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3 Angular Momentum and Spin 3 Angular Momentum and Spin In this chapter we review the notions surrounding the different forms of angular momenta in quantum mechanics, including the spin angular momentum, which is entirely quantum mechanical in nature. Some of the material presented in this chapter is taken from Auletta, Fortunato and Parisi, Chap. 6 and Cohen-Tannoudji, Diu and Laloë, Vol. II, Chaps. IX and X. 3.1 Orbital Angular Momentum Orbital angular momentum is as fundamental in quantum mechanics as it is in classical mechanics. In quantum mechanics, when applied to the realms of atoms and molecules, it can come in many different forms and flavours. For example, in molecular quan- tum mechanics the electronic angular momentum (for the motion of electrons about the atomic nuclei) and the rotational angular momentum (for the motions of nuclei about the molecular centre of mass) arise in the analysis of such systems. We define the quantum mechanical orbital angular momentum in the same manner as its classical counterpart L^ = ^r × p^; (3.1) with ^r and p^ the position and linear momentum observables, respectively. It follows that in quantum mechanics, the orbital angular momentum is also an observable. If we introduce the components x^j and p^j for the position and linear momentum, where j = 1; 2; 3 (i.e., in Cartesian coordinates x^1 =x ^, x^2 =y ^ and x^3 =z ^, and similarly for p^j), such that X x^ = ejx^j (3.2) j X p^ = ejp^j; (3.3) j with ej the different unit basis vectors, then ^ X L = (ej × ek)x ^jp^k j;k ! X X = ei"ijk x^jp^k j;k i 61 3 Angular Momentum and Spin X = ei"ijkx^jp^k: (3.4) i;j;k In equation (3.4) we have introduced the elements of the very important Levi-Civita tensor "ijk, which are defined as follows 8 < +1 ; even permutation of ijk (123; 231; 312) "ijk = −1 ; odd permutation of ijk (213; 132; 321) (3.5) : 0 ; if i = j; j = k or j = k: This tensor is particularly useful for expressing the elements of a vectorial cross-product P (i.e., (a × b)i = j;k "ijkajbk), and it is also related to the Kronecker delta (and thus the scalar product) through X "ijk"imn = δjmδkn − δjnδkm: (3.6) i It is straightforward through equation (3.6) to verify that [a × (b × c)]j = (a · c) bj − (a · b) cj. Coming back to equation (3.4), since it is also the case that X L^ = eiL^i; (3.7) i it then follows that X L^i = "ijkx^jp^k: (3.8) j;k There are several important and useful commutation relations involving the (orbital) angular momentum, we derive some here. First, the commutations with the position and linear momentum components h i X x^j; L^k = "kmn [^xj; x^mp^n] m;n X = "kmn (^xm [^xj; p^n] + [^xj; x^m]p ^n) m;n X = i~ "kmnx^mδjn m;n X = i~ "jkmx^m; (3.9) m and 62 3 Angular Momentum and Spin h i X p^j; L^k = "kmn [^pj; x^mp^n] m;n X = "kmn (^xm [^pj; p^n] + [^pj; x^m]p ^n) m;n X = −i~ "kmnp^nδjm m;n X = i~ "jknp^n (3.10) n where [^xj; p^k] = i~δjk was used for both derivations. For the commutation between components of the angular momentum h i X h i L^j; L^k = "jmnx^mp^n; L^k m;n X h i h i = "jmn x^m p^n; L^k + x^m; L^k p^n m;n X = i~ "jmn (^xm"nkrp^r + "mkrx^rp^n) m;n;r X = i~ (−"njm"nrkx^mp^r + "mnj"mkrx^rp^n) m;n;r " # X X = i~ − (δjrδmk − δjkδmr)x ^mp^r + (δnkδjr − δnrδjk)x ^rp^n m;r n;r ! X X = i~ − δjrδmkx^mp^r + δnkδjrx^rp^n m;r n;r X = i~ (δjrδkn − δjnδkr)x ^rp^n n;r X = i~ "ijk"irnx^rp^n n;r X ^ = i~ "ijkLi; (3.11) i where equations (3.6) and (3.9)-(3.10) were used. Considering the square of the angular momentum h 2 i X h i L^ ; L^j = L^kL^k; L^j k X h i h i = L^k L^k; L^j + L^k; L^j L^k k 63 3 Angular Momentum and Spin X ^ ^ ^ ^ = i~ "ikj LkLi + LiLk i;k X ^ ^ ^ ^ = i~ "ikjLkLi + "kijLkLi i;k X ^ ^ ^ ^ = i~ "ikj LkLi − LkLi i;k = 0^: (3.12) and h 2 i X h i L^ ; x^j = L^kL^k; x^j k X h i h i = L^k L^k; x^j + L^k; x^j L^k k X ^ ^ = −i~ Lk"ijkx^i + "ijkx^iLk i;k X ^ h ^ i = −i~ "ijk 2^xiLk − x^i; Lk i;k ! X ^ 2 X = 2i~"ijkx^kLi − "ijk~ "ikmx^m i;k m 0 1 X ^ 2 = 2 @i~ "jkix^kLi + ~ x^jA (3.13) i;k P P since k;i "ijk"ikm = k (δjkδkm − δjmδkk) = −2δjm, while similarly 0 1 h^2 i X ^ 2 L ; p^j = 2 @i~ "jkip^kLi + ~ p^jA ; (3.14) i;k A little more work will establish that (see the Second Problem List) " # h^2 i X ^ 2 2 L ; x^jx^k = 2 i~ (^xj"kmn +x ^k"jmn)x ^mLn − ~ x^ δjk − 3^xjx^k (3.15) m;n " # h^2 i X ^ 2 2 L ; p^jp^k = 2 i~ (^pj"kmn +p ^k"jmn)p ^mLn − ~ p^ δjk − 3^pjp^k ; (3.16) m;n h i h i which also imply that L^2; x^2 = L^2; p^2 = 0. Finally, we find using equations (3.9)- (3.10) 64 3 Angular Momentum and Spin h 2i L^j; x^ = 0 (3.17) h 2i L^j; p^ = 0: (3.18) 3.2 Eigenvalues of the Angular Momentum The fact that the three components of the angular momentum L^x, L^y, L^z commute with its square L^2, from equation (3.12), implies that we can find a common set of eigenvectors fj ig for L^2 and one component of L^ (the three components cannot share the same eigenvectors since they do not commute with each other; see equation (3.11)). This can be verified as follows. If a is an eigenvector of L^2 and j i one of its eigenvector, then 2 2 L^ L^j j i = L^j L^ j i = aL^j j i (3.19) 2 implies that the ket L^j j i is also a eigenvector of L^ . If this eigenvalue is non-degenerate, then L^j j i = b j i, with b a constant, and j i is also an eigenvector of L^j (and b its 2 corresponding eigenvalue). Similarly, we can assert that L^ j i is an eigenvector of L^j by swapping the order of the two operators in the first line of equation (3.19); they, therefore, share the same set of eigenvectors. It is common to choose L^j = L^z and we 2 2 denote by l ; m for the set of eigenvectors, with l and m the quantum numbers ^2 ^ 2 2 associated to L and Lz, respectively. As a matter of fact, we set l ~ as an eigenvalue ^2 ^ 2 of L and m~ one for Lz (although both l and m still need to be determined). We now introduce the raising and lowering operators L^± = L^x ± iL^y (3.20) for which the following commutation relations can easily be established h i h i h i L^z; L^± = L^z; L^x ± i L^z; L^y ^ ^ = i~ Ly ∓ iLx ^ = ±~L± (3.21) h 2 i L^ ; L^± = 0; (3.22) from equations (3.11) and (3.13). Using equation (3.21) we can calculate that ^ ^ 2 ^ ^ ^ 2 LzL± l ; m = L±Lz ± ~L± l ; m ^ 2 = (m ± 1) L± l ; m (3.23) 65 3 Angular Momentum and Spin which implies that whenever L^± acts on a eigenvector it transforms it to another eigen- ^ ^ vector of eigenvalue (m ± 1) ~ for Lz. It is therefore apparent that L± are ladders operators that will allow us to span the whole set of eigenvectors associated to a given 2 2 ^ ^2 eigenvalue l ~ (i.e., since L± commutes with L it shares the same eigenvectors). More- over, we deduce that m must have lower and upper bounds, since the following relation must be obeyed p m~ ≤ l2~2: (3.24) If we denote the upper and lower bounds of m as m+ and m−, respectively, then we can write ^ 2 L± l ; m± = 0: (3.25) Furthermore, we have ^ ^ 2 ^2 ^2 h^ ^ i 2 L∓L± l ; m± = Lx + Ly ± i Lx; Ly l ; m± ^2 ^2 ^ 2 = L − Lz ∓ ~Lz l ; m± 2 2 2 2 = l − m± ∓ m± ~ l ; m± = 0: (3.26) From this result we obtain the two equations 2 2 l − m− + m− = 0 (3.27) 2 2 l − m+ − m+ = 0 (3.28) from which we can take the difference and transform it to (m+ + m−)(m+ − m− + 1) = 0: (3.29) The term in the second set of parentheses must be greater than zero because m+ ≥ m−, which implies that the term within the first set of parentheses must cancel and m+ = −m− (3.30) m+ − m− = 2`; (3.31) where 2` has to be some positive integer number, since m+ − m− corresponds to the ^ 2 number of times (i.e., 2` times) L± is applied to l ; m∓ to span the whole set of kets.
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