A particle moves in a rotationally invariant three-dimensional potential

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Symmetry, angular momentum and spin · Symmetry, angular momentum and spin

The Structure of Quantum Mechanics, Ch Angular momentum from the rotation group

Problem

A particle moves in a rotationally invariant three-dimensional potential V(r)V(r), so that H^\hat H commutes with every component of J⃗^\hat{\vec J}: [H^,J^x]=[H^,J^y]=[H^,J^z]=0[\hat H,\hat J_x]=[\hat H,\hat J_y]=[\hat H,\hat J_z]=0. Let ∣E,j,m⟩|E,j,m\rangle be a simultaneous eigenstate of H^\hat H, J^2\hat J^2 and J^z\hat J_z with H^∣E,j,m⟩=E∣E,j,m⟩\hat H|E,j,m\rangle=E|E,j,m\rangle, and let J^±=J^x±iJ^y\hat J_\pm=\hat J_x\pm i\hat J_y. ((a)) Using only [H^,J^±]=0[\hat H,\hat J_\pm]=0 and the fact that J^±\hat J_\pm raises or lowers mm by one unit without changing jj, show that J^±∣E,j,m⟩\hat J_\pm|E,j,m\rangle (when non-zero) is again an eigenstate of H^\hat H with the same eigenvalue EE. Hence state how many linearly independent states with energy EE a given (E,j)(E,j) multiplet contains. ((b)) The one-dimensional oscillator Hamiltonian is H^osc=ℏω(N^+12)\hat H_{\rm osc}=\hbar\omega(\hat N+\tfrac12) with N^=a^†a^\hat N=\hat a^\dagger\hat a, and the raising operator obeys [H^osc,a^†]=ℏωa^†[\hat H_{\rm osc},\hat a^\dagger]=\hbar\omega\hat a^\dagger. Explain, using this commutator and the general connection between a commuting generator and a conserved quantity, why stepping up the oscillator ladder with a^†\hat a^\dagger changes the energy while stepping up an angular momentum multiplet with J^+\hat J_+ does not.

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