Rigid Rotor and Angular Momentum - Cobalt

CHEMISTRY 373 - TUTORIAL 4PRACTICE FOR ANGULAR MOMENTUM20001. Show that the three operators representing the three components of angular momentum,Lˆ x,Lˆy, Lˆzare Hermitian and therefore represent observables.(ANS. Use the fact that the operators xˆ , ŷ, ẑ,and pˆx, pˆy, pˆz, are Hermitian. Therefore, wehave by definition of the adjoint of an operator3 3∫ drL ( ˆ +X ψ )* φ = ∫ dr ψ * L ˆ x φ ,3= ∫ drψ*(ˆ ypˆ z − zp ˆ ˆ y) φ,3 3= ∫ drψ * yp ˆ ˆ zφ− ∫ drψ * zp ˆ ˆ yφ,3 3= ∫ dry (ˆ ψ)*ˆ pzφ − ∫ drz (ˆ ψ)*ˆ pyφ,3= drpy (ˆ zˆ 3∫ ψ ) * φ − ∫ drpz (ˆ yˆ ψ)* φ,3(ˆ ˆ3= ∫ drypzψ)* φ − ∫ drzp (ˆ ˆ yψ)* φ,3= ∫ dryp (ˆ ˆ z − zp ˆ ˆ yψ)* φ,3= ∫ drL ( ˆ xψ)* φ.Therefore, ˆ +Lx= L ˆ x,and Lˆ x is Hermitian. The proofs for Lˆ yand Lˆ are similar. In thezthird last line, we have made use of the fact that[ xˆ , pˆy] = [ xˆ , pˆz] = [ ŷ, pˆx] = [ ŷ, pˆz] = [ ẑ, pˆx] = [ ẑ, pˆy] = 0This completes the problem.)2. Show that the two operators Lˆ = Lˆ ±± xiLˆy, are Hermitian conjugates or adjoints. Use theresults of the previous question.(ANS. Use the results from Question. Beginning with the definition of adjoint, we have3 3drL ( ˆ ?∫ − ψ )* φ = ∫ dr ψ * L ˆ − φ ,3= ∫ dr ψ *(ˆ Lx−iLˆ y) φ ,3 3= ∫ drψ * L ˆ xφ−idr ∫ ψ * L ˆ yφ,3 3= ∫ drL ( ˆ xψ)* φ−idrL∫ (ˆ yψ)* φ,3 3= ∫ drL ( ˆ xψ)* φ+ ∫ driL ( ˆ yψ)* φ,= ∫ dr 3 (( L ˆ x+iL ˆ y ) ψ )* φ ,= ∫ drL3 ( ˆ + + ψ )* φ .Therefore, we have Lˆ ✝ −= Lˆ and similarly, we can show that+Lˆ ✝+= Lˆ , or simply note that−✝ ✝✝Lˆ ) = ( Lˆ ) Lˆ ).(− +=−3. Let Lˆx,Lˆy, Lˆ , be the three components of angular momentum. Then show thatz[ Lˆ ,Lˆ ] = ih Lˆ ,[ Lˆ ,Lˆ ] = ihLˆ,[ Lˆ ,Lˆ ] = ihLˆ.xyzyzxzxy

2 2 2 2Also, if Lˆ = Lˆx+ Lˆy+ Lˆz, then show that222[ Lˆ ,Lˆx] = [ Lˆ ,Lˆy] = [ Lˆ ,Lˆz] = 0 .(ANS. Express Lˆ x,Lˆy, Lˆz, in terms of three components of position xˆ , ŷ, ẑ , and momentum,pˆx, pˆy, pˆz ,Lˆx= ŷpˆz− ẑpˆy,Lˆy= ẑpˆx− xˆpˆz,Lˆz= xˆpˆy− ŷpˆx.Making use of the commutators[ xˆ , pˆx] = [ ŷ, pˆy] = [ ẑ, pˆz] = ih ,and[ xˆ , pˆy] = [ xˆ , pˆz] = [ ŷ, pˆx] = [ ŷ, pˆz] = [ ẑ, pˆx] = [ ẑ, pˆy] = 0,[ pˆx, pˆy] = [ pˆx, pˆz] = [ pˆy, pˆz] = 0along with[ Â,Bˆ Ĉ ] = [ Â,Bˆ ]Ĉ + Bˆ [ Â,Ĉ ], [ ÂBˆ ,Ĉ ] = [ Â,Ĉ ]Bˆ + Â[ Bˆ ,Ĉ ]we see that[ Lˆx,Lˆy] = [ ŷpˆz− ẑpˆy,ẑpˆx− xˆpˆz],= [ ŷpˆz− ẑpˆx] + ẑpˆy,xˆpˆz],= ŷ[ pˆz,ẑ ] pˆx+ xˆ [ ẑ, pˆx] pˆy,= ih(xˆpˆy− ŷpˆx),= ihLˆz.In arriving at this result, I have made use of what I call "eyeballing" techniques. Themissing commutators such as [ ŷpˆz,xˆpˆz] and [ ẑpˆy,ẑpˆx] clearly vanish. Why? Similarly,we can show that[ Lˆy,Lˆz] = ih Lˆx, [ Lˆz,Lˆx] = ihLˆ.yWe can use these three commutators to solve the final one. Hence, we get222[ Lˆ ,Lˆx] = [ Lˆx,Lˆx] + [ Lˆy,Lˆx] + [ Lˆz,Lx],= [ Lˆy,Lˆx]Lˆy+ Lˆy[ Lˆy,Lˆx] + [ Lˆz,Lˆx]Lˆz+ Lˆz[ Lˆz,Lˆx],= − ihLˆzLy− ihLˆyLˆz+ ihLˆyLz+ ihLˆzLˆy,= 0.The proofs for the remaining two commutators are analogous. This completes the problem.)4. Prove the following commutation relations:[ Lˆz,Lˆ±] = ± hLˆ±, [ Lˆ+,Lˆ−] = 2hLˆz.(ANS. Begin with the first commutator and use the results of the previous question.

[ Lˆz,Lˆ−] = [ Lˆ= [ Lˆzz= ihLˆy= − h(Lˆ= − hLˆ,Lˆ,Lˆ− i( −ihLˆ−xxx.−iLˆ−i[ Lˆ−iLˆThe proof for [ Lˆz,Lˆ−] is analogous. Next we show the second commutator[ Lˆ ,Lˆ ] = [ Lˆ ,iLˆ ,Lˆ − iLˆ ],We are done.)5. Prove thatThen, show thatLˆLˆLˆ+−+ Lˆz= − 2i[ Lˆ= − 2i( ihLˆ= 2hLˆ.[ Lˆzzy,Lˆ2=+ − z z− +zy),yxy],],zx),,Lˆ),yy],− h ) = Lˆ Lˆ + Lˆ ( Lˆ + h ).22[±] =z]Lˆ,Lˆ(ANS. From the definitions of Lˆ and+Lˆ in the previous problem, we can write−1iLˆx= ( Lˆ−+ Lˆ+), Lˆy= ( Lˆ−− Lˆ+),222 2 2 2and substituting these expressions into Lˆ = Lˆ + Lˆ + Lˆ we obtain[ Lˆ2 12 12 2Lˆ = ( Lˆ−+ Lˆ+) − ( Lˆ−− Lˆ+) + Lˆz,441 22= ( Lˆ−+ Lˆ−Lˆ++ Lˆ+Lˆ−+ Lˆz),41 22 2− ( Lˆ−− Lˆ−Lˆ+− Lˆ+Lˆ−+ Lˆz) + Lˆz,412= ( Lˆ−Lˆ++ Lˆ+Lˆ−) + Lˆz).2Substituting the commutator [ Lˆ ,Lˆ ] = 2hLˆin either the form+ −zLˆ Lˆ = Lˆ Lˆ + 2hLˆor the+ − − + zform Lˆ Lˆ = Lˆ Lˆ − 2hLˆinto the above equation, yields the desired results− + + − z2Lˆ = Lˆ+Lˆ−+ Lˆz( Lˆz( Lˆz− h ) = Lˆ−Lˆ++ Lˆz( Lˆz+ h ).22The operator Lˆ commutes with Lˆ +, Lˆ , and− Lˆ commutes with all components of angularmomentum. See Question 3.)x,Lˆy= 0zzz

6. Prove that2 k2 kLˆ ( Lˆ ψ ) = j( j 1)h ( Lˆ ψ −− j , j+for any nonnegative integer k such that ( j − k ) h ⎣ S( Lˆz) . Also, let s be a nonnegativeinteger such that ( j−s) h ∈S(ˆ L z ) but ( j− ( s+ 1)) h ∈S( L ˆ z ). Then show thatˆ 2 ( ˆ s, ) ( )( ) 2L L j s j s (ˆ s−ψjj= − − −1h L−ψjj, ),Now, using Eq. (1) and (2), show that the only values for j are 0, 1/2, 1, 3/2, 2, … .2(ANS. To do this Question, we need the results from the previous Question. Since Lˆ commutes Lˆ −, it commutes with any power of Lˆ −. Therefore, we can writeˆ 2 ( ˆ k) ˆ k( ˆ 2L L−ψjj, = L−L ψjj, ),ˆ k= L−( L ˆ − L ˆ + ψjj , + L ˆ z + h) ψjj, ),2 k= jj ( + 1) h (ˆ L−ψjj , ).2For the second equality, we choose the other expression for Lˆ . In addition, we shall needS + 1sss ssthe results (from class) that Lˆ −ψj , j= 0 , and [ Lˆz,Lˆ−] = − shLˆ−or LˆzLˆ−= Lˆ−Lˆz− khLˆ,−where s is a nonnegative interger. Therefore we can write2 sS + 1sLˆ ( L ψ ) = Lˆ Lˆ ψ + Lˆ ( Lˆ − h )( Lˆ ψ )−j , j= Lˆ= Lˆzz+−( Lˆ( Lˆs−Lˆs= ( j − s ) hLˆ( Lˆ ψ= ( j − s ) h(Lˆzzj , j( Lˆ ψs−− shLˆs−Lˆ)) − hLˆ) ψ− shLˆ= ( j − s )( j − s −1)hzj , jzzs−j , jzj , j= ( j − s ) h(j − s ) h(Lˆ ψ−− h(Lˆ) ψj , jLˆ) −(j − s ) hs−z).),s( Lˆ ( Lˆ ψs−j , j( Lˆ ψ2s−zj , j),− shLˆ−(j − s ) h) −(j − s ) hj , j−s−−j , jzj , j( Lˆ ψIn the first equality, set k = s, and then we have2 k2 sj(j + 1)h ( Lˆ ψ ) = ( j − s )( j − s −1)h ( Lˆ ψ −−j , j2s−s−j , j) ψ),s−j , js−( Lˆ ψ2,( Lˆ ψwhich means thatj(j + 1)= ( j − s )( j − s −1).This equation yields the quadratic equation in s2s −(2 j −1)s− 2 j = 0,which has two solutions s = 2j and s = -1. Since s must be a nonnegative integer, the secondroot is extraneous. Since s must be a nonnegative integer, then 2j must be a nonnegativeinteger. Therefore, j = 0, 1/2, 1, 3/2, 2, …)7. (Optional) Prove thatψ1⎡ 1 ( j+m)!⎤21jm , = ⎢( j)!( j−m)!⎥ j−m⎣ 2 ⎦ h− −jj ,(ANS. We start from the result given in classLˆψj mLˆ ψ .2= [( j + m )( j − m )] hψ− j ,m+1j ,m−11.2j , j),j , jj , j),),

If m = j, then we getOperating with Lˆ −again, we find thatLˆ2−ψj , j2Lˆ−ψ, j= [( 2 j )( 1)]hψj , j−j 1= [( 2 j )( 1)]Repeating this k times, we observe that12hLˆψ= [( 2 j )( 2 j −1)(2〈1)]−1j , j−1,12.hLˆψ−j , j−2ˆ kkL ψ = [( 2j)( 2j−1)...( 2j− k+ 1)( k• ... 2•1) 2 h ψ ,−jj , jj , −k1⎡ ( 2j)!( )! ! ,j k k ⎤2k= ⎢jj−k⎣ −⎥ h ψ2 ⎦.Inverting the formula, we haveψ1( j k)!jj , − k = 1 ⎡ 1 2 − ⎤2k ⎢jj ,⎣( 2j)!k!⎥h⎦Now, let k = j - m. Then, the result follows.)1kLˆ h ψ .8. Given the normalized spherical harmonic for l = 2 and m = 2,115 2 2 2Y22e i ϕ, ( θϕ , ) = ⎛ ⎞sin θ ,⎝ 32π⎠derive the other four spherical harmonics for l = 2 using the equation in the previousquestion. What is the difference between j and l?(ANS. It is easier to apply the operator Lˆ , expressed in spherical polar coordinates,−repeatedly to obtain the desired functions. So, sinceˆ −iϕ⎛ ∂L−=− e i cot∂ − ∂ ⎞h ⎜ θ ⎟⎝ θ ∂ ϕ ⎠we haveˆ −i⎛ ∂iL−Y , ( , ) =− e icot sin e ,∂ − ∂ ⎞⎜⎟ ⎛ 1ϕ15 ⎞ 2 2 2 ϕ22 θϕ hθ θ⎝ θ ∂ ϕ ⎠ ⎝ 32 π ⎠1−ii=− ⎛ 15 ⎞ 2 ϕ⎛∂e icot sin e ,⎝ ⎠ ∂ − ∂ ⎞ 2 2 ϕh ⎜ θ ⎟ θ32π⎝ θ ∂ ϕ ⎠1−ii=− ⎛ 15 ⎞ 2 ϕ 2 ϕ2 2iϕhe( 2sinθe − icotθsin θ( 2i) e ),⎝ 32π⎠1=− ⎛ 15 ⎞ 2iϕh4sinθcosθe⎝ 32π⎠= 2hY21, ( θφ , ).Therefore, solving for Y2 , 1(θ , φ ) , we obtain.

115 2Y21, ( θφ , ) =− ⎛ ⎞sinθcosθe iκ⎝ 8π⎠Following the same process, we getˆ −i⎛ ∂iL−Y , ( , ) =− e icot sin cos e ,∂ − ∂ ⎞⎛⎜⎟ − ⎛ 1⎞ϕ15⎞ 2 2 ϕ21 θϕ hθ θ⎝ θ ∂ ϕ ⎠⎜⎝ 32 π ⎠⎟⎝⎠1−iϕ i i=− e⎛ 15 ⎞ 2 2 2 ϕ ϕh ((cos θ−sin θ) e −icotθsinθcos θ( i) e ),⎝ 8π⎠1= ⎛ 15 ⎞ 2 2 2h ( 2cos θ− 1+cos θ),⎝ 8π⎠115 2 2= h⎛ ⎞( 3cos θ −1),⎝ 8π⎠= 6hY20, ( θϕ , ).Solving for Y2 , 0(θ , ϕ ) , we have15 2 2Y 20 , ( θφ , ) = ⎛ ⎞( 3cos θ −1).⎝16π⎠For Y2 , 1(θ , ϕ ) , we proceedˆ −i⎛ ∂L−Y , ( , ) =− e icot ( cos )∂ − ∂ ⎞⎜⎟ ⎛ 1ϕ5 ⎞ 2 220 θϕ hθ 3 θ −1⎝ θ ∂ ϕ ⎠⎝16π ⎠1i=− ⎛ 5 ⎞ 2 ϕh e ( −6sinθcos θ),⎝16π⎠1−i= ⎛ 5 ⎞ 2ϕh 6sinθcos θe,⎝16π⎠= 6hY2,−1( θϕ , ).Solving for Y2 , 1(θ , ϕ ) , we have15 2Y21, ( θφ , ) sin cos e i ψ= ⎛ ⎞−θ θ⎝ 8π⎠Finally, for Y2 , − 2(θ , ϕ ) , we find that

Hence,ˆ −iϕ⎛∂−iL−Y , ( , ) =− e icot sin cos e ,∂ − ∂ ⎞⎜⎟ ⎛ 5 ⎞ 2ϕ21 θϕ hθ θ θ⎝ θ ∂ ϕ ⎠⎝8 π ⎠=− ⎛ 5h⎞⎝ 8π⎠112 2 2iϕ −iϕ −iϕe ((cos θ −sin θ) e icotθsinθcos θ( −i) e ),= ⎛ 15 ⎞ 2 2−iϕh (cos θ−sinθ−cosθ) e ,⎝ 8π⎠= ⎛ 15h⎞⎝ 8π⎠= 2hY2,−212 2 2− iϕsin θ e ,( θϕ , ).15 2 2 −2iY2, −2( , ) = ⎛ ⎞ϕθϕ sin θ e⎝ 32π⎠This process is somewhat laborious but handy if tables of functions are unavailable.)9. In general, it can be shown that for two operators Âand Bˆ , the uncertainty product is1 3( ∆A)( ∆B) ≥ ∫ d rΨ*[ A,ˆ B] Ψ .2What is the uncertainty product for ∆ L ∆L? What is the uncertainty in Lz ?xy(ANS. Suppose that we choose Ψ to be an eigenstate of L ,Y ( θ , ϕ ). Then, we see that1 3 *∆Lx∆Ly≥∫ d rYjm , ( θϕ , )[ˆ Lx, L ˆ y] Yjm, ( θϕ , ),21 3 *≥ ∫ drYjm , ( θϕ , ) iLY h ˆx jm , ( θϕ , ),21 2≥ i m =m 2hh2 2We also have ∆Lz= 0.)11zj , m