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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 + ( r 2 3 z 2 ) , where is a small parameter. Neglect the spin of the electron. (a) Compute the shifts of the n = 2 energy levels (you may neglect finestructure 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 m = 1 4 Ry , where 1 Ry = me 4 / (2 planckover2pi1 2 ) = 13 . 6 eV. The corresponding unperturbed energy eigenfunc tions are: 200 ( vector r ) = 1 parenleftbigg 1 2 a parenrightbigg 3 / 2 parenleftbigg 1 r 2 a parenrightbigg e r/ (2 a ) , 210 ( vector r ) = 1 2 parenleftbigg 1 2 a parenrightbigg 3 / 2 r a e r/ (2 a ) cos , 21 1 ( vector r ) = 1 8 parenleftbigg 1 a parenrightbigg 3 / 2 r a e r/ (2 a ) sin e i , where a = planckover2pi1 2 / ( me 2 ). The perturbing Hamiltonian is: H (1) = ( r 2 3 z 2 ) = r 2 (1 3 cos 2 ) . To compute the firstorder energy shifts of degenerate states, we must construct the matrix elements of the matrix W , which is defined as: W m ; m ( 2 m  H (1)  2 m ) . (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 and H (1) is independent of . It follows that: 0 = ( 2 m  [ H (1) , L z ]  2 m ) = ( 2 m  ( H (1) L z L z H (1) )  2 m ) = ( m m ) ( 2 m  H (1)  2 m ) , which implies that ( 2 m  H (1)  2 m ) = 0 if m negationslash = m . 1 The only other nondiagonal 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 (1 3 cos 2 ) = 2 integraldisplay 1 1 d cos (1 3 cos 2 ) = (cos cos 3 ) vextendsingle vextendsingle vextendsingle +1 1 = 0 . (2) It therefore follows that the firstorder energy shifts of the n = 2 states are given by: E (1) 2 m = ( 2 m  H (1)  2 m ) . First we compute the firstorder energy shift of the state  2 0 0 ) : E (1) 200 = ( 2 0 0  r 2 (1 3 cos 2 )  2 0 0 ) = 8 a 3 integraldisplay d 3 r parenleftbigg 1 r 2 a parenrightbigg 2 r 2 e r/a (1 3 cos 2 ) . Using eq. (2), it follows that: E (1) 200 = 0 Turning to the n = 2, = 1 states, E (1) 210 = 32 a 5 integraldisplay d 3 r r 4 e r/a cos 2 (1 3 cos 2 ) , (3) E (1) 21 1 = 64 a 5 integraldisplay d 3 r r 4 e r/a sin 2 (1 3 cos 2 ) , (4) After writing d 3 r = r 2 drd , the radial integrals are all of the form: integraldisplay r n e r/a dr...
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This document was uploaded on 10/31/2011.
 Spring '09
 Energy

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