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Prerequisites Part 1

The EPR Paper

Resolution

Summary

Judgement DayQuantum Mechanics - Right or wrong!

Sunip Kumar Mukherjeesunipkmukherjee@gmail.com

19-March-2015

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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

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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

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.

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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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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.

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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>

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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

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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

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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

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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

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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

.

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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.

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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.

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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.

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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.

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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.

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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 .

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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

http://find/

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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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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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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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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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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

http://find/

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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

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

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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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