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This is just a pair of simultaneous linear equations. To solve for E we set the determinant (product of diagonal terms minus product of antidiagonal terms) to zero:

E2 + JE + (-Av2 - J2 + J2/4) = 0; E2 + JE + (-Av2 - 3J2/4) = 0

The solution to this quadratic equation in E is

We can visualize Av' as the hypotenuse of a right triangle with sides equal to J and Av (Fig. 10.38, left). If Av > J (weak coupling limit), Av' becomes nearly equal to Av and the energies are just the diagonal elements H22 and H33 of the Hamiltonian. If Av = 0, we have Av' = J and the two energies are +J/2 and -3J/2. The line positions are just the transition frequencies or the energy differences between the pairs of energy levels involved in the transition, so we can diagram the effect of strong coupling on the spectrum

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