Einstein's Recoiling-Slit Experiment: Uncertainty and ...qipa13/QIPA/docu/4.12.2013/evening...

Einstein’s Recoiling-Slit Experiment:Uncertainty and Complementarity

Radhika Vathsan

BITS Pilani K K Birla Goa Campus

QIPA 2013, 4th Dec...

Collaborator: Tabish QureshiCenter for Theoretical PhysicsJamia Millia Islamia, New Delhi

Radhika Vathsan (BITS Goa) Einstein’s Recoiling-Slit Experiment QIPA 2013, 4th Dec, Allahabad 1 / 43

Outline

1 Two-Slit Experiment and ComplementarityTwo Slit Experiment with Quantum ParticlesComplementarityEinstein’s Recoiling Slit Experiment...and Bohr’s Reply

2 Complementarity and Entanglementvon Neuman MeasurementsWhich-way Information and InterferencePath Distinguishability and Fringe Visibility

3 Complementarity and UncertaintyDuality and Uncertainty

4 Conclusions

Outline

4 Conclusions

The Two-Slit Experiment with Quantum particles

Slit 1 open

Slit 2 open

Both slits open

Two-slit experiment with electronsTonomura, Endo, Matsuda, Kawasaki, Ezawa, Am. J. Phys. 57(2) (1989).

Outline

4 Conclusions

Which slit did the electron pass through?Getting the “Welcher-Weg" (which-way) information

No Interference!

Which slit did the electron pass through?Getting the “Welcher-Weg" (which-way) information

Which-way Detector

No Interference!

Bohr’s Complementarity Principle

Niels Bohr in 1928

In describing the results ofquantum mechanical experiments,certain physical concepts arecomplementary. If two conceptsare complementary, an experimentthat clearly illustrates one conceptwill obscure the othercomplementary one.. . .("The Quantum Postulate and the Recent Development ofAtomic Theory," Supplement to Nature, April 14, 1928, p.580)

In the two-slit experiment: the “which-way" information vsexistence of interference pattern.

They can NEVER be observed at the same time, in the sameexperiment.

Niels Bohr in 1928

5th Solvay Conference (1927)

A Piccard, E Henriot, P Ehrenfest, E Herzen, T de Donder, E Schrodinger, J-E Verschaffelt,W Pauli, W Heisenberg, R H Fowler, LBrillouin,P Debye, M Knudsen, W L Bragg, H A Kramers, P Dirac, A Compton, L de Broglie, M Born, N Bohr,

I Langmuir, M Planck, M Sklodowska Curie, H Lorentz,A Einstein, P Langevin, C Guye, C T R Wilson, O W Richardson

5th Solvay Conference (1927)

A Piccard, E Henriot, P Ehrenfest, E Herzen, T de Donder, E Schrodinger, J-E Verschaffelt, W Pauli, W Heisenberg, R H Fowler,L Brillouin,P Debye, M Knudsen, W L Bragg, H A Kramers, P Dirac, A Compton, L de Broglie, M Born, N Bohr,

I Langmuir, M Planck, M Sklodowska Curie, H Lorentz, A Einstein, P Langevin, C Guye, C T R Wilson, O W Richardson

Outline

4 Conclusions

Einstein’s Recoiling-Slit Gedanken Experiment

... Einstein thought he had found a counterexample tothe uncertainty principle. "It was quite a shock forBohr .... he did not see the solution at once. Duringthe whole evening he was extremely unhappy, goingfrom one to the other and trying to persuade them thatit couldn’t be true, that it would be the end of physics ifEinstein were right; but he couldn’t produce anyrefutation. I shall never forget the vision of the twoantagonists leaving the club [of the FondationUniversitaire]: Einstein a tall majestic figure, walkingquietly, with a somewhat ironical smile, and Bohrtrotting near him, very excited ....

The next morningcame Bohr’s triumph."

ROSENFELD (1968)Fundamental Problems in Elementary Particle Physics

Proceedings of the Fourteenth Solvay Conference, Interscience, New York, p. 232.

... Einstein thought he had found a counterexample tothe uncertainty principle. "It was quite a shock forBohr .... he did not see the solution at once. Duringthe whole evening he was extremely unhappy, goingfrom one to the other and trying to persuade them thatit couldn’t be true, that it would be the end of physics ifEinstein were right; but he couldn’t produce anyrefutation. I shall never forget the vision of the twoantagonists leaving the club [of the FondationUniversitaire]: Einstein a tall majestic figure, walkingquietly, with a somewhat ironical smile, and Bohrtrotting near him, very excited ....

The next morningcame Bohr’s triumph."

... Einstein thought he had found a counterexample tothe uncertainty principle. "It was quite a shock forBohr .... he did not see the solution at once. Duringthe whole evening he was extremely unhappy, goingfrom one to the other and trying to persuade them thatit couldn’t be true, that it would be the end of physics ifEinstein were right; but he couldn’t produce anyrefutation. I shall never forget the vision of the twoantagonists leaving the club [of the FondationUniversitaire]: Einstein a tall majestic figure, walkingquietly, with a somewhat ironical smile, and Bohrtrotting near him, very excited .... The next morningcame Bohr’s triumph."

Replace the static source slit

by a movable slitto obtain which-way information without disturbing the particle

Figures after Bohr

by a movable slit

to obtain which-way information without disturbing the particle

Figures after Bohr

by a movable slitto obtain which-way information without disturbing the particle

Figures after Bohr

Particle going through upper/lower slit has momentum ±p0

Momentum conservation =⇒ recoil ∓p0 of slitMomentum of slit→ which-way information

Momentum conservation =⇒ recoil ∓p0 of slit

Momentum of slit→ which-way information

Outline

4 Conclusions

Bohr’s reply

For particles passing through Slit Aand those through slit B:

∆px = 2p sin(θ/2)

≈ pθ =hλθ =

This is the limit on accuracy ofmeasuring recoil momentum.

Min uncertainty in position of source slit: ∆x =~

2∆px=

λL4πd

This is the uncertainty in position of a fringe.Fringe separation = λL

d .Interference pattern is lost!

Bohr’s reply

∆px = 2p sin(θ/2) ≈ pθ =hλθ

2∆px=

λL4πd

Bohr’s reply

∆px = 2p sin(θ/2) ≈ pθ =hλθ =

2∆px=

λL4πd

Bohr’s reply

2∆px=

λL4πd

Bohr’s reply

2∆px=

λL4πd

Bohr’s reply

2∆px=

λL4πd

This is the uncertainty in position of a fringe.

Fringe separation = λLd .

Interference pattern is lost!

Bohr’s reply

2∆px=

λL4πd

Interference pattern is lost!

Bohr’s reply

2∆px=

λL4πd

Implication of Bohr’s resolution

Complementarity enforced by Uncertainty Principle?

Getting which-way information will necessarily disturb the state ofthe particle.Disturbance will be enough to wash out interference.This viewed as a restatement of Uncertainty Principle

Complementarity enforced by Uncertainty Principle?Getting which-way information will necessarily disturb the state ofthe particle.

Disturbance will be enough to wash out interference.This viewed as a restatement of Uncertainty Principle

Complementarity enforced by Uncertainty Principle?Getting which-way information will necessarily disturb the state ofthe particle.Disturbance will be enough to wash out interference.

This viewed as a restatement of Uncertainty Principle

Complementarity enforced by Uncertainty Principle?Getting which-way information will necessarily disturb the state ofthe particle.Disturbance will be enough to wash out interference.This viewed as a restatement of Uncertainty Principle

Realization of Recoiling-Slit Experiment

Is uncertainty a requirement for Complementarity?

Now it turns out that the concept of Uncertainty is not necessary forexplaining complementarity!

Obtaining information about a quantum system is throughMeasurement, which yields classical result.

Is uncertainty a requirement for Complementarity?

Now it turns out that the concept of Uncertainty is not necessary forexplaining complementarity!

Obtaining information about a quantum system is throughMeasurement, which yields classical result.

Outline

4 Conclusions

Quantum measurementAccording to von Neumann

A quantum measurement consists of two processes:

1 Process 1: Unitary operation establishes correlation betweensystem & detector.

Initial states: System:∑n

i=1 ci|ψi〉; Detector:|d0〉

|d0〉n∑

ci|ψi〉Unitary evolution−−−−−−−−−→

Process 1

n∑i=1

ci|di〉|ψi〉

2 Process 2: Non-unitary selection of a single state |ψk〉 withprobability |ck|2:

n∑i=1

ci|di〉|ψi〉 −−−−−→Process 2

|dk〉|ψk〉

"The Measurement Problem".

A quantum measurement consists of two processes:1 Process 1: Unitary operation establishes correlation between

system & detector.

Initial states: System:∑n

i=1 ci|ψi〉; Detector:|d0〉

|d0〉n∑

Process 1

n∑i=1

ci|di〉|ψi〉

n∑i=1

|dk〉|ψk〉

system & detector.Initial states: System:

∑ni=1 ci|ψi〉; Detector:|d0〉

|d0〉n∑

Process 1

n∑i=1

ci|di〉|ψi〉

n∑i=1

|dk〉|ψk〉

|d0〉n∑

Process 1

n∑i=1

ci|di〉|ψi〉

n∑i=1

|dk〉|ψk〉

|d0〉n∑

Process 1

n∑i=1

ci|di〉|ψi〉

n∑i=1

|dk〉|ψk〉

|d0〉n∑

Process 1

n∑i=1

ci|di〉|ψi〉

n∑i=1

|dk〉|ψk〉

Which-way Detection in Einstein’s experimentUsing von Neumann’s process 1

Two orthogonal states of the particle depending on the path:slit 1: |ψ1〉 slit 2: |ψ2〉

Two momentum states of the recoiling slit: |p1〉 and |p2〉.

(a) Final state of particle+slit: necessary entanglement :

|Ψ〉 = |ψ1〉|p1〉+ |ψ2〉|p2〉

(b) Reading out of which-way information: correlation of “readout"states with detector states without affecting the states of theparticle

Point (a) was not part of Bohr’s reply.

and is enough to rule out interference!

|Ψ〉 = |ψ1〉|p1〉+ |ψ2〉|p2〉

Outline

4 Conclusions

Which-way Information and Interference

Without which-way informationAmplitude for finding the particle at point x on the screen is

Ψ(x) = ψ1(x) + ψ2(x).

Probability (intensity):

|Ψ(x)|2 = |ψ1(x)|2 + |ψ2(x)|2 + ψ∗1(x)ψ2(x) + ψ∗

2(x)ψ1(x)︸︷︷︸.interference

WITH which-way information

Ψ(x) = ψ1(x)|p1〉+ ψ2(x)|p2〉

|Ψ(x)|2 = |ψ1(x)|2 + |ψ2(x)|2 + ψ∗1(x)ψ2(x)〈p1|p2〉+ ψ∗

2(x)ψ1(x)〈p2|p1〉

= |ψ1(x)|2 + |ψ2(x)|2, since 〈p1|p2〉 = 0

Ψ(x) = ψ1(x) + ψ2(x).

|Ψ(x)|2 = |ψ1(x)|2 + |ψ2(x)|2 + ψ∗1(x)ψ2(x) + ψ∗

2(x)ψ1(x)︸︷︷︸.

interference

Ψ(x) = ψ1(x)|p1〉+ ψ2(x)|p2〉

|Ψ(x)|2 = |ψ1(x)|2 + |ψ2(x)|2 + ψ∗1(x)ψ2(x)〈p1|p2〉+ ψ∗

2(x)ψ1(x)〈p2|p1〉

= |ψ1(x)|2 + |ψ2(x)|2, since 〈p1|p2〉 = 0

Ψ(x) = ψ1(x) + ψ2(x).

|Ψ(x)|2 = |ψ1(x)|2 + |ψ2(x)|2 + ψ∗1(x)ψ2(x) + ψ∗

Ψ(x) = ψ1(x)|p1〉+ ψ2(x)|p2〉

|Ψ(x)|2 = |ψ1(x)|2 + |ψ2(x)|2 + ψ∗1(x)ψ2(x)〈p1|p2〉+ ψ∗

2(x)ψ1(x)〈p2|p1〉

= |ψ1(x)|2 + |ψ2(x)|2, since 〈p1|p2〉 = 0

Ψ(x) = ψ1(x) + ψ2(x).

|Ψ(x)|2 = |ψ1(x)|2 + |ψ2(x)|2 + ψ∗1(x)ψ2(x) + ψ∗

Ψ(x) = ψ1(x)|p1〉+ ψ2(x)|p2〉

|Ψ(x)|2 = |ψ1(x)|2 + |ψ2(x)|2 + ψ∗1(x)ψ2(x)〈p1|p2〉+ ψ∗

2(x)ψ1(x)〈p2|p1〉

= |ψ1(x)|2 + |ψ2(x)|2, since 〈p1|p2〉 = 0

Ψ(x) = ψ1(x) + ψ2(x).

|Ψ(x)|2 = |ψ1(x)|2 + |ψ2(x)|2 + ψ∗1(x)ψ2(x) + ψ∗

Ψ(x) = ψ1(x)|p1〉+ ψ2(x)|p2〉

|Ψ(x)|2 = |ψ1(x)|2 + |ψ2(x)|2 + ψ∗1(x)ψ2(x)〈p1|p2〉+ ψ∗

2(x)ψ1(x)〈p2|p1〉

= |ψ1(x)|2 + |ψ2(x)|2, since 〈p1|p2〉 = 0

Ψ(x) = ψ1(x) + ψ2(x).

|Ψ(x)|2 = |ψ1(x)|2 + |ψ2(x)|2 + ψ∗1(x)ψ2(x) + ψ∗

Ψ(x) = ψ1(x)|p1〉+ ψ2(x)|p2〉

|Ψ(x)|2 = |ψ1(x)|2 + |ψ2(x)|2 + ψ∗1(x)ψ2(x)〈p1|p2〉+ ψ∗

2(x)ψ1(x)〈p2|p1〉= |ψ1(x)|2 + |ψ2(x)|2,

since 〈p1|p2〉 = 0

Ψ(x) = ψ1(x) + ψ2(x).

|Ψ(x)|2 = |ψ1(x)|2 + |ψ2(x)|2 + ψ∗1(x)ψ2(x) + ψ∗

Ψ(x) = ψ1(x)|p1〉+ ψ2(x)|p2〉

|Ψ(x)|2 = |ψ1(x)|2 + |ψ2(x)|2 + ψ∗1(x)ψ2(x)〈p1|p2〉+ ψ∗

2(x)ψ1(x)〈p2|p1〉= |ψ1(x)|2 + |ψ2(x)|2, since 〈p1|p2〉 = 0

Interference vanishes if which-way information is obtained.

Another interpretation: the recoil of the slit stores which-wayinformation.No need to invoke uncertainty!

If this entanglement between the particle and the recoiling-slit hadbeen recognized and its implications understood

Bohr could have provided a simpler rebuttal to Einstein!

Can this argument be made more quantitative?

Interference vanishes if which-way information is obtained.Another interpretation: the recoil of the slit stores which-wayinformation.

No need to invoke uncertainty!

Interference vanishes if which-way information is obtained.Another interpretation: the recoil of the slit stores which-wayinformation.No need to invoke uncertainty!

Outline

4 Conclusions

Path-distinguishability and Interference

Suppose our detector distinguishes the two paths inaccurately.

This means “which-way" states 〈d1|d2〉 6= 0.

Define Distinguishability:

D =√

1− |〈d1|d2〉|2,

Amplitude that the paths are perfectly distinguishedDefine Visibility:

V ≡ Imax − Imin

Imax + Imin,

measure of the interference observed.

Is there a relationship between them to capture complementarity?

D =√

1− |〈d1|d2〉|2,

V ≡ Imax − Imin

Imax + Imin,

D =√

1− |〈d1|d2〉|2,

Amplitude that the paths are perfectly distinguished

Define Visibility:

V ≡ Imax − Imin

Imax + Imin,

D =√

1− |〈d1|d2〉|2,

V ≡ Imax − Imin

Imax + Imin,

D =√

1− |〈d1|d2〉|2,

V ≡ Imax − Imin

Imax + Imin,

Path-distinguishability and InterferenceGaussian Wave-packet Model

t = 0: particle emerges from the double-slit with amplitude

Ψ(x, 0) =

A(|d1〉e−

(x−d/2)2

4ε2 + |d2〉e−(x+d/2)2

), A =

8πε2

After time t, traveling a distance L, amplitude for particle to arrive at xon screen:

Ψ(x, t) = At

(|d1〉e

− (x−d/2)2

4ε2+2i~t/m + |d2〉e− (x+d/2)2

4ε2+2i~t/m

where At =1√2

2π(ε+ i~t/2mε)]−1/2

Ψ(x, 0) = A(|d1〉e−

(x−d/2)2

4ε2 + |d2〉e−(x+d/2)2

8πε2

Ψ(x, t) = At

(|d1〉e

− (x−d/2)2

4ε2+2i~t/m + |d2〉e− (x+d/2)2

4ε2+2i~t/m

where At =1√2

2π(ε+ i~t/2mε)]−1/2

Ψ(x, 0) = A(|d1〉e−

(x−d/2)2

4ε2 + |d2〉e−(x+d/2)2

), A =

8πε2

Ψ(x, t) = At

(|d1〉e

− (x−d/2)2

4ε2+2i~t/m + |d2〉e− (x+d/2)2

4ε2+2i~t/m

where At =1√2

2π(ε+ i~t/2mε)]−1/2

Ψ(x, 0) = A(|d1〉e−

(x−d/2)2

4ε2 + |d2〉e−(x+d/2)2

), A =

8πε2

Ψ(x, t) = At

(|d1〉e

− (x−d/2)2

4ε2+2i~t/m + |d2〉e− (x+d/2)2

4ε2+2i~t/m

where At =1√2

2π(ε+ i~t/2mε)]−1/2

Ψ(x, 0) = A(|d1〉e−

(x−d/2)2

4ε2 + |d2〉e−(x+d/2)2

), A =

8πε2

Ψ(x, t) = At

(|d1〉e

− (x−d/2)2

4ε2+2i~t/m + |d2〉e− (x+d/2)2

4ε2+2i~t/m

where At =1√2

2π(ε+ i~t/2mε)]−1/2

Probability of finding particle at point x on the screen

|Ψ(x, t)|2 = 2|At|2e− x2+d2/4

2σ2t cosh(xd/2σ2

1 + |〈d1|d2〉|cos(

xdλL/2π4ε4+(λL/2π)2 + θ

)cosh(xd/2σ2

|Ψ(x, t)|2 = 2|At|2e− x2+d2/4

2σ2t cosh(xd/2σ2

1 + |〈d1|d2〉|cos(

)cosh(xd/2σ2

〈d1|d2〉 = |〈d1|d2〉|eiθ

p0 = h/λ =⇒ ~t/m = λL/2π,

σ2t = ε2 +

|Ψ(x, t)|2 = 2|At|2e− x2+d2/4

2σ2t cosh(xd/2σ2

1 + |〈d1|d2〉|cos(

)cosh(xd/2σ2

Fringe width =

+16π2ε4

Visibility of Interference

Visibility V ≡ Imax − Imin

Imax + Imin

=|〈d1|d2〉|

cosh(xd/2σ2t )

cosh(y) ≥ 1 =⇒ V ≤ |〈d1|d2〉|.

UsingD2 = 1− |〈d1|d2〉|2,

we get

V2 +D2 ≤ 1.

Englert-Greenberger-Yasin duality relation

A quantitative statement of complementarity

Imax + Imin=

|〈d1|d2〉|cosh(xd/2σ2

cosh(y) ≥ 1 =⇒ V ≤ |〈d1|d2〉|.

UsingD2 = 1− |〈d1|d2〉|2,

we get

V2 +D2 ≤ 1.

Imax + Imin=

cosh(y) ≥ 1 =⇒ V ≤ |〈d1|d2〉|.

UsingD2 = 1− |〈d1|d2〉|2,

we get

V2 +D2 ≤ 1.

Imax + Imin=

cosh(y) ≥ 1 =⇒ V ≤ |〈d1|d2〉|.

UsingD2 = 1− |〈d1|d2〉|2,

we get

V2 +D2 ≤ 1.

Imax + Imin=

cosh(y) ≥ 1 =⇒ V ≤ |〈d1|d2〉|.

UsingD2 = 1− |〈d1|d2〉|2,

we get

V2 +D2 ≤ 1.

Imax + Imin=

cosh(y) ≥ 1 =⇒ V ≤ |〈d1|d2〉|.

UsingD2 = 1− |〈d1|d2〉|2,

we get

V2 +D2 ≤ 1.

Origin of Complementarity?

Quantum correlations ?D.M. Greenberger, A. Yasin, Phys. Lett. A 128, 391 (1988),“Simultaneous wave and particle knowledge in a neutron interferometer",B-G. Englert, Phys. Rev. Lett. 77, 2154 (1996),“Fringe visibility and which-way information: an inequality"M.O. Scully, B.G. Englert, H. Walther, Nature 375, 367 (1995),

“Complementarity and uncertainty."

Uncertainty principleS.M. Tan, D.F. Walls,Phys. Rev. A 47, 4663-4676 (1993),“Loss of coherence in interferometry".E.P. Storey, S.M. Tan, M.J. Collett, D.F. Walls, Nature 367, 626 (1994).H. Wiseman, F. Harrison, Nature 377, 584 (1995),“Uncertainty over complementarity?"H. Wiseman, Phys. Lett. A 311, 285 (2003),

“Directly observing momentum transfer in twin-slit which-way experiments"

Does the particle really receive a “momentum kick"?S. Durr, T. Nonn, G. Rempe, Nature 395, 33 (1998),“Origin of quantum-mechanical complementarity probed by a which-way experiment in an atom interferometer."C.S. Unnikrishnan, Phys. Rev. A 62, 015601 (2000),

“Origin of quantum-mechanical complementarity without momentum back action in atom-interferometry

experiments".

Uncertainty principle and complementarityOther work

G. Bjork, J. Soderholm, A. Trifonov, T. Tsegaye, A. Karlsson, Phys. Rev.A 60, 1874 (1999), “Complementarity and the uncertainty relations".

K-P Marzlin, B.C. Sanders, P.L. Knight, Phys. Rev. A 78, 062107 (2008),“Complementarity and uncertainty relations for matter-waveinterferometry",

J-H Huang, S-Y Zhu, arXiv:1011.5273 [physics.optics],“Complementarity and uncertainty in a two-way interferometer".

G.M. Bosyk, M. Portesi, F. Holik, A. Plastino, arXiv:1206.2992 [quant-ph]“On the connection between complementarity and uncertainty principlesin the Mach-Zehnder interferometric setting".

Paul Busch, Christopher R. Shilladay. arXiv:quant-ph/0609048, PhysRep 435, 1-31 (2006)

Outline

4 Conclusions

Complementarity and UncertaintyUncertainty and duality

“Which-way" states of the recoiling slit: |d1〉 and |d2〉(normalized, not necessarily orthogonal)

Orthonormal basis for recoiling slit: {|p1〉, |p2〉}Eigenstates of some observable P̂ with eigenvalues ±1.

Which-way states in the P-basis:

|d1〉 = c1|p1〉+ c2|p2〉,|d2〉 = c∗2|p1〉+ c∗1|p2〉.

|c1| = 1, c2 = 0→ full which-way information|c1| = |c2| = 1/

√2→ no which-way information

Uncertainty: ∆P2 = 〈P̂2〉 − 〈P̂〉2 = 4|c1|2|c2|2.Distinguishability:

D2 = 1− |〈d1|d2〉|2 = 1− 4|c1|2|c2|2

= 1−∆P2

Orthonormal basis for recoiling slit: {|p1〉, |p2〉}

Eigenstates of some observable P̂ with eigenvalues ±1.Which-way states in the P-basis:

|d1〉 = c1|p1〉+ c2|p2〉,|d2〉 = c∗2|p1〉+ c∗1|p2〉.

D2 = 1− |〈d1|d2〉|2 = 1− 4|c1|2|c2|2

= 1−∆P2

|d1〉 = c1|p1〉+ c2|p2〉,|d2〉 = c∗2|p1〉+ c∗1|p2〉.

D2 = 1− |〈d1|d2〉|2 = 1− 4|c1|2|c2|2

= 1−∆P2

|d1〉 = c1|p1〉+ c2|p2〉,|d2〉 = c∗2|p1〉+ c∗1|p2〉.

D2 = 1− |〈d1|d2〉|2 = 1− 4|c1|2|c2|2

= 1−∆P2

|d1〉 = c1|p1〉+ c2|p2〉,|d2〉 = c∗2|p1〉+ c∗1|p2〉.

D2 = 1− |〈d1|d2〉|2 = 1− 4|c1|2|c2|2

= 1−∆P2

|d1〉 = c1|p1〉+ c2|p2〉,|d2〉 = c∗2|p1〉+ c∗1|p2〉.

Uncertainty: ∆P2 = 〈P̂2〉 − 〈P̂〉2 = 4|c1|2|c2|2.

Distinguishability:

D2 = 1− |〈d1|d2〉|2 = 1− 4|c1|2|c2|2

= 1−∆P2

|d1〉 = c1|p1〉+ c2|p2〉,|d2〉 = c∗2|p1〉+ c∗1|p2〉.

D2 = 1− |〈d1|d2〉|2 = 1− 4|c1|2|c2|2

= 1−∆P2

Uncertainty and Duality

Correlation of detector states with particle states:

Ψ(x) = ψ1(x)|p1〉+ ψ2(x)|p2〉.

Consider a basis change:

|p1〉+ |p2〉 → ψ1(x) + ψ2(x)

|p1〉 − |p2〉 → ψ1(x)− ψ2(x)

=⇒ ∃ another observable Q̂ with eigenvalues ±1 andcorresponding eigenstates

|q1〉 = (|p1〉+ |p2〉)/√

|q2〉 = (|p1〉 − |p2〉)/√

The particle states can be correlated with these states:

Ψ(x) =c1√

2[ψ1(x) + ψ2(x)]|q1〉+

c2√2

[ψ1(x)− ψ2(x)]|q2〉

Ψ(x) = ψ1(x)|p1〉+ ψ2(x)|p2〉.

|p1〉+ |p2〉 → ψ1(x) + ψ2(x)

|p1〉 − |p2〉 → ψ1(x)− ψ2(x)

|q1〉 = (|p1〉+ |p2〉)/√

|q2〉 = (|p1〉 − |p2〉)/√

Ψ(x) =c1√

2[ψ1(x) + ψ2(x)]|q1〉+

c2√2

[ψ1(x)− ψ2(x)]|q2〉

Ψ(x) = ψ1(x)|p1〉+ ψ2(x)|p2〉.

|p1〉+ |p2〉 → ψ1(x) + ψ2(x)

|p1〉 − |p2〉 → ψ1(x)− ψ2(x)

|q1〉 = (|p1〉+ |p2〉)/√

|q2〉 = (|p1〉 − |p2〉)/√

Ψ(x) =c1√

2[ψ1(x) + ψ2(x)]|q1〉+

c2√2

[ψ1(x)− ψ2(x)]|q2〉

Ψ(x) = ψ1(x)|p1〉+ ψ2(x)|p2〉.

|p1〉+ |p2〉 → ψ1(x) + ψ2(x)

|p1〉 − |p2〉 → ψ1(x)− ψ2(x)

|q1〉 = (|p1〉+ |p2〉)/√

|q2〉 = (|p1〉 − |p2〉)/√

Ψ(x) =c1√

2[ψ1(x) + ψ2(x)]|q1〉+

c2√2

[ψ1(x)− ψ2(x)]|q2〉

Correlate the detected particles on the screen with the measuredeigenstate of Q̂ (c1 = c2 case)

Two complementary interference patterns corresponding to |q1〉 and|q2〉.

0-1-2-3-4-5 1 2 3 4 5

For any c1, c2,

|Ψ(x)|2 =|ψ1(x)|2 + |ψ2(x)|2

2+|c1|2 − |c2|2

2[ψ∗

1(x)ψ2(x) + ψ∗2(x)ψ1(x)] .

Fringe visibility: V2 ≤ (|c1|2 − |c2|2)2.

The uncertainty in Q̂, in this entangled state:

∆Q2 = 1− (|c1|2 − |c2|2)2.

ThusV2 ≤ 1−∆Q2.

Combining with the earlier result D2 = 1−∆P2, we get

D2 + V2 ≤ 2− [∆P2 + ∆Q2].

For any c1, c2,

|Ψ(x)|2 =|ψ1(x)|2 + |ψ2(x)|2

2+|c1|2 − |c2|2

2[ψ∗

1(x)ψ2(x) + ψ∗2(x)ψ1(x)] .

∆Q2 = 1− (|c1|2 − |c2|2)2.

D2 + V2 ≤ 2− [∆P2 + ∆Q2].

For any c1, c2,

|Ψ(x)|2 =|ψ1(x)|2 + |ψ2(x)|2

2+|c1|2 − |c2|2

2[ψ∗

1(x)ψ2(x) + ψ∗2(x)ψ1(x)] .

∆Q2 = 1− (|c1|2 − |c2|2)2.

D2 + V2 ≤ 2− [∆P2 + ∆Q2].

For any c1, c2,

|Ψ(x)|2 =|ψ1(x)|2 + |ψ2(x)|2

2+|c1|2 − |c2|2

2[ψ∗

1(x)ψ2(x) + ψ∗2(x)ψ1(x)] .

∆Q2 = 1− (|c1|2 − |c2|2)2.

D2 + V2 ≤ 2− [∆P2 + ∆Q2].

For any c1, c2,

|Ψ(x)|2 =|ψ1(x)|2 + |ψ2(x)|2

2+|c1|2 − |c2|2

2[ψ∗

1(x)ψ2(x) + ψ∗2(x)ψ1(x)] .

∆Q2 = 1− (|c1|2 − |c2|2)2.

D2 + V2 ≤ 2− [∆P2 + ∆Q2].

Uncertainty and DualityThe Sum Uncertainty Relation

Sum uncertainty relation for angular momenta 1

∆L2x + ∆L2

y + ∆L2z ≥ `

Implication for Pauli spin matrices

∆σ2x + ∆σ2

y + ∆σ2z ≥ 2, ∆σ2

x + ∆σ2y ≥ 1.

In our case, P̂ = σ̂z, Q̂ = σ̂x. So, ∆P2 + ∆Q2 ≥ 1 .Using this on

D2 + V2 ≤ 2− [∆P2 + ∆Q2].

we getD2 + V2 ≤ 1.

The duality relation also emerges from the sum uncertainty relation.

1Hoffmann, Takeuchi, Phys. Rev. A 68, 032103 (2003).Radhika Vathsan (BITS Goa) Einstein’s Recoiling-Slit Experiment QIPA 2013, 4th Dec, Allahabad 41 / 43

∆L2x + ∆L2

y + ∆L2z ≥ `

∆σ2x + ∆σ2

y + ∆σ2z ≥ 2, ∆σ2

x + ∆σ2y ≥ 1.

D2 + V2 ≤ 2− [∆P2 + ∆Q2].

we getD2 + V2 ≤ 1.

∆L2x + ∆L2

y + ∆L2z ≥ `

∆σ2x + ∆σ2

y + ∆σ2z ≥ 2, ∆σ2

x + ∆σ2y ≥ 1.

In our case, P̂ = σ̂z, Q̂ = σ̂x. So, ∆P2 + ∆Q2 ≥ 1 .

Using this onD2 + V2 ≤ 2− [∆P2 + ∆Q2].

we getD2 + V2 ≤ 1.

∆L2x + ∆L2

y + ∆L2z ≥ `

∆σ2x + ∆σ2

y + ∆σ2z ≥ 2, ∆σ2

x + ∆σ2y ≥ 1.

D2 + V2 ≤ 2− [∆P2 + ∆Q2].

we getD2 + V2 ≤ 1.

∆L2x + ∆L2

y + ∆L2z ≥ `

∆σ2x + ∆σ2

y + ∆σ2z ≥ 2, ∆σ2

x + ∆σ2y ≥ 1.

D2 + V2 ≤ 2− [∆P2 + ∆Q2].

we getD2 + V2 ≤ 1.

The duality relation also emerges from the sum uncertainty relation.1Hoffmann, Takeuchi, Phys. Rev. A 68, 032103 (2003).

Conclusions

For any two orthogonal states of the recoiling slit (say) |ξ1〉 and|ξ2〉, one can always find operators P̂ and Q̂ whose uncertaintiesenforce complementarity.

P̂ = |ξ1〉〈ξ1| − |ξ2〉〈ξ2| Q̂ = |ξ1〉〈ξ2|+ |ξ2〉〈ξ1|

Englert-Greenberger-Yasin duality relation emerges fromcorrelations and also from the sum uncertainty relation.Complementarity enforced by correlations and the uncertaintyrelations are two sides of a coin (provided the observables arecorrectly identified).Momentum back-action of the recoiling slit on the particle plays norole in complementarity.

Conclusions

For any two orthogonal states of the recoiling slit (say) |ξ1〉 and|ξ2〉, one can always find operators P̂ and Q̂ whose uncertaintiesenforce complementarity.P̂ = |ξ1〉〈ξ1| − |ξ2〉〈ξ2| Q̂ = |ξ1〉〈ξ2|+ |ξ2〉〈ξ1|

Englert-Greenberger-Yasin duality relation emerges fromcorrelations and also from the sum uncertainty relation.Complementarity enforced by correlations and the uncertaintyrelations are two sides of a coin (provided the observables arecorrectly identified).Momentum back-action of the recoiling slit on the particle plays norole in complementarity.

Conclusions

For any two orthogonal states of the recoiling slit (say) |ξ1〉 and|ξ2〉, one can always find operators P̂ and Q̂ whose uncertaintiesenforce complementarity.P̂ = |ξ1〉〈ξ1| − |ξ2〉〈ξ2| Q̂ = |ξ1〉〈ξ2|+ |ξ2〉〈ξ1|Englert-Greenberger-Yasin duality relation emerges fromcorrelations and also from the sum uncertainty relation.

Complementarity enforced by correlations and the uncertaintyrelations are two sides of a coin (provided the observables arecorrectly identified).Momentum back-action of the recoiling slit on the particle plays norole in complementarity.

Conclusions

For any two orthogonal states of the recoiling slit (say) |ξ1〉 and|ξ2〉, one can always find operators P̂ and Q̂ whose uncertaintiesenforce complementarity.P̂ = |ξ1〉〈ξ1| − |ξ2〉〈ξ2| Q̂ = |ξ1〉〈ξ2|+ |ξ2〉〈ξ1|Englert-Greenberger-Yasin duality relation emerges fromcorrelations and also from the sum uncertainty relation.Complementarity enforced by correlations and the uncertaintyrelations are two sides of a coin (provided the observables arecorrectly identified).

Momentum back-action of the recoiling slit on the particle plays norole in complementarity.

Conclusions

For any two orthogonal states of the recoiling slit (say) |ξ1〉 and|ξ2〉, one can always find operators P̂ and Q̂ whose uncertaintiesenforce complementarity.P̂ = |ξ1〉〈ξ1| − |ξ2〉〈ξ2| Q̂ = |ξ1〉〈ξ2|+ |ξ2〉〈ξ1|Englert-Greenberger-Yasin duality relation emerges fromcorrelations and also from the sum uncertainty relation.Complementarity enforced by correlations and the uncertaintyrelations are two sides of a coin (provided the observables arecorrectly identified).Momentum back-action of the recoiling slit on the particle plays norole in complementarity.

Tabish Qureshi, Radhika VathsanEinstein’s Recoiling Slit Experiment, Complementarity andUncertaintyArxiv: 1210.4248 [quant-ph]Quanta Vol. 2 (April 2013)

THANK YOU!

Tabish Qureshi, Radhika VathsanEinstein’s Recoiling Slit Experiment, Complementarity andUncertaintyArxiv: 1210.4248 [quant-ph]Quanta Vol. 2 (April 2013)

THANK YOU!

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