Problem 8.2: Intermediate field Zeeman. (40 points) 212 4, = (3121(-1);) +(31210-2), (12) Just like as in the previous problem, consider only the n = 2 level of a hydrogen atom placed in a uniform static magnetic field B = BZ, and ignore the proton spin and the hyperfine (13) coupling. (8 = 121 7 2) 1=-(3121(-1);)+131210-1. In this problem, we assume that the Zeeman and fine structure Hamiltonians are of the same order, but both are therefore much smaller than the unperturbed hydrogen Hamilto- b) What are the first-order corrections to the energies of the 8 states in the intermediate nian. Thus we take as the perturbation field limit, in terms of y and B? H' = Hfs + Hz . (3) c) Using a software package, plot all of the first-order corrections to the energy on the same plot. For numerical convenience in making this plot, instead of E(), plot & = E()/y as a a) Using the basis In ( jm;), construct the 8 x 8 H'-matrix for the perturbation in the n = 2 function of B = B/y, and plot out to around B = 15. As B gets large, what regime are we level of hydrogen. Show that it is (note the overall minus sign) entering? Compare what you see to your results for both parts a) and b) of problem 8.1. Are suitable degeneracies arising in the right patterns? Include the plot in your writeup. 57 - 3 0 0 0 57 + 3 0 d) Plot the same first-order energy corrections, but only out to B = 3 or so. As B goes to Y - 23 0 O 0 zero, what regime are we entering? Compare what you see to your results for both parts 0 H = - 0 O O Y + 26 0 0 0 Y - 23/3 V28/3 , (4) c) and d) of problem 8.1. Are suitable degeneracies arising in the right patterns? Again, include the plot in your writeup. OO 0 O V23/3 57 - B/3 0 0 0 y + 23/3 V28/3 0 0 0 V23/3 5y+ 3/3 where we have defined a*mec2 128 B= ABB, (5) and where the basis states (nejm;) are given in the order indicated below. For your con- venience, the first ket on each line is the basis vector given in terms of quantum numbers [nejm;), and the rest of the line describes how they are written in the (n(mms) basis. 41= 20,7) = 12005) (6) 1 - (2 = 20,) = 1200-), (7) (3 = 1212 2) = 12112>, 2 2' (8) 24 = 218 3) =121 (-1) > (9) 31 2)= 1312105 + 31211-3, (10) 26 = 121 1 1, -V31210;) + (312117) (11)