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Unformatted text preview: 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 FeynmanHellmann 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 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 r . The exact eigenvalues in terms of e and ‘ are E n = Me 4 32 π 2 2 ~ 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 FeynmanHellman theorem with λ = e to derive: h n‘ (0)  R 1  n‘ (0) i = 1 n 2 a . 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 . According to FeynmanHellman h n‘m ‘  ∂H ∂e  n‘m ‘ i = ∂E n ∂e now ∂H ∂e = 2 e 4 π 1 R and we have E n = ~ 2 2 m e a 2 n 2 with a = 4 π ~ 2 m e e 2 this works out to E n = m e e 4 (4 π ) 2 2 ~ 2 n 2 so that ∂E n ∂e = 2 m e e 3 (4 π ) 2 ~ 2 n 2 Thus FeynmanHellman gives 2 e 4 π h R 1 i = 2 m e e 3 (4 π ) 2 ~ 2 n 2 Solving for h R 1 i thus gives h R 1 i = m e e 2 4 π ~ 2 n 2 = 1 n 2 a . 2 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 n 2 ( ‘ ) = ~ 2 m e a 3 n 3 where we have used ∂ ∂‘ n ( ‘ ) = 1. Putting these results into FeynmanHellman gives ~ 2 ( ‘ + 1 / 2) m e h R 2 i = ~ 2 m e a 2 n 3 Solving for h R 2 i gives h R 2 i = 1 ( ‘ + 1 / 2) n 3 a 2 3 3. [20 pts] Deriving Kramer’s relation: a.) First, show via integration by parts that R dr ur s u = s 2 R dr ur s 1 u . Then use this result to show that R dr u r s u = 2 s +1 R dr u r s +1 u 00 . Let: f = ur 2 dg = u dr df = ( u r 2 + 2 r 2 1 ) dr g = u Thus we have Z dr ur s u = Z f dg = fg Z gdf = ur s u Z dr ( u r s sur s 1 ) u Assuming the boundary term vanishes, this gives Z dr ur s u = Z dr u r s u s Z dr ur s 1 u Collecting like terms then gives Z dr ur s u = s 2 Z dr ur s 1 u This allows us to eliminate a firstderivative at the cost of reducing the power of r by one....
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This note was uploaded on 12/21/2011 for the course PHYS 852 taught by Professor Moore during the Spring '11 term at Michigan State University.
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