C3
fta aft fta aft
Multiplying the bottom row of the matrix by the column vector, we have
Jc2 + (Av — J/2)c3 = Ec3 J sin© + (Av — J/2)cos© = (Av' — J/2)cos© J sin© = (Av' — J/2 — Av + J/2)cos© tan© = sin©/cos© = (Av' — Av)/J
In the weak coupling limit, Av' becomes Av, the ratio on the right side becomes zero and the angle © goes to zero. The wave function becomes simply aft since c3 = cos © = 1 and c2 = sin © = 0. As Av gets smaller and approaches J, the difference Av' — Av gets larger relative to J and the angle © becomes positive. The wave function starts to be a mixture of aft and fta. When Av reaches zero (va = vb), Av' = J and we have tan © = 1 and © =
45°. The wave function is now an equal mixture of pa and ap:
with energy +J/2. Like aa and pp, this wave function is symmetric: if we switch the labels on Ha and Hb the wave function is the same.
A similar method shows that the other energy solution, —Av' — J/2, gives a wave function ^ = cos ©pa — sin © ap with © defined in the same way in terms of Av', Av and J. With Av > J, the angle is close to zero and we have ^ ~ pa. As Av gets smaller we get more of the ap state mixed into the wave function, but now with a negative coefficient. When Av = 0 we have the wave function
with energy —3J/2. This wave function is antisymmetric: switching the labels on Ha and Hb makes the pa state into the ap state and viceversa, and this changes the sign of the wave function. The two outer lines of the spectrum (lines 1 and 4) involve transitions from this antisymmetric state The frequencies are linel: E4 — E2 = ±[(2v + J/2) — (—3 J/2)] = v + J
line4: E2 — E1 = 2[(—3J/2) — (—2v + J/2)] = v — J
The inner lines (lines 2 and 3) involve transitions from the symmetric state ^3, line3: E4 — E3 = 2[(2v + J/2) — (J/2)] = v line2: E3 — E1 = ^[(J/2) — (—2v + J/2)] = v
A rule of quantum mechanics states that transitions between states of opposite symmetry are forbidden; this is why the intensity of the outer lines falls to zero in the limit of Av = 0. In between, in the strong coupling "zone," the outer lines are diminished in intensity and this gives the "leaning" or "house shape" of the AB system.
The intensity of the lines turns out to be: [^„(I^+Ib)^m]2 for a transition between state m and state n, so we can calculate them using matrix math. For line 1 (^2 ^ transition) we have for [^4(1* +I*)^2]:

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