1. Angular Momentum STM-Image Cs atoms on GaAs surface (NIST)
W. Udo Schröder, 2004

2. Motion of N-Body System
Separation of overall (center-of-mass “com”) and intrinsic motion: z x y
Associated  momenta omitted 
com = particle at with total mass M=S mi, total momentum plane wave, wave packet,…
Intrinsic motion: Superposition of normal modes
Vibrations
and associated momenta
Angular Momentum
Rotations
W. Udo Schröder, 2004

3. Rotational Motion: Angular Momentumz f m2 q
Separate rotational from other dof: rigid rotor
No external potential for rot motion: Uqf =0
Angles q, f describe spatial orientation of rotator (dumbbell)
Recap: circular motion around z-axis  found op Lz and: Lz quantized m
Angular Momentum
Generalize from z-component to total
Different for different axes!
must also be quantized
W. Udo Schröder, 2004

4. Rotational Motion: Polar Coordinatesz f m2 q
Polar coordinates appropriate for rotational motion about an axis. Transformation:
Rigid rotation: r = const. z x y q f
Angular Momentum
Similar for y, z
W. Udo Schröder, 2004

5. Component Representation of Angular Momentumz f m2 q
Are all L components simultaneously measurable? z x y q f
No, any two are incommensurable
Angular Momentum
Obviously
L2 and 1 component (Lz) commensurable
W. Udo Schröder, 2004

6. Instant Problem: Calculate Commutator [L2,Lz]
Given
and
Angular Momentum
W. Udo Schröder, 2004

7. Angular Momentum Eigen Value Equation
 only L2 and Lz sharp, look for simultaneous eigen functions Y(q,f) L r R m1 z f m2 q
Separate system variables q, f
solves Schrödinger Equ.
Angular Momentum
divide by eimf
W. Udo Schröder, 2004

8. Angular Momentum Eigen Value Equationz f m2 q
Variable transform
Legendre’s Equation
Simplest case: m = 0 (L perp to z-axis)
Angular Momentum
Trial: Power series expansion
W. Udo Schröder, 2004

9. Solving DEqu. with Power Seriesm1 z f m2 q
Case: m = 0 (L perp to z-axis), trial:
Angular Momentum
W. Udo Schröder, 2004

10. Asymptotic Boundary ConditionL r R m1 z f m2 q
Case: m = 0 (L perp to z-axis), trial:
Parity conservation: PL has either only even powers of x or only odd
Angular Momentum
If highest order is n=0 (p =+1) a0 ≠0.
If highest order is n=1 (p = -1) a1 ≠0.
W. Udo Schröder, 2004

11. Quantization of Angular Momentum
If highest order of x in power series for PL is n>1:
Recursion relations for an: Even n for even p, odd n for odd p
Recursion relation must stop at a certain n=L (L = 0,1,2,..) for PL
Require
Similar to E0 and E1 already found
Angular Momentum
Discrete rotational energy eigen values
Quantization of orbital angular momentum
W. Udo Schröder, 2004

12. Legendre Polynomials Quantization of angular momentum x:=cosq
Polynomial of finite order L
x:=cosq
orthogonal basis set
Angular Momentum
Even L  even functions PL Odd L  odd functions PL
W. Udo Schröder, 2004

13. Angular Momentum Eigen Functionsz f m2 q
General case: m ≠ 0
Legendre’s Equation
 Solutions depend on m2 or |m|
DEq. solved (check by inserting) by “associated Legendre Polynomials”
Angular Momentum
Total (normalized) wave function of rigid rotor:
“Spherical Harmonics”
W. Udo Schröder, 2004

14. Angular Momentum Eigen Value Spectrumz f m2 q
Quantization
Energy eigen values
Energy does not depend on orientation (mL) relative to z-axis (“quantization axis”)
Parity
DL = 2L+1 (-L ≤ mL ≤ +L) Degeneracy DL
L=5-
(11)
Spectroscopy: Electric dipole (E1) transitions between rot levels “selection rules”
L=4+
(9)
Angular Momentum
Electric dipole (E1) radiation emitted only, if molecule has a permanent electric dipole moment
L=3-
(7)
L=2+
(5)
L=1-
(3)
L=0+
(1)
W. Udo Schröder, 2004

15. Rotational Energy Level Scheme
Convention
(11)
Diatomic molecules: microwave
(9)
Transitions J+1J, energies hn or wave numbers..
(7)
(5)
(3)
Rotational Absorption Spectrum HCl
Angular Momentum
Ch. Gerthsen et al., 1963
W. Udo Schröder, 2004

16. Instant Problem: Deduce HCl Bond Lengthz f m2 q
Rotational Absorption Spectrum HCl
Angular Momentum
W. Udo Schröder, 2004

17. Legendre Polynomials: Polar Plotsq
Angular Momentum
Different L: out of phase except 00. Polar plot sign sensitive  |PL|
W. Udo Schröder, 2004

18. Orthogonality of Legendre Polynomials
Larger L for PL  smaller wave length l  larger momenta p  larger angular momenta L.
Angular Momentum
Different L: out of phase. Overlap=antisymmetric
q integral in polar coordinate system
W. Udo Schröder, 2004

19. Angular-momentum eigen functions form basis for all angular functions
Spherical Harmonics
Stationary wave functions of rigid rotor for fixed L, fixed m
Orthonormality
Angular-momentum eigen functions
form basis for all angular functions
Angular Momentum
Arbitrary function
W. Udo Schröder, 2004

20. Angular Wave Packets
Stationary WF (sharp L, m) : extended in q,f. Wave packets (LC over L,m): localized in q,f.
Angular Momentum
W. Udo Schröder, 2004

21. Legendre Polynomials P1|m|
Plot PLm=0 z x y q f
Angular Momentum
W. Udo Schröder, 2004

22. Spherical Harmonics P1|m|cos (mf)
Plot Re YLm z x y q f
Angular Momentum
W. Udo Schröder, 2004

23. Legendre Polynomials P2|m|
Angular Momentum
W. Udo Schröder, 2004

24. Spherical Harmonics P2|m|cos(mf)
Angular Momentum
W. Udo Schröder, 2004

25. Angular Momentum Orientation
Schrödinger Equ. solved by L r R m1 z f m2 a
Quantization of angular momentum z a
L=4 +1 +2 +3 +4 m -1 -2 -3 -4
Since
Angular Momentum
Angular momentum L can never be perfectly aligned with any direction
Normally, rotor has arbitrary spatial orientation Meaning of quantization axis? How to align ?
W. Udo Schröder, 2004

26. Magnetic Interactions: Zeeman Effect
Energy in homogeneous B-field || z axis
Bohr magneton mB:
Gyro-magnetic (Landé) g-factor: (e-) gL = -1 z
L and m cannot completely align with B field
Perpendicular component produces torque on L
Angular Momentum y
Angular momentum precesses on a cone around B field (because always ) x
W. Udo Schröder, 2004

27. Magnetic Interactions: Zeeman Effect
Energy in homogeneous B-field || z axis
Gyro-magnetic (Landé) g-factor: (e-) gL = -1
Magnetic splitting of energy levels in Bz
L=0
L=1 n +1 -1 mL
B≠0  breaks spatial symmetry, z||B is preferred (natural) for simplest description of system: degeneracy of energy levels is lifted  splitting of levels and spectroscopic lines.
For e-: mL > 0 higher energy
Angular Momentum
W. Udo Schröder, 2004

28. Electronic m: The Einstein-deHaas Experiment
Conjecture: magnetism due to electronic ring currents
e me A
Current loop
Measure m/L NO!
Torsion Wire
Laser
Mirror
Lini=0
Lfin=0
Angular Momentum e-
Fe rod
Scale
Angular Momentum
Battery Capacitor
Fe Rod
W. Udo Schröder, 2004

29. Conclusions on Electronic Spin and m
Modern Einstein-deHaas experiment setup
PAS: U NM
Conclusions:
m and mechanical angular momentum related
Measured ang. mom. of Fe rod not due to orbital L of e-
e- have intrinsic spin S = mechanical ang. mom. about e- “axis.”
e- have spin magnetic moment ms || S
Angular Momentum
Spin = intrinsic mechanical angular momentum (rotation), but Re = 0 (?)
W. Udo Schröder, 2004

30. Measuring Electronic Spin
Stern and Gerlach (1922): Splitting of a Ag beam in inhomogeneous B. Inh. B field exerts force on e- magnetic dipole
Ag (e-) s1/2
Magnet B
Oven
Angular Momentum
Ag atom: outer s1/2 electron. Experiment: 2 orientations of e- spin/dipole  2S+1=  S = 1/2.
Prof. Gerlach’s postcard with the announcement to Bohr in Copenhagen.
W. Udo Schröder, 2004

31. Spin Quantization z z Quantization of spin angular momentum
ms= +1/2
Quantization of spin angular momentum
Precise structure/coordinates of spin ?? assume similar to z
ms= -1/2
Angular Momentum
Spin=1/2 particles : Spin=1 particles : (Fermions) e-, e+, n, p, n,.. (Bosons)photons, pions,….
W. Udo Schröder, 2004

32. Enough spinning for today--
But: We'll be back!
Angular Momentum
W. Udo Schröder, 2004

33. Force in inhomogeneous B-field || z axis
Angular Momentum
W. Udo Schröder, 2004

34. Magnetic Dipole Moments
Moving charge e  current density j  vector potential A, influences particles at via magnetic field =0
Angular Momentum
current loop:
mLoop = j x A= current x Area
W. Udo Schröder, 2004

35. End of
Section
Angular Momentum
W. Udo Schröder, 2004

36. Particle Reflection and Transmission at Potential Step
Total energy E=K(x)+U(x)=const E >0  unbound, free particle Piece-wise constant potential step: x
U,E
U(x)>0 E K 1 2
Continuous wf matching at U discontinuities
Transmitted
matching point x
Reflected
Incoming 1 2
General solution: construct from p-EFs
Angular Momentum
Effects of potential step (not surprising): > Partial reflection of incoming wave B1 ≠ 0 > Slows transmitted wave > No reflection right of step (X>0): B2 =0
W. Udo Schröder, 2004

37. Drawing Elements z a L r R z f q L=4 a L r R z f L r R z f q L r R z f+1 +2 +3 +4 m -1 -2 -3 -4 a L r R m1 z f m2 L r R m1 z f m2 q L r R m1 z f m2 z x y L r R f m z
Angular Momentum
W. Udo Schröder, 2004