Physics 2D Lecture Slides Lecture 29: March 8th 2005 Vivek Sharma - PDF document

Physics 2D Lecture Slides Lecture 29: March 8th 2005 Vivek Sharma UCSD Physics The Hydrogen Atom In Its Full Quantum Mechanical Glory � � � r 2 2 2 � = 2 + + Instead of writing Laplacian , � 2 � 2 � 2 x y z � 2 write for spherical polar coordinates: � � � � � � � � � 2 1 1 1 � + � + 2 = r 2 sin � � � � � � � � � � � � � � 2 � � 2 � � 2 2 2 r r r r s n i r sin � � � � Thus the T.I.S.Eq. for (x,y,z) = (r, , ) be come s � � � � � � � � � � 1 � (r, , ) � 1 � (r, , ) � 2 + � + r sin � � � � � � � � � � � r 2 r � r � r 2 sin � � � 2 � � � 1 (r, , ) 2m � � � + (E-U(r)) (r, , ) = 0 2 2 � � 2 � 2 r si n � 1 1 � = with U r ( ) r + + x 2 y 2 z 2 1

So What do we have after all the shuffling! 2 � d + � = 2 m 0.. ..... ............(1) l � 2 d � � � 2 1 d � d � + m � + � � � = l sin l l ( 1) ( ) 0.....(2) � � � � � � � � 2 si n d � d � sin � � � � � 2 2 + 1 d � R � + 2m r ke l l ( 1) = 2 r (E+ )- R r ( ) 0....(3) � � � � 2 � 2 2 r d r � r � � r r � � These 3 "simple" diff. eqn describe the physics of the Hydrogen atom. All we need to do now is guess the solutions of the diff. equations Each of them, clearly, has a different functional form And Now the Solutions of The S. Eqns for Hydrogen Atom Φ 2 � + d � = The Azimuthal Diff. Equation : m 2 0 � 2 l d � � im � Solution : ( ) = A e but need to check "Good Wavefunction Condition" l � � � � � � + � Wave Function must be Single Valued for all ( )= ( 2 ) � � + � � � � im = im ( 2 ) � = ± ± ± ( ) = A e A e m 0, 1, 2, 3....( Magneti c Q uantum # ) l l l � � � � � + 2 1 d d m � + � � � = The Polar Diff. Eq: sin l ( l 1) l ( ) 0 � � � � � � � � sin d � d � sin 2 � � Solutions : go by the name of "Associated Legendr e Functions" only exist when the integers and l m are related as follows l = ± ± ± ± = m 0, 1, 2, 3.... l ; l positive number l l : Orbital Quantum Number 2

Wavefunction Along Azimuthal Angle φ and Polar Angle θ 1 = � � � For l 0, m =0 ( ) = ; l 2 = ± � For l 1, m =0, 1 Three Possibilities for the Orbital part of wavefunction l 6 3 = = � � � � = = ± � � � � [ l 1, m 0 ] ( ) = cos [ l 1, m 1 ] ( ) = sin l l 2 2 10 = = � � � 2 � � [ l 2, m 0 ] ( ) = (3cos 1) l 4 .... and so on and so forth (see book for more Functions) Radial Differential Equations and Its Solutions � � � 2 2 + 1 d � R � + 2m r ke l l ( 1) = The Radial Diff. Eqn : r 2 (E+ ) - R r ( ) 0 � � � � � r d 2 r � r � � 2 r r 2 � � Solu tio ns : Associated Laguerre Functions R(r ), Solutions exist only i f : 1. E>0 or has negtive values given by 2 � � 2 ke 1 � = = E=- 2a ; with a Bohr Radius � � 0 2 2 � n � mke 0 = � 2. And when n = integer such that l 0,1,2,3,4,.......( n 1) n = principal Quantum # or the "big daddy" quantum # 3

The Hydrogen Wavefunction: ψ (r, θ , φ ) and Ψ (r, θ , φ ,t) To Summarize : The hydrogen atom is brought to you by the letters: � n = 1,2,3,4,5,.... = � l 0, 1,2,3,,4....( n 1) Quantum # appear onl y in Trapped sys tems = ± ± ± ± m 0, 1, 2, 3,... l l The Spatial part of the Hydrog en Atom Wave Function is : � � � = � � � � = m ( , , ) r R ( ) . r ( ) . ( ) R Y l nl lm m nl l l l m Y are kn own as Sphe rical Harmonics. They define the angu lar stru cture l l in the Hydrogen-like atoms. iEt � � � � = � � � The Full wavefunction is (r, , , ) t ( , r , ) e � Radial Wave Functions For n=1,2,3 n l m R(r)= l 2 e -r/a 1 0 0 3/2 a 0 r 1 r -2a 2 0 0 (2- )e 0 3/2 a 2 2a 0 0 r 2 2 r r � � + 3 a 3 0 0 (27 18 2 ) e 0 2 3/2 a a 81 3a 0 0 0 n=1  K shell l=0  s(harp) sub shell l=1  p(rincipal) sub shell n=2  L Shell l=2  d(iffuse) sub shell n=3  M shell l=3  f(undamental) ss n=4  N Shell l=4  g sub shell …… …….. 4

Symbolic Notation of Atomic States in Hydrogen � = = = = = l ( s l 0) ( p l 1) ( d l 2) ( f l 3) ( g l 4 ) ..... n � 1 1 s 2 2 s 2 p 3 3 3 s p 3 d 4 4 s 4 p 4 d 4 f 5 5 5 5 s p d 5 f 5 g Note that: •n =1 is a non-degenerate system 2 � � ke 1 •n>1 are all degenerate in l and m l. E=- 2a � � 2 � n � All states have same energy 0 But different angular configuration Energy States, Degeneracy & Transitions 5

Facts About Ground State of H Atom 2 1 1 = = = � = -r/a � � = � � = n 1, l 0, m 0 R ( ) r e ; ( ) ; ( ) l a 3/2 � 2 2 0 1 � � � = -r/a ( , , ) r e ......look at it caref ully 100 � a 0 � � � 1. Spherically s ymmetric no , dependence (structure) 2 r 1 � 2 � � � = 2. Probab i lity Per Unit Volume : ( , r , ) e a 100 � 3 a 0 � � Likelihood of finding the electron is same at all , and depends only on the radial seperation (r) between elect ron & the nucleus. ke 2 = � 3 Energy of Ground ta S te =- 13.6 eV 2a 0 Overall The Ground state wavefunction of the hydrogen atom is quite boring Not much chemistry or Biology could develop if there was only the ground state of the Hydrogen Ato m ! We ne e d structure, we need variety, we need some curves! Interpreting Orbital Quantum Number (l) � + � 1 d � dR � + 2m ke 2 l l ( 1) 2 = Radia l part of S.Eqn: r ( + E )- R r ( ) 0 � � � � r dr 2 � dr � � 2 r r 2 � � L 2 k e + � For H Atom: E = K + U = K K ; substitute this i n E RADIAL ORBITAL r p � + � 1 d � dR � + 2m � 2 l l ( 1) 2 + = r K K - R ( ) r 0 � � � � r RADIAL ORBI TAL 2 � 2 2 r dr � dr � 2 m r � � 2 + � l l ( 1) = Examine the equation, if we set K then ORBITAL 2 2m r what remains is a differential equation in r � � + 1 d dR 2m [ ] 2 = r K R r ( ) 0 which depends only on radius r of orbit � � RAD AL I 2 2 r dr � dr � � � 2 1 � � L = � � = Further, we a s l o know t hat K mv 2 ; L= r p ; |L| =mv r K ORBITAL orbit orb ORBIT AL 2 2 2 mr 2 + 2 � l l ( 1) L = = � = + Putting it all togat her: K magnitude of Ang . Mom | L | l l ( 1) � ORBITAL 2 2 2m r 2 mr = � = + = Since l positive integer =0,1,2, 3 ...(n-1) angular momentum | L | l l ( 1) � discrete values = + | L | l ( l 1) : QUANTIZATION OF � E lect ron's Angular Mom entu m 6

Magnetic Quantum Number m l � � � = � L r p (Right Hand Rule) � Classically, direction & Magnitud e of L always well defi n ed � QM: Can/Does L have a definite direction ? Proof by Negat io n : � � ˆ Suppose L was precisely known/defined (L || z) � � � = � � S ince L r p Electron MUST be in x-y orbit plane 2 p � � � � � � � = � z = 0 ; p z � � p � ; E � !!! z z 2 m � So , in Hydrogen atom, L can not have precise measurable value � � � � � Uncertainty Principle & An gular Momentum : L z Magnetic Quantum Number m l Consider � = 2 | L | = � ( � + 1) = 6 � � In Hydrogen atom, L can not have precise measurable value Arbitararily picking Z axis as a reference direction : � L vector spins around Z axis (precesses). � = = ± ± ± ± The Z component of L : | L | m � ; m 1 , 2, 3... l Z l l < Note : since | L | | L | (always) Z < + = = + sin c e m � l l ( 1) It can never be that |L | � m � l ( l 1) � l Z l (break s Uncertainty Pri ncip le) So......the Electron's dance has be gun ! 7