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NRQM10finsol

# NRQM10finsol - Physics 216 Solutions to the Final Exam...

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Unformatted text preview: Physics 216 Solutions to the Final Exam Spring 2010 1. An electron is placed in a potential V ( vector r ) = − e 2 r + β ( r 2 − 3 z 2 ) , where β is a small parameter. Neglect the spin of the electron. (a) Compute the shifts of the n = 2 energy levels (you may neglect fine-structure effects) using first order perturbation theory. Indicate the relative positions of the energy levels. The unperturbed n = 2 state of hydrogen are degenerate states with energy: E (0) 2 ℓ m ℓ = − 1 4 Ry , where 1 Ry = me 4 / (2 planckover2pi1 2 ) = 13 . 6 eV. The corresponding unperturbed energy eigenfunc- tions are: ψ 200 ( vector r ) = 1 √ π parenleftbigg 1 2 a parenrightbigg 3 / 2 parenleftbigg 1 − r 2 a parenrightbigg e − r/ (2 a ) , ψ 210 ( vector r ) = 1 √ 2 π parenleftbigg 1 2 a parenrightbigg 3 / 2 r a e − r/ (2 a ) cos θ , ψ 21 ± 1 ( vector r ) = ∓ 1 √ 8 π parenleftbigg 1 a parenrightbigg 3 / 2 r a e − r/ (2 a ) sin θ e ∓ iφ , where a = planckover2pi1 2 / ( me 2 ). The perturbing Hamiltonian is: H (1) = β ( r 2 − 3 z 2 ) = βr 2 (1 − 3 cos 2 θ ) . To compute the first-order energy shifts of degenerate states, we must construct the matrix elements of the matrix W , which is defined as: W ℓ ′ m ′ ℓ ; ℓm ℓ ≡ ( 2 ℓ ′ m ′ ℓ | H (1) | 2 ℓ m ℓ ) . (1) However, it is easy to show that the matrix W is diagonal. First, note that H (1) com- mutes with L z , since L z = − i planckover2pi1 ∂ ∂φ and H (1) is independent of φ . It follows that: 0 = ( 2 ℓ ′ m ′ ℓ | [ H (1) , L z ] | 2 ℓ m ℓ ) = ( 2 ℓ ′ m ′ ℓ | ( H (1) L z − L z H (1) ) | 2 ℓ m ℓ ) = ( m ℓ − m ′ ℓ ) ( 2 ℓ ′ m ′ ℓ | H (1) | 2 ℓ m ℓ ) , which implies that ( 2 ℓ ′ m ′ ℓ | H (1) | 2 ℓ m ℓ ) = 0 if m ℓ negationslash = m ′ ℓ . 1 The only other non-diagonal elements to examine are ( 2 0 0 | H (1) | 2 1 0 ) and ( 2 1 0 | H (1) | 2 0 0 ) . When we evaluate both matrix elements, we find that the angular integral vanishes, since integraldisplay d Ω (1 − 3 cos 2 θ ) = 2 π integraldisplay 1 − 1 d cos θ (1 − 3 cos 2 θ ) = (cos θ − cos 3 θ ) vextendsingle vextendsingle vextendsingle +1 − 1 = 0 . (2) It therefore follows that the first-order energy shifts of the n = 2 states are given by: E (1) 2 ℓm ℓ = ( 2 ℓ m ℓ | H (1) | 2 ℓ m ℓ ) . First we compute the first-order energy shift of the state | 2 0 0 ) : E (1) 200 = β ( 2 0 0 | r 2 (1 − 3 cos 2 θ ) | 2 0 0 ) = β 8 πa 3 integraldisplay d 3 r parenleftbigg 1 − r 2 a parenrightbigg 2 r 2 e − r/a (1 − 3 cos 2 θ ) . Using eq. (2), it follows that: E (1) 200 = 0 Turning to the n = 2, ℓ = 1 states, E (1) 210 = β 32 πa 5 integraldisplay d 3 r r 4 e − r/a cos 2 θ (1 − 3 cos 2 θ ) , (3) E (1) 21 ± 1 = β 64 πa 5 integraldisplay d 3 r r 4 e − r/a sin 2 θ (1 − 3 cos 2 θ ) , (4) After writing d 3 r = r 2 drd Ω, the radial integrals are all of the form: integraldisplay ∞ r n e − r/a dr...
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NRQM10finsol - Physics 216 Solutions to the Final Exam...

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