1. Physics 3 for Electrical Engineering Ben Gurion University of the Negev www.bgu.ac.il/atomchipwww.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenterwww.bgu.ac.il/nanocenter Lecturers: Daniel Rohrlich, Ron Folman Teaching Assistants: Daniel Ariad, Barukh Dolgin Week 6. Quantum mechanics – probability current 1D scattering and tunneling simple 1D potentials parity general 1D potential 2D and 3D square wells and degeneracy Sources: Merzbacher (2 nd edition) Chap. 6; Merzbacher (3 rd edition) Chap. 6; Tipler and Llewellyn, Chap. 6 Sects. 1-3,6; פרקים בפיסיקה מודרנית, יחידה 7

2. Probability current When we solved the finite square well potential for bound states (E < V 0 ), we threw away the outside term e k′|x| because it is not normalizable. Recall x V(x)V(x) V0V0 0 L/2 −L/2 outside inside

3. Probability current But for free states (E > V 0 ) the solutions outside are linear combinations of e ik′x and e −ik′x ; and they are not normalizable either. Should we throw them away? x V(x)V(x) V0V0 0 L/2 −L/2 outside inside

4. Probability current We know why the wave e ik′x is not normalizable: its momentum is with zero uncertainty (Δp = 0), so the uncertainty Δx in its position must be infinite – the particle can be anywhere along the x-axis with equal probability. x V(x)V(x) V0V0 0 L/2 −L/2 outside inside

5. Probability current We know why the wave e ik′x is not normalizable: its momentum is with zero uncertainty (Δp = 0), so the uncertainty Δx in its position must be infinite – the particle can be anywhere along the x-axis with equal probability. Should we throw away the wave e ik′x ? x V(x)V(x) V0V0 0 L/2 −L/2 outside inside

6. Probability current If we ask what fraction of an incoming beam ψ(x) = A e ikx is reflected and what fraction is transmitted, then the normalization doesn ’ t matter.

7. Probability current If we ask what fraction of an incoming beam ψ(x) = A e ikx is reflected and what fraction is transmitted, then the normalization doesn ’ t matter. The flux density is ρv where ρ = |ψ(x)| 2 = |A| 2 and v =.

8. Probability current Consider the time derivative of the probability density ρ(x,t): where we have applied Schrödinger’s equation and its complex conjugate. (What happened to the potential term?)

9. Probability current We found which implies the 1D “continuity equation” if we define

10. Probability current Let ’ s check: If Ψ(x,t) = Ae ikx-iωt, then which is just what we found from our earlier calculation.

11. So we could look for an outside solution of the form in which the incoming flux /m splits into a reflected flux /m and a transmitted flux /m. x V(x)V(x) V0V0 0 L/2 −L/2 outside inside

12. 1D scattering and tunneling But let’s consider a simpler scattering problem:

13. x V(x)V(x) V0V0 0 1D scattering and tunneling But let’s consider a simpler scattering problem:

14. We take E > V 0 and in which the incoming flux /m splits into a reflected flux /m and a transmitted flux /m. The reflection probability is R = and the transmission probability is T = x V(x)V(x) V0V0 0

15. Continuity of ψ at x = 0: A + B = C. Continuity of dψ/dx at x = 0: ik(A – B) = ik′C. Now 2A = (1+k′/k) C and 2B = (1–k′/k) C; hence C/A = 2k/(k +k′) and B/A = (k–k′)/(k +k′). x V(x)V(x) V0V0 0

16. Continuity of ψ at x = 0: A + B = C. Continuity of dψ/dx at x = 0: ik(A – B) = ik′C. Now 2A = (1+k′/k) C and 2B = (1–k′/k) C; hence C/A = 2k/(k +k′) and B/A = (k–k′)/(k +k′). Does R + T = 1? x V(x)V(x) V0V0 0

17. Continuity of ψ at x = 0: A + B = C. Continuity of dψ/dx at x = 0: ik(A – B) = ik′C. Now 2A = (1+k′/k) C and 2B = (1–k′/k) C; hence C/A = 2k/(k +k′) and B/A = (k–k′)/(k +k′). Does R + T = 1? x V(x)V(x) V0V0 0

18. Isn’t something very odd – “quantumly” odd – going on here? x V(x)V(x) V0V0 0

19. A classical particle with E > V 0 would never reflect back from the potential step…but here there is reflection unless k = k′, i.e. unless V 0 = 0. x V(x)V(x) V0V0 0

20. Isn’t something very odd – “quantumly” odd – going on here? A classical particle with E > V 0 would never reflect back from the potential step…but here there is reflection unless k = k′, i.e. unless V 0 = 0. In fact, even if we replace V 0 by –V 0 so that k >> k′, there is still reflection at x = 0! x V(x)V(x) −V 0 0

21. The case E < V 0 : Now R= 1, T = 0: no transmission. How deep do particles penetrate the potential step? x V(x)V(x) V0V0 0

22. 0.5 1.0 Summary of scattering from the potential step:

23. Snapshots of the probability density, for an incident wave packet:

24. Another scattering problem: the “square” potential barrier: x V(x)V(x) V0V0 0 L

25. As before, x V(x)V(x) V0V0 0 L

26. Boundary conditions (ψ and dψ/dx continuous) at x = 0: Hence x V(x)V(x) V0V0 0 L

27. Boundary conditions (ψ and dψ/dx continuous) at x = L: Hence x V(x)V(x) V0V0 0 L

28. If we combine these equations we can show Hence x V(x)V(x) V0V0 0 L

29. For k′L >> 1, both cosh k′L and sinh k′L approach and the expression for T simplifies: x V(x)V(x) V0V0 0 L

30. Simulation of a Gaussian wave packet with kinetic energy 500 eV, incident on a square potential barrier with height 600 eV and thickness 25 pm. About 17% of the wave packet tunnels through the barrier. To see the probability density: http://www.youtube.com/watch?v=4-PO-RHQsFA&NR=1 To see the real (black) and imaginary (red) components of the wave function: http://www.youtube.com/watch?v=_3wFXHwRP4s

31. Three applications of quantum tunneling: 1. Light crosses a “total internal reflection barrier”.crosses 2. The NH 3 molecule: Distance of N atom from plane of 3 H atoms, at various energy levels E i

32. Three applications of quantum tunneling: 1. Light crosses a “total internal reflection barrier”.crosses 2. The NH 3 molecule: Distance of N atom from plane of 3 H atoms, at various energy levels E i

33. 3. α–decay of a U 238 nucleus:

34. Simple 1D potentials – summary

36. Parity When the potential term V(x) of the 1D Schrödinger equation is symmetric, i.e. V(x) = V(−x), then the Schrödinger equation itself is symmetric and the solutions must have the same physical symmetry. But what is physical, ψ(x) or |ψ(x)| 2 ?

37. Parity When the potential term V(x) of the 1D Schrödinger equation is symmetric, i.e. V(x) = V(−x), then the Schrödinger equation itself is symmetric and the solutions must have the same physical symmetry. But what is physical, ψ(x) or |ψ(x)| 2 ? Only |ψ(x)| 2 (and the probability current) are physical. Hence ψ(x) can be a solution whether ψ(−x) = ψ(x) or ψ(−x) = −ψ(x). But we give these cases different names: in the first case ψ(x) has even parity; in the second ψ(x) has odd parity.

38. Parity When the potential term V(x) of the 1D Schrödinger equation is symmetric, i.e. V(x) = V(−x), then the Schrödinger equation itself is symmetric and the solutions must have the same physical symmetry. But what is physical, ψ(x) or |ψ(x)| 2 ? Only |ψ(x)| 2 (and the probability current) are physical. Hence ψ(x) can be a solution whether ψ(−x) = ψ(x) or ψ(−x) = −ψ(x). But we give these cases different names: in the first case ψ(x) has even parity; in the second ψ(x) has odd parity. If two solutions are degenerate (have the same energy) then a linear combination of them need not have definite parity. But if the energies are nondegenerate (one solution for each energy) then the solutions will have definite parity (even or odd).

39. General 1D potential Now let’s consider a 1D potential without any special symmetry: What can we say about the solutions of the time-independent Schrödinger equation? V(x)V(x) x

40. General 1D potential If ψ(x) > 0 and E > V(x), then ψ(x) will turn down until it becomes negative. If ψ(x) V(x), then ψ will turn up until it becomes positive. The larger E − V(x), the more up-and- down curves in ψ(x) and the more nodes (zeros) in ψ(x). E V(x)V(x) x

41. General 1D potential Hence the lowest-energy solution will have no node. E0E0 ψ(x)ψ(x)V(x)V(x) x

42. General 1D potential The next-lowest-energy solution will have one node. E1E1 ψ(x)ψ(x)V(x)V(x) x

43. General 1D potential As the energy increases, so does the number of nodes. E5E5 ψ(x)ψ(x)V(x)V(x) x

44. 2D and 3D square wells and degeneracy The time-independent Schrödinger equation for a particle in two dimensions: Suppose V(x,y) is an infinite square well potential, vanishing for |x| < L x /2 and |y| < L y /2, and infinite otherwise. What are the five lowest energy levels? Are the energies nondegenerate? If not, what are their degeneracies? How do we normalize the solutions in 2D? What are the normalized solutions? Does it matter whether or not L x and L y are equal?

45. What are the corresponding questions and answers in 3D? The time-independent Schrödinger equation for a particle in three dimensions is where r = (x,y,z). How do we define the probability current J(r,t) in 3D so that the continuity condition, holds?