Lecture 14 Angular momentum operator algebra In this lecture we present the theory of angular momentum operator algebra in quantum mechanics. 14.1 Basic relations ^ ^ ^ Consider the three Hermitian angular momentum operators Jx; Jy and Jz, which satisfy the commutation relations ^ ^ ^ ^ ^ ^ ^ ^ ^ Jx; Jy = i~ Jz; Jz; Jx = i~ Jy; Jy; Jz = i~ Jx: (14.1) The operator ^2 ^2 ^2 ^2 J = Jx + Jy + Jz ; (14.2) ^ ^ ^ is also Hermitian and it commutes with Jx; Jy and Jz: ^2 ^ ^2 ^ ^2 ^ J ; Jx = J ; Jy = J ; Jz = 0: (14.3) These relations are not difficult to prove using the operator identity A;^ B^C^ = A;^ B^C^ + B^A;^ C^: (14.4) 1 2 LECTURE 14. ANGULAR MOMENTUM OPERATOR ALGEBRA For example, ^2 ^ ^2 ^2 ^2 ^ J ; Jz = Jx + Jy + Jz ; Jz ^2 ^ ^2 ^ = Jx ; Jz + Jy ; Jz ; (14.5) ^ because by definition Jz commutes with itself and, using (14.4), ^2 ^ ^ ^ ^ ^ ^ ^ Jz ; Jz = − Jz; Jz Jz − Jz Jz; Jz = 0: (14.6) The remaining two commutators in the last row of (14.5) can be calculated using again (14.4): ^2 ^ ^ ^ ^ ^ ^ ^ Jx ; Jz = Jx; Jz Jx + Jx Jx; Jz ^ ^ ^ ^ = − i~ JyJx + JxJy ; (14.7) and ^2 ^ ^ ^ ^ ^ ^ ^ Jy ; Jz = Jy; Jz Jy + Jy Jy; Jz ^ ^ ^ ^ = i~ JxJy + JyJx : (14.8) ^2 ^ ^2 ^ The sum of Jx ; Jz and Jy ; Jz is therefore zero and from (14.5) it follows ^2 ^ that J ; Jz = 0. ^ ^ Rather then working with the Hermitian operators Jx and Jy, it is more convenient to work with the non-Hermitian linear combinations, ^ ^ ^ J+ = Jx + iJy; (14.9a) ^ ^ ^ J− = Jx − iJy; (14.9b) ^ y ^ where, by definition, (J−) = J+. For reasons that will become clear later, ^ ^ J+ and J− are called ladder operators. Using (14.1) and (14.3), it is straight- 14.1. BASIC RELATIONS 3 forward to show that ^ ^ ^ Jz; J+ = ~ J+; (14.10a) ^ ^ ^ Jz; J− = − ~ J−; (14.10b) ^ ^ ^ J+; J− = 2~ Jz; (14.10c) ^2 ^ ^2 ^ J ; J+ = J ; J− = 0: (14.10d) ^ ^ ^ ^ ^ ^ The operators J+ and J− often appear in the products J+J− and J−J+, which are equal to ^ ^ ^ ^ ^ ^ J+J− = Jx + iJy Jx − iJy ^2 ^2 ^ ^ = Jx + Jy − i Jx; Jy ^2 ^2 ^ = Jx + Jy + ~ Jz; (14.11) and ^ ^ ^ ^ ^ ^ J−J+ = Jx − iJy Jx + iJy ^2 ^2 ^ ^ = Jx + Jy + i Jx; Jy ^2 ^2 ^ = Jx + Jy − ~ Jz; (14.12) ^2 ^2 ^2 ^2 respectively. After noticing that (14.2) implies Jx + Jy = J − Jz , we can ^ ^ ^ ^ straightforwardly rewrite J+J− and J−J+ as ^ ^ ^2 ^2 ^ J+J− = J − Jz + ~ Jz; (14.13a) ^ ^ ^2 ^2 ^ J−J+ = J − Jz − ~ Jz: (14.13b) Adding equations (14.13) side-by-side and rearranging the terms, we obtain 1 J^2 = J^ J^ + J^ J^ + J^2: (14.14) 2 + − − + z 4 LECTURE 14. ANGULAR MOMENTUM OPERATOR ALGEBRA Two additional useful relations are: ^ ^ n ^ n Jz; J+ = n~ J+ ; (14.15a) ^ ^ n ^ n Jz; J− = − n~ J− ; (14.15b) where n = 0; 1; 2;:::, is an integer number. We demonstrate, for example, ^ 0 ^ the first relation (14.15a). The proof is by iteration. If n = 0, then (J+) = I and (14.15a) becomes a trivial identity. If n = 1, we simply recover (14.10a). If n > 1, we first rewrite ^ ^ n ^ ^ ^ n−1 Jz(J+) = JzJ+(J+) ; (14.16) ^ ^ ^ ^ ^ and then use (14.10a) to write JzJ+ = J+Jz + ~J+. This permits us to find the following recursion relation: ^ ^ n ^ ^ ^ ^ n−1 Jz J+ = J+Jz + ~J+ J+ ^ n ^ h ^ ^ n−1i = ~ J+ + J+ Jz J+ : (14.17) If we rewrite this equation replacing n with n − 1, we easily find ^ ^ n−1 ^ n−1 ^ h ^ ^ n−2i Jz J+ = ~ J+ + J+ Jz J+ : (14.18) Substituting this expression into (14.17), we obtain ^ ^ n ^ n ^ n ^ n−1 ^ h ^ ^ n−2io Jz J+ = ~ J+ + J+ ~ J+ + J+ Jz J+ ^ n ^ 2 h ^ ^ n−2i = 2~ J+ + J+ Jz J+ : (14.19) This procedure can be iterated again and again. After p iterations we find ^ ^ n ^ n ^ p h ^ ^ n−pi Jz J+ = p~ J+ + J+ Jz J+ ; (14.20) which becomes, for p = n, ^ ^ n ^ n ^ n ^ Jz J+ = n~ J+ + J+ Jz: (14.21) 14.2. COMPATIBLE AND INCOMPATIBLE OBSERVABLES 5 This proves (14.15a). Equation (14.15b) can be obtained in the same way using (14.10b). ^2 ^ 1 Finally, from J ; J± = 0 and the equation A;^ B^n = A;^ B^B^n−1 + B^A;^ B^B^n−2 + ::: + B^n−1A;^ B^; (14.22) where n is a positive integer number, it follows that ^2 ^ n J ; (J±) = 0: (14.23) 14.2 Compatible and incompatible observables In quantum mechanics, two observables A and B are said to be compatible when the corresponding operators A^ and B^ commute, A;^ B^ = 0; (14.24) and incompatible when they do not A;^ B^ 6= 0: (14.25) If A and B are compatible and j i is an eigenvector of A^ associated with the eigenvalue a, A^j i = aj i; (14.26) then B^j i is also an eigenvector of A^ associated with the same eigenvalue a. This can be easily seen applying B^ to both sides of (14.26): B^A^j i = aB^j i: (14.27) Since, by hypothesis, A;^ B^ = 0, we can replace B^A^ on the left side of this equation with A^B^, thus obtaining A^ B^j i = a B^j i : (14.28) Now, there are two possibilities: 1This equation can be easily proved by induction using (14.4). 6 LECTURE 14. ANGULAR MOMENTUM OPERATOR ALGEBRA 1) If a is a non-degenerate eigenvalue, then all vectors j i satisfying (14.26) are parallel2 and B^j i is necessarily proportional to j i, that is B^j i = bj i: (14.29) Therefore, j i is also an eigenvector of B^. 2) If a is a degenerate eigenvalue, then the set of all vectors j i satisfying (14.26) spans a subspace Ea associated with the eigenvalue a. Then, if ^ j i 2 Ea, all what we can say is that Bj i belongs to Ea: ^ Bj i 2 Ea: (14.30) In the first case, the knowledge of the common eigenvector j i of A^ and B^, uniquely determines the values of both a and b. This means that a and b are not independent variables and the knowledge of a fixes the value of b and vice versa. In the second case, instead, we need to specify both values a and b to uniquely identify the common eigenvectors of A^ and B^. A fundamental theorem of quantum mechanics states that3: When two observables A and B are compatible, it is always pos- sible to find an orthonormal basis of the vector space spanned by the eigenvectors common to the corresponding operators A^ and B^. We denote with fja; big such a basis, where the eigenvectors ja; bi are defined by the properties A^ja; bi = aja; bi; (14.31a) B^ja; bi = bja; bi; (14.31b) 2Here and hereafter, the statement that two vectors j i and jφi are parallel, signifies that j i = γjφi, where γ is any nonzero complex number. 3For a demonstration see, e.g., C. Cohen-Tannoudji, B. Diu, F. Lalo¨e, Quantum Me- chanics, Vol. I (Wiley-Vch, 2005), Chapter II, x D-3-a. 14.2. COMPATIBLE AND INCOMPATIBLE OBSERVABLES 7 and 0 0 ha ; b ja; bi = δaa0 δbb 0 : (14.32) If the ordered pair of eigenvalues (a; b) determine a unique common eigenvec- tor ja; bi of A^ and B^, then the set of observables fA; Bg is called a Complete Set of Commuting Observables (CSCO), or a maximal set of commuting observables; that is there is not a third observable C such that A;^ B^ = A;^ C^ = B;^ C^ = 0: (14.33) The eigenvalues of operators A^ and B^ may still be degenerate, but if we specify a pair (a; b), then the corresponding eigenvector ja; bi common to A^ and B^ is uniquely specified. The Hermitian operator A^ possess at least one degenerate eigen- value when there are two observables B and C compatible with A but incompatible each other. To prove this statement, consider three observables A; B and C such that fA; Bg is a CSCO, with A;^ B^ = 0 = A;^ C^; and B;^ C^ 6= 0: (14.34) According to our nomenclature, equations (14.34) imply that A is compatible with both B and C, but B and C are incompatible. Let ja; bi be a common eigenvector of A^ and B^ associated with the eigenvalues a and b, respectively. From A;^ C^ = 0 it follows that C^ja; bi is also eigenvectors of A^ associated with the same eigenvalue a. Therefore, we have A^ B^ja; bi = a B^ja; bi ; (14.35a) A^ C^ja; bi = a C^ja; bi ; (14.35b) Suppose that a is a non-degenerate eigenvalue. Then C^ja; bi must be proportional to B^ja; bi, otherwise according to (14.35) C^ja; bi and B^ja; bi 8 LECTURE 14. ANGULAR MOMENTUM OPERATOR ALGEBRA would be two different eigenvectors associated with the same eigenvalue a, which is incompatible with the hypothesis that a is a non-degenerate eigenvalue. Therefore, we can write C^ja; bi = γ B^ja; bi = γ bja; bi; (14.36) because B^ja; bi = bja; bi, and γ is a numerical constant. Multiplying by B^ both sides of this equation, we obtain: B^C^ja; bi = γ b B^ja; bi = γ b2ja; bi: (14.37) On the other hand, applying C^ to both sides of B^ja; bi = bja; bi; (14.38) we find C^B^ja; bi = b C^ja; bi = γ b2ja; bi; (14.39) where (14.36) has been used.