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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.
((Penro
In th1/2 parttravel awa great d
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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)
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(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
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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
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)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
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e
e
e
e
e
e
e
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Supposespins is
this case
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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
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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.
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 N2.
a
b
a
b
a
b
a
b
25
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 N3.
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 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{+{+{-{-
measuremenof B's meas
Pa
article-1+z, +x}+z, -x}-z, +x}-z, -x}
nts of S1x andurements wi
article-1+z, +x}+z, -x}-z, +x}-z, -x}
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
{+{+{-{-
___________
measuremenof B's meas
of S2x ( 2/ )
Pa{+{+{-{-
measuremenof B's meas
of S2x ( 2/ )
Pa{+{+{-{-
+z, +x}+z, -x}-z, +x}-z, -x}
__________
nts of S1z andsurements w and E2 for B
article-1+z, +x}+z, -x}-z, +x}-z, -x}
nts of S1z andsurements w and E3 for B
article-1+z, +x}+z, -x}-z, +x}-z, -x}
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
, and B makwill yield
2/ ).
__________
kes measurem2/ . E1 for
kes measurem2/ . E4 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
___________
measuremenof B's meas
of S2z ( 2/ )
Pa{+{+{-{-
measuremenof B's meas
of S2z ( 2/ )
Pa{+{+{-{-
___________mechanics (n
chanics, we e
__________
nts of S1x andsurements w
and E1 for B
article-1+z, +x}+z, -x}-z, +x}-z, -x}
nts of S1x andsurements w
and E2 for B
article-1+z, +x}+z, -x}-z, +x}-z, -x}
__________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
___________
, and B makwill yield
2/ ) and
, and B makwill yield
2/ ).
___________
__________
kes measurem2/ . E3 for
kes measurem2/ . E4 for
__________
_____
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
particle-1 particle-2
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 _____________________________________________________________________________
Fig. Left: configuration for the particle-1 (Alice). Right: configuration for the particle-2
(Bob). {a+, b+, c+} (particle-1). {a-, b-, c-} (particle-2) __________________________________________________________________________
Fig. 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
a
b c
a
b c
a
b c
34
Fig. Left: configuration for the particle-1 (Alice). Right: configuration for the particle-2
(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
Probability: particle 1 (1
a ) and particle 2 (2
c ),
8
1
42),(
iiN
NNP ca
Probability: particle 1 (1
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
particle-1 particle-2
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
Transpose1.K1.1 Simplify
14—2 Cos
Transpose2.K1.2 Simplify
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
2KroneckerProduct1, 2
KroneckerProduct2, 1 MatrixForm
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
2KroneckerProduct1, 1
KroneckerProduct2, 2 MatrixForm
1200
12
B4
1
2KroneckerProduct1, 2
KroneckerProduct2, 1 MatrixForm
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`";
exp_ : exp . Complexre_, im_ Complexre, im;
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
2KroneckerProductR, L KroneckerProductL, R;
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
and the joint probability
)(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
and the joint probability
)(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
and the joint probability
)(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`";
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 1
2KroneckerProductR, L KroneckerProductL, R;
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;
Transpose2.1 Simplify
Cos 2
3 KroneckerProductX1, Y2 Simplify;
Transpose3.1 Simplify
Sin 2
4 KroneckerProductY1, X2 Simplify;
Transpose4.1 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;
Transpose1.1 Simplify
Sin 2
2 KroneckerProductY1, Y2 Simplify;
Transpose2.1 Simplify
Sin 2
3 KroneckerProductX1, Y2 Simplify;
Transpose3.1 Simplify
Cos 2
4 KroneckerProductY1, X2 Simplify;
Transpose4.1 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
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)
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)
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
117
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,
123
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