Av

Figure 10.38

(Fig. 10.38, right). The doublets retain their separation of J Hz but they move apart from the positions we would expect based on the weak coupling assumption. Av' is the distance between lines 1 and 3 (or between lines 2 and 4). Av' can be viewed as the hypotenuse of a right triangle with sides of length Av and J, and the angle 2© opposite J. As Av decreases relative to J, Av' approaches J and 2© approaches 90°. In the extreme case of Av = 0, Av' = J and the two central lines (lines 2 and 3) coincide in the spectrum. In this case, the outer lines at v + J and v — J have zero intensity.

Once we have the energies we can solve for the wave functions; that is, for the coefficients c2 and c3. First try the solution E = Av' — J/2, which in the weak coupling limit is the energy of the aft state (c3 coefficient). We know that the probability of being in the Pa state plus the probability of being in the aft state has to be 1, since c1 and c4 are zero. This means that c^ + c| = 1. If we restrict c2 and c3 to real numbers, we can set c3 = cos© and c2 = sin © and we know that cos2 © + sin2 © = 1. Now we only have one variable, which along with E makes two variables to extract from our two linear equations. Setting E to the solution Av' — J/2 we can solve for the "angle" ©,

-Av-J/2 J J Av-J/2
0 0

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