1. Splitting the wavefunctions of two particles in two boxes* OUTLINE 1. Introduction: a QM problem; Probability of finding 2 particles(both bosonic or fermionic) in the box after splitting and merging; 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 ge B. Fermions: can be ge only 3. Discussion: the correct answer from 4 choices 4. 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. . : * S.J. van Enk, Dept. of Physics, Oregon Center for Optics and Institute for Theoretical Sciences University of Oregon, Eugene, Oregon. AJP,77,140(2009)

3. the interaction is turned off,

4. systems I and II remain correlated(coherent or entangled).

5. measurement of observable A on I is done

6. If measurement of observable B on I is done:

7. What if I and II are far apart? Measuring A or B on I changes II instantly.

8. EPR paradox: QM is either incomplete or action at a distance(no causality)is possible*A. Einstein, B. Podolski and N. Rosen, PRV,47,777(1935)

10. No non-local effect or entanglement involved

11. This is a local effect problem

13. Probability of 2 particles in WL and SR, Pws=? Particle 2 Note: 0<x1,x2<2L before splitting 0<x1,x2<L and L<x1,x2<2L after splitting

14. For Classical distinguishable particles 25% 2 WS 25% 2 SW 50% 1 WS + 1 SW Pws =1/4

15. Identical quantum particlesbosonic and fermionic? The answer is …… P = ¼ because only particle numbers are concerned, no other Q.#s 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.

16. 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!

17. Yg Ye Y+ Y- However, excited state and ground state are degenerate after splitting the box. Take the 2nd term of Eq. (3) of the splitted boxes

20. Can you tell actually which one it is?Prepare a 2nd box, say Y+. Then merge two halves on each side . We will have ……….

22. We can tell which one,Y+ or Y-, is given to us if wemeasure the states of the 2 particles.

23. conclusion Summary and some key points References