852_2010hw8_Solutions - PHYS852 Quantum Mechanics II Spring...

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PHYS852 Quantum Mechanics II, Spring 2010 HOMEWORK ASSIGNMENT 8: Solutions Topics covered: hydrogen fine structure 1. [10 pts] Let the Hamiltonian H depend on the parameter λ , so that H = H ( λ ). The eigenstates and eigenvalues of H are then also functions of λ , i.e. E n = E n ( λ ) and | n i = | n ( λ ) i . Use the property H | n i = E n | n i to prove the Feynman-Hellmann theorem: ∂E n ( λ ) ∂λ = h n ( λ ) | ∂H ( λ ) ∂λ | n ( λ ) i Start from the definition E n = h n | H | n i Differentiation gives ∂λ E n = ∂λ h n | H | n i + h n | ∂λ H | n i + h n | H ∂λ | n i Since H | n i = E n | n i and h n | H = h n | E n , we can rewrite this as ∂λ E n = E n ∂λ h n | | n i + h n | ∂λ | n i E n + h n | ∂H ∂λ | n i = E n ∂λ h n | n i + h n | ∂H ∂λ | n i Since h n | n i = 1, it follows that ∂E n ( λ ) ∂λ = h n ( λ ) | ∂H ( λ ) ∂λ | n ( λ ) i 1
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2. [15 pts] The effective Hamiltonian which governs the radial wave equation is H = - ~ 2 2 M 2 ∂r 2 + ~ 2 ( + 1) 2 Mr 2 - e 2 4 π 0 r . The exact eigenvalues in terms of e and are E n = - Me 4 32 π 2 2 0 ~ 2 n ( ) 2 , where n ( ) = n r + + 1, with n r being the highest power in the series expansion or R n‘ ( r ). Apply the Feynman-Hellman theorem with λ = e to derive: h n‘ (0) | R - 1 | n‘ (0) i = 1 n 2 a 0 . Then use λ = , with treated as a continuous parameter, to derive: h n‘ (0) | R - 2 | n‘ (0) i = 1 ( + 1 / 2) n 3 a 2 0 . According to Feynman-Hellman h n‘m | ∂H ∂e | n‘m i = ∂E n ∂e now ∂H ∂e = - 2 e 4 π 0 1 R and we have E n = - ~ 2 2 m e a 2 0 n 2 with a 0 = 4 π 0 ~ 2 m e e 2 this works out to E n = - m e e 4 (4 π 0 ) 2 2 ~ 2 n 2 so that ∂E n ∂e = - 2 m e e 3 (4 π 0 ) 2 ~ 2 n 2 Thus Feynman-Hellman gives - 2 e 4 π 0 h R - 1 i = - 2 m e e 3 (4 π 0 ) 2 ~ 2 n 2 Solving for h R - 1 i thus gives h R - 1 i = m e e 2 4 π 0 ~ 2 n 2 = 1 n 2 a 0 . 2
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Now ∂H ∂‘ = ∂‘ ~ 2 ( + 1) 2 m e R 2 = ~ 2 ( + 1 / 2) m e R 2 and ∂E n ∂‘ = - ∂‘ ~ 2 2 m e a 2 0 n 2 ( ) = ~ 2 m e a 3 0 n 3 where we have used ∂‘ n ( ) = 1. Putting these results into Feynman-Hellman gives ~ 2 ( + 1 / 2) m e h R - 2 i = ~ 2 m e a 2 0 n 3 Solving for h R - 2 i gives h R - 2 i = 1 ( + 1 / 2) n 3 a 2 0 3
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3. [20 pts] Deriving Kramer’s relation: a.) First, show via integration by parts that R dr ur s u 0 = - s 2 R dr ur s - 1 u . Then use this result to show that R dr u 0 r s u 0 = - 2 s +1 R dr u 0 r s +1 u 00 . Let: f = ur 2 dg = u 0 dr df = ( u 0 r 2 + 2 r 2 - 1 ) dr g = u Thus we have Z dr ur s u 0 = Z f dg = fg - Z gdf = ur s u - Z dr ( u 0 r s - sur s - 1 ) u Assuming the boundary term vanishes, this gives Z dr ur s u 0 = - Z dr u 0 r s u - s Z dr ur s - 1 u Collecting like terms then gives Z dr ur s u 0 = - s 2 Z dr ur s - 1 u This allows us to eliminate a first-derivative at the cost of reducing the power of r by one. This result can be rewritten as Z dr ur s - 1 u = - 2 s Z dr ur s u 0 with u u 0 and s s + 1 this becomes Z dr u 0 r s u 0 = - 2 s + 1 Z dr u 0 r s +1 u 00 b.) With R n‘ ( r ) = u ( r ) /r , the radial eigenvalue equation of the hydrogen atom becomes u 00 = ( + 1) r 2 - 2 a 0 r + 1 n 2 a 2 0 u.
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