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Unformatted text preview: HOMEWORK ASSIGNMENT 7 PHYS852 Quantum Mechanics II, Spring 2008 New topics covered: Green’s functions, Tmatrix . 1. ‘Weakfield Zeeman Effect’ : Consider a hydrogen atom in a uniform magnetic field. Assume that the Zeeman shift is large compared to the hyperfine splitting, but smaller than the finestructure shift. Thus we can ignore nuclear spin (i.e. the hyperfine structure) and consider the magnetic field as a perturbation on the fine structure. The finestructure lifts the degeneracy of states with different j , so the bare eigenstates are  nℓsjm j ) states. Use your previous work, and/or reference materials, to find the bare energy eigenvalues, which should depend on n and j only. Show that the states with the same n , s and j , but different ℓ and m j are degenerate. Now add the magnetic field interaction V = − eB 2 M ( L z + 2 S z ). What are the good eigenstates? (hint: Use Clebsch Gordan coefficients to expand the  nℓsjm j ) states onto  nℓsm ℓ m z ) states. Then use the selection rule m j = m ℓ + m s to show that ( nℓsjm j  L z + 2 S z  nℓ ′ sjm ′ j ) ∝ δ ℓ,ℓ ′ δ m j ,m ′ j . This result should allow you to deduce the good basis). Now use firstorder perturbation theory to compute the Zeeman sublevels of a state with arbitrary n , ℓ , and j . Answer : From Eq. (138) in the TIPT notes, we see that the bareenergies of the finestructure eigenstates are given by E n,j = − α 2 Mc 2 2 bracketleftbigg 1 n 2 + α 2 j + 1 / 2 1 n 3 − 3 α 2 4 1 n 4 bracketrightbigg , (1) where α = 1 / 137, M = 9 . 11 × 10 − 31 kg, and c = 3 . 00 × 110 8 m/s. Clearly states with different ℓ and m j , but the same n and j are degenerate, since the bare energies depend on n and j only. To find the ‘good eigenstates’ of V = − eB 2 M ( L z + 2 S z ), we need to diagonalize V in the degenerate subspace. The degenerate subspace consists of all states with the same n , s , and j , but different ℓ and m j . The matrix elements of V D = P D V P D are thus ( nℓsjm j  L z + 2 S z  nℓ ′ sjm ′ j ) . (2) If we can verify that they are proportional to δ ℓ,ℓ ′ δ m j ,m ′ j , then we will have proven that V D is diagonal, so that the ‘good states’ are just the  nℓsjm j ) states themselves. To prove the diagonality of V D , we insert the projector onto the  nℓsm ℓ m s ) basis into Eq. (1). This gives ( nℓsjm j  L z + 2 S z  nℓ ′ sjm ′ j ) = summationdisplay ℓ ′′ m ′′ ℓ m ′′ s ( nℓsjm j  ( L z + 2 S z )  nℓ ′′ sm ′′ ℓ m ′′ s )( nℓ ′′ sm ′′ ℓ m ′′ s  nℓ ′ sjm ′ j ) = summationdisplay ℓ ′′ m ′′ ℓ m ′′ s ( nℓsjm j  ( m ′′ ℓ + 2 m ′′ s )  nℓ ′′ sm ′′ ℓ m ′′ s ) δ ℓ ′′ ,ℓ ′ ( m ′′ ℓ m ′′ s  jm ′ j ) = δ ℓ,ℓ ′ summationdisplay m ′′ ℓ m ′′ s ( m ′′ ℓ + 2 m ′′ s ) δ ℓ,ℓ ′ ( jm j  m ′′ ℓ m ′′ s )( m ′′ ℓ...
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 Spring '10
 MichaelMoore
 mechanics, Work, U0, Mj, Fermi golden rule, Clebsch Gordan Coeﬃcients

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