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Quantum Physics 2005 Notes-8 Three-dimensional Schrodinger Equation Notes 8 Quantum Physics F2005 1 The Schrodinger Equation Here’s what we have been working with: Hˆ ! = Eˆ! h2 " 2 "# $ # + V ( x, t ) # = i h 2 "t 2m "x Here’s what the 3D equation is: Hˆ ! = Eˆ! h2 r 2 "# r $ % # + V ( r , t ) # = ih "t 2m Notes 8 Quantum Physics F2005 2 A quick review of coordinate systems Cartesian coordinates: iˆ, ˆj , kˆ form an orthogonal (rectangular) system with each axis at right angles to the other two. r ! Line element: dl = dxiˆ + dyˆj + dzkˆ ! Volume element: dV = dxdydz rˆ " r " " % = iˆ + ˆj + k "x "y "z z 2 2 2 " " " %2 = 2 + 2 + 2 "x "y "z y x Notes 8 Quantum Physics F2005 3 Cylindrical Converting to Cartesian coordinates: ! Line element: ! Length ds: x = ' cos& , y = ' sin & , z = z ' = ( x 2 + y 2 )1 / 2 , tan & = y / x r dl = d ' uˆ ' + ' d& uˆ& + dzuˆ z ds 2 = dr 2 + r 2 d& 2 + dz 2 ! Volume element: dV = ' d ' d& dz 'ˆ ,&ˆ, ẑ r " 1 " " % = uˆ ' + uˆ& + uˆ z "' ' "& "z 2 2 1 1 " " " " ' %2 = + 2 + 2 2 ' "' "' ' "& "z Notes 8 Quantum Physics F2005 4 Spherical x = r sin & cos ( ; y = r sin & sin ( ; z = r cos& 2 2 2 r = x + y + z ; tan ( = y / x; tan& = r dl = drrˆ + rd&&ˆ + r sin& d((ˆ x2 + y 2 z dV = r 2 sin& drd& d( r " 1 " 1 " % = uˆr + uˆ& + uˆ( "r r "& r sin& "( 1 " 2 " 1 " " 1 "2 % = 2 + 2 sin& + 2 r "& r sin2 & "( 2 r "r "r r sin& "& 2 Notes 8 Quantum Physics F2005 5 The Rectangular 3D Box – Quantum well Notes 8 Quantum Physics F2005 6 5-11 The 3-D Box a a x in $ , 0 2 2 b b V ( x, y , z ) = y in $ , 2 2 c c ) otherwise x in $ 2 , 2 The 3D Box problem is a straightforward extension of the 1D infinite well or 1D box problem where $ a , a 0 x in 2 2 V ( x) = ) x > a 2 Notes 8 Quantum Physics F2005 7 Px 1- D V (x) Px * h" i"x E * ih " "t 2 Px + V ( x) = E 2m Notes 8 2 h" # i "x +V ( x)# = ih " # 2m "t 2 "2 h $ # +V ( x)# = ih" # 2m "x2 "t Quantum Physics F2005 8 3 $D Px, Py , Pz V ( x, y , z ) Px * h " , Py * h " , Pz * h " i "z i "x i "y E * ih 2 2 Px + Py + Pz 2m " "t 2 + V ( x, y , z ) = E " h 2 " 2 "2 " 2 $ #+ #+ # + V ( x , y , z ) # = ih # "t 2m "x 2 "y 2 "z 2 Notes 8 h2 2 " )# =F2005 $ % # + V (Quantum x, y, zPhysics ih # 2m "t 9 1- D • Volume element d+ = dx • Probability of finding a particle in d+ at time t 2 ) P( x, t ) = ∫$) # ( x, t ) dx • Normalization condition 2 ) ∫$) # ( x, t ) dx = 1 • Stationary State wave function $ iEt #( x,t) =! ( x)e h &&dinger equation • Time independen t Schro 2 2 $ h " ! +V ( x)! = E! 2m "x2 • Eigenfunction Notes 8 ! (x ) Quantum n Physics F2005 10 • Volume d+ = dx dy dz 3 -D • Probabilit y of finding a particle in d+ at time t 2 P( x, y, z,t) = #( x, y, z,t) dx dy dz • Normalization conditon ) $) ) $) 2 ) $) ∫ dx ∫ dy ∫ dz # ( x, y, z , t ) = 1 • Stationary state wave function # ( x, y , z , t ) = ! ( x , y , z ) e $ iEt h • Eigenfunction ! n( x) && • Time independent Schrodinger equation 2 $ h %2! ( x, y, z) +V ( x, y, z)! = E! 2m • Eigenfunction ! n ( x), ! n ( y), ! n ( z) Notes 8 1 Quantum2 Physics F20053 11 1D ! n ( x) a x, 2 ! n ( x) = 2 cos n- x a sin a x> a 2 ! n ( x) = 0 Subsitute ! back into time independent Schr&o&dinger eq. to obtain allowed energies for the particle 2 2 h2-F2005 Notes 8 Quantum En = nPhysics 2ma2 12 3D ! n n n ( x, y, z) =! n ( x) ! n ( x) ! n ( z) 1 2 3 1 a x, , 2 ! n n n ( x, y, z) = 1 2 3 2 b c y , ,z , 2 2 2 cos n1- x a sin a 3 n 2 cos • 2 y b sin b n 2 • c cos 3c z sin Notes 8 Quantum Physics F2005 13 x > a, y > b, z > c 2 2 2 ! n n n ( x, y, z) = 0 1 2 3 2 2 2 n n n h 2- 2 En n n = 1 + 2 + 3 1 2 3 a b c 2m Notes 8 Quantum Physics F2005 14 Degeneracy & symmetry A degeneracy always reflects the existence of some symmetry in a given problem. E.g. 1 - 3D box with a = b . c 2 2 2 + n1 n2 n3 h2- 2 + En n n = a2 c2 2m 1 2 3 Exchange n1 & n2, En1n2n3 stays the SAME. But ! n n n &! n n n are DISTINCT. 1 2 3 because !n n n 213 ( x, y, z,t ) & !n n n ( x, y, z,t ) 1 2 3 213 Notes 8 Quantum Physics F2005 differ in their dependence on x & y. 15 • Symmetry here is interchang e coordinate s x & y. •Degeneracy here is two DISTINCT eigenfunctions having the SAME energy value. See Fig. 5-27 112 a = b . c box case $ $211,121 E.g. 2 Fig. 5-27 a=b=c box case $ 211,121,112 See Figure 5 - 27 a . b . c box Notes 8 $ 112 $ 211 $ 121Quantum Physics F2005 16 Notes 8 Quantum Physics F2005 17 Solutions for Central Potentials The 3D time-independent Schrodinger Eq: h2 2 ˆ % # + V # = E# H# = $ 2m " "# " 2# 1 1 h 2 1 " 2 "# $ + (V $ E ) # = 0 r + sin & + "& r 2 sin 2 & "( 2 2m r 2 "r "r r 2 sin & "& 2 2 h2 " 2 ( r# ) " ( r# ) 1 " ( r# ) h2 " ( r# ) $ + cot & $ + (V $ E ) ( r # ) = 0 + 2 2 2 2 2m "r 2 " " " & & & ( 2 mr sin Multiply both sides by r 2 , letting V = V (r ) and separating variables: 2 2 2 " ( r# ) 1 " ( r# ) h2 " ( r# ) 2 h2 " ( r# ) $r + r (V $ E ) ( r # ) = + cot & + 2 2 2m "r 2 2m "& 2 sin " " & & ( r # = u (r ) F (& , ( ) 2 2 2 2 "(F ) h2 1 " ( u (r ) ) 2 1 1 " (F ) h2 " ( F ) $r + r (V $ E ) = + cot & + 2 2 2m u (r ) "r 2 & & ( " " F (& , ( ) 2m "& 2 sin 2 =Notes = l (l + 1) a constant 8 Quantum Physics F2005 18 The ( equation Notes 8 Quantum Physics F2005 19 ( solution for central potential 1 2 1 1 " 2 "# 1 1 " "# 1 1 " 2# 2mE & % #= r + sin + = $ h2 # # r 2 "r "r # r 2 sin & "& "& # r 2 sin 2 & "( 2 # = R/0 1 2 " 2 "R 1 " "/ 2 2 2m( E $ V (r )) 1 " 20 2 & & & $ sin & r $ $ = = $ sin sin r sin m l h2 "r "r # "& "& 0 "( 2 R 0 = e± iml( Because the wavefunction must be single valued ml must be an integer. Notes 8 Quantum Physics F2005 20 The & equation Notes 8 Quantum Physics F2005 21 & solution for central potential 1 1 " 2 "R 2 2m( E $ V (r )) ml2 1 1 " "/ & r $ r = + $ sin = l (l + 1) 2 2 R "r "r sin & / sin & "& "& h The angular part of Laplace's equation is called the Legendre Equation. ml2 1 1 " "/ & sin + 2 $ = l (l + 1) sin & / sin & "& "& Making the substitution x=cos& d dx d d d = = $ sin & = $ 1 $ x2 , so d& d& dx dx dx d m2 2 d /lm (1 $ x ) )/lm = 0 + (l (l + 1) $ 2 1$ x dx dx Notes 8 Quantum Physics F2005 22 & solution for central potential 2 We will start by solving the special case when m=0. d 2 dPl (1 $ x ) + (l (l + 1)) Pl = 0 dx dx To find the solution we use the power series method, ) assume a solution of the form, y ( x) = ∑ an x n n =0 and substitution into gives ) ∑ an n( n $ 1) x n=2 n$2 ) ) n ) $ ∑ an n(n $ 1) x $ 2 ∑ an nx + l (l + 1) ∑ an x n = 0 n=2 n n =1 n =0 from which we get the recursion relation, (n $ l )(n + l + 1) an + 2 = an (n + 1)(n + 2) Notes 8 Quantum Physics F2005 23 & solution for central potential 3 We can deduce the solution for n=0 and use with the recursion relation. (2n $ 2 j )! x n $ 2 j Pn ( x) = ∑ ( $1) n j =0 2 l !(n $ 2 j )!(n $ 1)! where N=n/2 for n even and N=(n-1)/2 for n odd. N j A second set of possible solutions yields unphysical results. The first few Legendre polynomials (Pn ) are: P0 ( x) = 1, P1 ( x) = x, P2 ( x) = ( 3 x 2 $ 1) / 2 or P0 = 1, P1 (& ) = cos & , P2 (& ) = ( 3cos 2 & $ 1) / 2 Notes 8 Quantum Physics F2005 24 & solution for central potential 4 • Example plots of the first few Legendre functions (Legendre_plots.mws) 1 2 3 Notes 8 Quantum Physics F2005 25 & solution for central potential 5 Orthogonality and normalization of Legendre Polynomials 1 2 ∫ [ Pn ( x) Pm ( x) ] dx = 1 nm 2n + 1 $1 Orthogonality means that we can express the angular part of any wavefunction using a sum of Legendre polynomials. Notes 8 Quantum Physics F2005 26 & solution for central potential 6 We will not solve the m . 0 case now, but we will state the relation between the Legendre functions (m=0) and the full solutions (the associated Legendre functions). m /lm ( x) = (1 $ x 2 ) m /2 d Pl m dx from which we can find /00 = 1, /10 = x, /1±1 = (1 $ x 2 )1/ 2 , /02 = 1 $ 3 x 2 , / ±2 1 = (1 $ x 2 )1/ 2 x Notes 8 Quantum Physics F2005 27 & and ( together Sperical harmonics Notes 8 Quantum Physics F2005 28 Associated Legendre functions from Gasiorowicz Notes 8 Quantum Physics F2005 29 Angular solutions put together The & and ( solutions can now be combined Flm (& , ( ) = /lm (& )0 m (( ) and when normalized, yields the Spherical Harmonics: 2l + 1 (l $ m)! m im( / Yl = (cos ) e & l + 4 ( l m )! - The orthonormality relation is: m m m' Y ∫∫ l Yl ' d& d( = 1 ll '1 mm ' Notes 8 Quantum Physics F2005 30 Spherical Harmonic Functions Notes 8 l m 0 0 1 0 1 1 2 0 Ylm (( ,& ) ( 4- ) $1/ 2 1/ 2 ( 3 / 4- ) 1/ 2 $ ( 3 / 8- ) 1/ 2 ( 5 /16- ) cos & sin & ei( ( 3cos & $ 1) Quantum Physics F2005 2 31 The radial equation Notes 8 Quantum Physics F2005 32 The r part of the 3D TISE " 2 ( u (r ) ) l (l + 1) 2m $ + u (r ) + 2 (V $ E )u ( r ) = 0 2 2 "r r h # is finite, continuous, and smooth, so u(r) must go to zero at r=0. First, let's look at u(r) solutions for l = 0. 2 h 2 " ( u (r ) ) $ + (V $ E )u ( r ) = 0 2 2m "r Notes 8 Quantum Physics F2005 33 u(r) example for l=0: particle in a box 0 r < a assume V = ) r 2 a 2 h 2 " ( u (r ) ) $ $ Eu (r ) = 0 inside 2 2m "r u (r ) = A sin kr + B cos kr = A sin kr because u must = 0 at r=0 Boundary condition at a gives: kn = na n- r 2 2 2 2 2 2 2 2 k n nh h h a ; E = n = = n r 2m 2ma 2 8md 2 A sin # (r ) = Notes 8 Quantum Physics F2005 34 The radial equation for the Coulomb potential $ " 2 ( u (r ) ) "r 2 l (l + 1) 2m e2 + u (r ) + 2 ($ )u (r ) = Eu (r ) 2 h r 4-3 0 r h2 r Let ' = where a0 = 4-3 0 a0 me 2 E me 4 Let 3 = where ER = 2 ER ( 4-3 0 ) 2h 2 and so: $ Notes 8 " 2 (u( ' ) ) "' 2 + l (l + 1) ' 2 u(' ) + Quantum Physics F2005 2 ' u( ' ) = 3 u( ' ) 35 l=0 + Coulomb potential Let's look at forms of the radial equation for l = 0 in the large r limit: d 2u $ 3' ' *) 3 + = 0 ⇒ 4 u u e d'2 Remember that 3 is negative because these are bound states and the potential is negative. We now search for solutions to the full equation using the polynomial expansion approach. The lowest order polynomial with the right behavior as ' * 0 is u ( ' ) = re $ 3' . (Think about the number of roots for a polynomial.) Notes 8 Quantum Physics F2005 36 l=0 Coulomb solutions Substituting back into the original diff eq and solving for 3 . ( ) $ $ 31 $ 31 ' $ 1 = 31 ' ⇒ 31 = $1 E1 = me 4 ( 4-3 0 ) 2 2h 2 = $13.6eV (Rydberg) Notes 8 Quantum Physics F2005 37 l=0 Coulomb solutions We look for solution of increasing polynomial order in similar manner and can find: u2 4 2 ' $ ' 2 e $ ' / 2 u3 4 27 ' $ 18' 2 + 2 ' 3 e $ ' / 3 and we find that the energies are: E1 En = 2 n Notes 8 Quantum Physics F2005 38 Energy levels for solutions so far (1) $1 En = 2 n n=4 n=3 n=2 n =1 3 r-nodes 2 r-nodes 1 r-node 0 r-nodes $ 1 Ry 4 $1 Ry l =0 Notes 8 Quantum Physics F2005 39 The l=0 wavefunctions !1 := e ( $' ) !2 := ( 2 $ ' ) e 2 !3 := ( 27 $ 18 ' + 2 ' ) e Notes 8 Quantum Physics F2005 1 $ ' 2 1 $ ' 3 40 Probability of finding electron at r: radial probability # *# is the probability of finding the electron in a particular position. If we want the probability of finding the electron at a distance between r and r+dr from the nucleus, then we have to integrate # *# around the sphere. 2- - w( ' ) = ∫ ∫ r 2 sin & R 2 ( ' )Ylm* (& , ( ) Ylm (& , ( ) d& d( 0 0 For l=0, w( ' ) = 4- r 2 R 2 = 4- u 2 Notes 8 Quantum Physics F2005 41 Radial probability for l=0 solutions w1 w2 w3 Notes 8 Quantum Physics F2005 42 Radial probability At what distance is the electron most likely to be found? For the n = 1, l = 0 state: d ( ' 2 R2 ) 2 du du = =2 =2 d' d' d' ' = 1!!! r = a0 = 0.5 Angstroms Notes 8 d ( ' e$ ' ) d' = 2 ( e$ ' $ ' e$ ' ) = 0 Quantum Physics F2005 43 Solutions to Coulomb potential with l>0 " 2 ( u ( ' ) ) l (l + 1) 2 $ + $ u(' ) = 3 u(' ) 2 2 "' ' ' This is equivalent to an effective potential of l (l + 1) 2 $ Veff = 2 ' ' On the next page we plot effective potential for various l. Notes 8 Quantum Physics F2005 44 Effective potential with l>0 So higher l will push the wavefunctions out and push the energies up. Notes 8 Quantum Physics F2005 45 Radial solutions for l>0 using r limits d 2u $ 5 $ 3 ⇒ 4 u u e For large ': d'2 d 2u For small ' : d'2 l (l + 1) '2 3' u ⇒ u 4 ' l +1 or ' $ l 6Look for solutions of the form: u 4 ' l +1e $ 3' If we plug this form as it stands back into the 1 full equation, we get 3 = $ . 2 (l + 1) The solution we have found corresponds to the l =0 solutions for l = n -1. Notes 8 Quantum Physics F2005 46 Radial solutions with l>0 1 . 2 (l + 1) Note that this for l > 0 has the same energy as the l =0 solutions for l = n -1. Remember these are solutions for u (r ) with no nodes. (Radially, they actually look like the function for l = 0, n = 1. In & they have increasing number of nodes.) For a given l we have 3 = $ Notes 8 Quantum Physics F2005 47 Energy levels for solutions so far $1 En = 2 n n=4 n=3 n=2 l =3 l=2 3 r-nodes 2 r-nodes 1 r-node $ 1 Ry 4 l =1 0 r-nodes Because the energies line up, n =1 0 r-nodes l =0 Notes 8 $1 Ry these states are labeled by the corresponding n number. Quantum Physics F2005 48 What about solutions with more rnodes? Let's look at a solution with one node: Guess: u ( ' ) = ( c0 $ c1 ' ) ' l +1e $ ' 3 We find that this can be a solution if: 3 = $ 1 (l + 2) 2 . This is the same energy as our nodeless solution with l = n - 2. Notes 8 Quantum Physics F2005 49 Energy levels for solutions so far $1 En = 2 n n=4 n=3 n=2 l =3 l=2 3 r-nodes 2 r-nodes l =1 1 r-node 0 r-nodes n =1 l=2 l =1 1 r-node 0 r-nodes l =0 Notes 8 Quantum Physics F2005 50 A table of the radial wavefunctions n 1 2 l=0 e$ ' (1 $ ' 2 )e $ ' / 2 3 ( 27 $ 18 ' + 2 ' ) e 2 Notes 8 $' /3 l=1 no solution l=2 no solution ' e$ ' / 2 no solution ' (6 $ ' )e$ ' / 3 Quantum Physics F2005 ' 2e$ ' / 3 51 Atomic energy structure • • • • • • This should begin to look familiar. The energy levels correspond to the energies of the Bohr model. For n=1 energy level, only l=0 is found. For the n=2 energy level, only l=0 and l=1 solutions are found. For the n=3 level, only l=0,1,2 are found. In chemistry, we designate the l=0 case as s, l=1 as p, l=2 as d, and l=3 as f. Note the ml does not affect the energy of a state because it does not appear in the radial equation. Notes 8 Quantum Physics F2005 52 A big summary of energy levels and quantum numbers (from Semat 1972) Notes 8 Quantum Physics F2005 53 A big summary of hydrogen wavefunctions (from Semat 1972) Notes 8 Quantum Physics F2005 54 A big summary of pictures of wavefunctions (from Semat 1972) Notes 8 Quantum Physics F2005 55 What we observe: emission from transitions between states E photon Notes 8 1 1 = 1Ry × 2 $ 2 n n f i Quantum Physics F2005 56 Calculating transition probabilities The probability for an transition between two atomic levels with the emission of a photon polarized along the z axis is proportional to the matrix element: M fi7 ! f z! i = ! f r cos &! i . It can be shown that M fi is non-zero only when l f $ li = ±1 and 8ml = 0 or ± 1 Notes 8 Quantum Physics F2005 57 Angular momentum and magnetic quantum numbers: l and ml Notes 8 Quantum Physics F2005 58 Another look at the 3D Schrodinger equation Another way to construct the 3D Schrodinger equation is to compare it to the classical equation: p2 +V = E 2m Breaking p up into components: p 2 = px2 + p y2 + pz2 = pr2 + pt2 L and in terms of angular momentum: pt = . r L2 pr2 + +V = E 2 2m 2mr Notes 8 Quantum Physics F2005 59 Angular momentum If we now compare this expression to the Laplacian version we see that L2 should be an operator and we can test if it has eigenvalues and functions. We already know from our prior math that the angular part of the Laplacian has the same eigenfunctions as the Hamiltonian. Let's look more closely at the L operator. We think: Lˆ2! = L2! . r r r Classically: L = r × p, so let's do the math. Notes 8 Quantum Physics F2005 60 Angular momentum operators In Cartesian coordinates: i r r r×p= x px j y py k z pz " h " Lx = ypz $ zp y = y $ z i "z "y " h " Ly = zpx $ xpz = z $ x i "x "z " h " Lz = xp y $ ypx = x $ y i "y "x Notes 8 Quantum Physics F2005 61 Angular momentum in spherical coordinates h " " Lx = $sin( $ cot& cos( i "& "( h " " Ly = $cos( $ cot& sin( i "& "( h " Lz = i "( Notes 8 Quantum Physics F2005 62 What can we tell about L? We can learn quite a bit about angular momentum without doing integrals. We can show that Lz , and Lx cannot have a complete set of common eigenstates by considering the commutator [ Lz , Lx ] . Lz = xp y $ ypx ; Lx = ypz $ zp y ; Ly = zpx $ xpz L2 = L2x + L2y + L2z [ Lx , Lz ] = ypz $ zp y , xp y $ ypx = ypz , xp y $ [ ypz , ypx ] $ zp y , xp y + zp y , ypx = ypz , xp y + 0 + 0 + zp y , ypx = x ypz , p y + [ ypz , x ] p y + zp y , ypx = $ x p y , ypz $ [ x, ypz ] p y + zp y , ypx = $ xy p y , pz $ x p y , y pz $ y [ x, pz ] p y $ [ x, y ] pz p y + zp y , ypx = = 0 $ x p y , y pz + 0 $ 0 + zp y , ypx = x y, p y pz + zp y , ypx = x y, p y pz + y zp y , px + zp y , y px = x y, p y pz $ y px , zp y $ y, zp y px = x y, p y pz $ yz px , p y $ y [ px , z ] p y $ z y, p y px $ [ y, z ] p y px = x y, p y pz $ 0 $ 0 $ z y, p y px $ 0 = ih { xpz $ zpx } = $ihLy Similarly: Lx , Ly = ihLz and Ly , Lz = ihLx Notes 8 Quantum Physics F2005 63 What can we tell about L? • We can also show (homework!) that [L2,Lz]=0. • This tells us that there exists a complete set of eigenfunctions for both operators, or that we can measure both quantities simultaneously with infinite accruracy. • (We already knew that because we did the work to find the eigenfunctions and values, but we could have avoided that work if all we wanted was to know whether it could be done.) Notes 8 Quantum Physics F2005 64 Useful commutator/operator relations [ A, B ] = AB $ BA [ A, B ] = $ [ B, A] [ A, B + C ] = [ A, B ] + [ A, C ] [ A, BC ] = [ A, B ] C + B [ A, C ] A, [ B, C ] + B, [C , A] + C , [ A, B ] = 0 Notes 8 Quantum Physics F2005 65 Angular Momentum in Spherical coordinates 2 2 2 " " " 1 2 2 ˆ + 2 L = $ h 2 + cot & 2 & & & ( " " " sin • We have already found the eigenfunctions of this operator. They are the spherical harmonics. • We have previously found that the eigenvalues of L2 are l(l+1) with l=integers 0, 1, 2, 3... if the potential is central. • Angular momentum manifests itself as a magnetic dipole moment when the particle with L has charge. • It is most useful to know the projection of the dipole onto an applied magnetic field. (Let’s say in the z direction.) Notes 8 Quantum Physics F2005 66 Angular momentum projection We want to know now whether the projection of the angular momentum onto the z-axis can be an eigenvalue. h " ˆ Lz! = ! = Lz! i "( In the previous section we found: ! = R/0 with 0 = eiml( , thus Lˆ ! = hm ! . z l ! is simultaneously an eigenstate of Lˆ2 and Lˆ z . For L2 = hl (l + 1), ml can take on integer values from -l to + l. Notes 8 Quantum Physics F2005 67 Pictorial representation of L and ml Notes 8 Quantum Physics F2005 68 Angular momentum and magnetic moment An electron moving in a plane orbit of area A e is equivalent to a current given by I = cT where T is the period of the orbit. Such a plane current has a magnetic moment given by: µ = IA. So the magnetic moment of a classical orbiting eA electron is: µ = . cT We can relate the area and period to the angular momentum. Notes 8 Quantum Physics F2005 69 Angular momentum and magnetic moment 1 2- 2 Area can be expressed as: A = ∫ r d( 2 0 2 d( Angular momentum is: L = mr dt r r LT So A = 2m e r eh r L=$ l (l + 1) Lˆ. Therefore: µ = $ 2mc 2mc Notes 8 Quantum Physics F2005 70 Angular momentum and magnetic moment r If an atom has a magnetic moment µ, then it will r experience a force when it is placed in a magnetic field B. If the magnetic field is uniform, then the atom will r r r experience a torque M = µ × B and precess about the direction of the field. The shift in energy due to eh Bml . precession is: 8E = µ B cos & = 2mc Notes 8 Quantum Physics F2005 71 Electron spin • It is observed that a free electron exhibits a magnetic moment which is quantized. • We have associated this moment with a new quantum number which we call spin s so that µs = 2s eh 1 with s = 2mc 2 • When we compute the total magnetic moment of an atom, we must include both electron spin and orbital angular momentum as a vector sum. Notes 8 Quantum Physics F2005 72 Appendix – general radial wavefunctions for V~1/r using series solutions (from Eisberg and Resnick) d 2 F 2 dF 9 $ 1 l (l + 1) F = 0. + $ 1 + $ 2 2 d' ' d' ' ' a0 . 0 and s 2 0. Assume F ( ' ) = ' s ∑ ak ' k k =0 This form is used because it assures that F will be finite at ' =0. ) ∑ {[ ( s + k )( s + k + 1) $ l (l + 1) ] ak ' k =0 s+k $2 $ ( s + k + 1 $ 9 ) ak ' s + k $1} = 0 It can be shown that: [( s)( s + 1) $ l (l + 1)] a0 ' s $2 ) + ∑ {[ ( s + j + 1)( s + j + 2) $ l (l + 1) ] a j +1 $ ( s + j + 1 $ 9 )a j } ' s + j $1 = 0 j =0 Notes 8 Quantum Physics F2005 73 General radial solutions Collecting terms in powers of ' , we see that the sum above will only be zero for all ' if: [( s)( s + 1) $ l (l + 1)] = 0 (This relation requires that s = l or s = -(l + 1). Only the s = l solution can be physically allowed.) and (l + j + 1 $ 9 ) a j +1 = aj (l + j + 1)(l + j + 2) $ l (l + 1) Notes 8 Quantum Physics F2005 74