137A_HW10_solutions.pdf

# 137A_HW10_solutions.pdf - Physics 137A Problem Set 10...

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Physics 137A: Problem Set 10 solutions April 27, 2017 Problem 1 (a) Recall from homework 9 that for a nucleus of charge Ze , we have the energies: E n = - m 2 ~ 2 Ze 2 4 π 0 2 1 n 2 ! = - 13 . 6 Z 2 n 2 So for Z = 3, the 2 s state, corresponding to n = 2, has energy -30.6 eV. The 2 p state has the same energy; indeed, in terms of the angular momentum number l , a state with the letter s means l = 0, while p means l = 1, but the energies above are degenerate in the letter l , since they only depend on n . (b) If Z eff = 1, the energies are nothing but the usual energies of hydrogen. So we get at once that the 2 s state has energy - 13 . 6 / 2 2 = - 3 . 4 eV. The energy for the 2 p state is once again the same. (c) P ( r < a 0 ) = a 0 Z 0 r 2 | R 20 | 2 dr = a 0 Z 0 r 2 1 2 a 3 0 1 - r 2 a 0 2 e - r/a 0 dr = 1 - 21 8 e 0 . 034 1

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(d) P ( r < a 0 ) = a 0 Z 0 r 2 | R 21 | 2 dr = a 0 Z 0 r 2 1 24 a 3 0 r 2 a 2 0 e - r/a 0 dr = 1 - 65 24 e 0 . 0036 NOTE: we only cared about the radial part of the wavefunction, because the angular part is automatically normalized; indeed, P ( r < a 0 ) = 2 π Z 0 π Z 0 a 0 Z 0 | R nl | 2 | Y lm | 2 r 2 sin( θ ) drdθdφ = 2 π Z 0 π Z 0 | Y lm | 2 sin( θ ) dθdφ a 0 Z 0 | R nl | 2 r 2 dr = 1 · a 0 Z 0 | R nl | 2 r 2 dr So it was fine to ignore the angular part here. (e) Here we won’t have to compute anything new, since P ( r > a 0 ) = 1 - P ( r < a 0 ). For the 2 s state, we get: Z eff = 3 P ( r < a 0 ) + P ( r > a 0 ) = 3 1 - 21 8 e + 1 - 1 - 21 8 e 1 . 069 For the 2 p state, we get: Z eff = 3 P ( r < a 0 ) + P ( r > a 0 ) = 3 1 - 65 24 e + 1 - 1 - 65 24 e 1 . 007 2
(f) For the 2 s state, we get: E 2 = - 13 . 6 Z 2 eff 2 2 = - 3 . 88 eV For the 2 p state, we get: E 2 = - 13 . 6 Z 2 eff 2 2 = - 3 . 45 eV Problem 2 (a) We need to recall that [ AB, C ] = A [ B, C ] + [ A, C ] B (you can prove that easily by expanding both sides). Then we get: [ ˆ L 2 , ˆ L z ] = [ ˆ L 2 x + ˆ L 2 y + ˆ L 2 z , ˆ L z ] = [ ˆ L 2 x , L z ] + [ ˆ L 2 y , ˆ L z ] + 0 = ˆ L x [ ˆ L x , ˆ L z ] + [ ˆ L x , ˆ L z ] ˆ L x + ˆ L y [ ˆ L y , ˆ L z ] + [ ˆ L y , ˆ L z ] ˆ L y = ˆ L x ( - i ~ ˆ L y ) + ( - i ~ ˆ L y ) ˆ L x + ˆ L y ( i ~ ˆ L x ) + ( i ~ ˆ L x ) ˆ L y = 0 (b) We use the fact that ˆ L x and ˆ L y are hermitian operators (since they represent physical observables).

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