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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 , so that H = H ( ). The eigenstates and eigenvalues of H are then also functions of , i.e. E n = E n ( ) and  n i =  n ( ) i . Use the property H  n i = E n  n i to prove the FeynmanHellmann theorem: E n ( ) = h n ( )  H ( )  n ( ) i Start from the definition E n = h n  H  n i Differentiation gives E n = h n  H  n i + h n  H  n i + h n  H  n i Since H  n i = E n  n i and h n  H = h n  E n , we can rewrite this as E n = E n h n   n i + h n   n i E n + h n  H  n i = E n h n  n i + h n  H  n i Since h n  n i = 1, it follows that E n ( ) = h n ( )  H ( )  n ( ) 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 r . The exact eigenvalues in terms of e and are E n = Me 4 32 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 = e to derive: h n (0)  R 1  n (0) i = 1 n 2 a . Then use = , 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 nm  H e  nm i = E n e now H e = 2 e 4 1 R and we have E n = ~ 2 2 m e a 2 n 2 with a = 4 ~ 2 m e e 2 this works out to E n = m e e 4 (4 ) 2 2 ~ 2 n 2 so that E n e = 2 m e e 3 (4 ) 2 ~ 2 n 2 Thus FeynmanHellman gives 2 e 4 h R 1 i = 2 m e e 3 (4 ) 2 ~ 2 n 2 Solving for h R 1 i thus gives h R 1 i = m e e 2 4 ~ 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 Kramers 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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 Spring '11
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