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J. BroidaUCSD Fall 2009Phys 130BQM IIAngular Momentum1Angular momentum in Quantum MechanicsAs is the case with most operators in quantum mechanics, we start from the classical definition and make the transition to quantum mechanical operators via thestandard substitution x x and p i . Be aware that I will not distinguisha classical quantity such as x from the corresponding quantum mechanical operatorx. One frequently sees a new notation such as x̂ used to denote the operator, butfor the most part I will take it as clear from the context what is meant. I will alsogenerally use x and r interchangeably; sometimes I feel that one is preferable overthe other for clarity purposes.Classically, angular momentum is defined byL r p.Since in QM we have[xi , pj ] i δijit follows that [Li , Lj ] 6 0. To find out just what this commutation relation is, firstrecall that components of the vector cross product can be written (see the handoutSupplementary Notes on Mathematics)(a b)i εijk aj bk .Here I am using a sloppy summation convention where repeated indices are summedover even if they are both in the lower position, but this is standard when it comesto angular momentum. The Levi-Civita permutation symbol has the extremelyuseful property thatεijk εklm δil δjm δim δjl .Also recall the elementary commutator identities[ab, c] a[b, c] [a, c]band[a, bc] b[a, c] [a, b]c .Using these results together with [xi , xj ] [pi , pj ] 0, we can evaluate the commutator as follows:[Li , Lj ] [(r p)i , (r p)j ] [εikl xk pl , εjrs xr ps ] εikl εjrs [xk pl , xr ps ] εikl εjrs (xk [pl , xr ps ] [xk , xr ps ]pl )1

εikl εjrs (xk [pl , xr ]ps xr [xk , ps ]pl ) εikl εjrs ( i δlr xk ps i δks xr pl ) i εikl εjls xk ps i εikl εjrk xr pl i εikl εljs xk ps i εjrk εkil xr pl i (δij δks δis δjk )xk ps i (δji δrl δjl δri )xr pl i (δij xk pk xj pi ) i (δij xl pl xi pj ) i (xi pj xj pi ) .But it is easy to see thatεijk Lk εijk (r p)k εijk εkrs xr ps (δir δjs δis δjr )xr ps xi pj xj piand hence we have the fundamental angular momentum commutation relation[Li , Lj ] i εijk Lk .(1.1a)Written out, this says that[Lx , Ly ] i Lz[Ly , Lz ] i Lx[Lz , Lx ] i Ly .Note that these are just cyclic permutations of the indices x y z x.Now the total angular momentum squared is L2 L · L Li Li , and therefore[L2 , Lj ] [Li Li , Lj ] Li [Li , Lj ] [Li , Lj ]Li i εijk Li Lk i εijk Lk Li .Butεijk Lk Li εkji Li Lk εijk Li Lkwhere the first step follows by relabeling i and k, and the second step follows by theantisymmetry of the Levi-Civita symbol. This leaves us with the important relation[L2 , Lj ] 0 .(1.1b)Because of these commutation relations, we can simultaneously diagonalize L2and any one (and only one) of the components of L, which by convention is takento be L3 Lz . The construction of these eigenfunctions by solving the differentialequations is at least outined in almost every decent QM text. (The old book Introduction to Quantum Mechanics by Pauling and Wilson has an excellent detaileddescription of the power series solution.) Here I will follow the algebraic approachthat is both simpler and lends itself to many more advanced applications. Themain reason for this is that many particles have an intrinsic angular momentum(called spin) that is without a classical analogue, but nonetheless can be describedmathematically exactly the same way as the above “orbital” angular momentum.2

In view of this generality, from now on we will denote a general (Hermitian)angular momentum operator by J. All we know is that it obeys the commutationrelations[Ji , Jj ] i εijk Jk(1.2a)and, as a consequence,[J 2 , Ji ] 0 .(1.2b)Remarkably, this is all we need to compute the most useful properties of angularmomentum.To begin with, let us define the ladder (or raising and lowering) operatorsJ Jx iJyJ (J )† Jx iJy .(1.3a)Then we also haveJx 1(J J )2andJy 1(J J ) .2i(1.3b)Because of (1.2b), it is clear that[J 2 , J ] 0 .(1.4)In addtion, we have[Jz , J ] [Jz , Jx ] i[Jz , Jy ] i Jy Jxso that[Jz , J ] J .(1.5a)Furthermore,22[Jz , J ] J [Jz , J ] [Jz , J ]J 2 J and it is easy to see inductively thatkk.[Jz , J ] k J (1.5b)It will also be useful to noteJ J (Jx iJy )(Jx iJy ) Jx2 Jy2 i[Jx , Jy ] Jx2 Jy2 Jzand hence (since Jx2 Jy2 J 2 Jz2 )J 2 J J Jz2 Jz .(1.6a)Similarly, it is easy to see that we also haveJ 2 J J Jz2 Jz .3(1.6b)

Because J 2 and Jz commute they may be simultaneously diagonalized, and wedenote their (un-normalized) simultaneous eigenfunctions by Yαβ whereJ 2 Yαβ 2 αYαβandJz Yαβ βYαβ .Since Ji is Hermitian we have the general result2hJi2 i hψ Ji2 ψi hJi ψ Ji ψi kJi ψk 0and hence hJ 2 i hJz2 i hJx2 i hJy2 i 0. But Jz2 Yαβ 2 β 2 Yαβ and hence we musthaveβ2 α .(1.7)Now we can investigate the effect of J on these eigenfunctions. From (1.4) wehaveJ 2 (J Yαβ ) J (J 2 Yαβ ) 2 α(J Yαβ )so that J doesn’t affect the eigenvalue of J 2 . On the other hand, from (1.5a) wealso haveJz (J Yαβ ) (J Jz J )Yαβ (β 1)J Yαβand hence J raises or lowers the eigenvalue β by one unit of . And in general,from (1.5b) we see thatJz ((J )k Yαβ ) (J )k (Jz Yαβ ) k (J )k Yαβ (β k)(J )k Yαβso the k-fold application of J raises or lowers the eigenvalue of Jz by k units of .This shows that (J )k Yαβ is a simultaneous eigenfunction of both J 2 and Jz withcorresponding eigenvalues 2 α and (β k), and hence we can write(J )k Yαβ Yαβ k(1.8)where the normalization is again unspecified.Thus, starting from a state Yαβ with a J 2 eigenvalue 2 α and a Jz eigenvalue β,we can repeatedly apply J to construct an ascending sequence of eigenstates withJz eigenvalues β, (β 1), (β 2), . . . , all of which have the same J 2 eigenvalue 2 α. Similarly, we can apply J to construct a descending sequence β, (β 1), (β 2), . . . , all of which also have the same J 2 eigenvalue 2 α. However, becauseof (1.7), both of these sequences must terminate.Let the upper Jz eigenvalue be βu and the lower eigenvalue be βl . Thus, bydefinition,andJz Yαβl βl Yαβl(1.9a)Jz Yαβu βu YαβuwithJ Yαβu 0andJ Yαβl 0and where, by (1.7), we must haveβu2 αand4βl2 α .(1.9b)

By construction, there must be an integral number n of steps from βl to βu , sothatβl βu n .(1.10)(In other words, the eigenvalues of Jz range over the n intervals βl , βl 1, βl 2, . . . , βl (βl βu ) βu .)Now, using (1.6b) we haveJ 2 Yαβu J J Yαβu (Jz2 Jz )Yαβu .Then by (1.9b) and the definition of Yαβ , this becomes 2 αYαβu 2 βu (βu 1)Yαβuso thatα βu (βu 1) .In a similar manner, using (1.6a) we haveJ 2 Yαβl J J Yαβl (Jz2 Jz )Yαβlor 2 αYαβl 2 βl (βl 1)Yαβlso alsoα βl (βl 1) .Equating both of these equations for α and recalling (1.10) we conclude thatβ u βl n: j2where j is either integral or half-integral, depending on whether n is even or odd.In either case, we finally arrive atα j(j 1)(1.11)and the eigenvalues of Jz range from j to j in integral steps of .We can now label the eigenvalues of Jz by m instead of β, where the integer orhalf-integer m ranges from j to j in integral steps. Thus our eigenvalue equationsmay be writtenJ 2 Yjm j(j 1) 2 YjmJz Yjm m Yjm .(1.12)We say that the states Yjm are angular momentum eigenstates with angular momentum j and z-component of angular momentum m. Note that (1.9b) is nowwrittenandJ Yj j 0 .(1.13)J Yjj 05

Since (J )† J , using equations (1.6) we havehJ Yjm J Yjm i hYjm J J Yjm i hYjm (J 2 Jz2 Jz )Yjm i 2 [j(j 1) m2 m]hYjm Yjm i 2 [j(j 1) m(m 1)]hYjm Yjm i .We know that J Yjm is proportional to Yjm 1 . So if we assume that the Yjm arenormalized, then this equation implies thatpJ Yjm j(j 1) m(m 1) Yjm 1 .(1.14)If we start at the top state Yjj , then by repeatedly applying J , we can construct allof the states Yjm . Alternatively, we could equally well start from Yj j and repeatedlyapply J to also construct the states.Let us see if we can find a relation that defines the Yjm . Since Yjj is definedby J Yjj 0, we will only define our states up to an overall normalization factor.Using (1.14), we haveppJ Yjj j(j 1) j(j 1) Yjj 1 2j Yjj 1or1Yjj 1 1 J Yjj .2jNext we have(J )2 Yjj 2orpp p2j j(j 1) (j 1)(j 2) Yjj 2 2 (2j)2(2j 1) Yjj 21Yjj 2 2 p(J )2 Yjj .(2j)(2j 1)2And once more should do it:pp(J )3 Yjj 3 (2j)(2j 1)2 j(j 1) (j 2)(j 3) Yjj 3p 3 (2j)(2j 1)(2)(3)(2j 2) Yjj 3or1Yjj 3 3 p(J )3 Yjj .2j(2j 1)(2j 2)3!Noting that m j 3 so that 3! (j m)! and 2j 3 2j (j m) j m, itis easy to see we have shown thats(j m)!mm jYj (J )j m Yjj .(1.15a)(2j)!(j m)!6

And an exactly analogous argument starting with Yj j and applying J repeatedlyshows that we could also writes(j m)!Yjm m j(J )j m Yj j .(1.15b)(2j)!(j m)!It is extremely important to realize that everything we have done up to thispoint depended only on the commutation relation (1.2a), and hence applies to bothinteger and half-integer angular momenta. While we will return in a later section todiscuss spin (including the half-integer case), for the rest of this section we restrictourselves to integer values of angular momentum, and hence we will be discussingorbital angular momentum.The next thing we need to do is to actually construct the angular momentumwave functions Ylm (θ, φ). (Since we are now dealing with orbital angular momentum, we replace j by l.) To do this, we first need to write L in spherical coordinates.One way to do this is to start from Li (r p)i εijk xj pk where pk i ( / xk ),and then use the chain rule to convert from Cartesian coordinates xi to sphericalcoordinates (r, θ, φ). Usingx r sin θ cos φy r sin θ sin φz r cos θso thatr (x2 y 2 z 2 )1/2θ cos 1 z/rφ tan 1 y/xwe have, for example, r θ φ x x r x θ x φ x xz y 3 2 cos2 φr r r sin θ θ x φ sin θ cos φcos θ cos φ sin φ rr θ r sin θ φwith similar expressions for / y and / z. Then using terms such as Lx ypz zpy i y z z ywe eventually arrive at cot θ cos φLx i sin φ θ φ cot θ sin φLy i cos φ θ φLz i . φ(1.16a)(1.16b)(1.16c)7

However, another way is to start from the gradient in spherical coordinates (seethe section on vector calculus in the handout Supplementary Notes on Mathematics) r̂ 1 1 θ̂ φ̂. rr θr sin θ φThen L r p i r i r (r̂ ) so that (since r̂, θ̂ and φ̂ areorthonormal) 1 1L i r r̂ r̂ r̂ θ̂ r̂ φ̂ rr θr sin θ φ 1 θ̂ i φ̂ θsin θ φIf we write the unit vectors in terms of their Cartesian components (again, see thehandout on vector calculus)θ̂ (cos θ cos φ, cos θ sin φ, sin θ)φ̂ ( sin φ, cos φ, 0)then ŷ cos φ ẑ cot θ cos φ cot θ sin φL i x̂ sin φ θ φ θ φ φwhich is the same as we had in (1.16).Using these results, it is now easy to write the ladder operators L Lx iLyin spherical coordinates: .(1.17) i cot θL e iφ θ φTo find the eigenfunctions Ylm (θ, φ), we start from the definition L Yll 0. Thisyields the equation Yll Y l i cot θ l 0 . θ φWe can solve this by the usual approach of separation of variables if we writeYll (θ, φ) T (θ)F (φ). Substituting this and dividing by T F we obtain1 T1 F i.T cot θ θF φFollowing the standard argument, the left side of this is a function of θ only, andthe right side is a function of φ only. Since varying θ won’t affect the right side,and varying φ won’t affect the left side, it must be that both sides are equal to aconstant, which I will call k. Now the φ equation becomesdF ikdφF8

which has the solution F (φ) eikφ (up to normalization). But Yll is an eigenfunction of Lz i ( / φ) with eigenvalue l , and hence so is F (φ) (since T (θ) justcancels out). This means that i eikφ k eikφ : l eikφ φand therefore we must have k l, so that (up to normalization)Yll eilφ T (θ) .With k l, the θ equation becomesdTcos θd sin θ l cot θ dθ ldθ l.Tsin θsin θThis is also easily integrated to yield (again, up to normalization)T (θ) sinl θ .Thus, we can writeYll cll (sin θ)l eilφwhere cll is a normalization constant, fixed by the requirement thatZZ π22Yll dΩ 2π cll(sin θ)2l sin θ dθ 1 .(1.18)0I will go through all the details involved in doing this integral. You are free to skipdown to the result if you wish (equation (1.21)), but this result is also used in otherphysical applications.First I want to prove the relationZZn 11n 1nx cos x sinn 2 x dx .(1.19)sin x dx sinnnRby parts (remember the formula u dv uv RThis is done as an integrationv du) letting u sinn 1 x and dv sin x dx so that v cos x and du (n 1) sinn 2 x cos x dx. Then (using cos2 x 1 sin2 x in the third line)ZZnsin x dx sinn 1 x sin x dxn 1 sinx cos x (n 1) sinn 1 x cos x (n 1)ZZsinn 2 x cos2 x dxsinn 2 x dx (n 1)Zsinn x dx .Now move the last term on the right over to the left, divide by n, and the result is(1.19).9

We need to evaluate (1.19) for the case where n 2l 1. To get the final resultin the form we want, we will need the basically simple algebraic result(2l 1)!! (2l 1)!2l l!l 1, 2, 3, . . .(1.20)where the double factorial is defined byn!! n(n 2)(n 4)(n 6) · · · .There is nothing fancy about the proof of this fact. Noting that n 2l 1 is alwaysodd, we haven!! 1 · 3 · 5 · 7 · 9 · · · (n 4) · (n 2) · n 1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · · · (n 4) · (n 3) · (n 2) · (n 1) · n2 · 4 · 6 · 8 · · · (n 3) · (n 1) 1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · · · (n 4) · (n 3) · (n 2) · (n 1) · nn 1(2 · 1)(2 · 2)(2 · 3)(2 · 4) · · · (2 · n 32 )(2 · 2 ) n!2n 12( n 12 )!.Substituting n

recall that components of the vector cross product can be written (see the handout Supplementary Notes on Mathematics) (a b) i ε ijka jb k. Here I am using a sloppy summation convention where repeated indices are summed over even if they are both in the lower position, but this is standard when it comes to angular momentum. The Levi-Civita .