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I think the equation (10A-7) is wrong. In (10A-6), it said L_Ylm = [(l- m +1)(l +m)]^(1/2) hYlm-1. But in (10A-7), it gives [(l+ m +1)(l -m)]^(1/2) hYlmX+ at the last term. Shouldn't it be [(l- m +1)(l +m)]^(1/2) hYlmX+? If this is the case, please derive the following equations 10A-9 to 10A-14 correctly. I don't have much time. I would appreciate if you could give me an answer within 6 hours. Orbital Angular Momentum (Details) W-44 Supplement 10-A: The Addition of Spin 1/2 and Orbital Angular Momentum (Details) This will be of the form h(j + 1)+1/2 = hj( + 1)(Yx, + Y,+-) (10A-8) provided that (I + //=[I + + (r )]g + [+ + (I + IN] 1+/=[I++]+[=+1+1] (10A-9). This requires that [w 1 +F 1 +/)]=1 +m+1r 1 [( + 1) (I + 1) + m + 1] which evidently has two solutions. Of great importance for future applications is the combination of a spin with an orbital angular momentum. Since L depends on spatial coordinates and S does not, they commute [L,S]=0 (10A-1). It is therefore evident that the components of the total angular momentum J, defined by J = L + S (10A-2), will satisfy the angular momentum commutation relations. In asking for linear combinations of the Y_ and the that are eigenstates of J, = L, + S, (10A-3) and J^2 = L^2 + S^2 + 2LS (-v01) = L^2 + S^2 + 2LS + ZS_ + ZS, we are again looking for the expansion coefficients of one complete set of eigenfunctions in terms of another set of eigenfunctions. Let us consider the linear combination (j + 1) (/ + 1) = (01-v01) that is, (10A-11), * = Y, + Y;*+:X- (-vo1). It is, by construction, an eigenfunction of J, with eigenvalue (m + ). We now determine and such that it is also an eigenfunction of.3. We shall make use of the fact that L,Y = [(/+ 1) m(m +1)]1Y1 = [(1+m+1/m)]Y+1 (9-V01) L,Y,= [(m + 1)(/+m)]3Y=-1 Sx=S--=0Sx:=hx Then J^2,+1/2 = h^2 ((7 + 1) Y, + Y, + 2m() Y For/ = / + 1/2, we get, after a little algebra +m+1 = (10A-12) (Actually we just get the ratio; these are already normalized forms.) Thus +1/2,+1/2= +m Yim+1X (10A-13) We can guess that the/ = / 1/2 solution must have the form )1/2,+12 = I+m+1 Y+X (10A-14) + Y,+- + 2(m + 1 Y;+1X + (m)(/ + m + 1)]12 Y} in order to be orthogonal to the / = / + 1/2 solution. General Rules for Addition of Angular Momenta, and Implications for Identical Particles W-43

          I think the equation (10A-7) is wrong. In (10A-6), it said L_Ylm = [(l- m +1)(l +m)]^(1/2) hYlm-1. But in (10A-7), it gives [(l+ m +1)(l -m)]^(1/2) hYlmX+ at the last term. Shouldn't it be [(l- m +1)(l +m)]^(1/2) hYlmX+? If this is the case, please derive the following equations 10A-9 to 10A-14 correctly. I don't have much time. I would appreciate if you could give me an answer within 6 hours.

Orbital Angular Momentum (Details)
W-44 Supplement 10-A: The Addition of Spin 1/2 and Orbital Angular Momentum (Details)

This will be of the form h(j + 1)+1/2 = hj( + 1)(Yx, + Y,+-) (10A-8) provided that (I + //=[I + + (r )]g + [+ + (I + IN] 1+/=[I++]+[=+1+1] (10A-9). This requires that [w 1 +F 1 +/)]=1 +m+1r 1 [( + 1) (I + 1) + m + 1] which evidently has two solutions.

Of great importance for future applications is the combination of a spin with an orbital angular momentum. Since L depends on spatial coordinates and S does not, they commute [L,S]=0 (10A-1). It is therefore evident that the components of the total angular momentum J, defined by J = L + S (10A-2), will satisfy the angular momentum commutation relations. In asking for linear combinations of the Y_ and the that are eigenstates of J, = L, + S, (10A-3) and J^2 = L^2 + S^2 + 2LS (-v01) = L^2 + S^2 + 2LS + ZS_ + ZS, we are again looking for the expansion coefficients of one complete set of eigenfunctions in terms of another set of eigenfunctions. Let us consider the linear combination (j + 1) (/ + 1) = (01-v01) that is, (10A-11), * = Y, + Y;*+:X- (-vo1). It is, by construction, an eigenfunction of J, with eigenvalue (m + ). We now determine and such that it is also an eigenfunction of.3. We shall make use of the fact that L,Y = [(/+ 1) m(m +1)]1Y1 = [(1+m+1/m)]Y+1 (9-V01) L,Y,= [(m + 1)(/+m)]3Y=-1 Sx=S--=0Sx:=hx Then J^2,+1/2 = h^2 ((7 + 1) Y, + Y, + 2m() Y

For/ = / + 1/2, we get, after a little algebra +m+1 = (10A-12) (Actually we just get the ratio; these are already normalized forms.) Thus +1/2,+1/2= +m Yim+1X (10A-13)

We can guess that the/ = / 1/2 solution must have the form )1/2,+12 = I+m+1 Y+X (10A-14) + Y,+- + 2(m + 1 Y;+1X + (m)(/ + m + 1)]12 Y} in order to be orthogonal to the / = / + 1/2 solution.

General Rules for Addition of Angular Momenta, and Implications for Identical Particles W-43

i think the equation 10a 7 is wrong in 10a 6 it said lylm l m 1l m12 hylm 1 but in 10a 7 it gives l m 1l m12 hylmx at the last term shouldnt it be l m 1l m12 hylmx if this is the case pleas 33516

Added by Rosa S.

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