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Lecture 14: Angular Momentum-II. The material in this lecture covers the following in Atkins. Rotational Motion Section 12.7 Rotation in three dimensions Lecture on-line Angular Momentum-II (3- D) (PDF) Angular Momentum-II (3-D) (PowerPoint) Handout for this lecture (PDF) Tutorials on-line Vector concepts Basic Vectors More Vectors (PowerPoint) More Vectors (PDF) Basic concepts Observables are Operators - Postulates of Quantum Mechanics Expectation Values - More Postulates Forming Operators Hermitian Operators Dirac Notation Use of Matricies Basic math background Differential Equations Operator Algebra Eigenvalue Equations Extensive account of Operators Extensive account of Operators Audio-visuals on-line Rigid Rotor (PowerPoint) (Good account from the Wilson Group,****) Rigid Rotor (PDF)(Good account from the Wilson Group,****) Slides from the text book (From the CD included in Atkins ,**) Classical Angular Momentum Angular momentum in classical physics Consider a particle at the position r Review of classical physics Position and velocity in 3D v r k i j Where r = ix + jy + kz The velocity of this particle is given by dr dx dy dz v = dt = i dt + j dt + k dt Classical Angular Momentum The linear momentum of the particle with mass m is given by dx p = mv where e.i px = mvx = m dt The angular momentum is defined as Review of classical physics Angular Momentum in 3D L = rXp L |r| |p| sin p r L =rXp The angular momentum is perpendicular to the plane defined by r and p. Classical Angular Momentum Review of classical physics We have in addition Angular Momentum in 3D L = rXp = (ix +jy + kz)X (ipx + jpy +kpz) L = (r ypz -rzpy)i + (rzpx -rxpz)j + (rxpy - rypx)k or i rXp = rx px j k ry rz py pz Classical Angular Momentum Why are we interested in the angular momentum Review of classical physics ? Angular Momentum in 3D Consider the change of L with time F dL dr dp dt = dt Xp + rX dt dL dp dt = mvXv + rX dt r dp = rX dt dL d dr d2 r dt = rX dt [m dt ] = rXm dt 2 m d2r dt 2 F (Newtons Law) Classical Angular Momentum Review of classical physics Angular momentum and central force in 3D dL r F dt F r For centro-symmetric systems in which the force works in the same direction as r we must have dL dt = 0 : THE ANGULAR MOMENTUM IS CONSERVED Classical Angular Momentum Review of classical physics Angular momentum and r Examples : F central force in 3D movement of electron around nuclei movement of planets around sun For such systems L is a constant of motion, e.g. does not change with time since dL dt = 0 In quantum mechanics an operator O representing a constant of motion will commute with the Hamiltonian which means that we can find eigenfunctions that are both eigenfunctions to H and O Quantum mechanical representation of angular momentum operator Rotation..Quantum Mechanics 3D We have L = rXp = iLx + jLy + kLz where Angular momentum operators of quantum mechanics in 3D Lx = ry pz - rzpy ; Ly = rzpx - rx py ; Lz = rx py - ry px In going to quantum mechanics we have x --> x ; y --> y ; z --> z px --> -i x ; py --> -i y ; pz --> -i z Thus : Lx = -i (y z - zy ) ; Ly = -i (z x - xz ) L z = -i (x y - yx ) Rotation..Quantum Mechanics 3D We have Angular momentum operators of quantum mechanics in 3D L = iLx + jLy + kLz thus L.L = L 2 =(iLx + jLy + kLz).(iLx + jLy + kLz) L2 = Lx 2 + Ly 2 + Lz2 Rotation..Quantum Mechanics 3D Can we find common eigenfunctions to L 2 , Lx , Ly , Lz ? Commutation relations for angular momentum operators of quantum mechanics in 3D Only if all four operators commute We must now look at the commutation relations The two operators L1 and L2 will commute if [L1,L2 ] f(x,y,z) =(L1L2 - L2L1) f(x,y,z) = 0 Rotation..Quantum Mechanics 3D Commutation relations for angular momentum operators of quantum mechanics in 3D ˆ2 L ˆ L ˆ representing For the quantum mechanical operators L the square of the length of the angular momentum as well as the operators representing the three Cartesian ˆ x ;L ˆ y; L ˆz components of the angular momentum vector L we have [Lˆ2 , Lˆ x ] [Lˆ2 , Lˆ y ] [Lˆ2 , Lˆ z ] 0 [Lˆ , Lˆ ] i Lˆ x y [Lˆ y , Lˆ z ] i [Lˆ , Lˆ ] i z x z Lˆ x Lˆ y Rotation..Quantum Mechanics 3D Common eigenfunctions for ˆ and L ˆ2 . L z How do we find the eigenfunctions ? The eigenfunctions f must satisfy L zf = af and L 2f = bf The function f must in other words satisfy the differential equations L zf = af as well as L 2f = bf Rotation..Quantum Mechanics 3D It is more convenient to solve the equations in Angular momentum operators spherical coordinates (x,y,z) (r, of quantum mechanics in spherical coordinates in 3D r We find after some tedious but straight forward manipulations d Lz = -i ] d d2 d 1 d2 L 2 = - 2[ +cot sin2 ] 2 d d d Rotation..Quantum Mechanics 3D Common eigenfunctions for ˆ z and L ˆ2 . L We must solve : ˆ z(,) b(,) and L ˆ 2 (,) c(,) L The eigenfunctions to L2 and Lz are given by (,) = Yl,m ((,) 2l+ 1 (l | m!| |m| = Pl (cos) exp[im] 4 (l | m!|) Eigenfunctions are orthonormal 2 sindd * Y (,)Y (,) r lm l'm' l,l' m,m' Rotation..Quantum Mechanics 3D Common eigenfunctions for ˆ and L ˆ2 . L z 2l+ 1 (l | m!| |m| (,) = Yl,m ((,) = Pl (cos ) exp[im] 4 (l | m!|) We have that l can take the values : l = 0,1, 2, 3,4.. and the possible eigenvalues for L2 are 2l(l 1) ˆ 2 (,) 2 l(l 1) (,) L lm lm for a given l value m can take the 2l+ 1 values - l, - l + 1,...,-1,0,1,...,l - 1,l and the possible eigenvalues for Lz are m ˆ z lm (,) m lm (, ) L Rotation..Quantum Mechanics 3D ˆz Common eigenfunctions for L ˆ 2 . Spherical harmonics and L 2l+ 1 (l | m!| |m| (,) = Yl,m ((,) = Pl (cos ) exp[im] 4 (l | m!|) l m 0 0 1 1 0 1 Y lm (,) 1 4 3 cos 4 3 sin exp[i] 4 Rotation..Quantum Mechanics 3D ˆz Common eigenfunctions for L ˆ 2 . Spherical harmonics and L 2l+ 1 (l | m!| |m| (,) = Yl,m ((,) = Pl (cos ) exp[im] 4 (l | m!|) l m Y lm (,) 2 0 5 (3cos2 1) 16 2 1 15 cossin[i] 8 2 15 sin 2 exp[2i] 32 2 ˆz Common eigenfunctions for L ˆ 2 . Spherical harmonics and L 2l+ 1 (l | m!| |m| (,) = Yl,m ((,) = Pl (cos ) exp[im] 4 (l | m!|) Rotation..Quantum Mechanics 3D l 3 3 3 3 m Y lm (,) 0 7 (5cos3 3 cos) 16 1 21 (5cos 2 1)sin [i] 64 2 105 sin2 cos exp[2i] 32 3 35 sin3 exp[2i] 64 What you should learn from this lecture 1. you should know the definition of angular momentum as L = rxp. 2. You should be aware of the commutation relations [Lˆ2 , Lˆ ] [Lˆ2 , Lˆ ] [Lˆ2 , Lˆ ] 0 x y z [Lˆ x , Lˆ y ] i Lˆ z ;[Lˆ y , Lˆ z ] i Lˆ x;[Lˆ z , Lˆ x ] i Lˆ y 3. You should realize that the above commutation has the consequence that we only can find ˆ 2 and one of the find common eigenfunctions to L ˆ . Thus we can components , normally taken as L z only know L2 and Lz precisely. What you should learn from this lecture 4. You are not required to know the exact form of the eigenfunctions 2l+ 1 (l | m!| |m| (,) = Yl,m ((,) = Pl (cos ) exp[im] 4 (l | m!|) ˆ z and L ˆ2 to L 5. You should know that l can take the values : l = 0,1, 2, 3,4.. and the possible eigenvalues for L2 are 2l(l 1) ˆ 2 (,) 2 l(l 1) (,) L lm lm 6. You should know that for a given l value m can take the 2l+ 1 values - l, - l + 1,...,-1,0,1,...,l - 1,l and the possible eigenvalues for Lz are m ˆ z lm (,) m lm (, ) L We have Appendix : Commutator relations for angular momentum components Lx ;L y ;Lz ;L2 . f f Lxf = -i ( yz - zy ) = -i u x f f L yf = -i ( zx - xz ) = -i u y Next LxLyf = -i Lxu y u y u y LxLyf = -i [ -i ( y z - z y ) ] u y u y LxLyf = - 2 [ y z - z y ] Appendix : Commutator relations for angular momentum components Lx ;L y ;Lz ;L2 . We have uy f f z = z (zx - xz ) uy f 2f 2f z = x + zzx - xz Further uy f f = y (zx - xz ) y uy y = 2f 2f z yx - x yz combining terms Thus Appendix : Commutator relations for angular momentum components Lx ;L y ;Lz ;L2 . f 2f 2f 2f 2f LxLyf = - 2[ yx + yzzx - yxz - z2yx +zxyz ] f 2f 2f 2f 2f LxLyf = - 2[ yx + yzzx - yxz - z2yx +zxyz ] It is clear that L xLyf can be evaluated by interchanging x and y We get: f 2f 2f 2f 2f LyLxf = - 2[ xy + xzzy - xyz - z2xy +zyxz ] Appendix : Commutator relations for angular momentum components Lx ;L y ;Lz ;L2 using the relations 2f 2f zy = yz etc. We have f f [ LxLy - LyLx] f = - 2[ yx - xy ] = - 2[ yx - xy ] f We have: L z = -i [ x y - yx Thus: [ LxLy - LyLx] f ] = i Lz f ; [Lx,Ly] = i Lz We have shown [L x,Ly] = i Lz Appendix : Commutator relations for angular momentum components Lx ;L y ;Lz ;L2 . By a cyclic permutation Z X C3 z Y Y z X X Y [ Ly,L z] = i Lx [ Lz,Lx ] = i Ly We have shown that the three operators L x ,L y ,L z are non commuting What about the commutation between L x ,Ly ,Lz and L2 Appendix : Commutator relations for angular momentum components Lx ;L y ;Lz ;L2 . Let us examine the commutation relation between L2 and L x We have : [ L2 ,L x ] [L2x L2y L2z ,L x ] [ L2 ,L x ] [L2x ,L x ] [L2y ,Lx ] [L2z ,Lx ] For the first term [L2x ,L x ] L2xL x L xL2x L3x L3x 0 Appendix : Commutator relations for angular momentum components Lx ;L y ;Lz ;L2 . For the second term [L2y ,L x ] L2yLx LxL2y 2 2 L yLx LyL xLy L yLxL y L xLy L y [L yL x L xL y ] [L yLx LxL y ]L y Z i LyL z i LzL y X Y Appendix : Commutator relations for angular momentum components Lx ;L y ;Lz ;L2 . For the third term [L2z,Lx ] L2zL x L xL2z 2 2 L zL x LzLxL z LzL xLz L xLz L z[LzL x L xLz ] [LzL x L xLz ]L z i L zL y LyL z Z X Y Appendix : Commutator relations for angular momentum components Lx ;L y ;Lz ;L2 . In total [ L2 ,L x ] [L2x L2y L2z ,L x ] 0 i L yLz i LzLy i L zL y LyL z Z X Y 0 Appendix : Commutator relations for angular momentum components Lx ;L y ;Lz ;L2 . We have shown [L2,Lx] = [L x 2+Ly 2+Lz2,Lx ] = O now by cyclic permutation Z X Y [Ly 2+Lz2+Lx 2,Ly ] = [L2,Ly ] = 0 [Lz2+Lx 2+Ly 2,Lz] = [L2,Lz] = 0 Thus Lx ,Ly ,Lz all commutes with L 2 and we can find common L2 and Lx or L 2 and Ly eigenfunctions for or L 2 and Lz the normal convention is to obtain eigenfunctions that are at the same time eigenfunctions to L z and L2. How do we find the eigenfunctions ? Rotation..Quantum Mechanics 3D We have -i S()T() = b S()T() or -i S() T()= b S()T() multiplying with 1/ S( ) from left T() = ib T() The general solution is T() = AExp[ ib ] A general point in 3-D space is given by ( r,) Z rcos (x,y,z) (r, r Y X rsin We have the following relation x= r sincos y= r sinsin z= r cos The same point is represented by (r, +2) We must thus have ib ib ib ib Exp[ ] = Exp[ ( 2) = Exp[ ] Exp[ 2] Thus Exp[ ib 2b 2b 2] = cos + isin =1 This equation is only satisfied if b = m with m = 0,±1,±2,...... Thus the eigenvalue b is quantized as b = m m = 0,±1,±2,...... The possible eigenfunctions are T() = AExp[im] , m = 0,±1,±2,......