Angular Momentum 1 Angular Momentum In Quantum Mechanics

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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 2l 1 we arrive at (1.20).Now we are ready to do the integral in (1.18). Since the limits of integrationare 0 and π, the first term on the right side of (1.19) always vanishes, and we canignore it. Then we haveZ πZ π2l(sin x)2l 1 dx (sin x)2l 1 dx2l 1 00 Z π2l2l 2 (sin x)2l 3 dx2l 12l 10 Z π2l2l 22l 4 (sin x)2l 5 dx2l 12l 12l 30 2l 22l 42l ··· 2l 12l 12l 3 Z π 2l (2l 2)sin x dx ··· 2l (2l 3)0 2l l(l 1)(l 2) · · · (l (l 1))2(2l 1)!! 2(2l l!)22l l! 2(2l 1)!!(2l 1)!10(1.21)

Rπwhere we used 0 sin x dx 2 and (1.20).Using this result, (1.18) becomes4π clland hencecll ( 1)l2(2l l!)2 1(2l 1)! (2l 1)!4π 1/212l l!(1.22)where we included a conventional arbitrary phase factor ( 1)l . Putting this alltogether, we have the top orbital angular momentum state 1/2 1ll (2l 1)!(sin θ)l eilφ .(1.23)Yl (θ, φ) ( 1)4π2l l!To construct the rest of the states Ylm (θ, φ), we repeatedly apply L from equation (1.17) to finally obtain 1/2 1/2 1(l m)!(2l 1)!Ylm (θ, φ) ( 1)l4π2l l! (2l)!(l m)! eimφ (sin θ) mdl m(sin θ)2l .d(cos θ)l m(1.24)It’s just not worth going through this algebra also.2SpinIt is an experimental fact that many particles, and the electron in particular, havean intrinsic angular momentum. This was originally deduced by Goudsmit and Uhlenbeck in their analysis of the famous sodium D line, which arises by the transitionfrom the 1s2 2s2 2p6 3p excited state to the ground state. What initially appears as astrong single line is slightly split in the presence of a magnetic field into two closelyspaced lines (Zeeman effect). This (and other lines in the Na spectrum) indicates adoubling of the number of states available to the valence electron.To explain this “fine structure” of atomic spectra, Goudsmit and Uhlenbeckproposed in 1925 that the electron possesses an intrinsic angular momentum inaddition to the orbital angular momentum due to its motion about the nucleus.Since magnetic moments are the result of current loops, it was originally thoughtthat this was due to the electron spinning on its axis, and hence this intrinsic angularmomentum was called spin. However, a number of arguments can be put forth todisprove that classical model, and the result is that we must assume that spin is apurely quantum phenomena without a classical analogue.As I show at the end of this section, the classical model says that the magneticmoment µ of a particle of charge q and mass m moving in a circle is given byqµ L2mc11

where L is the angular momentum with magnitude L mvr. Furthermore, theenergy of such a charged particle in a magnetic field B is µ · B. Goudsmit andUhlenbeck showed that the ratio of magnetic moment to angular momentum of theelectron was in fact twice as large as it would be for an orbital angular momentum.(This factor of 2 is explained by the relativistic Dirac theory.) And since we knowthat a state with angular momentum l is (2l 1)-fold degenerate, the splittingimplies that the electron has an angular momentum /2.From now on, we will assume that spin is described by the usual angular momentum theory, and hence we postulate a spin operator S and corresponding eigenstates s ms i such that[Si , Sj ] iεijk Sk2(2.1a)2S s ms i s(s 1) s ms i(2.1b)Sz s ms i ms s ms ipS s ms i s(s 1) ms (ms 1) s ms 1i .(2.1c)(2.1d) χi c ẑ i c ẑ i .(2.2a)(I am switching to the more abstract notation for the eigenstates because the statesthemselves are a rather abstract concept.) We will sometimes drop the subscript son ms if there is no danger of confusing this with the eigenvalue of Lz , which wewill also sometimes write as ml . Be sure to realize that particles can have a spin sthat is any integer multiple of 1/2, and there are particles in nature that have spin0 (e.g., the pion), spin 1/2 (e.g., the electron, neutron, proton), spin 1 (e.g., thephoton, but this is a little bit subtle), spin 3/2 (the ’s), spin 2 (the hypothesizedgraviton) and so forth.Since the z component of electron spin can take only one of two values /2, wewill frequently denote the corresponding orthonormal eigenstates simply by ẑ i,where ẑ i is called the spin up state, and ẑ i is called the spin down state.An arbitrary spin state χi is of the formIf we wish to think in terms of explicit matrix representations, then we will writethis in the form(z)(z)χ c χ c χ .(2.2b)Be sure to remember that the normalized states s ms i belong to distinct eigenvaluesof a Hermitian operator, and hence they are in fact orthonormal.To construct the matrix representation of spin operators, we need to first choosea basis for the space of spin states. Since for the electron there are only two possiblestates for the z component, we need to pick a basis for a two-dimensional space,and the obvious choice is the standard basis" #" #10(z)(z)χ andχ .(2.3)0112

With this choice of basis, we can now construct the 2 2 matrix representation ofthe spin operator S.Note that the existence of spin has now led us to describe the electron by amulti-component state vector (in this case, two components), as opposed to thescalar wave functions we used up to this point. These two-component states arefrequently called spinors.(z)The states χ were specifically constructed to be eigenstates of Sz (recall thatwe simultaneously diagonalized J 2 and Jz ), and hence the matrix representationof Sz is diagonal with diagonal entries that are precisely the eigenvalues /2. Inother words, we have hẑ Sz ẑ i 2so that"#0 1Sz .(2.4)2 0 1(We are being somewhat sloppy with notation and using the same symbol Sz todenote both the operator and its matrix representation.) That the vectors definedin (2.3) are indeed eigenvectors of Sz is easy to verify:"" #"" ##" ##" #0010 1 1 1 0and. 2 0 12 02 0 12 101To find the matrix representations of Sx and Sy , we use (2.1d) together withS Sx iSy so that Sx (S S )/2 an

Angular Momentum 1 Angular momentum in Quantum Mechanics As is the case with most operators in quantum mechanics, we start from the clas-sical definition and make the transition to quantum mechanical operators via the standard substitution x x and p i . Be aware that I will not distinguish

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