1 Bound States in 3 Dimensions

2 From 1-d to 3-d

3 3-d Hamiltonian

4 Bass-Ackwards z y x   r

5 For a particle of mass, m,

6 Continuing Only on r Only on  and 

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( ,  )=P(  )  (  ) Multiply by

11 And Voila! Let the separation constant = -m 2

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

13 Back to P(  )

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(  )=P(  )  (  ) 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 The Rigid Rotor m2 m1 z y x   r 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 r1 r2

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

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

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 V i is the interaction potential appropriate to the transition

28 For most atomic and molecular systems, V i looks like If M fi vanishes, the transition is said to be “forbidden” by dipole selection rules If M fi <>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  l=+/- 1  m=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(  )=P(  )  (  ) 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}U l =U l+1 Proof:

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

39 Conclusions R is a “sort-of” raising operator If U l is a sol’n of the SE then RU l is a sol’n with l’=l+1 When l=0 then U 0 ” +U 0 =0 U 0 = A sin x + B cos x For the spherical region which contains the origin, it is necessary B=0 since U l /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 x l,n denote the n th root of j l (x)

42 Energy Levels of Hollow Sphere s 1p 1d 1f 1g 1h 2s 2p 2d 2f 2g 3s 3p Units Of H-bar^2/(2  R 2 )

43 Conclusions Does not depend on m but l Degeneracy = (2l +1) Recall S = sharpl=0 P = principle l=1 D = diffusel=2 F = fundamentall=3

44 Normalizin’ Selection rules are  L=+/-1  m=0, +/-1 There is no selection rules depending on n

45 What if V=V 1 when r R Inside Outside where h l 1 = Spherical Hankel Functions

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

47 The Hydrogen Atom  defines the electrical units. For SI, the units  = {4  0 } -1 For electrostatic units,  = 1 So the SE becomes

48 Change of Variables Bound states mean that E<0 and thus, is a real number We now introduce  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  goes to ∞ A must = 0 in order to normalize U.

51 As  goes to 0 And p=-l and p=l+1 but  -l cannot be normalized so

52 The Hydrogen Atom Radial Wave function And if we substitute this into the Whittaker Equation, we have Which is called Laguerre’s Equation And it’s solutions are called Laguerre’s polynomials

53 Our friend, the confluent hypergeometric equation If we set a=-n+l+1 and b=2(l+1), then the equation takes the form of the confluent hypergeometric equation: Pochmanner symbol

54 In our case, Series expands infinitely… which is bad in the whole “good / bad” thing And thus, we force a=0 or –n+l+1=0 n=l+1, l+2, l+3 … n is an integer and n>l

55 The point of the matter is to solve for E, remember?

56 An aside One problem with Laguerre polynomials is that there is not an unique definition For example from Abramowitz and Stegun denote L (   )=(-1) n n! 1 F 1 (-n,  +1;  ) Where Schiff, Saxon, and Pauling use L (   )=(-1) n (n+  )! 1 F 1 (-n-1,  ;  ) Yuck! But all agree on the formula of Rodriguez for Laguerre’s And if  =0 then associated Laguerre polys become Laguerre polys

57 Normalizin’ Use And going back to Then

58 Finally! where

59 Some Hydrogen Wavefunctions

60 If n=1 and l=m=0, then gnd state And 1/ is called a 0 a 0 is called the Bohr radius

61 Bound States in 3 Dimensions Radial Solutions Part 2

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

63 Angular Solution Y(  )=P(  )  (  ) Called Spherical Harmonics Explicit Form Phase factor does not affect normalization Which is chosen to agree with Condon & Shortley, Blatt & Weisskopf, Particle Data Group

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

65 The Isotropic Harmonic Oscillator Let then

66 Defining k and lambda

67 As r goes to infinity

68 As r goes to 0 Just like the Hydrogen atom

69 Our main bud, the confluent hypergeometric equation Let We need a new definition of rho so that the confluent hypergeometric equation will pop out 1 st Let

70 And thus becomes

71 Since the confluent hypergeometric equation looks like Then

72 The W functions must converge just like the L functions We find that to make the series converge n r =0, 1, 2, 3, … However, the fact that –n r is the coefficient of W/rho has the following implications

73 But we know s=E/h-bar/omega This is so that n looks like the quantization of the Hydrogen atom l=s, p, d, f Note: In the H-atom, n>l was boundary condition Here 1p, 1d, 1f allowed

74 Final Wave Function The problem that we just solved is of considerable importance in a variety of subjects. It has an analytic solution with equal spacing of states. Most famously, it is the beginning point for a shell model of the nucleus: assuming that each nucleon is bound to other nucleons by a quadratic potential. Adding a perturbation of the spin coupled with the orbital angular momentum will further refine the theory. Meson theory with a quark-anti-quark pair