Report

OUTLINE1. Introduction:

a QM problem; Probability of finding 2 particles(both bosonicor fermionic) in the box after ; similar to EPR paradox.

2. Where Do The Particles End up?Y ←→ Y~ ; ground state(g) and excited state(e); Yg ←→ Y+, Ye ←→ Y-

A. Bosons: can be gg , ee and geB. Fermions: can be ge only

3. Discussion: the correct answer from 4 choices4. Distinguishing Y+ and Y-: a way to find relative phase between

w.f.s.5. Conclusion: A better way to understand is to use the second

quantization formulation.

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* S.J. van Enk, Dept. of Physics, Oregon Center for Optics and Institute for Theoretical SciencesUniversity of Oregon, Eugene, Oregon.AJP,77,140(2009)

What if I and II are far apart? Measuring A or B on I changes II instantly. EPR paradox: QM is either incomplete or action at a distance(no causality)is

possible

two interacting systems, I and II the interaction is turned off, systems I and II remain correlated(coherent or entangled). measurement of observable A on I is done

If measurement of observable B on I is done:

*A. EINSTEIN, B. PODOLSKI AND N. ROSEN, PRV,47,777(1935)

+ Dozens of experiments after 1970 support QM predictions.

+ In the following:

Problem we considered is a simpler version of EPR paradoxNo non-local effect or entanglement involvedThis is a local effect problemCan be solved with QM using 1st quantization and 2nd quantization

methods.

2 particles(1&2) in the groud

state of two boxes(S&W); each box can be splitted into 2 halves(L&R); SR and WR exchanegd and boxes merged.

Probability of 2 particles in WL and SR, Pws=?

Note: 0<x1,x2<2L before splitting0<x1,x2<L and L<x1,x2<2L after splitting

Particle 1

Particle 2

+ 25% 2 WS

+ 25% 2 SW

+ 50% 1 WS + 1 SW

Pws =1/4

The answer is ……

1. P = ¼ because only particle numbers are concerned, no other Q.#s

2. Bosons together, P = ½

Fermions expells , P = 0

3. Exchange particles, same as color repainted, no actual change, P= 0 for B and F.

4. P=0 for Boson as in 3;

P=1/2 for Fermion because minus sign of w.f.

W.F. describing 2 particles in 2 boxes where S is symmetrization operator or anti-symmetrization operator.

After splitting the boxes, w.f. is :

Multiply terms,

So, you can see P=1/4 is the correct answer!

However, excited state and ground state are degenerate after splitting the box.

YgYe

Y+ Y-

Take the 2nd term of Eq. (3) of the splitted boxes

For Bosons, 2nd and 3rd terms cancelled out. Bosons like to be together(gg andd ee)

For Fermions, 1st and 4th terms cancelled out.Fermions expelled each other(only eg).

After merging, the 2nd term becomes

Prepare a 2nd box, say Y+.

Given a box which could be either one on the left.

Can you tell actually which one it is?

Then merge two halves on each side . We will have ……….

For Y+ and Y- states respectively,

Difference only appears in the 1st term which says , after merging two halves, that the 2 particles are either both in the ground state as shown in Eq.(8) or 1 in the ground state and another in the excited state as shown in Eq.(9).We can tell which one,Y+ or Y-, is given to us if we measure the states of the 2 particles.

+ Summary and some key points

+ References