《无机化学 Chemical Theory》课程教学资源（讲稿）Part I Basic concepts of Quantum mechanics Chapter 4 Angular Momentum

4.1Angular MomentumClassicallytheangularmomentumofanAPisolatedsystemis a constant ofthemotionrsineQuantummechanicallythismeansthatweYexpecttobeabletofind statesofdefiniteangular momentum.In three dimensions, the angular momentum about a point is the magnitude p of themomentum multiplied by theperpendiculardistance of the momentumvectorfromthe point (r sin inthediagram)The angularmomentum describesrotation about an axis perpendicular totheplane containing r and p, so in vector notation it is J= r X p

4.1 Angular Momentum Classically, the angular momentum of an isolated system is a constant of the motion. Quantum mechanically, this means that we expect to be able to find states of definite angular momentum. In three dimensions, the angular momentum about a point is the magnitude p of the momentum multiplied by the perpendicular distance of the momentum vector from the point (r sin θ in the diagram). The angular momentum describes rotation about an axis perpendicular to the plane containing r and p, so in vector notation it is J = r × p

4.2AngularmomentumoperatorsThe angular momentum is the vector product J= r X p. That is,J, = yp. -zPyJ,= zP-XPJ. = xPy - yPxMaking the usual substitutions yields the operatorsaaJ, =-in(vo, -2o,)=-ihayCj, = -in(za, - xa.)FC-tj, =-in(xa, - yo.)y

4.2 Angular momentum operators The angular momentum is the vector product J = r × p. That is, Making the usual substitutions yields the operators x z y y x z z y x J yp zp J zp xp J xp yp       ˆ ˆ ˆ x y z J i y z z y J i z x x z J i x y y x                                              ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ x z y y x z z y x J i y z J i z x J i x y               

Angular momentum operators don't commuteThe components of the angular momentum operator do not commute with each otherWriting a/ax = a,for brevity, and remembering that a, x = 1 + xax, but a, y = y, anda,o,=a,ox,j,j,w=(-in)(va. -z,)(-in)(zo, -xo.)y--h2 (vo,z0, -yo.xo. -2z0,20+ +z0,xo.)y= -h? (vo, + yza,a, - xya.a. -2a,a + xza,a.)yj,j=(-in)(-o, -xa.)(-in)(vo. -z,)y=-h2 (2o.yo. -xo.yo. -z0,20, +xo,20,)y=-h2(yz0,a. -xya.a,-2a,a, +xa, + xza.a,)ywe find thatJx,j,=jj,-j,j,=-h2(vor-xa,)=inj

Angular momentum operators don’t commute The components of the angular momentum operator do not commute with each other. Writing 𝜕/𝜕𝑥 = 𝜕መ 𝑥 for brevity, and remembering that 𝜕መ 𝑥 𝑥 = 1 + 𝑥𝜕መ 𝑥 , but 𝜕መ 𝑥 𝑦 = 𝑦𝜕መ 𝑥 and 𝜕መ 𝑥𝜕መ 𝑦 = 𝜕መ 𝑦𝜕መ 𝑥 ,          2 2 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ x y z y x z z x z z y x y z x z x z z y x y z J J i y z i z x y z y x z z z x y yz xy z xz                                                   2 2 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ y x x z z y x z z z x y z y x z z z x y y z y J J i z x i y z z y x y z z x z yz xy z x xz                                            2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ , x y x y y x x y z   J J J J J J y x i J           we find that

Commutation relationsforangularmomentumWehavefoundthat7[J,j,]=jj,-j,j,=-ha1[,j,]-injSimilarly,[3.,j,]=inj,[3,j,]-inj,(Note that x, y and z appear in cyclic order in these equations.)The uncertainty principle tells us that, for example,N,N,([.j,J)-.)so in general we cannot find wavefunctions that are simultaneously eigenfunctions of two ormore of Jx , J, and Jz . The only exception is that it is possible to find wavefunctions forwhich Jx, J, and J, are all exactly zero

Commutation relations for angular momentum We have found that Similarly, (Note that x, y and z appear in cyclic order in these equations.) The uncertainty principle tells us that, for example, so in general we cannot find wavefunctions that are simultaneously eigenfunctions of two or more of 𝐽መ 𝑥 , 𝐽መ 𝑦 and 𝐽መ 𝑧 . The only exception is that it is possible to find wavefunctions for which Jx , Jy and Jz are all exactly zero. 2 ˆ ˆ ˆ ˆ ˆ ˆ , ˆ ˆ ˆ , x y x y y x x y z J J J J J J y x x y J J i J                         ˆ ˆ ˆ , y z x   J J i J    ˆ ˆ ˆ , z x y   J J i J    1 1 ˆ ˆ , 2 2 x y x y z     J J J J J    

However, Jx, J, and J, all commute with j? = J,2 + j,2 + j2. For example,[.,J3]-J.3?-j2j-JSS-JJJ+JJ-JS[3,j]j,+j[,j]=in(j,j,+j,j)and similarly [z,J,2] = -in(JJx +J,J,), while [Jz,J2] = 0Adding these results together shows that [1 z, J2] = 0Therefore we can find wavefunctions that are eigenfunctions of both f2 and oneonly ofjx , J, and Jz . It is customary to choose Jz

However, 𝐽መ 𝑥 , 𝐽መ 𝑦 and 𝐽መ 𝑧 all commute with 𝑱෠2 = 𝐽መ 𝑥 2 + 𝐽መ 𝑦 2 + 𝐽መ 𝑧 2 . For example, and similarly 𝐽መ 𝑧 ,𝐽መ 𝑦 2 = −𝑖ℏ 𝐽መ 𝑦 𝐽መ 𝑥 + 𝐽መ 𝑥 𝐽መ 𝑦 , while 𝐽መ 𝑧 ,𝐽መ 𝑧 2 = 0. Adding these results together shows that 𝐽መ 𝑧 , 𝑱෠2 = 0. Therefore we can find wavefunctions that are eigenfunctions of both 𝑱෠2 and one only of 𝐽መ 𝑥 , 𝐽መ 𝑦 and 𝐽መ 𝑧 . It is customary to choose 𝐽መ 𝑧 .   2 2 2 ˆ ˆ ˆ ˆ ˆ ˆ , ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ , , ˆ ˆ ˆ ˆ z x z x x z z x x x z x x z x x x z z x x x z x y x x y J J J J J J J J J J J J J J J J J J J J J J J J i J J J J                      

SphericalpolarcoordinatesBefore proceeding, we recall the definition of spherical polar coordinates. They aredefined byx = r sindcos(py=rsinosingpz = r cosoand converselyr =/x?+y+??Q = arccos(= / r)β = arctan(y / x)The volume element for integration over spherical polar coordinates isdV=r?sinodrdodpForgetting the r?sino is a very common source of mistakes

Spherical polar coordinates Before proceeding, we recall the definition of spherical polar coordinates. They are defined by and conversely The volume element for integration over spherical polar coordinates is Forgetting the r 2 sinθ is a very common source of mistakes. sin cos sin sin cos x r y r z r             2 2 2 arccos / arctan / r x y z z r y x        2 d sin d d d V r r    

aaaOra00adaxaxa0axadaraxOrOr2x=2rsin0cosd=sinOcosΦOxOxaa(sind+cotecosda0adaCiicoscotsin00Oda=-inInvolved only 0 and Φada2a1+cote-h000sin?a

y z y z y z , , , r x x r x x                                        ˆ (sin cot cos ) x J i            , 2 2 2 sin cos , sin cos y z r r r x r x x                        ˆ (cos cot sin ) y J i             ˆ z J i      2 2 2 2 2 2 1 ˆ ( cot ) sin J                Involved only  and 

V=-V(r)Acentral forceh?v? +V(r)2ma2P21f11Or.2forfrr sin'ofof0a22a1↑2Or2r?h?r Or[H,3,]=0H,j2=01aj2aj.+-,H)=0I.12,H)=0dtatatindtinh

A central force V =V(r) 2 2 ˆ ˆ ˆ ( ) 2 H T V V r m       Ñ 2 = ¶ 2 ¶r 2 + 2 r ¶ ¶r + 1 r 2 ¶ 2 ¶q 2 + 1 r 2 cotq ¶ ¶q + 1 r 2 sin 2 q ¶ ¶f 2 2 2 2 2 2 2 1 ˆJ r r r r        2 ˆ ˆ   H J, 0    z ˆ ˆ   H J, 0    2 2 2 ˆ 1 ˆ ˆ [ , ] 0 d J J J H dt i t      ˆ ˆ 1 ˆ ˆ [ , ] 0 z z z d J J J H dt i t     

4.3SphericalharmonicsIn spherical polar coordinates, j, = -in %. J, commutes with j2, so we can find functions thatare eigenfunctions of both. Eigenfunctions of J, satisfy the eigenvalue equationj.y=-ih-y=kyaThe unnormalised solutions are of the form exp(ik/h), but the value of k is restricted by therequirement that the wavefunction is single-valued - that is, ( + 2r) must be the same as(p). This means thatexp(ik(β + 2元) / h) = exp(ikp / h)Thus k = Mh, where M is an integer (positive, negative or zero), and the wavefunctionsbecome(afternormalisation)1exp(iMp)VM2元Theeigenvalue is Mh :the angular momentum is an integer multiple ofh

In spherical polar coordinates, 𝐽መ 𝑧 = −𝑖ℏ 𝜕 𝜕𝜑. 𝐽መ 𝑧 commutes with 𝑱෠2 , so we can find functions that are eigenfunctions of both. Eigenfunctions of 𝐽መ 𝑧 satisfy the eigenvalue equation The unnormalised solutions are of the form exp(𝑖𝑘𝜑/ℏ), but the value of k is restricted by the requirement that the wavefunction is single-valued — that is, 𝜓(𝜑 + 2𝜋) must be the same as 𝜓(𝜑). This means that Thus 𝑘 = 𝑀ℏ, where M is an integer (positive, negative or zero), and the wavefunctions become (after normalisation) The eigenvalue is 𝑀ℏ : the angular momentum is an integer multiple of ℏ. ˆ z J i k          exp 2 / exp / ik ik            1 exp 2 M   iM   4.3 Spherical harmonics

Eigenfunctions of f2J2 is more complicated:aa2a112 -一sinsin? osingaeaoTo obtain eigenfunctions of f2 we have to multiply the functions eiMo by suitable functions of0. WriteYM = OM(0)M(β)where Φm(p) = eiMs, Then the eigenvalue equation J2 Yjm = AY jm becomesM?aaJ'Yjm =-h?sine0M(0)ΦM(p)=a0JM(0)Φ(p)singaoaosin?0We can cancel out Φm(p) to get an eigenvalue equation in 0. The eigenvalues turn out to beh?JU +1)for integerJ.The functions Y μm(0, β) are spherical harmonics

Eigenfunctions of 𝑱 ෠𝟐 𝑱෠2 is more complicated: To obtain eigenfunctions of 𝑱෠2 we have to multiply the functions e 𝑖𝑀𝜑 by suitable functions of θ. Write where Φ𝑀(𝜑) = e 𝑖𝑀𝜑 . Then the eigenvalue equation 𝑱෠2𝑌𝐽𝑀 = 𝜆𝑌𝐽𝑀 becomes We can cancel out Φ𝑀(𝜑) to get an eigenvalue equation in θ. The eigenvalues λ turn out to be ℏ 2 𝐽(𝐽 + 1) for integer J. The functions 𝑌𝐽𝑀(𝜃,𝜑) are spherical harmonics. 2 2 2 2 2 1 1 ˆ sin sin sin                      J ( ) ( ) YJM JM M      2 2 2 2 1 ˆ sin ( ) ( ) ( ) ( ) sin sin JM JM M JM M M Y                             J