PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Symmetry, angular momentum and spin · Symmetry, angular momentum and spin
Quantum Mechanics: From the Photon to the Hydrogen Atom, Ch Orbital angular momentum, the spherical harmonics and rotations
Problem
Earlier in this chapter, a common eigenfunction flml of L^2 and L^z cannot also be sharp in Lx or Ly, because [L^x,L^y]=iℏL^z forcesΔLxΔLy≥2ℏ⟨L^z⟩.A worked calculation elsewhere in this chapter checked this bound only at the very top of a ladder, where it turned out to be an equality. This problem asks whether the top rung is special, or whether every rung reaches the bound. Let flml be a normalised eigenfunction of L^2 and L^z, with l any value the algebra allows (a non-negative integer or half-integer) and ml running from −l to l in unit steps. Such a state has ⟨Lx⟩=⟨Ly⟩=0, and, by the symmetry of the state about the z axis, ⟨Lx2⟩=⟨Ly2⟩.
((a)) Using L^2=L^x2+L^y2+L^z2, show that ⟨Lx2⟩=⟨Ly2⟩=2ℏ2[l(l+1)−ml2].
((b)) For ml=0, define the ratio R=ΔLxΔLy/[2ℏℏ∣ml∣] of the actual product to the bound. Show that R=1+∣ml∣(l−∣ml∣)(l+∣ml∣+1), and hence that R≥1, with equality only at the top and bottom rungs, ∣ml∣=l.
((c)) Evaluate R for l=4, ml=1, to three significant figures.
((d)) In one or two sentences, say whether the conclusion of (b) depends on l being a whole number or a half, and say what, if not this uncertainty relation, keeps orbital motion from ever reaching l=23.