CLEBSCH-GORDAN COEFFICIENTS - EXAMPLES

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1 CLEBSCH-GORDAN COEFFICIENTS - EXAMPLES Link to: physicspages home page. To leave a comment or report an error, please use the auxiliary blog. Shankar, R. (994), Principles of Quantum Mechanics, Plenum Press. Section 5.; Exercises [If some equations are too small to read easily, use your browser s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).] We ve seen a detailed example of calculating Clebsch-Gordan coefficients by using the angular momentum lowering operator, where we calculated the coefficients for the case of spin- combined with spin-. Here we ll give a slightly more involved example by combining two spin- systems. Shankar gives the conditions satisfied by the CG coefficients, so we ll apply these here. We start with the combined system with the maximum values of j (the total angular momentum number) and m (the z component of the total angular momentum), which here means we have j = m =. As there is only one member of the product space (m = m = ) satisfying this condition, we have () t = p As in the previous post, we give the ket in the total- j space as jm t (subscript t for total- j ) and in the product space as m m p (subscript p for product ). The two individual spins j and j are always the same in all cases, so we omit them from the notation. We now apply the lowering operator to both sides to generate the next state. This operator is, in the total- j space: () J jm t = h ( j + m)( j m + ) j (m ) t () In the product space, we have (J + J ) m m p = h ( j + m )( j m + ) (m )m p + (4) h ( j + m )( j m + ) m (m ) p In what follows, we ll omit the h since it always occurs in every term on both sides of the equation, so it always cancels out in the final result.

2 Starting from we have CLEBSCH-GORDAN COEFFICIENTS - EXAMPLES (5) () (7) J t = t (J + J ) p = 0 p + 0 p t = 0 p + 0 p For the next state, we have (8) J t = 0 t (9) ) (J + J ) ( 0 p + 0 p = p + ) 00 p + (0) () () 00 p + ), p = p + 00 p +, p 0 t = p + 00 p +, p [The last line equates the first line with the fourth line.] To get the states with negative m, we can use equation 5.. in Shankar: () m m jm = ( ) j + j j m, m j, m Here, the bracket m m jm is the CG coefficient that multiplies m m in the expansion of jm. For example, in (4) (5) () 0 = 00 0 =, 0 = Using and 7, we have j + j j = + = 0, so (7), t = 0, p + 0 p The final ket in the column with j = is

3 CLEBSCH-GORDAN COEFFICIENTS - EXAMPLES (8), t =, p For the next column, we have j = and the top entry therefore has m =. This total- j state t must be a combination of the product states 0 p and 0 p, must be orthogonal to 7 and the coefficient of the term with m = j is by convention positive. By inspection, we have (9) t = 0 p + 0 p Note that m = j = in the second term, so it s the positive one. To get the next ket, we apply the lowering operator again: (0) () ) (J + J ) ( 0 p 0 p () J t = 0 t = p + 00 p )+ 00 p + ), p () (4) = p +, p 0 t = p +, p We apply to get the final entry in this column. This time j + j j = so (5), t = 0, p 0 p Finally, there is one total- j ket in the third column, where j = m = 0. This time, the ket 00 t must be a combination of p, 00 p and, p, and must be orthogonal to both and 4. Suppose () 00 t = a p + b 00 p + c, p Then the orthogonality conditions tell us that (7) (8) a + b + c = 0 a + c = 0

4 We can solve these to find CLEBSCH-GORDAN COEFFICIENTS - EXAMPLES 4 (9) c = a (0) a + b = 0 () b = a We can find a from the normalization condition () () Thus we have (4) (5) a = c = b = a + b + c = a = () 00 t = p 00 p +, p These CG coefficients agree with those given in Griffiths s Table 4.8, for example. A final comment on the dimensionality of the various spaces. If we combine two single spins j and j, then the dimensionality of the product space j j is ( j + )( j + ), since there are j i + possible m i values for spin j i. In the above example, both j and j are, so the dimensionality of is = 9. The dimensionality of the corresponding total- j space is the sum of the dimensions for each possible value of j within this space. For, there are 5 states with j =, states with j = and state with j = 0, for a total of = 9. If we combine more than spins, we can apply the same argument, provided we count the number of total- j states properly. For, the product space contains = 8 dimensions, so the total- j space must also contain 8 dimensions. In the total- j space, j can be or. We have 4 states with j =. For j =, we can have m = ±. Consider the ket t. It must be a combination of product states where two spins are up and one is down (that is, two of the m i are + and one is ), so we have (7) = a + b + c t

5 CLEBSCH-GORDAN COEFFICIENTS - EXAMPLES 5 The only constraints we have on a, b and c are () the state t must be orthogonal to and () the state must be normalized. As we have t only constraints on unknowns, the subspace occupied by is twodimensional. The same argument applies to t, so it, too, is twodimensional. Thus the total dimensionality of the total- j space is 4++ = t 8, or in other words, =.