Bound States in 3 Dimensions

1. Bound States in 3 Dimensions

2. From 1-d to 3-d

3. 3-d Hamiltonian

4. z q r y f x Bass-Ackwards

5. For a particle of mass, m,

6. Continuing Only on w and f Only on r

7. Appropriate Choice of Separation Constant 1-d Sch. Eq. Called centrifugal potential

8. Radial Solution • What happens next depends on V(r). • We will leave this part of the Sch. Equation alone and concentrate on the angular part.

9. Angular Solution

10. Another fortuitous choice! Let Y(q,f)=P(q)F(f) Multiply by

11. And Voila! Let the separation constant = -m2

12. The Azimuthal Dependence • m is quantized • m carries the plus/minus sign so we write a general expression • f is azimuth angle and goes from 0 to 2p . • Demand that F(f+2p)=F(f) • Therefore

13. Back to P(q)

14. Some cleanin’ up

15. And now When m=0 then called Legendre Equation When m<>0 then called Associated Legendre Equation Solutions called Legendre Functions

16. Finally, Y(q,f)=P(q)F(f) Called Spherical Harmonics Explicit Form Phase factor does not affect normalization Which is chosen to agree with Condon & Shortley, Blatt & Weisskopf, Particle Data Group

17. Properties of Spherical Harmonics

18. Connection Between Spherical Harmonics and Angular Momentum

19. And thus

20. Recall

21. Back Substituting

22. Using these, we can find new relationships for the various components of angular momentum

23. The Importance of these Results • Particle in a central force field will execute a motion such that orbital angular momentum about the central force is conserved • The QM result is that in a central force field, the eigenvalues of H can ALWAYS be written as Eigenfunctions of the orbital angular momentum. • It is often said that the orbital angular momentum is a “good quantum number”

24. z q m1 r m2 y f x The Rigid Rotor r1 Consider a molecule with two masses, m1 and m2, attached by an inflexible rod of length, r0, as shown in the figure to the left. The distance from m1 to the origin is r1 and the distance from the origin to m1 is r2. r1+r2=r0 And m1*r1=m2*r2 r2

25. The rigid rotor Hamiltonian • Let the molecule rotate freely but there is no potential energy • Therefore the Hamiltonian is H=L2/2I

26. Solving for the Energy eigenvalues Thus, there is a 2L+1 degeneracy of eigenvalues of energy. |2 2> |2 1> |2 0> |2 -1> |2 -2> 3E1 |1 1> |1 0> |1 -1> E1 |0 0> 0

27. Dipole Selection Rules Assumption 15 – If an initial state, | i >, can make a transition to a final state, |f>, by means of the emission or absorption of a single photon, probability of the occurrence is proportional to Where Vi is the interaction potential appropriate to the transition

28. For most atomic and molecular systems, Vi looks like If Mfi vanishes, the transition is said to be “forbidden” by dipole selection rules If Mfi <>0, the transition is allowed by dipole selection rules

29. Spherical Coordinates Again

30. Skipping to the result • The radial dependence is not considered in the dipole problem; only the angular dependence • x,y, and z depend on l=1 and thus have m value of -1, 0, 1 • If should be no surprise that the allowed transitions between angular momentum states with a central potential must follow the rules that • Dl=+/- 1 • Dm=0, +/- 1 • This means that only 1 unit of h-bar may be carried away from or given to a state.

31. Bound States in 3 Dimensions Radial Solutions

32. For a particle of mass, m,

33. Appropriate Choice of Separation Constant 1-d Sch. Eq. Called centrifugal potential

34. Angular Solution Y(q,f)=P(q)F(f) Called Spherical Harmonics Explicit Form Phase factor does not affect normalization Which is chosen to agree with Condon & Shortley, Blatt & Weisskopf, Particle Data Group

35. Radial Solutions • The potential V(r) will determine the radial solution to the Schroedinger Equation described below:

36. The Hollow Sphere • If V=0, then free particle • However, imagine a hollow sphere with a hard wall at r=R that cannot be penetrated. • Let Is called a spherical Bessel function

37. Thm: {(l+1)/x - d/dx}Ul=Ul+1 Proof:

38. Proof cont’d Then can be written We can update this equation by the transformation of l->l+1 and RUl is a sol’n to the SE

39. Conclusions • R is a “sort-of” raising operator • If Ul is a sol’n of the SE then RUl is a sol’n with l’=l+1 • When l=0 then U0” +U0=0 • U0 = A sin x + B cos x • For the spherical region which contains the origin, it is necessary B=0 since Ul/x = (cos x)/x and cos x/ x is undefined at x=0

40. Raisin’ it and the spherical Bessel functions are then

41. The Solution This means that we need to find the “roots” of the Bessel function i.e. where the Bessel functions equal 0 Let xl,n denote the nth root of jl(x)

42. Energy Levels of Hollow Sphere 13 2g 3p 2f 11 1h 3s 2d 1g 9 2p 1f Units Of H-bar^2/(2mR2) 7 1d 2s 1p 5 1s 3

43. Conclusions • Does not depend on m but l • Degeneracy = (2l +1) • Recall • S = sharp l=0 • P = principle l=1 • D = diffuse l=2 • F = fundamental l=3

44. Normalizin’ Selection rules are DL=+/-1 Dm=0, +/-1 There is no selection rules depending on n

45. What if V=V1 when r<Rand V=V2 when r>R Inside Outside where hl1 = Spherical Hankel Functions

46. Spherical Hankel Functions We know this is a sol’n of the SE For l>1, hl1 can be generated by R Using the following boundary conditions We can generate energy levels similar to hollow sphere

47. The Hydrogen Atom k defines the electrical units. For SI, the units k= {4pe0}-1 For electrostatic units,k= 1 So the SE becomes

48. Change of Variables Bound states mean that E<0 and thus, l is a real number We now introduce r and n

49. Re-writing the SE This called Whittaker’s Differential Equation In order to solve it, we look at the endpoints over which r can vary: (0,∞ )

50. As r goes to ∞ A must = 0 in order to normalize U.