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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 spin1E 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 sconiponents simply
add SIXLXZ = {39’+ﬁililxiri=t5§1ixilxr+xt391nl
= 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 denialaw“
it: m = D: Iii9's
it: in = —l. 3‘tpntuterii'uitlw.garmentstatesotherewon'theanyetatitres 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 :_ ﬁfeo
' 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} = EiTlHl: 5=ﬂiﬁillsiell [41T3] {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” ~Smlﬂl = ﬁgéo H — ti +2 ti — to = Elms E4150]
and 2 s
It 1 3h
gt”. :2: =__ _ _ =_._
3 . Iﬂﬂl 4 ﬁt: «LT H EliL + H] 4 lﬂﬂl {4.1311
Returning to Equation 4.1”}? (and again using Equation 4.142}. we conclude that
n2 it1 s2 1
sﬂioi=(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! MOMENTI.
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 [Bi}, 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 {BI5] and [BrI5}: .' i w'" st = s; + s: + zslzsl, + as. + sis1+ [315] F 9x 1 L53, The fear vectors  e1, ez } are eigenveeters et' 5%, 5:, SI, and Sh [fennulas {Ii2}], and the action of the operators 31: and SH can be derived i'rem fennulas [iiiT} of
chapter IX. We therefore ﬁnd: ' _5r:§{£p+5)
2 3 32 I . a 
Si+’+>=(Ttﬁz+Eﬁ)l+*+>+§ﬁ2i+'+} :55"??ng
=2a3+,+} [Bl?—a}_
ssl+_>=(gﬁl+§ﬁ2)i+_)—lﬁji+}+ﬁ1]—+}
' 4. 4 ' 2 ’ *
zﬁlff+,—}+i—,+}] t'BITb] Slim +§=Gﬁz +ghz)f. +>*%ﬁ3!—. +} +a= +.—) = a1[[ —, + ) +i +. — }] iBITCJ
S’J,—>=G&= +§ﬁ1)[. > +és1I,—>
= Eli _, _ > {BITHI” 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 giten in {BI}, is therefore: 2tl 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 nonzero matrix elements only,r between eigentreetors ofS, associated withtllesameeigenvaiue.lteeordingtotheretultsof §2,the
onlynondiagonalelementsofszwhiehmuldbediﬁ'erentfromzeromthoae
whjehrelatei+.—}to—,+>. b. EIEEHVALLIES AND EIGEHVECTIJIIS DF 3: Matrix {318} ean be broken down into three submatriees {as shown by the
dottedtinesLTwoot'themareonedimenaionai: rite veernrsl+. +}anrl —. —}
are eigenveetors of 5", as is also shown by relations {Ii11:1] and [ll«1741). The
minted eigenvalues are both equal to 231. Wemustaowdiagonaiizethel x isubmattilt: ts’le =a=G D ' {1319) 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—Zla}
in +.  > —  —. + }] tor the eigenvalue tl {B—llh] 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 nondegenerate and corresponds to vector (32 lb]. The sooonrl one
ii threefold degenerate, nntl tlre treetors+,+},1—,— ) and lBEla] form an
Wotma] 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 {Elll}. The quantum number 5 of {Bi[243} can take on two values: it and l. The
ﬁrst one is associated with a single vector, iEllb}, 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)+—.+>] IL1>=1—.—> [Mu It can easii},r be shown that the four vectors  S, M } given in {3'22} and {323}
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 coercedlite with S. COMMENTS: it} The l'arnii},r [ll23] 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 [323] remain invariant, while
ill22] 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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 Spring '10
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