The Rigid Rotor Problem: A quantum par7cle confined to a sphere Spherical Harmonics. Reading: McIntyre 7.6

Transcription

1 The Rigid Rotor Problem: A quantum par7cle confined to a sphere Spherical Harmonics Reading: McIntyre 7.6

2 Legendre s equa7on (m = 0) The series must be finite! If the series is not finite, the polynomial blows up (check ra7o for large n) ( ) ( ) A = n max n max + 1 n(n + 1)! A a A =!! + 1 n+2 = (n + 2)(n + 1) a n An7cipated this! These special values of A give Legendre polynomials. P 0 (z) = 1 P 1 (z) = z P 2 (z) = 1 2 (3z 2! 1) P 3 (z) = 1 2 (5z3! 3z) P 4 (z) = 1 8 (35z 4! 30z 2 + 3) P 5 (z) = 1 8 (63z5! 70z z)

3 The energy for the rigid rotor problem Go back to the original statement of the rigid rotor problem to see how A is related to E. A =!! + 1 ( )" ( )! E! =!! + 1 el is an integer (energy is quan7zed) 2I 2

4 Legendre polynomials Rodrigues formula P! (z) = 1 2!!! d! ( )! dz! z 2! 1 Orthogonal (Herm op)! Of degree! 1 " P * (z)p k! (z)dz = # k!! + 1!1 2 Either even or odd

5 Legendre polynomials Orthogonal means we can express any func7on defined on that interval as a superposi7on of Legendre polynomials: HW (! + 1 ) 2 1 " P k* (z)p! (z)dz = # k!!1 This means we can project and find probabili7es just as with sines, cosines and exp(imφ)

6 Associated Legendre polynomials m " (1! z 2 ) d 2 dz! 2z d 2 dz +!(! +1)! m 2 % $ # (1! z 2 ) ' & P(z) = 0 P! m (z) = P!! m (z) = (1! z 2 ) m/2 d m dz m ( P (z) )! = 1 2!!! (1! z 2 ) m/2 d m+! dz m+! (z 2!1)! ( )

7 Associated Legendre polynomials m, integer! P 0 0 = 1 P 0 1 = cos! P 1 1 = sin! P 0 2 = 1 2 (3cos2! " 1) P 1 2 = 3sin! cos! P 2 2 = 3sin 2! P 0 3 = 1 2 (5cos3! " 3cos!) P 1 3 = 3 2 sin!(5cos2! " 1) P 2 3 = 15sin 2! cos! P 3 3 = 15sin 3!

8 Associated Legendre polynomials P! m (z) = 0 if m >! restricts m to ± el determines # of orbitals P!! m (z) = P! m (z) opposite projec7ons have same spa7al form yz plane is a node P m! (±1) = 0 for m! 0 parity (selec7on rules) P m! (!z) = (!1)!! m P m! (z) Orthogonal (for given m) 1 " P m (z)p m! q (z)dz =!1 2 (! + m)! (2! + 1) (!! m)! #!q

9 Finally.. The normalized polynomial that solves the theta equa7on is (typo in book Eq ) :! m! (") = (#1) m (2! +1) 2 (!# m )! (!+ m )! P m (cos")!

10 Summary The fact that the series must be finite polynomial of degree el is what (mathema7cally) leads to a maximum (and integer) value of el! The fact that the series is finite means that we can differen7ate it only a finite number of 7mes (el) and that number turns out to be m!! Thus m is limited to el el. Legendre func7ons are orthogonal on [ 1,1] Legendre and associated Legendre func7ons.

11 Spherical harmonics: Original problem was separated into angular and radial parts. We further separated angular parts into theta and phi. Y m (!!,") = # m (!!)$ ( m ") These func7ons are the angular part of the solu7on to the hydrogen atom problem, and indeed ANY central force problem!!

12 Do you know what this is?

13 The cosmic microwave background is analyzed with spherical harmonics!

14 Spherical Harmonics Y! m (!,") = (#1) (m+ m )/ 2 (2! + 1) 4$! m Y m! (!,") 0 0 Y 0 0 = 1 4# 1 0 Y 1 0 = 3 4# cos! ±1 Y 1 ±1 = " 3 8# sin!e±i" 2 0 Y 2 0 = 5 16# 3cos 2! $ 1 (!# m )! (!+ m )! P m (cos!)e im"! ( ) ±1 Y ±1 2 = " 15 8# sin! cos!e±i" ±2 Y 2 ±2 = 15 32# sin2!e ±i2" ( ) 3 0 Y 3 0 = 7 16# 5cos 3! $ 3cos! ( )e ±i" ±1 Y 3 ±1 = " 21 64# sin! 5cos2! $ 1 ±2 Y 3 ±2 = # sin2! cos!e ±i2" ±3 Y ± = 64# sin3 ±i 3"!e

15 Spherical Harmonics Y! m (!,") = (#1) (m+ m )/ 2 (2! + 1) 4$ (!# m )! (!+ m )! P m (cos!)e im"! What is the ac7on of these operators on the spherical harmonics? L 2 Y! m (!,") =? L z Y! m (!,") =? Do the two operators commute?

16 Spherical Harmonics!m What is the ac7on of these operators on the spherical harmonics? L 2!m =? L z!m =? Do the two operators commute?

17 Spherical Harmonics Y! m (!,") = (#1) (m+ m )/ 2 (2! + 1) 4$ Orthonormal on the sphere (!# m )! (!+ m )! P m (cos!)e im"! 2!! ( )* " 0 0 ( ) " Y! 1 m 1 #,$! 1 m 1! 2 m 2 =!!1! 2! m1 m 2 m Y 2 (!2 #,$)sin#d#d$ " $ # $ % = % %!! m1 m Now we ll spend some 7me visualizing probability distribu7ons of eigenstates and superposi7ons.? Do you remember where this comes from?

18 Spherical Harmonics

19 Spherical Harmonics

20 Spherical Harmonics

21 Spherical Harmonic Expansion! (",#) = % $ & ' ( cos3 2" + sin 2 # ) *! # m="! P E! =!m! 2

22 About the spherical harmonics ( )! Y! m ($ #!," + $ ) parity ( even or Y m! (!,") = #1 odd. This will be helpful when study selec7on rules for transi7ons between states. opposite projec7ons have same spa7al form yz plane is a node Orthonormal

23 Spherical harmonics as a basis set The orthonormality of the spherical harmonics means that any func7on define on a sphere can be expressed as a superposi7on of spherical harmonics. Electromagne7sm mul7poles of a charge distribu7on Gravita7on mul7poles of a mass distribu7on The CMB example mul7poles of the temperature fluctua7ons of the early universe

24 Spherical harmonics as a basis set Find the coefficients of the func7on 15 f (!,") = sin2! sin" 16# expanded as a Laplace series (series of sph. har.) %! $ m=#! f (!,") = $ c!m Y m!!,"!=0 ( )

25 The real (as opposed to complex) spherical harmonics In gravita7on problems, the complex numbers in the spherical harmonics are rarely useful, so other linear combina7ons are used. d yz (!,")! Y 1 ( 2!,") + e i# Y $1 ( 2!,") You ll recognize these as the p, d, f orbital forms from chemistry

26 Summary The fact that the series must be finite polynomial of degree el is what (mathema7cally) leads to a maximum (and integer) value of el! The fact that the series is finite means that we can differen7ate it only a finite number of 7mes (el) and that number turns out to be m!! Thus m is limited to el el. Legendre func7ons are orthonormal on [ 1,1] Legendre and associated Legendre func7ons.