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7/21/2019 A Presentation Introducing Bell's Inequality
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
Judgement DayQuantum Mechanics - Right or wrong!
Sunip Kumar [email protected]
19-March-2015
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
Outline
1 Prerequisites Part 1Elementary Introduction to Quantum SpinSpin HalfTwo ParticleStatesA Problem!
2 The EPR PaperA little bit of philosophyThe Way Quantum Mechanics WorksImplications of the exampleIllustration of the paradoxConclusions of the EPR Paper
3 Resolution of the paradoxBegin from the end.Incompatibility of Bells inequality with QMDemise Of Non-locality
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
Outline
1 Prerequisites Part 1Elementary Introduction to Quantum SpinSpin HalfTwo ParticleStatesA Problem!
2 The EPR PaperA little bit of philosophyThe Way Quantum Mechanics WorksImplications of the exampleIllustration of the paradoxConclusions of the EPR Paper
3 Resolution of the paradoxBegin from the end.Incompatibility of Bells inequality with QMDemise Of Non-locality
Sunip K. Mukherjee EPR Paradox
http://find/7/21/2019 A Presentation Introducing Bell's Inequality
4/183
Prerequisites Part 1
The EPR Paper
Resolution
Summary
Outline
1 Prerequisites Part 1Elementary Introduction to Quantum SpinSpin HalfTwo ParticleStatesA Problem!
2 The EPR PaperA little bit of philosophyThe Way Quantum Mechanics WorksImplications of the exampleIllustration of the paradoxConclusions of the EPR Paper
3 Resolution of the paradoxBegin from the end.Incompatibility of Bells inequality with QMDemise Of Non-locality
Sunip K. Mukherjee EPR Paradox
http://find/7/21/2019 A Presentation Introducing Bell's Inequality
5/183
Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
Elementary Relations
S2, Si
= 0
Sj, Sk
=
3i=1
ijkSi
The commutations relations ensure two things.1 S2 is measurable for all spin states.2 You can measure only one of the three spin components.
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
Elementary Relations
S2, Si
= 0
Sj, Sk
=
3i=1
ijkSi
The commutations relations ensure two things.1 S2 is measurable for all spin states.2 You can measure only one of the three spin components.
Sunip K. Mukherjee EPR Paradox
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7/183
Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
Elementary Relations
S2, Si
= 0
Sj, Sk
=
3i=1
ijkSi
The commutations relations ensure two things.1 S2 is measurable for all spin states.2 You can measure only one of the three spin components.
Sunip K. Mukherjee EPR Paradox
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8/183
Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
Elementary Relations
S2, Si
= 0
Sj, Sk
=
3i=1
ijkSi
The commutations relations ensure two things.1 S2 is measurable for all spin states.2 You can measure only one of the three spin components.
Sunip K. Mukherjee EPR Paradox
http://find/7/21/2019 A Presentation Introducing Bell's Inequality
9/183
7/21/2019 A Presentation Introducing Bell's Inequality
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7/21/2019 A Presentation Introducing Bell's Inequality
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
Spin Eigenfunctions
We selectS2 andSzto be our measurable quantities.
We call |s, m>as our spin eigenstate (eigenfunction of S2 andSz),
With eigenvalues:
S2|s, m>=2s(s+1)|s, m>
Sz|s,m>= m|s,m>
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
Spin Eigenfunctions
We selectS2 andSzto be our measurable quantities.
We call |s, m>as our spin eigenstate (eigenfunction of S2 andSz),
With eigenvalues:
S2|s, m>=2s(s+1)|s, m>
Sz|s,m>= m|s,m>
Sunip K. Mukherjee EPR Paradox
S
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
Spin Eigenfunctions
We selectS2 andSzto be our measurable quantities.
We call |s, m>as our spin eigenstate (eigenfunction of S2 andSz),
With eigenvalues:
S2|s, m>=2s(s+1)|s, m>
Sz|s,m>= m|s,m>
Sunip K. Mukherjee EPR Paradox
P i it P t 1 S i
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
Spin Eigenfunctions
We selectS2 andSzto be our measurable quantities.
We call |s, m>as our spin eigenstate (eigenfunction of S2 andSz),
With eigenvalues:
S2|s, m>=2s(s+1)|s, m>
Sz|s,m>= m|s,m>
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1 Spin
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
Spin Eigenfunctions
We selectS2 andSzto be our measurable quantities.
We call |s, m>as our spin eigenstate (eigenfunction of S2 andSz),
With eigenvalues:
S2|s, m>=2s(s+1)|s, m>
Sz|s,m>= m|s,m>
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1 Spin
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
We deal with spin half particles specifically for our purposes.
Spin half particles are particles with s= 1
2
. So,m=
1
2
and 1
2
.
We denote by+ andthe two simultaneous eigenstates of
S2 andSz, which are equally represented by
1
0
and
0
1
respectively.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1 Spin
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
We deal with spin half particles specifically for our purposes.
Spin half particles are particles with s= 12 . So,m=
12 and
12 .
We denote by+ andthe two simultaneous eigenstates of
S2 andSz, which are equally represented by
1
0
and
0
1
respectively.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1 Spin
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
We deal with spin half particles specifically for our purposes.
Spin half particles are particles with s= 12 . So,m=
12 and
12 .
We denote by+ andthe two simultaneous eigenstates of
S2 andSz, which are equally represented by
1
0
and
0
1
respectively.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1 Spin
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
We deal with spin half particles specifically for our purposes.
Spin half particles are particles with s= 12 . So,m=
12 and
12 .
We denote by+ andthe two simultaneous eigenstates of
S2 andSz, which are equally represented by
1
0
and
0
1
respectively.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1 Spin
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q
The EPR Paper
Resolution
Summary
p
Spin Half
TPS
A Problem!
We deal with spin half particles specifically for our purposes.
Spin half particles are particles with s= 12 . So,m=
12 and
12 .
We denote by+ andthe two simultaneous eigenstates of
S2 andSz, which are equally represented by
1
0
and
0
1
respectively.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1 Spin
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q
The EPR Paper
Resolution
Summary
p
Spin Half
TPS
A Problem!
We deal with spin half particles specifically for our purposes.
Spin half particles are particles with s= 12 . So,m=
12 and
12 .
We denote by+ andthe two simultaneous eigenstates of
S2 andSz, which are equally represented by
1
0
and
0
1
respectively.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1 Spin
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The EPR Paper
Resolution
Summary
Spin Half
TPS
A Problem!
We deal with spin half particles specifically for our purposes.
Spin half particles are particles with s= 12 . So,m=
12 and
12 .
We denote by+ andthe two simultaneous eigenstates of
S2 andSz, which are equally represented by
1
0
and
0
1
respectively.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1 Spin
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The EPR Paper
Resolution
Summary
Spin Half
TPS
A Problem!
We deal with spin half particles specifically for our purposes.
Spin half particles are particles with s= 12 . So,m=
12 and
12 .
We denote by+ andthe two simultaneous eigenstates of
S2 andSz, which are equally represented by
1
0
and
0
1
respectively.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1 Spin
S f
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The EPR Paper
Resolution
Summary
Spin Half
TPS
A Problem!
We deal with spin half particles specifically for our purposes.
Spin half particles are particles with s= 12 . So,m=
12 and
12 .
We denote by+ andthe two simultaneous eigenstates of
S2 andSz, which are equally represented by
1
0
and
0
1
respectively.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
Th EPR P
Spin
S i H lf
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The EPR Paper
Resolution
Summary
Spin Half
TPS
A Problem!
We deal with spin half particles specifically for our purposes.
Spin half particles are particles with s= 12 . So,m=
12 and
12 .
We denote by+ andthe two simultaneous eigenstates of
S2 andSz, which are equally represented by
1
0
and
0
1
respectively.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Spin
Spin Half
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The EPR Paper
Resolution
Summary
Spin Half
TPS
A Problem!
We deal with spin half particles specifically for our purposes.
Spin half particles are particles with s= 12 . So,m=
12 and
12 .
We denote by+ andthe two simultaneous eigenstates of
S2 andSz, which are equally represented by
1
0
and
0
1
respectively.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Spin
Spin Half
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The EPR Paper
Resolution
Summary
Spin Half
TPS
A Problem!
We deal with spin half particles specifically for our purposes.
Spin half particles are particles with s= 12 . So,m=
12 and
12 .
We denote by+ andthe two simultaneous eigenstates of
S2 andSz, which are equally represented by
1
0
and
0
1
respectively.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Spin
Spin Half
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The EPR Paper
Resolution
Summary
Spin Half
TPS
A Problem!
Using these representations for eigenstates, we can arrive at
the following representations for the corresponding operators:
S2 = 32
4
1 00 1
Sz
=
2 1 0
0 1
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Spin
Spin Half
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The EPR Paper
Resolution
Summary
Spin Half
TPS
A Problem!
Using these representations for eigenstates, we can arrive at
the following representations for the corresponding operators:
S2 = 32
4
1 00 1
Sz
=
2 1 0
0 1
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Spin
Spin Half
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The EPR Paper
Resolution
Summary
Spin Half
TPS
A Problem!
Using these representations for eigenstates, we can arrive at
the following representations for the corresponding operators:
S2 = 32
4
1 00 1
Sz
=
2 1 0
0 1
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Spin
Spin Half
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p
Resolution
Summary
p
TPS
A Problem!
Using these representations for eigenstates, we can arrive at
the following representations for the corresponding operators:
S2 = 32
4
1 00 1
Sz=
2 1 0
0 1
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Spin
Spin Half
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p
Resolution
Summary
p
TPS
A Problem!
Raising and Lowering Operators
Some more operators and their functions: We define
S=Sx Sy.
SinceS + =0, S = +, 0
S are respectively called the raising and lowering operatorsbased on what they do to
. In matrix representation,
S+=
0 1
0 0
,
S=
0 0
1 0
.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Spin
Spin Half
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Resolution
Summary
TPS
A Problem!
Raising and Lowering Operators
Some more operators and their functions: We define
S=Sx Sy.
SinceS + =0, S = +, 0
S are respectively called the raising and lowering operatorsbased on what they do to
. In matrix representation,
S+=
0 1
0 0
,
S=
0 0
1 0
.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Spin
Spin Half
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Resolution
Summary
TPS
A Problem!
Raising and Lowering Operators
Some more operators and their functions: We define
S=Sx Sy.
SinceS + =0, S = +, 0
S are respectively called the raising and lowering operatorsbased on what they do to
. In matrix representation,
S+=
0 1
0 0
,
S=
0 0
1 0
.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Spin
Spin Half
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Resolution
Summary
TPS
A Problem!
Raising and Lowering Operators
Some more operators and their functions: We define
S=Sx Sy.
SinceS + =0, S = +, 0
S are respectively called the raising and lowering operatorsbased on what they do to
. In matrix representation,
S+=
0 1
0 0
,
S=
0 0
1 0
.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
R l ti
Spin
Spin Half
TPS
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Resolution
Summary
TPS
A Problem!
Raising and Lowering Operators
Some more operators and their functions: We define
S=Sx Sy.
SinceS + =0, S = +, 0
S are respectively called the raising and lowering operatorsbased on what they do to . In matrix representation,
S+=
0 1
0 0
,
S=
0 0
1 0
.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Spin
Spin Half
TPS
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Resolution
Summary
TPS
A Problem!
Raising and Lowering Operators
Some more operators and their functions: We define
S=Sx Sy.
SinceS + =0, S = +, 0
S are respectively called the raising and lowering operatorsbased on what they do to . In matrix representation,
S+=
0 1
0 0
,
S=
0 0
1 0
.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Spin
Spin Half
TPS
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Resolution
Summary
TPS
A Problem!
Raising and Lowering Operators
Some more operators and their functions: We define
S=Sx Sy.
SinceS + =0, S = +, 0
S are respectively called the raising and lowering operatorsbased on what they do to . In matrix representation,
S+=
0 1
0 0
,
S=
0 0
1 0
.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Spin
Spin Half
TPS
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Resolution
Summary
TPS
A Problem!
Raising and Lowering Operators
Some more operators and their functions: We define
S=Sx Sy.
SinceS + =0, S = +, 0
S are respectively called the raising and lowering operatorsbased on what they do to . In matrix representation,
S+=
0 1
0 0
,
S=
0 0
1 0
.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Spin
Spin Half
TPS
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Resolution
Summary
TPS
A Problem!
SxandSy
From the representations ofS, we can deduce the matrixrepresentations ofSx &Sy:
Sx=2
0 1
1 0
,
Sy =2
0 0
.
TheSx andSyeigenstates (common withS2, but not among
themselves) are:(x) =
12
1
1,
(y) =
12
1
.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Spin
Spin Half
TPS
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Resolution
Summary
TPS
A Problem!
SxandSy
From the representations ofS, we can deduce the matrixrepresentations ofSx &Sy:
Sx=2
0 1
1 0
,
Sy =2
0 0
.
TheSx andSyeigenstates (common withS2, but not among
themselves) are:(x) =
12
1
1,
(y) =
12
1
.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Spin
Spin Half
TPS
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Summary A Problem!
SxandSy
From the representations ofS, we can deduce the matrixrepresentations ofSx &Sy:
Sx=2
0 1
1 0
,
Sy =2
0 0
.
TheSx andSyeigenstates (common withS2, but not among
themselves) are:(x) =
12
1
1,
(y) =
12
1
.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Spin
Spin Half
TPS
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Summary A Problem!
SxandSy
From the representations ofS, we can deduce the matrixrepresentations ofSx &Sy:
Sx=2
0 1
1 0
,
Sy =2
0 0
.
TheSx andSyeigenstates (common withS2, but not among
themselves) are:(x) =
12
1
1,
(y) =
12
1
.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Spin
Spin Half
TPS
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Summary A Problem!
SxandSy
From the representations ofS, we can deduce the matrixrepresentations ofSx &Sy:
Sx=2
0 1
1 0
,
Sy =2
0 0
.
TheSx andSyeigenstates (common withS2, but not among
themselves) are:(x) =
12
1
1,
(y) =
12
1
.
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1
The EPR Paper
Resolution
Spin
Spin Half
TPS
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Summary A Problem!
Addition of Angular Momenta
If we have two spin-half particles, they can be arranged in the
following manner:
, , , .
Of course,+ and in our example. Let us nowconsider what is the possible value of thezcomponent of the
spin for such two particles: Notice that, first, Sz=S(1)
z +S(2)
z .Also,S
(i)z operates only on theith particle.
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1
The EPR Paper
Resolution
S mmar
Spin
Spin Half
TPS
A Problem!
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Summary A Problem!
Addition of Angular Momenta
If we have two spin-half particles, they can be arranged in the
following manner:
, , , .
Of course,+ and in our example. Let us nowconsider what is the possible value of thezcomponent of the
spin for such two particles: Notice that, first, Sz=S(1)
z +S(2)
z .Also,S
(i)z operates only on theith particle.
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
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Summary A Problem!
Addition of Angular Momenta
If we have two spin-half particles, they can be arranged in the
following manner:
, , , .
Of course,+ and in our example. Let us nowconsider what is the possible value of thezcomponent of the
spin for such two particles: Notice that, first, Sz=S(1)
z +S(2)
z .Also,S
(i)z operates only on theith particle.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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Summary A Problem!
Addition of Angular Momenta
If we have two spin-half particles, they can be arranged in the
following manner:
, , , .
Of course,+ and in our example. Let us nowconsider what is the possible value of thezcomponent of the
spin for such two particles:Notice that, first, Sz=S(1)
z +S(2)
z .Also,S
(i)z operates only on theith particle.
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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Then,
Sz12 = (S(1)z +S
(2)z )12
= (S(1)
z 1)2+1(S(2)
z 2)=m112+1(m22)
=(m1+m2)12.
So, we see that thezcomponents of the spinadd up.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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Then,
Sz12 = (S(1)z +S
(2)z )12
= (S(1)
z 1)2+1(S(2)
z 2)=m112+1(m22)
=(m1+m2)12.
So, we see that thezcomponents of the spinadd up.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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Spin values of two spin-half particles
So, the totalzcomponent of spin for the four combinations
discussed are:
: m=1;
: m=0;
: m=0;
: m= 1.Something is fishy!
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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Spin values of two spin-half particles
So, the totalzcomponent of spin for the four combinations
discussed are:
: m=1;
: m=0;
: m=0;
: m= 1.Something is fishy!
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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Spin values of two spin-half particles
So, the totalzcomponent of spin for the four combinations
discussed are:
: m=1;
: m=0;
: m=0;
: m= 1.Something is fishy!
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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Spin values of two spin-half particles
So, the totalzcomponent of spin for the four combinations
discussed are:
: m=1;
: m=0;
: m=0;
: m= 1.Something is fishy!
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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The Triplet States
If we apply the lowering operator to the state, we will obtain:
S = (S(1) +S(2) )
= (S(1) ) + (S(2) )= () + ()=( + )
If we apply the lowering operator to the (normalized) state1
2( + ), we will obtain .
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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The Triplet States
If we apply the lowering operator to the state, we will obtain:
S = (S(1) +S(2) )
= (S(1) ) + (S(
2) )
= () + ()=( + )
If we apply the lowering operator to the (normalized) state1
2( + ), we will obtain .
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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The Triplet States
If we apply the lowering operator to the state, we will obtain:
S = (S(1) +S(2) )
= (S(1) ) + (S(
2) )
= () + ()=( + )
If we apply the lowering operator to the (normalized) state1
2( + ), we will obtain .
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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The Triplet States
If we apply the lowering operator to the state, we will obtain:
S = (S(1) +S(2) )
= (S(1) ) + (S
(2) )
= () + ()=( + )
If we apply the lowering operator to the (normalized) state1
2( + ), we will obtain .
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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The Triplet States
If we apply the lowering operator to the state, we will obtain:
S = (S(1) +S(2) )
= (S(1) ) + (S
(2) )
= () + ()=( + )
If we apply the lowering operator to the (normalized) state1
2( + ), we will obtain .
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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The Triplet States
If we apply the lowering operator to the state, we will obtain:
S = (S(1) +S(2) )
= (S(1) ) + (S
(2) )
= () + ()=( + )
If we apply the lowering operator to the (normalized) state1
2( + ), we will obtain .
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
Th T i l S
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The Triplet States
So, we have a set of three:
|1, 1>=|1, 0>= 1
2( + )
|1,1>=
These states haves=1, and since there are three such states,they are collectively calledthe triplet states.
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
Th T i l t St t
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The Triplet States
So, we have a set of three:
|1, 1>=|1, 0>= 1
2( + )
|1,1>=
These states haves=1, and since there are three such states,they are collectively calledthe triplet states.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
Th T i l t St t
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The Triplet States
So, we have a set of three:
|1, 1>=|1, 0>= 1
2( + )
|1,1>=
These states haves=1, and since there are three such states,they are collectively calledthe triplet states.
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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We discussed three out of four combinations, what happened to
the fourth?
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
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We discussed three out of four combinations, what happened to
the fourth?
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
The Singlet State
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The Singlet State
If we consider the state
|0, 0>= 1
2 (),We will find that any raising or lowering operation on this state
produces zero, and measurement ofS2 does the same.
This is the missing fourth, and since this one comes alone, this
one is called the singlet state.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
THE Problem
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THE Problem
Suppose two spin half particles are known to be in the singlet
configuration. LetS(1)a be the component of the spin angular
momentum of particle 1 in the direction specified by the unit
vectora, and similarly letS(2)
b be the component of the spin
angular momentum of particle 2 in the direction specified by the
unit vectorb. Then, determine the value of
S(1)
a S
(2)
b .
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
Solution
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Solution
Without loss in generality, we can alignzaxis alonga, andconsideraandbto lie on theZX plane.So, we can decomposeSbasSzcos +Sxsin , wherecos =a b.Now,
S(1)a S
(2)
b =
0, 0 S(1)a S(2)b0,0 for our problem.
So,S(1)a S
(2)b |0, 0>=S
(1)z (cos S
(2)z +sin S
(2)x )|0,0>. Upon
expansion,
S(1)z (cos S
(2)z +sin S
(2)x )|0,0>
= 12
(S(1)z )(S(2)z cos +S(2)x sin )
(S(1)z )(S(2)z cos +S(2)x sin ).
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
Solution
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Solution
Without loss in generality, we can alignzaxis alonga, andconsideraandbto lie on theZX plane.So, we can decomposeSbasSzcos +Sxsin , wherecos =a b.Now,
S(1)a S
(2)
b =
0, 0 S(1)a S(2)b0,0 for our problem.
So,S(1)a S
(2)b |0, 0>=S
(1)z (cos S
(2)z +sin S
(2)x )|0,0>. Upon
expansion,
S(1)z (cos S
(2)z +sin S
(2)x )|0,0>
= 12
(S(1)z )(S(2)z cos +S(2)x sin )
(S(1)z )(S(2)z cos +S(2)x sin ).
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
Solution
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Solution
Without loss in generality, we can alignzaxis alonga, andconsideraandbto lie on theZX plane.So, we can decomposeSbasSzcos +Sxsin , wherecos =a b.Now,
S(1)a S
(2)
b =
0, 0 S(1)a S(2)b0,0 for our problem.
So,S(1)a S
(2)b |0, 0>=S
(1)z (cos S
(2)z +sin S
(2)x )|0,0>. Upon
expansion,
S(1)z (cos S
(2)z +sin S
(2)x )|0,0>
= 12
(S(1)z )(S(2)z cos +S(2)x sin )
(S(1)z )(S(2)z cos +S(2)x sin ).
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1The EPR Paper
Resolution
Summary
SpinSpin Half
TPS
A Problem!
Solution
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Solution
Without loss in generality, we can alignzaxis alonga, andconsideraandbto lie on theZX plane.So, we can decomposeSbasSzcos +Sxsin , wherecos =a b.Now,
S(1)a S
(2)
b =
0, 0 S(1)a S(2)b0,0
for our problem.
So,S(1)a S
(2)b |0, 0>=S
(1)z (cos S
(2)z +sin S
(2)x )|0,0>. Upon
expansion,
S(1)z (cos S
(2)z +sin S
(2)x )|0,0>
= 12
(S(1)z )(S(2)z cos +S(2)x sin )
(S(1)z )(S(2)z cos +S(2)x sin ).
Sunip K Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
Solution
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Solution
Without loss in generality, we can alignzaxis alonga, andconsideraandbto lie on theZX plane.So, we can decomposeSbasSzcos +Sxsin , wherecos =a b.Now,
S(1)a S
(2)
b =
0, 0 S(1)a S(2)b0,0
for our problem.
So,S(1)a S
(2)b |0, 0>=S
(1)z (cos S
(2)z +sin S
(2)x )|0,0>. Upon
expansion,
S(1)z (cos S
(2)z +sin S
(2)x )|0,0>
= 12
(S(1)z )(S(2)z cos +S(2)x sin )
(S(1)z )(S(2)z cos +S(2)x sin ).
Sunip K Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
Solution
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Solution
Without loss in generality, we can alignzaxis alonga, andconsideraandbto lie on theZX plane.So, we can decomposeSbasSzcos +Sxsin , wherecos =a b.Now,
S(1)a S
(2)b
=
0, 0 S(1)a S(2)b 0,0for our problem.
So,S(1)a S
(2)b |0, 0>=S
(1)z (cos S
(2)z +sin S
(2)x )|0,0>. Upon
expansion,
S(1)z (cos S
(2)z +sin S
(2)x )|0,0>
= 12
(S(1)z )(S(2)z cos +S(2)x sin )
(S(1)z )(S(2)z cos +S(2)x sin ).
Sunip K Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
Spin
Spin Half
TPS
A Problem!
Solution
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Finally, after all the manipulations, we will be left with:
24
a b.
You can try your hand at the algebra if you want, as our time here is abit less than I would have liked.
Sunip K Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Parts of Physical Theory
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y y
Objective Reality.Physical Concepts.
Sunip K Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Parts of Physical Theory
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y y
Objective Reality.Physical Concepts.
Sunip K Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Parts of Physical Theory
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y y
Objective Reality.Physical Concepts.
Sunip K Mukherjee EPR Paradox
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Testing Physical Theory
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The two questions:
1 Is the theory correct?
2 Is the description given by the theory complete?
Sunip K Mukherjee EPR Paradox
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Testing Physical Theory
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The two questions:
1 Is the theory correct?
2 Is the description given by the theory complete?
Sunip K Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Correctness of Theory
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Only when positive answers can be given to both of the
questions, we can say that the concepts of the theory are
satisfactory. -EPR
The correctness of the theory is judged by the comparison of
the predictions of the theory and human experience, which
takes the form of experiments and measurements in physics.
This experience alone enables us to make inferences about
reality.
Sunip K Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Correctness of Theory
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Only when positive answers can be given to both of the
questions, we can say that the concepts of the theory are
satisfactory. -EPR
The correctness of the theory is judged by the comparison of
the predictions of the theory and human experience, which
takes the form of experiments and measurements in physics.
This experience alone enables us to make inferences about
reality.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Correctness of Theory
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Only when positive answers can be given to both of the
questions, we can say that the concepts of the theory are
satisfactory. -EPR
The correctness of the theory is judged by the comparison of
the predictions of the theory and human experience, which
takes the form of experiments and measurements in physics.
This experience alone enables us to make inferences about
reality.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Consideration for Quantum Mechanics
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It is the second question that we wish to consider here, as
applied to Quantum Mechanics.EPR Criterion for completeness of a physical theory:
Every element of the physical reality must have a counterpart in
the physical theory.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Consideration for Quantum Mechanics
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It is the second question that we wish to consider here, as
applied to Quantum Mechanics.EPR Criterion for completeness of a physical theory:
Every element of the physical reality must have a counterpart in
the physical theory.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Consideration for Quantum Mechanics
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It is the second question that we wish to consider here, as
applied to Quantum Mechanics.EPR Criterion for completeness of a physical theory:
Every element of the physical reality must have a counterpart in
the physical theory.
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Solution
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The question is then answered if we can decide what are theelements of physical reality.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Physical Reality
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EPR Criterion for Physical Reality:
If, without in any way disturbing a system, we can predict with
certainty (ie. with probability equal to unity) the value of aphysical quantity, then there exists an element of physical
reality corresponding to that quantity. This condition can merely
be regarded as sufficient, not necessary, which is in agreement
with both quantum mechanical and classical ideas of reality.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Physical Reality
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EPR Criterion for Physical Reality:
If, without in any way disturbing a system, we can predict with
certainty (ie. with probability equal to unity) the value of aphysical quantity, then there exists an element of physical
reality corresponding to that quantity. This condition can merely
be regarded as sufficient, not necessary, which is in agreement
with both quantum mechanical and classical ideas of reality.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Physical Reality
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EPR Criterion for Physical Reality:
If, without in any way disturbing a system, we can predict with
certainty (ie. with probability equal to unity) the value of aphysical quantity, then there exists an element of physical
reality corresponding to that quantity. This condition can merely
be regarded as sufficient, not necessary, which is in agreement
with both quantum mechanical and classical ideas of reality.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
Physical Reality
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106/183
EPR Criterion for Physical Reality:
If, without in any way disturbing a system, we can predict with
certainty (ie. with probability equal to unity) the value of aphysical quantity, then there exists an element of physical
reality corresponding to that quantity. This condition can merely
be regarded as sufficient, not necessary, which is in agreement
with both quantum mechanical and classical ideas of reality.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
RecapImplications
Paradox
Conclusions
The way Quantum Mechanics Works
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Quantum mechanical state, which is supposed to be
completely characterized by the wave function.
There is a corresponding operator to every physically
observable quantity.
We can say with certainty that the physically observablequantityAhas the valueaonly if
A=a
whereais a number. In accordance with the EPR criterion,then, we have an element of physical reality corresponding
to Afor a particle in the state.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
Implications
Paradox
Conclusions
The way Quantum Mechanics Works
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108/183
Quantum mechanical state, which is supposed to be
completely characterized by the wave function.
There is a corresponding operator to every physically
observable quantity.
We can say with certainty that the physically observablequantityAhas the valueaonly if
A=a
whereais a number. In accordance with the EPR criterion,then, we have an element of physical reality corresponding
to Afor a particle in the state.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
Implications
Paradox
Conclusions
The way Quantum Mechanics Works
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109/183
Quantum mechanical state, which is supposed to be
completely characterized by the wave function.
There is a corresponding operator to every physically
observable quantity.
We can say with certainty that the physically observablequantityAhas the valueaonly if
A=a
whereais a number. In accordance with the EPR criterion,then, we have an element of physical reality corresponding
to Afor a particle in the state.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
Implications
Paradox
Conclusions
The way Quantum Mechanics Works
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110/183
Quantum mechanical state, which is supposed to be
completely characterized by the wave function.
There is a corresponding operator to every physically
observable quantity.
We can say with certainty that the physically observablequantityAhas the valueaonly if
A=a
whereais a number. In accordance with the EPR criterion,then, we have an element of physical reality corresponding
to Afor a particle in the state.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
Implications
Paradox
Conclusions
The way Quantum Mechanics Works
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111/183
Quantum mechanical state, which is supposed to be
completely characterized by the wave function.
There is a corresponding operator to every physically
observable quantity.
We can say with certainty that the physically observablequantityAhas the valueaonly if
A=a
whereais a number. In accordance with the EPR criterion,then, we have an element of physical reality corresponding
to Afor a particle in the state.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
Implications
Paradox
Conclusions
An Example
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Let =exp (p0x/).
Momentum operator is:
p=
x.
So,
p=
x =p0.Thus, for such a state, the momentum has certain value, p0.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
Implications
Paradox
Conclusions
An Example
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113/183
Let =exp (p0x/).
Momentum operator is:
p=
x.
So,
p=
x =p0.Thus, for such a state, the momentum has certain value, p0.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
Implications
Paradox
Conclusions
An Example
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Let =exp (p0x/).
Momentum operator is:
p=
x.
So,
p=
x =p0.Thus, for such a state, the momentum has certain value, p0.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
Implications
Paradox
Conclusions
An Example
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Let =exp (p0x/).
Momentum operator is:
p=
x.
So,
p=
x =p0.Thus, for such a state, the momentum has certain value, p0.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
Implications
Paradox
Conclusions
Continued
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On the other hand, if the eigenvalue equation does not hold, wecan not speak of the physical quantity in question having a
particular value.Like,
x=a,whereais a definite number. In accordance with quantum
mechanics, we can only say that te relative probability that ameasurement of co-ordinate will give a result lying between a
andbis given by:
P(a, b) = b
a
dx= b
a
dx=b
a,
which is independent ofaand depends only on the difference
b a. So, we can conclude that all the values of co-ordinatesare equally probable.
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
Implications
Paradox
Conclusions
Continued
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On the other hand, if the eigenvalue equation does not hold, wecan not speak of the physical quantity in question having a
particular value. Like,
x=a,whereais a definite number. In accordance with quantum
mechanics, we can only say that te relative probability that ameasurement of co-ordinate will give a result lying between a
andbis given by:
P(a, b) = b
a
dx= b
a
dx=b
a,
which is independent ofaand depends only on the difference
b a. So, we can conclude that all the values of co-ordinatesare equally probable.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
Implications
Paradox
Conclusions
Continued
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On the other hand, if the eigenvalue equation does not hold, wecan not speak of the physical quantity in question having a
particular value. Like,
x=a,whereais a definite number. In accordance with quantum
mechanics, we can only say that te relative probability that ameasurement of co-ordinate will give a result lying between a
andbis given by:
P(a, b) = b
a
dx= b
a
dx=b
a,
which is independent ofaand depends only on the difference
b a. So, we can conclude that all the values of co-ordinatesare equally probable.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
Implications
Paradox
Conclusions
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Thus, a definite value of the co-ordinate can only be obtained
by a direct measurement.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
Implications
Paradox
Conclusions
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But such a measurement will disturb the particle and thus alter
its state.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Conclusion of the example
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When the momentum of a particle is known, its co-ordinates
have no physical reality.
We can extend this statement to any two non-commutating
operators corresponding to physically observable quantities.
Then,
Any knowledge of one such quantity precludes the other.
Any attempt to determine the latter experimentally alters
the state of the system in such a way as to destroy the
knowledge of the first.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Conclusion of the example
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When the momentum of a particle is known, its co-ordinates
have no physical reality.
We can extend this statement to any two non-commutating
operators corresponding to physically observable quantities.
Then,
Any knowledge of one such quantity precludes the other.
Any attempt to determine the latter experimentally alters
the state of the system in such a way as to destroy the
knowledge of the first.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Conclusion of the example
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When the momentum of a particle is known, its co-ordinates
have no physical reality.
We can extend this statement to any two non-commutating
operators corresponding to physically observable quantities.
Then,
Any knowledge of one such quantity precludes the other.
Any attempt to determine the latter experimentally alters
the state of the system in such a way as to destroy the
knowledge of the first.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Conclusion of the example
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When the momentum of a particle is known, its co-ordinates
have no physical reality.
We can extend this statement to any two non-commutating
operators corresponding to physically observable quantities.
Then,
Any knowledge of one such quantity precludes the other.
Any attempt to determine the latter experimentally alters
the state of the system in such a way as to destroy the
knowledge of the first.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Conclusion of the example
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When the momentum of a particle is known, its co-ordinates
have no physical reality.
We can extend this statement to any two non-commutating
operators corresponding to physically observable quantities.
Then,
Any knowledge of one such quantity precludes the other.
Any attempt to determine the latter experimentally alters
the state of the system in such a way as to destroy the
knowledge of the first.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Implications of the example
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From the previous discussion it follows that one of the following
statements is true about quantum mechanics:
1 The quantum mechanical description of reality given by the
wave function is not complete.
2 When the operators corresponding to two physical
quantities do not commute, the two quantities can not have
simultaneous reality, for if both of them had simultaneous
realities, they would have definite values and those values
would have entered into the complete description,according to the condition of completeness.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Implications of the example
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From the previous discussion it follows that one of the following
statements is true about quantum mechanics:
1 The quantum mechanical description of reality given by the
wave function is not complete.
2 When the operators corresponding to two physical
quantities do not commute, the two quantities can not have
simultaneous reality, for if both of them had simultaneous
realities, they would have definite values and those values
would have entered into the complete description,according to the condition of completeness.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Implications of the example
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From the previous discussion it follows that one of the following
statements is true about quantum mechanics:
1 The quantum mechanical description of reality given by the
wave function is not complete.
2 When the operators corresponding to two physical
quantities do not commute, the two quantities can not have
simultaneous reality,for if both of them had simultaneous
realities, they would have definite values and those values
would have entered into the complete description,according to the condition of completeness.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR PaperResolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Implications of the example
F h i di i i f ll h f h f ll i
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From the previous discussion it follows that one of the following
statements is true about quantum mechanics:
1 The quantum mechanical description of reality given by the
wave function is not complete.
2 When the operators corresponding to two physical
quantities do not commute, the two quantities can not have
simultaneous reality, for if both of them had simultaneous
realities, they would have definite values and those values
would have entered into the complete description,according to the condition of completeness.
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Goal of the EPR Paper
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In quantum mechanics it is usually assumed that the wave
functiondoescontain a complete description of the physical
reality of the system to which it corresponds. At first sight this
assmption is entirely reasonable, for the information obtainablefrom the wave function seems to correspond exactly to what
can be measured without altering the state of the system. We
shall show, however, that this assumption, together with the
criterion of reality given above, leads to a contradiction.
Sunip K. Mukherjee EPR Paradox
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Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
The Famous Example
The pion decay (David Bohm):
A neutral meson decays into an electron and a positron
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A neutral-meson decays into an electron and a positron.
0 e + e+.Outcomes:
Assuming that the pion was at rest prior to the decay, theelectron and the positron will move in opposite directions to
conserve the momentum.
Since a pion has spinzero, conservation of angular
momentum dictates that the electron and the positron are
insinglet configuration:1
2(+ +).
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
The Famous Example
The pion decay (David Bohm):
A neutral -meson decays into an electron and a positron
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A neutral-meson decays into an electron and a positron.
0 e + e+.Outcomes:
Assuming that the pion was at rest prior to the decay, theelectron and the positron will move in opposite directions to
conserve the momentum.
Since a pion has spinzero, conservation of angular
momentum dictates that the electron and the positron are
insinglet configuration:1
2(+ +).
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
The Famous Example
The pion decay (David Bohm):
A neutral -meson decays into an electron and a positron
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A neutral meson decays into an electron and a positron.
0 e + e+.Outcomes:
Assuming that the pion was at rest prior to the decay, theelectron and the positron will move in opposite directions to
conserve the momentum.
Since a pion has spinzero, conservation of angular
momentum dictates that the electron and the positron are
insinglet configuration:1
2(+ +).
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
The Famous Example
The pion decay (David Bohm):
A neutral -meson decays into an electron and a positron
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A neutral meson decays into an electron and a positron.
0 e + e+.Outcomes:
Assuming that the pion was at rest prior to the decay, theelectron and the positron will move in opposite directions to
conserve the momentum.
Since a pion has spinzero, conservation of angular
momentum dictates that the electron and the positron are
insinglet configuration:1
2(+ +).
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Measurement of spin just after creation
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Now, if we measure the spin of the electron, it will be either spin
up or spin down.
Correspondingly, the positron will be either spin down or spin
up.Quantum mechanics can not tell you which combination you
will get for a particular pion decay, but it does say that the
measurements arecorrelated, and that you will get each
combination half the time (on average).
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Measurement of spin just after creation
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Now, if we measure the spin of the electron, it will be either spin
up or spin down.
Correspondingly, the positron will be either spin down or spin
up.Quantum mechanics can not tell you which combination you
will get for a particular pion decay, but it does say that the
measurements arecorrelated, and that you will get each
combination half the time (on average).
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Measurement of spin just after creation
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Now, if we measure the spin of the electron, it will be either spin
up or spin down.
Correspondingly, the positron will be either spin down or spin
up.Quantum mechanics can not tell you which combination you
will get for a particular pion decay, but it does say that the
measurements arecorrelated, and that you will get each
combination half the time (on average).
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Measurement of spin long after creation
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Now, if the electron and the positron are let to fly away and
separated by a large distance, and someone measures the spin
of the electron and gets spin up or down (provided the
electron-positron system has not been interacted with in any
other way prior to this measurement), it will be known
instantaneously that the positron has spin down or spin up
respectively,without ever affecting the state of the positron.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Aftermath
At this point, we are very happy to conclude that this
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At this point, we are very happy to conclude that this
gedankenexperimentallows superluminal motion by allowing
the person measuring the spin of the electron complete
knowledge of the spin of the positroninstantaneouslywithout
disturbing the positron at all. But, this is not the whole story. Wecould have in this way found out some other observable that
does not commute with the spin operator, starting from the
same two particle state.
This implies a graver conclusion: Two non commutating
operators can have simultaneous reality, provided the wavefunctiondoes notprovide a complete description of a state.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Aftermath
At this point, we are very happy to conclude that this
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p , y ppy
gedankenexperimentallows superluminal motion by allowing
the person measuring the spin of the electron complete
knowledge of the spin of the positroninstantaneouslywithout
disturbing the positron at all. But, this is not the whole story. Wecould have in this way found out some other observable that
does not commute with the spin operator, starting from the
same two particle state.
This implies a graver conclusion: Two non commutating
operators can have simultaneous reality, provided the wavefunctiondoes notprovide a complete description of a state.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Aftermath
At this point, we are very happy to conclude that this
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p , y ppy
gedankenexperimentallows superluminal motion by allowing
the person measuring the spin of the electron complete
knowledge of the spin of the positroninstantaneouslywithout
disturbing the positron at all. But, this is not the whole story. Wecould have in this way found out some other observable that
does not commute with the spin operator, starting from the
same two particle state.
This implies a graver conclusion: Two non commutating
operators can have simultaneous reality, provided the wavefunctiondoes notprovide a complete description of a state.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
Resolution
Summary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Implications
Previously we had shown that:
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Previously we had shown that:
1 The quantum-mechanical description of reality given by the
wave function is not complete. (Hidden variables theory)
2
Or, when the operators corresponding to two observabesdo not commute, theycan nothave simultaneous reality.
But in the previous example, we negated the first statement,
and came up with the negation of the second. So,
We are forced to conclude that the description of physical
reality provided by quantum mechanics is incomplete.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
ResolutionSummary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Implications
Previously we had shown that:
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Previously we had shown that:
1 The quantum-mechanical description of reality given by the
wave function is not complete. (Hidden variables theory)
2
Or, when the operators corresponding to two observabesdo not commute, theycan nothave simultaneous reality.
But in the previous example, we negated the first statement,
and came up with the negation of the second. So,
We are forced to conclude that the description of physical
reality provided by quantum mechanics is incomplete.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
ResolutionSummary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Implications
Previously we had shown that:
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147/183
Previously we had shown that:
1 The quantum-mechanical description of reality given by the
wave function is not complete. (Hidden variables theory)
2
Or, when the operators corresponding to two observabesdo not commute, theycan nothave simultaneous reality.
But in the previous example, we negated the first statement,
and came up with the negation of the second. So,
We are forced to conclude that the description of physical
reality provided by quantum mechanics is incomplete.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
ResolutionSummary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Implications
Previously we had shown that:
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Previously we had shown that:
1 The quantum-mechanical description of reality given by the
wave function is not complete. (Hidden variables theory)
2
Or, when the operators corresponding to two observabesdo not commute, theycan nothave simultaneous reality.
But in the previous example, we negated the first statement,
and came up with the negation of the second. So,
We are forced to conclude that the description of physical
reality provided by quantum mechanics is incomplete.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
ResolutionSummary
The Philosophy
Recap
ImplicationsParadox
Conclusions
A counter argument
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Once could object to this conclusion on the grounds that our
criterion of reality is not sufficiently restrictive. Indeed, one
would not arrive at our conclusion if one insisted that two or
more physical quantities can be regarded as simultaneouselements of realityonly when they can be simultaneously
measured or predicted. On this point of view, since either one
or the other of the two non-commutating quantities can be
predicted, they are not simultaneously real.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
ResolutionSummary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Countering the counter argument
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But the argument makes the reality of the two quantities in
question depend upon the process of measurement carried out
on the first system, which does not disturb the second systemin any way. No reasonable definition of reality could be
expected to permit this.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
ResolutionSummary
The Philosophy
Recap
ImplicationsParadox
Conclusions
Concluding note of the EPR Paper
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While we have shown that the wave function does not provide
a complete description of the physical reality, we left open the
question of whether or not such a description exists. We
believe, however, that such a theory is possible.
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
ResolutionSummary
Begin from the end.
Error!Demise Of Non-locality
Bells Design of the Bohm experiment
Instead of measuring the electron and positron spins along thesame direction Bell proposed that we rotate the detectors
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same direction, Bell proposed that we rotate the detectors
independently, and measure the spins alongafor electron, andbfor positron.The results can be as follows (in units of
2
):
electron positron product+1 1 11 1 +11 +1 1
... ... ...
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
ResolutionSummary
Begin from the end.
Error!Demise Of Non-locality
Bells Design of the Bohm experiment
Instead of measuring the electron and positron spins along thesame direction Bell proposed that we rotate the detectors
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same direction, Bell proposed that we rotate the detectors
independently, and measure the spins alongafor electron, andbfor positron.The results can be as follows (in units of
2):
electron positron product+1 1 11 1 +11 +1 1
... ... ...
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
ResolutionSummary
Begin from the end.
Error!Demise Of Non-locality
He further proposed that we calculate the averagevalue of the
product of the spins, for a given set of detector orientations,
P( b)
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P(a,b).Ifb=a, we recover the actual EPRB configuration, andP(a, a) = 1.By the same token,P(a,a) = +1.For arbitrary orientations, as we proved long time ago,
P(a,b) = a b,
in units of 2 .
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
ResolutionSummary
Begin from the end.
Error!Demise Of Non-locality
He further proposed that we calculate the averagevalue of the
product of the spins, for a given set of detector orientations,
P( b)
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P(a,b).Ifb=a, we recover the actual EPRB configuration, andP(a, a) = 1.By the same token,P(a,a) = +1.For arbitrary orientations, as we proved long time ago,
P(a,b) = a b,
in units of
2 .
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
ResolutionSummary
Begin from the end.
Error!
Demise Of Non-locality
He further proposed that we calculate the averagevalue of the
product of the spins, for a given set of detector orientations,
P(a b)
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P(a,b).Ifb=a, we recover the actual EPRB configuration, andP(a, a) = 1.By the same token,P(a,a) = +1.For arbitrary orientations, as we proved long time ago,
P(a,b) = a b,
in units of
2 .
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
ResolutionSummary
Begin from the end.
Error!
Demise Of Non-locality
He further proposed that we calculate the averagevalue of the
product of the spins, for a given set of detector orientations,
P(a b)
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P(a,b).Ifb=a, we recover the actual EPRB configuration, andP(a, a) = 1.By the same token,P(a,a) = +1.For arbitrary orientations, as we proved long time ago,
P(a,b) = a b,
in units of
2 .
Sunip K. Mukherjee EPR Paradox
Prerequisites Part 1
The EPR Paper
ResolutionSummary
Begin from the end.
Error!
Demise Of Non-locality
Bells Inequality
Let quantum mechanics be incomplete description and let
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Let quantum mechanics be