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Unformatted text preview: éq‘fflé 4.4.3 Addition of Angular Memento F" I; p Suppose nonI that we have two spin-1E particles—for example. the electron and the proton hi the ground state“ of hydrogen. Each can have spin up orspin down. so there are four possibilities in all“: it ti. it. ti. [4.1m where the first arrow refers to the electron and the second to the proton. Question: What is the total angular momentum of the atom? Let s g s“J + sat. _ [4.11s] Each of the four composite states is an eigenstate of SI—tlte s-coniponents simply add SIXLXZ = {39’+fiililxiri=t5§1ixilxr+x|t391nl = ifimixilxz+xiifimrxfl =fiIEMi +milxlx:. [note that 5‘” acts only on in. and 5'” acts only on xi]. So in (the quantum number for the com ites stem is‘ustm in: 13'5"S 3" l J l+ 1 TT=/+EJ+3> TTIM = l: Ham = it; self denial-aw“ it: m = D: Iii-9's it: in =- —l. 3‘tpntuterii'uitlw.garmentstatesotherewon'theanyeta-titres angular ramhentantowenyahuot. At first glance. this doesn't look right: this supposed to advance in integer steps. from —.r to +3. so it appears that s = l—but there is an extraatatewithni = II}. One way to untangle this problem is to apply the lowering operator 3. = SE” + SE} tuthestate T't. using Equation 4.145: 5 :-_ fife-o '- t 9 3—HT} = iii” n t + t is?“ ii a! = oni+toti=ttit+tn Leila» Evidently the three states withs = l are {in thenotation Intel}: 5* 5 5‘: 1:51} [Hit tt no II—ll fittt + it} s =]{trlp1ot}. [4.111] it {As a check. try applying the lowering operator to II it}; what should you get'E' See Problem 4.?- This is called the triplet combination. for the obvious reason. Meanv while.theort ogonalstatewithm = [Icarriess =0: 1 {run} = EiTl-Hl: 5=flifiillsiell [4-1T3] {If you apply the raising or lowering operator to this state. you'll get zero. See Prob— 1em'435.) . . I claim. then. thatthe combination of two spin—HZ particles cancerry a total spin of l or (l. depending on whether they ooeupy the triplet or the singlet oonfiguratlnn. _ To confirm this. I need to prose that the triplet states are eigenvectors of 59 with eigenvalue Eli2 and the singlet is an eigenveotor of 5'2 with eigenvalue (J. New s1 = {silt + at”; ~ is“: + salt I tsflh2 + remit + as“:I - st1t . [4.1T9] q. Using Equations stifled 4.14.5, weliave fit =- if: {ff} etc.) 5‘” ~Smtttl = in“ no}? ti + tit” tltSf’ ti + tit” no? ti = (it)(it)+(it)(%it)+(it)(‘it) i = ”In it - iii. Similarly. It! . sf‘t smut} = It: it — tti- ltfollows that 2 1:1 st” ~Sm|lfll = figéo H — ti +2 ti — to = Elms E4150] and 2 s It 1 3h gt”. :2: =__ _ _ =_._ 3 . Iflfll 4 fit: «LT H Eli-L + H] 4 lflfll- {4.1311 Returning to Equation 4.1”}? (and again using Equation 4.142}. we conclude that n2 it1 s2 1 sfl|ioi=(T+—4—+2E)|iui=2i no}. [4.132] so :1 u} is indeed an eigenstate of s2 with eigenvalue 2F; and is2 set n2 Salflfl} —( 4 + 4 2 4 )Iflfl} — I], [4.133] so ll} I1} is an eigenstate of .5"1 with eigenvalue I]. {I will leave it for you to confirm that J] I} and ”—11 areeigenstates old“. with the appropriate eigenvalue—see Proh- lero 4.35.} What we have just done {combining spin In with spin ME to get spin 1 and spin Ellis the simplest example of a larger problem: 11' you combine spin :1 with spin .93. what total spins s can you get?“ The answer” is that you get every spin from [31 +ng down tots. — Sal—mist; - s1}. ifs; :- si—in integer steps: t=tai+iii. tsi+e— ll, {31+52—2]. _ n. fist. [4.134] {Roughly speaking. thehigbest total spin occurswhenthe individual spins are aligned parallelto one another. andthe lowest occurs when they areantiparallel.i Qbh“—rm J ‘1‘: CHAPTER I ADDI'I'IDH 'EIIF ANGELA! MOMENT-I. These results appear clearlyr in the matrix which represents 3: in the { i s, . a: > } basis. Choosing the basis vectors in the order indicated in [B-i}, that. matrix can be written : is.) = 1: DEED GDDD 99:} Fax“ 3:“ LN" 3'" ”MM" ‘7‘ \3 '1“ W W I I} [I U —1 fifis’a fSéfi 3. Diagen alization of 53 - All that remains to he done is in find and then diagonalize the matrix which represents 81 in the { fe,. s2 } } basis. We knew in advance that it is net diagonal. since 52 does not commute with S 1, and 31,. 2. L. 5‘ 45: +3.) a. CALCULATION OF THE MATH”! HEFHESEHTIHG 5‘ We are going to app];-r S] to each of the basis 1tweeters. To do this, we shall use = 5- :2 females {BI-5] and [Br-I5}: .' i w'" st = s; + s: + zslzsl, + as. + sis1+ [3-15] F 9x 1 L53, The fear vectors | e1, ez } are eigenveeters et' 5%, 5:, SI, and Sh [fennulas {Ii-2}], and the action of the operators 31: and SH can be derived i'rem fen-nulas [iii-T} of chapter IX. We therefore find: ' _5r:§{£p+5-) 2 3 32 I .- a - Si+’+>=(Ttfiz+Efi)l+*+>+§fi2i+'+} :55"??ng =2a3|+,+} [B-l?—a}_ ssl+_>=(gfil+§fi2)i+_)—lfiji+-}+fi1]—+} ' 4. 4 ' 2 ’ * zfilff+,—}+i—-,+}] t'B-IT-b] Slim +§=Gfiz +ghz)f-. +>*%fi3!—. +} +a=| +.—) = a1[[ —, + ) +i +. — }] iB-IT-CJ S’J-,—>=G&= +§fi1)[-.- > +és1I-,—> = Eli _, _ > {B-ITHI” iflflfi I. ADDI‘HOH “IF THE IP'IH 'I-‘I'I The matrix representing 5’ in the basis of the four vectors Ie,. it2 ), arranged in the order git-en in {BI}, is therefore: 2-tl tl tt o'l tie {51}an i : [n.ls] {1:1 1:0 tl tl ail CDHHENTI The zeros appearing in this matrilt were to he expected. 51 commutes With 3,, and therefore has non-zero matrix elements only,r between eigentreetors ofS, associated withtllesameeigenvaiue.lteeordingtotheretultsof §2,the onlynon-diagonalelementsofszwhiehmuldbedifi'erentfromzeromthoae whjehrelatei+.—}to|—,+>. b. EIEEHVALLIES AND EIGEHVECTIJIIS DF 3: Matrix {3-18} ean be broken down into three submatriees {as shown by the dottedtinesLTwoot'themareone-dimenaionai: rite veernrsl+. +}anrl| —. —} are eigenveetors of 5", as is also shown by relations {Ii-11:1] and [ll-«1741). The minted eigenvalues are both equal to 231. Wemustaowdiagonaiizethel x isubmattilt: ts’le =a=G D ' {13-19) Whidlrepresentssj insidethetwo—rlimensional subspaeeSpannedhy I +, - )and |-. + it, that is, the eigensubspaee or a, corresponding to M = tl. roe eigen- value lfiz of matrix (Bel?) can be obtained by solving the characteristic equation: [l—lF—l=tl (are; Themotsofthisequarionarel =flantii. :- 2.1'hisyieldstllelasttwoeigemraiues I151 : fl and 21?. Art elementary calculation yields the corresponding eigenvectors: 71—2‘[f+,—}+|—.+}] fortilleeige'nltaltleZfi2 [B—Zl-a} in +. - > — | —. + }] tor the eigenvalue tl {B—ll-h] Simon. theyaredefined onlytowithin a global phase tnetor; tneooetseteno l.- o": ”lure their normalization}. The Operator 5: therefore possesses two distinct eigenvalues : Ill and 21?. The 1318!: one is non-degenerate and corresponds to vector (3-2 l-b]. The sooonrl one ii threefold degenerate, nntl tlre treetors|+,+},1—,— ) and lB-El-a] form an Wot-ma] basis in the associated eigensttbepaee. 1W? CHAPTER ll ADDITION [IF AMBULRH MOMENT! —————'—''_'''''———‘—_—-"—"‘—‘——-—-—.__II 4. Results: triplet and singlet Thus we have obtained the eigenvalues of 33 and 5,. as well as a system of eigenvectors common to these two observables. We shall summarise these result; by expressing them in the notation of equations {El-ll}. The quantum number 5 of {Bi-[243} can take on two values: it and l. The first one is associated with a single vector, iE-ll-b}, which is also an eigenvectm of 5', with the eigenvalue fl, since it isa linear combination of | +. — j» and | —, + );i we shall therefore denote this vector by | ll, il b: l I}. I} = _: .ii.t _ _ _, + 342. I } J; [i 3* I P] i } Three vectors which difiet by their values of M are associated with the value .5' - l : [1.1) =i+,+} 11,0} =—12-[|+.e)+|—.+>] IL-1>=1—.—> [Mu It can easii},r be shown that the four vectors | S, M } given in {3'22} and {3-23} form an orthonormal basis. Specification of S and M suffices to define uniquely a vector of this basis. From this, it can be shown that 51 and 5', constitute a C.S.C.{J. [which could include Sf and 5:. although it is not necessary here}. Therefore, when two spits lii‘s isI = s, = [3'2]: are added, tire another 5' which characterizes the eigenvalues Sis + Hit] of the observable 5‘ can be ennui either to i or to I}. With sari: of these two voices of S is associated a fourth of (23 + i J orthogomri rectors [three for .5' =- i, one for 5' = Ill correspontihig to tile {23 + U utilities of M which are coerced-lite with S. COMMENTS: it} The l'arnii},r [ll-23] of the three vectors I l, M > [M = i, U, - ll constitutes what is called a triplet ;. the vector [0, I} ) is called a singlet state. {it} The triplet states are symusrrte with respect to an exchange of two spins. whereas the singlet state is antisymmetric. This means that ii‘each vector | 2,, s2 :- is replaced by the vector i Es: el }, expressions [3-23] remain invariant, while ill-22] changes sign. We shall see in chapter KW the importance of this property when the two particles whose spins are added are identical. Further- more, it enables us to find use right linear combination of i +, — ) and |—, + } which must be associated with J +. + J and f -. — } [elisarltr synt- mettic] in order to complete the triplet. The singlet state, on the other hand, is the antvsvrnrnetric linear combination of | +, — } and I —, + }, which is orthogonal to the preceding one. "Its ...
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