Splitting the wavefunctions of two particles in two boxes *

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)

2. What is EPR paradox?* • 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: • 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 *A. Einstein, B. Podolski and N. Rosen, PRV,47,777(1935)

3. EPR paradox type experiment • Dozens of experiments after 1970 support QM predictions. • In the following: • Problem we considered is a simpler version of EPR paradox • No non-local effect or entanglement involved • This is a local effect problem • Can be solved with QM using 1st quantization and 2nd quantization methods.

4. 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=? Particle 1 Particle 2 Note: 0<x1,x2<2L before splitting 0<x1,x2<L and L<x1,x2<2L after splitting

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

6. 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.

7. 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!

8. 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

9. After merging, the 2nd term becomes • For Bosons, 2nd and 3rd terms cancelled out. • Bosons like to be together(gganddee) • For Fermions, 1st and 4th terms cancelled out. • Fermions expelled each other(only eg).

10. Distinguishing Y+ and Y-, can we? Yes, we can. • Given a box which could be either one on the left. • Can you tell actually which one it is? Prepare a 2nd box, say Y+. Then merge two halves on each side . We will have ……….

11. 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 wemeasure the • states of the 2 particles.

12. conclusion • Summary and some key points • References