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E q . 30) is a n example of a Fredholm equation of t h e second kind [ 3 8 ] . T h e general m e t h o d for solving such a n equation is t o e x p a n d b o t h kernel a n d function in some series of orthonormal polynomials, thereby obtaining a series of equations for t h e coefficients in t h e expansion of t h e function. T h e n a t u r a l choice for our polynomials is t h e four-dimen sional spherical harmonics. If we specialize t o t h e hydrogen a t o m , we can set R = 0, corresponding t o a charge Ze a t t h e origin.

35) m We have now to interpret 0 in terms of 0ι, φχ, 0 2, and φ%. Suppose Ζ>(ωχ) takes a unit vector r 0 into Τχ; that is, r 0 — * r i . Then corresponds t o Μΐ corresponds to r 2 - + r i . This transformation is Γ Ο — » Γ 2 , and £ ( Ω Ι ) £ ( Ω » ) effected by Ώ(ω)} and thus depends on 0 and φ. However, φ does not appear on the left-hand side of Eq. 35), which implies that the entire rotation can be described in terms of 0 alone. From elementary geometry, cos 0 = Γ Ι · Γ 2 = cos 0i cos 0 2 + sin $% sin 0 2 οοβ(φι — φ%).

The two angular momenta \t and \t' can couple to produce the resultants *(« + 0 , i ( * + 0 - ι , * ( < + Ο -2, each of which occurs just once. Thus A(tf's) = 1 if t, t'y and s satisfy the triangular condition and if the sum t + t' + s is even: otherwise A(tt'$) = 0. The origin of the angular momentum it lies in the collapse of Ynim to C n- i when I = m « 0; for then η — 1 = 2k = tj and so \t = k. It might be worthwhile to illustrate this result with an example. (μ) = 1, CMM) 2 M, - C 2 ( M) C 4( M ) 4μ* - - 1, - 16μ< - 12μ' + 1, δ C .