master dissertation topic

1. Discussion on Quantum Entanglement
Project Report
Submitted in partial fulfilment of the requirements of
MPHYD-D405 Master Dissertation
By
Jatin Mahato (1801168007)
Under the supervision of
Dr. Suman Ghosh
Assistant Professor
Department of Physics
Indira Gandhi National Tribal University,
Amarkantak (M.P.)-484 887
(06/08/2020)

2. Contents
0.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
0.2 Entangled state . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
0.3 The EPR Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . 4
0.3.1 Faster than light ? . . . . . . . . . . . . . . . . . . . . . . 4
0.3.2 Einstein locality and hidden variables . . . . . . . . . . . . 5
0.4 Bell’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
0.4.1 Spin 1
2 along arbitrary direction . . . . . . . . . . . . . . . 6
0.4.2 What’s Wrong with Hidden Variables . . . . . . . . . . . . 7
0.5 CHSH Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
0.6 The CHSH Game . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
0.7 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
i

3. Acknowledgement
I would like to express my special thanks of gratitude to my guide Dr. Suman
Ghosh for their able guidance and support in completing the project.
I would like to extend my sincere and heartfelt obligation towards Mr.Ajithkumar
K and Ms. Tamanna Jain and all other persons who have helped me in this en-
deavour. Without their active cooperation, encouragement and guidance, i would
not have made headway in the project.
At last but not least gratitude goes to all of my friends who directly or in-
directly helped me to complete this project report. Any omission in this brief
acknowledgement does not mean lake of gratitude.
Thank you
Jatin Mahato
1

4. 0.1 Introduction
Quantum entanglement is one of the important and interesting phenomena in quantum mechanics.
If two or more quantum particles are inextricably linked and whatever happens to one of then
which immediately affects the other regardless of their distance of separation, then we can say
that they are entangled. Einstein described it as ’Spooky action at a distance’. There exist a
correlations between observable physical properties of the systems.
For example, if we measure the spin of one of the two particles, which are entangled and
observed to be spin up along a given direction, then the other one will be spin down along the
same direction and vice versa. Before doing the measurement, the state of a quantum system
will be in the superposition of all possible states. According to the special theory of relativity,
nothing can travel faster than light. Here the spooky action at a distance is because of the
instantaneous collapse of wave function. If their distance of separation is very large, this collapse
of wave function appears to be flat contradiction to the special theory of relativity. This is the
famous EPR paradox. But using entanglement it is not possible to transmit classical information
faster than light.
Einstein believed that, quantum mechanics was an incomplete description of reality and there
exist hidden variables which predetermines state of quantum systems. Hidden variable theory
says quantum system is assigned with variables those are hidden to us, since they are hidden the
probability comes to play in quantum mechanical measurements.
In 1964 John Stewart Bell explained the non-existence of hidden variables through Bell’s
theorem. The essence of this theorem is that quantum mechanical probability cannot arise from
the ignorance of local pre-existing variable.
Quantum entanglement has wide range of applications in the emerging technologies of quan-
tum computing and quantum cryptography, and has been used to realize quantum teleportation
experimentally.
2

5. 0.2 Entangled state
Consider two systems, with associated Hilbert spaces H1 and H2 respectively, then the composite
system has the Hilbert space H1 ⊗ H2. Let |ψ1i be the state of first system and |ψ2i be the state
of second one, then the state of composite system is
|Ψi = |ψ1i ⊗ |ψ2i (1)
States of the composite system that can be represented in the form Eq.(1) are called separable
states or product states. All states of composite system are not separable. More generally let |ii
and |ji be the basis for the Hilbert spaces H1 and H2 respectively, then the most general state
in the H1 ⊗ H2 is of the form:
|Ψi =
X
i,j
cij |ii ⊗ |ji (2)
This state is separable if there exist vectors |ψ1i =
P
i ci |ii and |ψ2i =
P
j cj |ji such that
cij = cicj. If cij 6= cicj then |Ψi is inseparable and it is called ’Entangled state’. Simply a state
of this combined system is said to be entangled if it cannot be written in the form of Eq.(1).
For example, consider a composite system of two quantum particles in which each one can
have one of two states (called ’quantum bit or qubit’ ) i.e., ’spin up’ (|↑i) or ’spin down’ (|↓i).
Here the spin of these particles are measured along z-axis and the corresponding eigenstates (|↑i)
and (|↓i) are chosen to be the basis of Hilbert space. Then we can write the state
|Ψi = |↑i ⊗ |↓i (3)
which is not entangled. But, the state
|Ψi =
1
√
2
(|↑i ⊗ |↓i − |↓i ⊗ |↑i) (4)
is entangled (called EPR pair, after Einstein, Podolsky and Rosen).
Consider another state |Ψ0
i which can also be express in the form of Eq.(2)
|Ψ0
i =
1
√
2
(|↑i ⊗ |↓i + |↓i ⊗ |↓i) (5)
But it is not an entangled state, since we can write Eq.(5) as |Ψi = |→i ⊗ |↓i, where |→i =
1
√
2
(|↑i + |↓i).
Here we are interested in electron spin based qubit realizations. Note that outside the context
of the spin, Eq.(4) is one member of a basis of four mutually orthogonal maximally entangled
3

6. bell’s states for the two qubits. The basis are

8. φ+
=
1
√
2
(|0i ⊗ |0i + |1i ⊗ |1i)

10. φ−
=
1
√
2
(|0i ⊗ |0i − |1i ⊗ |1i)

12. ψ+
=
1
√
2
(|0i ⊗ |1i + |1i ⊗ |0i)

14. ψ−
=
1
√
2
(|0i ⊗ |1i − |1i ⊗ |0i) (6)
Here [|ψi , |φi] is the pairity bit and [+ , −] is the phase bit.
0.3 The EPR Paradox
The Einstein-Podolsky-Rosen paradox (or EPR paradox) is a thought experiment proposed by
Albert Einstein, Boris Podolsky and Nathan Rosen, which is intended to demonstrate an inherent
paradox in the early formulation of quantum theory. In a 1935 paper titled ”Can Quantum
Mechanical Description of Physical Reality” they argued for the existence of ”elements of reality”
that were not part of quantum theory, and speculated that it should be possible to construct a
theory containing them.
0.3.1 Faster than light ?
The thought experiment involves a pair of particles in the state Eq.(4)(EPR pair). Consider Alice
takes one of the particles to the Andromeda galaxy and Bob keep another on earth. Let’s say
Alice measuring spin of her z-direction. She either finds spin up |↑i or spin down |↓i with 50%
probably of each. Similarly Bob’s measurement result also could be spin up or spin down, again
with probability 50%.
Since these particles are in the state
|Ψi =
1
√
2
(|↑i ⊗ |↓i − |↓i ⊗ |↑i) (7)
the possible outcomes of their measurement(as a whole) is limited to two not four. If Alice’s
measurement result spin up then the Bob’s measurement will results in spin down, because the
wave function collapses to |↑i ⊗ |↓i immediately after Alice measuring the spin up. Similar thing
will happen when Alice’s measurement results is spin down but now the wave function collapses
to |↓i ⊗ |↑i. Here we can see the perfect anti-correlation in their measurements.
4

15. The difficulty is coming in our mind when we think about their distance of separation. Since
these particles are separated by a large distance this instantaneous collapse of wave function is
seems to be violating the Special theory of relativity. This is the EPR paradox. There is one
more things which makes trouble here : Quantum physics says that, until the moment of the
measurement, the particles do not have a definite quantum spin but are in a superposition of
possible states, but Alice can predict spin of Bob’s particle with an absolute certainty after his
measurement.
[There is another point also getting importance here, even though this entanglement seems to
be violating the special theory of relativity it is not possible to use this perfect anti-correlations
to transmit classical information faster than the speed of light. There is no way for both of them
to know that they measured spin up ( or spin down ) because whether he/she collapsed the wave
function, or because of the other guy already did it.]
0.3.2 Einstein locality and hidden variables
Einstein called it as ”spooky action at a distance”. He believed in hidden variable theory, because
he felt that quantum mechanics was an incomplete description of reality and there exist hidden
variables which predetermines state of quantum systems. Hidden variable theory says quantum
system is assigned with some variables, because they are hidden to us the probability comes to
play in quantum mechanical measurements.
For example, perhaps when a spin is prepared in what quantum theory would describe as the
pure state |↑zi or |↓zi, there is actually a deeper theory in which the state prepared is parametrized
as (z, λ) where λ is the hidden variable. If we know λ, the outcome is deterministic, but if λ
is completely unknown, then the probability distribution governing the measurement will agree
with the predictions of quantum theory. In other words our ignorance of these hidden variables
is where the probability of quantum mechanics comes from, then the uncertainties of quantum
mechanics are entirely due to lack of knowledge.
Einstein locality sometimes known as local realism :
Suppose that A and B are spacelike separated systems. Then in a complete description
of physical reality an action performed on system A must not modify the description
of system B.
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16. 0.4 Bell’s Inequality
Bell disproved the existence hidden variable using the Bell’s Inequality. While measuring, Alice
and Bob need not to choose the same axis for measuring the spin. Suppose, Alice and Bob
choose to measure the spin in the x-direction. The eigenstates for a single spin are
1
√
2


1
1

 and
1
√
2


1
−1

, in terms of |↑i and |↓i these states can be expressed as
|→i =
1
√
2
(|↑i + |↓i) and |→i =
1
√
2
(|↑i − |↓i) (8)
because |↑i =


1
0

 and |↓i =


0
1

.
0.4.1 Spin 1
2 along arbitrary direction
Let us say unit vector b
a points in the direction (θ, φ),i.e.,
b
a = (sin θ cos φ, sin θ sin φ, cos θ) (9)
An operator that measures spin along the direction b
n is
S · b
a = b
axSz + b
aySy + b
azSz
=
~
2


b
az b
ax − ib
ay
b
ax + ib
ay −b
az


=
~
2


cos θ sin θe−iφ
sin θeiφ
− cos θ

 (10)
The matrix S · b
n has eigenvalues λ = ±~
2
and the eigenstates are
|θ+i =


cos θ
2
e
−iφ
2
sin θ
2
e
iφ
2

 and |θ−i =


− sin θ
2
e
−iφ
2
cos θ
2
e
iφ
2

 (11)
If we use −
→
σ instead of S , then the matrix −
→
σ ·b
n has eigenvalues λ = ±1 with the same eigenstates
(as Eq.10). Because
S =
~
2
−
→
σ
=
~
2
(σx
b
i + σx
b
j + σx
b
k)
6

17. where
σx =


0 1
1 0

 , σy =


0 −i
i 0

 , σz =


1 0
0 −1

 (12)
are Pauli matrices. If unit vector is in xz-plane then φ = 0 (i.e., b
a = (sin θ, 0, cos θ)), then Eq.(10)
can be expressed in terms of |↑i and |↓i as
|θ+i = cos
θ
2
|↑i + sin
θ
2
|↓i and |θ−i = − sin
θ
2
|↑i + cos
θ
2
|↓i (13)
We can write the EPR pair Eq.(4) as
|EPRi =
1
√
2
(|↑i |↓i − |↓i |↑i) =
1
√
2
(|θ+i |θ−i − |θ−i |θ+i) (14)
( ⊗ symbol is omitted ) for any θ. i.e., here the anti-correlation of the spins are guaranteed if
Alice and Bob measure the spin along the same direction. The hypothetical hidden variables that
determine the measurements of spin also has to explain this correlation.
0.4.2 What’s Wrong with Hidden Variables
Suppose Alice and Bob measures the spin of particles along z-axis and b
a-axis respectively(here
b
a = (sin θ, 0, cos θ) ).
x
z
b
a
θ
Figure 1: Vector b
a
Consider Alice’s measurement results the spin |↑i, then we can be sure that Bob’s measurement
has to result spin |↓i along z-axis. But he measures spin along b
a-axis. Then the probability that
Bob measuring either spin + or spin − is
P(θ+) = sin2 θ
2
and P(θ−) = cos2 θ
2
(15)
7

18. respectively( from Eq.(12)). Also probability of Alice measures the spin |↑i is 1
2
. Then the
probability of Alice measuring spin |↑i and Bob measuring spin + is given by
P(σA
z = +, σB
θ = +) =
1
2
sin2 θ
2
(15. i)
where σA
z −→Alice’s particle spin along z-axis and σB
θ −→Bob’s particle spin along θ-axis.
Eq.(15. i) is written using conditional probability, P(A ∩ V ) = P(A)P(B/A). Similarly Bob
measuring Spin − provided Alice measured |↑i is
P(σA
z = +, σB
θ = −) =
1
2
cos2 θ
2
(15. ii)
There are also two more probabilities exists
P(σA
z = −, σB
θ = +) =
1
2
cos2 θ
2
(15. iii)
P(σA
z = −, σB
θ = −) =
1
2
sin2 θ
2
(15. iv)
where we considered that Alice measured spin |↓i ( Note that for z-axis θ = 0, so σA
z = + or −
means Alice’s particle is in |↑i or in |↓i state respectively ).
According to Hidden variable theory each of these particle have the variables such that
it’s value determine the spin along different directions. Let sz and sθ be the hidden variables
associated both of the particles, which can have values +1 or −1 and responsible for results
obtained from the measurement spin along the z-axis and b
a axis respectively. If Alice’s particle is
assigned with sA
z = +1, then spin measurement results |↑i along z-axis. Also Bob’s particle must
be assigned with sB
z = −1 for perfect anti-correlation of spins. So sA
z = −sB
z . Since states |θ+i
and |θ−i are also entangled sA
θ = −sB
θ . These theory have to reproduce the above probabilities.
Whenever Bob measures his spin σB
θ to be say, +1 then this determines the spin of Alice’s particle
to be sA
θ = −1 even though Alice didn’t measure the spin in the direction b
a . In this way we can
talk about the probability distribution P(sA
z , sA
θ ) for the spins of Alice’s particle.
P(sA
z = +, sA
θ = −) =
1
2
sin2 θ
2
and P(sA
z = +, sA
θ = +) =
1
2
cos2 θ
2
P(sA
z = −, sA
θ = −) =
1
2
cos2 θ
2
and P(sA
z = −, sA
θ = +) =
1
2
sin2 θ
2
(16)
By applying common logic Hidden variable theory could reproduce the probabilities. But this is
against the spirit of quantum mechanics because these operators for spins in different directions
don’t commute. Since this theory is against to the flavour of Quantum Mechanics let’ s wait until
it fails.
8

19. Till now Alice and Bob can choose any one of two axes ( z and b
a = (sin θ, 0, cos θ) ) for
measuring spin of their particles, where Alice measures spin along z-axes and Bob measures along
b
a-axis. Let there exist another axis say b
b = (sinφ, 0, cos φ), along this also they can measure the
spin.
x
z
b
b
b
a
θ
φ
Figure 2: Vectors b
a and b
b
Now according to the hidden variable theory there must be hidden variable corresponding to
choice of this axes also. So let sA
z , sA
θ and sA
φ , be the variables associated with Alice particle, each
of which can take value ±1 and P(sA
z , sA
θ , sA
φ ) be the probability distribution. Then we can write
P(sA
θ = +, sA
φ = −) = P(+ + −) + P(− + −)
≤ [P(+ + −)P(+ + +)] + [P(− + −) + P(− − −)]
= P(sA
z = +, sA
φ = +) + P(sA
z = −, sA
φ = −) (17)
From Eq.(16) we can write
P(sA
θ = +, sA
φ = −) =
1
2
sin2 θ − φ
2
RHS of Eq.(17) is known from Eq.(16), then Eq.(17) becomes
sin2 θ − φ
2
≤ cos2 θ
2
+ cos2 φ
2
(18)
Eq.(21) is called Bell inequality. For the existence of hidden variables and the probability distri-
bution P(sA
z , sA
θ , sA
φ ), Bell inequality has to valid for all θ and φ.
Let’s take θ = 3π/2and φ = 3π/4. Then
sin2 3π
8
− cos2 3π
8
= − cos
3π
4
=
1
√
2
9

20. Meanwhile
cos2 3π
4
=
1
2
but 1/2 1/
√
2.
This violation of Bell’s inequality disproves the existence of hidden variables in the quantum
system. i.e., quantum mechanical probabilities cannot arise from the ignorance of local pre-
existing variable, which is the Bell’s theorem.
0.5 CHSH Inequality
CHSH inequality is another (or simpler) form of the Bell inequality, which applies to a situation
in which Alice can measure either one of two observables A1 and A2, while Bob can measure
either B1 or B2, where these four observables can have possible eigenvalues, ai = ±1 and bi = ±1
respectively.
Consider the observable
C = (A1 + A2)B1 + (A1 − A2)B2 (19)
here [Ai, Bj] = 0 where i,j = 1, 2 so that their measurements will not interfere with each other.
If the both eigenvalues a1, a2 are +1 or −1 then a1 + a2 = +2, a1 − a2 = 0 and a1 + a2 = −2,
a1 − a2 = 0 respectively. If a1 = +1 and a2 = −1, then a1 + a2 = 0, a1 − a2 = 2 and if a1 = −1
and a2 = +1, then a1 + a2 = 0, a1 − a2 = −2 respectively. Therefore expectation value of the
observable C is
Ch.v ≡ (a1 + a2)b1 + (a1 − a2)b2 = ±2bi (20)
(Here is where the local hidden-variable assumption sneaks in — we have imagined that values in
±1 can be assigned simultaneously to all four observables, even though it is impossible to measure
both of a1 and a2, or both of b1 and b2.) Since bi can only take values ±1, we have |hbii| ≤ 1,
then |hCh.vi| ≤ 2, and so
−2 ≤ hCh.vi ≤ 2 (21)
This is the CHSH inequality.
According to Quantum theory, we can’t take a1 and a2 simultaneously, so cannot write Eq.(20).
To determine the upper and lower bounds hCi we are going to manipulate Eq.(19) as an operator
10

21. equation. Here A2
1 = A2
2 = B2
1 = B2
2 = I (∵ the eigenvalues are ±1). Let’s find
C2
= (A1 + A2)2
B2
1 + (A1 − A2)2
B2
2 + (A1 + A2)B1(A1 − A2)B2 + (A1 − A2)B2(A1 + A2)B1
=
4I + A1B1A1B2 − A1B1A2B2 + A2B1A1B2 − A2B1A2B2
+A1B2A1B1 + A1B2A2B1 − A2B2A1B1 − A2B2A2B1
since [Ai, Bj] = 0, C2
becomes
C2
= 4I + B1B2 − A1B1A2B2 + A2B1A1B2 − B1B2 + B2B1 + A1B2A2B1 − A2B2A1B1 − B2B1
= 4I − A1B1A2B2 + A2B1A1B2 + A1B2A2B1 − A2B2A1B1 (22)
= 4I − [A1, A2][B1, B2] (23)
Now [A1, A2] = A1A2 − A1A2, then
h[A1, A2]i = hA1A2i − hA1A2i
i.e.,
|h[A1, A2]i| ≤ |hA1A2i| + |hA1A2i|
Each of our operator has eigenvalue ±1, then |hA1A2i|, |hA1A2i| ≤ 1, which implies that |h[A1, A2]i| ≤
2. Similarly |h[B1, B2]i| ≤ 2. Then using Eq.(23) we can write
hC2
i = 4hIi − h[A1, A2]ih[B1, B2]ii (24)

23. hC2
i

25. = 4|hIi| + |h[A1, A2]i||h[B1, B2]ii|
which implies

27. hC2
i

29. ≤ 8 (25)
From Eq.(24), we can say that hC2
i ≥ 0 and hence
hC2
i ≤ 8 (26)
We have hC2
i ≥ hCi2
( because there is a Quantum uncertainty 4C =
q
hC2i − hCi2
), so
hCi2
≤ 8. The range of hCi in quantum mechanics is
−2
√
2 ≤ hCi ≤ 2
√
2 (27)
called Cirel’son bound. It exceeds the CHSH inequality (Eq.(21)) and again proves the non-
existence of hidden variables.
11

30. Form our Eq.(15), we can write that
hEPR|σA
z ⊗ σB
θ |EPRi = sin2 θ
2
− cos2 θ
2
= − cos θ (28)
If we choose spin operators A1, A2, B1 and B2 measures the spin along four directions as shown
in the Fig.(3) ( i.e., A2 has θ = 0, B1 has θ = π
4
, A1 has θ = π
2
and B2 θ = 3π
4
), then using the
Eq.(28) we can write the expectation values as
x
z
A2
B1
A1
B2
3π/4
π/4
π/2
Figure 3: Vectors A1, A2, B1 and B2
hA1B1i = hA1B2i = hA2B1i = −
1
√
2
and hA2B2i = +
1
√
2
(29)
and hence we can see
hCi = −2
√
2 (30)
saturating the Cirel’son bound.
0.6 The CHSH Game
This is a game shows the usefulness of entanglement. Obviously the players are Alice and Bob.
Before starting the game, to both of them an envelope is given. Inside each of envelope there is
either a red card or a blue card.
It’s like a team game, but they can’t communicate with each other but can build a strategy
before starting the game. The game is that, after opening their respective envelop both of them
have to say a word, ”turtles” or ”cucumber”. The rules of the game are :
12

31. 1. Alice and Bob win if both cards are red and they said different words.
2. Alice and Bob win if at least one card was blue and they said the same word.
3. Otherwise, they lose
There are four possibilities for their cards : (Red,Red), (Red,Blue), (Blue,Red) and (Blue,Blue)
The best strategy of them is repeating the same word ”turtles” or ”cucumber”. Using this
strategy there is a winning probability of 75% ( because Probability of getting atleast one Blue
is 75% ). If they could share a number of EPR pairs, one for each time that the game is played,
Figure 4: Alice’s plan
Figure 5: Bob’s plan
they can increase the winning probability. The game plan explained in the given diagrams.
13

32. If Alice got Blue card, she measures A1 and if it is red card she measures A2. She says ”turtles”
whenever her measurement gives +1 and says ”cucumber” for −1. Bob does the similar thing
but he measures B1 when blue, B2 when red.
Consider that both of them get blue cards. For winning the game they have to say same
words. For that measurement A1 and B1 have to give same result. i.e., they will win if their
measurement gives A1B1 = +1 and lose when A1B1 = −1. We can write this as
P(win) − P(lose) = hA1B1i (31)
If both cards are red then for winning they have to say the different words, so they lose if both
A2 and B2 give +1 or −1 i.e., the same result and win otherwise, so that
P(win) − P(lose) = −hA2B2i (32)
There are two more possibilities Alice getting Red and Bob getting Blue and Alice getting Blue
and Bob getting Red. Since there is one Blue in the both cases they win if they say the same
word. So they win if measurement gives A2B1 = +1 and lose when A2B1 = −1 ( for Alice→ Blue
Bob→ Red ). Also win when A1B2 = +1 and A1B2 = +1 ( for Alice→ Red Bob→ Blue ).
Probability of getting each of the card combination P = 1
4
then the total probability is
P(win) − P(lose) =
1
4
hA1B1 + A1B2 + A2B1 − A2B2i
=
1
4
hCi (33)
( from Eq.(19) ). Using Cirel’son bound −2
√
2 ≤ hCi ≤ 2
√
2 we can write Eq.(33) as
P(win) − P(lose) ≤
1
√
2
(34)
Since P(win) + P(lose) = 1 we can write as
P(win) − P(lose) + P(win) + P(lose) =
1
√
2
+ 1 (35)
i.e.,
P(win) =
1
2
1
√
2
+ 1
≈ 0.854 (36)
Remember in our classical strategy winning probability was 75%. The game does reveal an
important point: the correlations in entangled states can be used to do things that wouldn’t
otherwise be possible. If we can harness this ability to perform tasks that we actually care about,
then we might genuinely be able to change the world.
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33. 0.7 References
1. David Tong, 2017, ”Lectures on Applications of Quantum Mechanics”, University of Cam-
bridge Part II Mathematical Tripos ( page 251-259).
2. John Preskill, 2001, ”Lecture Notes on Quantum Information and Computation” (Chapter
4) , California Institute of Technology ( page 4-18).
3. R. Shankar, 1994, ”Principles of Quantum Mechanics” (second edition), Yale University
(page 373-382).
4. A. Einstein, B. Podolsky and N. Rosen, 1935, ”Can Quantum Mechanical Description of
Physical Reality Be Considered Complete”, Institute for Advanced Study, Princeton. New
Jersey, Phys. Rev. 47 ,777(1935).
5. Guido Burkard, 2007, ”Spin-entangled electrons in solid-state systems”, University of Besel,
J. Phys.: Condens. Matter 19, 233202 (2007)
6. Ryszard Horodecki, Pawel Horodecki, Mickel Horodecki, Krol Horodecki, 2007, ”Quantum
entanglement” arXiv:quant-ph/0702225v2 (2007)
7. ”Quantum Entanglement”, (2020, April 15). In Wikipedia. Retrived from https://en.
wikipedia.org/wiki/Quantum_entanglement
8. ”EPR paradox”, (2020, April 15). In Wikipedia. Retrived from https://en.wikipedia.
org/wiki/EPR_paradox
9. ”Einstein-Podolsky-Rosen paradox observed in many-particle system for the first time”,
(26 April 2018), ScienceDaily, Retrived from https://www.sciencedaily.com/releases/
2018/04/180426141601.htm.
10. Lorenzo Maccone, ”A simple proof of Bell’s inequality”,(2013), American Journal of Physics
81,854(2013).
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