Ladder operators: The angular momentum eigenvalue equations (5) can also be solved by introducing ladder operators very similar to the one applied to SHO, L± ≡ Lx ± i Ly: (28) The commutation relations involving L± and components of angular momentum are derived using the relations (4), [Lz;L±] = [Lz;Lx ± iLy] = ± ~ L± 2 L ;L± = 0 [L±;L∓] = ±2~Lz: (29) The ladder operators L± also satisfy, 2 2 L± L∓ = (Lx ± iLy)(Lx ∓ iLy) = Lx + Ly ∓ i(LxLy − LyLx) 2 2 = L − Lz ± ~Lz (30) For the present purpose, let the angular momentum eigenvalue equations be written in the following form, 2 2 L f = λ~ f and Lz f = µ~ f (31) 2 Just as in SHO, if f is an eigenfunction of L and Lz, so also is L±f, 2 2 2 2 L (L±f) = L± (L f) = L± (λ~ f) = λ~ (L±f) Lz (L±f) = ±~ (L±f) + L±(Lz f) = (µ ± 1)~ (L±f) (32) Interestingly, L±f is indeed an eigenfunciton of Lz but with new eigenvalues, raised or lowered by one unit of ~. Hence, L+ (L−) operator is called raising (lowering) operator and together are called ladder operator. Therefore, each time we apply L+ or L−, the eigenvalue µ increases or decreases by ~. But can this continue indefinitely? The restriction on µ follows from the observation, 2 2 2 2 hL − Lzi = (λ − µ )~ ; (33) 2 2 2 2 hL − Lzi = hLx + Lyi 2 2 = hLxi + hLyi Z Z ? ? = dτ f LxLxf + dτ f LyLyf Z Z 2 2 = dτ (Lxf) + dτ (Lyf) using hermiticity of Lx;Ly ≥ 0 (34) where dτ is some appropriate volume element. Comparing (33) and (34), we get, λ ≥ µ2: (35) Therefore, raising or lowering of µ must stop to obey (35). Say raising must stop at some fmax and maximum value for µ be, say, l~, 2 2 L+fmax = 0 ) Lzfmax = l ~ fmax and L fmax = λ~ fmax (36) 1 Using (30), it follows that, 2 2 L fmax = λ~ fmax ) 2 2 2 2 (L−L+ + Lz + ~ Lz) fmax = (0 + l ~ + l~ ) fmax 2 and therefore, the eigenvalue λ of L in terms of maximum eigenvalue of Lz is, 2 λ = l(l + 1) ~ (37) which is exactly what we got by solving angular part of Schr¨odingerequation in (21). ¯ Similarly, lowering also should end at some fmin and minimum value for µ be l~, ¯ 2 2 L−fmin = 0 ) Lzfmin = l ~ fmin and L fmin = λ~ fmin (38) In the same way as with L+, using (30), it follows that, 2 2 L fmin = λ~ fmin ) 2 ¯2 2 ¯ 2 (L+L− + Lz − ~ Lz) fmin = (0 + l ~ − l~ ) fmin 2 and therefore, the eigenvalue λ of L in terms of minimum eigenvalue of Lz is, 2 λ = ¯l(¯l − 1) ~ : (39) 2 Since eigenvalue λ of L does not change with action of L±, comparing equations (37) and (39), we see that l(l + 1) = ¯l(¯l − 1) and solving for ¯l we get, ¯l = −l and ¯l = l + 1: But ¯l = l+1 is absurd since minimum eigenvalue cannot be larger than maximum eigenvalue, so the acceptable solution is ¯l = −l ) −l ≤ µ ≤ +l (40) which the limit of µ (the magnetic quantum number) we talked about before without actu- ally showing it. So for a given value of l, there are 2l + 1 different values of µ and, since µ changes by one unit of ~, it goes from −l to +l (or the other way round) in N (say) integer steps, l = −l + N ) l = N=2 i.e. l must be an integer or a half-integer, l = 0; 1=2; 1; 3=2; 2;:::. However, we have seen before when angular part of the Schr¨odingerequation was solved explicity that l is an integer. Thus here we have determined the eigenvalues of generic angular momentum operator without even knowing its eigenfunctions. 2 The ladder operator when acted upon the eigenfunctions of L and Lz changes the eigen- values of Lz by one unit which can be represented as, L+ fµ = cλµ fµ+1 and L− fµ = dλµ fµ−1: (41) 2 The coefficient cλµ can be determined as, L+ fµ = cλµ fµ+1 ) Z Z Z ? ? 2 2 2 dτ fµL−L+fµ = dτ (L+fµ) (L+fµ) = jcλµj dτ jfµ+1j = jcλµj Z Z ? ? 2 2 or; dτ fµL−L+fµ = dτ fµ(L − Lz − ~Lz)fµ Z 2 2 2 2 2 = l(l + 1)~ − µ ~ − µ~ dτ jfµj 2 = (l − µ)(l + µ + 1)~ 1=2 ) cλµ = [(l − µ)(l + µ + 1)] ~ (42) 1=2 In the same way we can get dλµ = [(l + µ)(l − µ + 1)] . In short, p cλµ fµ = (l − µ)(l + µ + 1) ~ fµ+1 (43) p dλµ fµ = (l + µ)(l − µ + 1) ~ fµ−1: (44) 3.