels the distance between them, a process limited by the speed of light. Nonlocality

In physics, nonlocality or action at a distance is the direct interaction of two objects that are separated in space with no intermediate agency or mechanism. Regarding the unexplained nature of gravity, Isaac Newton (1642-1727) considered action-at-a-distance "so great an Absurdity that I believe no Man who has in philosophical Matters a competent Faculty of thinking can ever fall into it". Quantum nonlocality refers to what Einstein called the "spooky action at a distance" of quantum entanglement.

129

(Local) hidden variable theory Historically, in physics, hidden variable theories were espoused by some physicists who

argued that the state of a physical system, as formulated by quantum mechanics, does not give a complete description for the system; i.e., that quantum mechanics is ultimately incomplete, and that a complete theory would provide descriptive categories to account for all observable behavior and thus avoid any indeterminism. The existence of indeterminacy for some measurements is a characteristic of prevalent interpretations of quantum mechanics; moreover, bounds for indeterminacy can be expressed in a quantitative form by the Heisenberg uncertainty principle. Principle of locality

In physics, the principle of locality states that an object is influenced directly only by its immediate surroundings. Experiments have shown that quantum mechanically entangled particles must either violate the principle of locality or engage in superluminal communication. Local realism

Local realism is a significant feature of classical mechanics, of general relativity, and of electrodynamics; but quantum mechanics largely rejects this principle due to the theory of distant quantum entanglements, an interpretation rejected by Einstein in the EPR paradox but subsequently proven by Bell's inequalities. Any theory, such as quantum mechanics, that violates Bell's inequalities must abandon either locality or realism; but some physicists dispute that experiments have demonstrated Bell's violations, on the grounds that the sub-class of inhomogeneous Bell inequalities has not been tested or due to experimental limitations in the tests. Different interpretations of quantum mechanics violate different parts of local realism and/or counterfactual definiteness. Qubit In quantum computing, a qubit or quantum bit is a unit of quantum information—the quantum analogue of the classical bit. A qubit is a two-state quantum-mechanical system, such as the polarization of a single photon: here the two states are vertical polarization and horizontal polarization. In a classical system, a bit would have to be in one state or the other, but quantum mechanics allows the qubit to be in a superposition of both states at the same time, a property which is fundamental to quantum computing. Quantum entanglement

Quantum entanglement is a physical phenomenon that occurs when pairs (or groups) of particles are generated or interact in ways such that the quantum state of each member must subsequently be described relative to each other. Quantum entanglement is a product of quantum superposition. However, the state of each member is indefinite in terms of physical properties such as position, momentum, spin, polarization, etc. in a manner distinct from the

130

intrinsic uncertainty of quantum superposition. When a measurement is made on one member of an entangled pair and the outcome is thus known (e.g., clockwise spin), the other member of the pair is at any subsequent time always found (when measured) to have taken the appropriately correlated value (e.g., counterclockwise spin). There is thus a correlation between the results of measurements performed on entangled pairs, and this correlation is observed even though the entangled pair may be separated by arbitrarily large distances. Repeated experiments have verified that this works even when the measurements are performed more quickly than light could travel between the sites of measurement: there is no light speed or slower influence that can pass between the entangled particles. Recent experiments have measured entangled particles within less than one part in 10,000 of the light travel time between them; according to the formalism of quantum theory, the effect of measurement happens instantly.

This behavior is consistent with quantum theory, and has been demonstrated experimentally with photons, electrons, molecules the size of buckyballs, and even small diamonds. It is an area of extremely active research by the physics community. However, there is some heated debate about whether a possible classical underlying mechanism could explain entanglement. The difference in opinion derives from espousal of various interpretations of quantum mechanics.

Research into quantum entanglement was initiated by a 1935 paper by Albert Einstein, Boris Podolsky, and Nathan Rosen describing the EPR paradox and several papers by Erwin Schrödinger shortly thereafter. Although these first studies focused on the counterintuitive properties of entanglement, with the aim of criticizing quantum mechanics, eventually entanglement was verified experimentally, and recognized as a valid, fundamental feature of quantum mechanics. The focus of the research has now changed to its utilization as a resource for communication and computation. Decoherence

Decoherence is an important effect – a physical system is never completely isolated, it interacts with an environment. Through this interaction, the number of degrees of freedom involved in the full description of the evolution increases rapidly. Roughly speaking, one can say that the information about the state, initially concentrated on the system under study, is rapidly diluted into the environment. Mathematically, it can be shown that such an interaction indeed happens very rapidly for large objects, and has the effect of destroying interferences. Locality loophole

The results of any Bell experiment can in principle be explained by a hidden-variable model if information about the measurement settings on one side can travel to and influence the measurement on the other side. To exclude this possibility, or at least make it incompatible with special relativity, the entangled particles should be sufficiently distant from each other and the measurement settings varied sufficiently rapidly, so that such information should necessarily travel at a speed greater than the speed of light to have any influence on the measurement results. Failure to satisfy this condition is known as the “locality loophole”

131

Detection loophole Current detectors detect less than half the photons that are sent. This is a fact, but it is

legitimate to ask ourselves whether or not the photons detected constitute a representative sample of all of the photons. It could be that it is not the case, that only certain photons, suitably ‘programmed’, activate our detectors. These photons could be programmed to violate Bell’s inequality, but if we detected all of the photons, we might see that Bell’s inequality is not violated. APPENDIX-B SG measurements

[2

1

[2

1

[2

1

[2

1

where these ket vectors are orthogonal to each other.

)ˆ(2

ˆ1111 nσnS

, )ˆ(

2ˆ

2222 nσnS

(a)

)sin()cos()ˆ)(ˆ( 21212211 nσnσ

leading to

)cos()ˆ)(ˆ( 212211 nσnσ

(b)

132

)sin()cos()ˆ)(ˆ( 21212211 nσnσ

leading to

)cos()ˆ)(ˆ( 212211 nσnσ

(c)

)cos()sin()ˆ)(ˆ( 21212211 nσnσ

leading to

)cos()ˆ)(ˆ( 212211 nσnσ

(i) When 021 ,

)ˆ)(ˆ( 2211 nσnσ

)ˆ)(ˆ( 2211 nσnσ

(ii) When 221

,

)ˆ)(ˆ( 2211 nσnσ

)ˆ)(ˆ( 2211 nσnσ

(iii) When 021

)ˆ)(ˆ( 2211 nσnσ

)ˆ)(ˆ( 2211 nσnσ

133

(iv) When 221

,

)ˆ)(ˆ( 2211 nσnσ

)ˆ)(ˆ( 2211 nσnσ

__________________________________________________________________________ APPENDIX-C Therefore, all three components of spin correspond to “elements of reality,” as defined by EPR, because a definite value will be predictable with certainty, for any one of them, if we measure the corresponding spin component of the other particle. This claim, however, is incompatible with quantum mechanics, which asserts that at most one spin component of each particle may be definite.

We consider the two spin particles, far apart from each other, in a singlet state'

][2

1212112

zzzz

We know that measurements of xx 21 ˆˆ , if performed, shall yield opposite values,

1212211221 )ˆˆ( xxxx mm (quantum mechanics)

Sincee 121 xxmm and 122

21 xx mm , we have

xx mm 21

Similarly, the measurements of yy 21 ˆˆ , if performed, shall yield opposite values,

1212211221 )ˆˆ( yyyy mm (quantum mechanics)

Since 121 yymm and 122

21 yy mm , we have

m1y = –m2y.

134

Furthermore, since x1 and y2 commute, and both correspond to elements of reality, their

product yx 21 ˆˆ also corresponds to an element of reality. The numerical value assigned to

the product yx 21 ˆˆ is the product of the individual numerical values, m1x m2y,such that

12211221 )ˆˆ( yxyx mm (EPR

Likewise, the numerical value of xy 21 ˆˆ is the product m1y m2x, such that

12211221 )ˆˆ( xyxy mm

These two products must be equal,

xyyxyx mmmmmm 211221 ))((

because m1x = – m2x and m1y = – m2y .

The quantum theory asserts that these products have opposite values, because the singlet state satisfies

0)ˆˆˆˆ(122121

xyyx (quantum mechanics)

From the EPR, element of reality, on the other hand, we should obtain the relation

02121 xyyx mmmm

which is inconsistent with the value

02 212121 yxxyyx mmmmmm

which can be derived from the above result. Then we are thus forced to the conclusion that the definition of elements of reality is incompatible with quantum theory. APPENDIX-C The experiments of Aspect et al. Aspect, P. Grangier, and G. Roger, Phys. Rev. Lett. 47, 460 (1981). Aspect, P. Grangier, and G. Roger, Phys. Rev. Lett. 49, 91 (1981). A. Aspect, J. Dalibard, and G. Roger, Phys. Rev. Lett. 49, 1804 (1982) A. Aspect, Nature 398, 189 (1991).

135

A.Aspect, Bell’s theorem: the naïve view of an experimentalist, Quantum [Un]speakables – From Bell to Quantum information, edited by R.A. Bertlmann and A. Zeilinger edited, (Springer, 2002).

((Experiment with one channels))

((Experiment with two channels))

Fig. ),(),( baba NN . ),(),( baba NN . ),(),( baba NN .

),(),( baba NN

S

1

1

ba

PMPM

PM

1

PM

( , ) , ( , )

( , ) , ( , )

N N

N N

a b a b

a b a b

1

136

),(),(),(),(

),(),(),(),(),(

babababa

bababababa

NNNN

NNNNE

When ),( baE is +1 or -1, there is a complete correlation between the results of measurements made by Alice and Bob; when ),( baE is 0, there is none.

Fig. Schematic Design of the Second Aspect Experiment testing Bell's inequality

)','(),'()',(),( babababa EEEES

137

Fig. Schematic Design of the third Aspect Experiment testing Bell's inequality.

)2cos(

)(sin)(cos

),(),(),(),(),(22

ab

abab

PPPPE

bababababa

Fig. Correlation of polarizations as a function of the relative angle of the polarimeters. The

indicated errors are ± 2 standard deviations. The dotted curve is not a fit to the data, but

138

quantum mechanical predictions for the actual experiment. For ideal polarizers, the curve would reach the values + 1. (Experimental result reported by Aspect et al.)

Fig. S vs the angle between the vectors a and b. )2cos( S with ab .

_____________________________________________________________________________ APPENDIX-D Invariance of the Bell’s state under the rotation (a) Two spin states with S = 1/2

0

1z ,

1

0z

0

1

1

0

2

1][

2

10,0 zzzz

We now consider the state

0,0)ˆˆ(

][2

1)ˆˆ(

]ˆˆˆˆ[2

1][

2

1

RR

zzzzRR

zRzRzRzR

nnnn

E

4 2

1.

0.5

0.5

1.

139

The rotation operator is given by R

2cos

2sin

2sin

2cos

ˆ22

22

ii

ii

ee

eeR

0,0

0

1

1

0

2

1

0

1

1

0

)cos1(2

1sin

2

1sin

2

1

2sin

sin2

1

2cos)cos1(

2

1sin

2

1

sin2

1)cos1(

2

1

2cossin

2

12

sinsin2

1sin

2

1)cos1(

2

1

2

10,0)ˆˆ(

2

2

2

2

iiii

iiii

eeee

eeee

RR

We consider the eigenvalue problem of RR ˆˆ (4 x 4 matrix). For simplicity we assume that = 0.

The eigenvalue The eigenket

1

1

0

0

1

2

11

1

0

1

1

0

2

12 (the Bell’s state)

140

ie

1

1

2

13 i

i

ie

1

1

2

14 i

i

Then we conclude that these states are invariant under the rotation. (b) Two spin states with S = Suppose that the system is a two spin state expressed by

0

0

1

0

1

0

1

0

0

3

1

0,10,11,11,11,11,1[3

1

The rotation matrix R with one spin state (3x3 matrix) is given by

2cossin

2

1

2sin

sin2

1cossin

2

12

sinsin2

1

2cos

)ˆexp()ˆexp(ˆ

22

22

iii

iii

yz

eee

eee

Si

Si

R

141

RR ˆˆ is given by

2cossin

2

1

2sin

sin2

1cossin

2

12

sinsin2

1

2cos

2cossin

2

1

2sin

sin2

1cossin

2

12

sinsin2

1

2cos

ˆˆ

22

22

22

22

iii

iii

iii

iii

eee

eee

eee

eee

RR

It is found that

0,10,11,11,11,11,1[3

1

]0,1ˆ0,1ˆ1,1ˆ1,1ˆ1,1ˆ1,1ˆ[3

1

]0,10,11,11,11,11,1[3

1)ˆˆ()ˆˆ(

RRRRRR

RRRR

So that the state is invariant under the rotation.

(c) Two photon states

0

1x ,

1

0y

0

1

1

0

2

1][

2

1xyyx

We now consider the state

142

)ˆˆ(

][2

1)ˆˆ(

]ˆˆˆˆ[2

1][

2

1

RR

xyyxRR

xRyRyRxRa

aaa

The rotation operator is given by R

cossin

sincosR

0

1

1

0

2

1

0

1

1

0

cos2sin2

12sin

2

1sin

2sin2

1cossin2sin

2

1

2sin2

1sincos2sin

2

1

sin2sin2

12sin

2

1cos

2

1)ˆˆ(

22

22

22

22

RR

We consider the eigenvalue problem of RR ˆˆ (4 x 4 matrix).

The eigenvalue The eigenket

1 ],,[2

1

1

0

0

1

2

11 yyxx

143

1 ],,[2

1

0

1

1

0

2

12 xyyx

2ie

1

1

2

13 i

i

2ie

1

1

2

14 i

i

Then we conclude that these states are invariant under the rotation. REFERENCES D.L. Hemmick and A.M. Shakur, Bell’s Theorem and Quantum Realism: Reassesment in Light of the Schrodinger Paradox (Springer, 2012). _____________________________________________________________________________ APPENDIX-E ((Problem-1)) The photon correlation is defined by

),(),(),(),(),( bababababa PPPPE

where ),( baP is defined by

2

,),( babaP

We call the angle that a makes with the x axis a with the x axis

yx aa sincos a , yx aa cossin a

144

We call the angle that b makes with the x axis b with the x axis

yx bb sincos b , yx bb cossin b ;

(a) First show the invariance of the Bell’s state under the rotation.

],,[2

1,

2

1,

2

1 aaaayyxx

],,[2

1,

2

1,

2

1 aaaaxyyx

(b) Calculate ),( baE for the two-photon entangled state as a function of baab .

yyxx ,2

1,

2

1

(c) Calculate ),( baE for the two-photon entangled state as a function of baab .

xyyx ,2

1,

2

1

((Solution)) (a) (i)

1

0

0

1

2

1

1

0

0

0

2

1

0

0

0

1

2

1

1

0

1

0

2

10

1

0

1

2

1,

2

1,

2

1yyxx

145

and

1

0

0

1

2

1

cos

cossin

cossin

sin

2

1

sin

cossin

cossin

cos

2

1

cos

sin

cos

sin

2

1sin

cos

sin

cos

2

1],,[

2

1

2

2

2

2

a

aa

aa

a

a

aa

aa

a

a

a

a

a

a

a

a

a

aaaa

Thus we have

],,[2

1,

2

1,

2

1 aaaayyxx

(ii)

0

1

1

0

2

1

0

1

0

0

2

1

0

0

1

0

2

1

0

1

1

0

2

11

0

0

1

2

1,

2

1,

2

1xyyx

and

146

0

1

1

0

2

1

cossin

cos

sin

cossin

2

1

cossin

sin

cos

cossin

2

1

sin

cos

cos

sin

2

1cos

sin

sin

cos

2

1],,[

2

1

2

2

2

2

aa

a

a

aa

aa

a

a

aa

a

a

a

a

a

a

a

a

aaaa

Thus we have

],,[2

1,

2

1,

2

1 aaaaxyyx

(b)

aaaa ,,[2

1,

2

1,

2

1yyxx

abba2

1,

abba2

1,

abba2

1,

abba2

1,

We use

a

a

sin

cosa ,

a

a

cos

sina

147

b

b

sin

cosb ,

b

b

cos

sinb

ab

ba

baba

a

abb

cos

)cos(

sinsincoscos

sin

cossincos

ab

ab

ba

baba

a

abb

sin

)sin(

cossinsincos

sin

coscossin

ab

ab

ba

baba

a

abb

sin

)sin(

sincoscossin

cos

sinsincos

ab

ab

ba

baab

a

abb

cos

)cos(

coscossinsin

cos

sincossin

ab

Then we get

)2cos(

)sin2cos2(2

1

),(),(),(),(),(

22

ab

abab

PPPPE

bababababa

(c)

148

],,[2

1,

2

1,

2

1 aaaaxyyx

abba2

1,

abba2

1,

abba2

1,

abba2

1,

We use

a

a

sin

cosa ,

a

a

cos

sina

b

b

sin

cosb ,

b

b

cos

sinb

abcos ab

absin ab

ab

ba

baba

a

abb

sin

)sin(

sincoscossin

cos

sinsincos

ab

abcos ab

149

Then we get

)2cos(

)cos2sin2(2

1

),(),(),(),(),(

22

ab

abab

PPPPE

bababababa

((Problem-II)) The spin correlation is defined by

),(),(),(),(),( bababababa PPPPE

where ),( baP is defined by

2

,),( babaP

We call the angle that a makes with the x axis a with the x axis

zz aa 2

sin2

cos

a , zz aa 2

cos2

sin

a

We call the angle that b makes with the x axis b with the x axis

zz bb 2

sin2

cos

b , zz bb 2

cos2

sin

b ;

(a) First show the invariance of the Bell’s state under the rotation.

],,[2

1),,(

2

1aaaa zzzz

(b) Calculate ),( baE for the two-photon entangled state as a function of baab .

zzzz ,2

1,

2

1

((Solution))

150

(a) (i)

0

1

1

0

2

1

0

1

0

0

2

1

0

0

1

0

2

1

0

1

1

0

2

11

0

0

1

2

1,

2

1,

2

1zzzz

and

0

1

1

0

2

1

2cos

2sin

2cos

2sin

2cos

2sin

2

1

2cos

2sin

2sin

2cos

2cos

2sin

2

1

2sin

2cos

2cos

2sin

2

1

2cos

2sin

2sin

2cos

2

1],,[

2

1

2

2

2

2

aa

a

a

aa

aa

a

a

aa

a

a

a

a

a

a

a

a

aaaa

Thus we have

],,[2

1,

2

1,

2

1aaaa zzzz

(b)

151

abba2

1,

abba2

1,

abba2

1,

abba2

1,

We use

2sin

2cos

a

a

a ,

2cos

2sin

a

a

a

2sin

2cos

b

b

b ,

2cos

2sin

b

b

b

2cos

)2

cos(

2sin

2sin

2cos

2cos

2sin

2cos

2sin

2cos

ab

ba

baba

a

a

bb

ab

152

2sin

)2

sin(

2cos

2sin

2sin

2cos

2sin

2cos

2cos

2sin

ab

ba

baba

a

a

bb

ab

2sin

)2

sin(

2sin

2cos

2cos

2sin

2cos

2sin

2sin

2cos

ab

ba

baba

a

a

bb

ab

2cos

)2

cos(

2cos

2cos

2sin

2sin

2cos

2sin

2cos

2sin

ab

ba

baab

a

a

bb

ab

Then we get

153

ba

abababab

bababababa

)cos(

)2

sin22

cos2(2

1

[2

1

),(),(),(),(),(

22

2222

ab

abab

PPPPE

Quantum entanglement Masatsugu Sei Suzuki Department of...

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