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0 Va-

1- Vb + J/2.

jj

The diagonal terms are just twice the energies of the four spin states aa (top), ja, aft and jj (bottom) given in hertz. Here we are using the convention that aligned pairs of spins (aa and jj) are higher in energy, meaning that the " j" components of doublets are downfield of the "a" components. Up till now in this book we have used the opposite convention, effectively assuming that the J value is negative. We will drop the 1/2 factor for simplicity, and bring it back when we need to calculate energies.

To find the stationary states ("eigenfunctions") and the energies ("eigenvalues") of the Hamiltonian we need to solve the Schrodinger equation:

This would be easy except for the off-diagonal terms. If the chemical-shift difference in hertz, Av = vb - Va, is much larger than the coupling constant J, we can ignore the off-diagonal terms because they are very small compared to the diagonal terms. Then the energies are just the diagonal terms and the stationary states are just the aa, ¡a, aft and ¡3 states. This "weak coupling" assumption is what we have been working under for most of this book:

10.8.1 Strong Coupling: The AB System

We saw in Chapter 2 that when the chemical-shift difference in Hz, Av, decreases and approaches the order of magnitude of J we get distortions of the peak heights ("leaning") and of the line positions. We are now in a position to derive this effect precisely for the AB system (Ha, Hb). The Hn and H44 terms of the Hamiltonian are enormous compared to the central portion of the matrix: for a 600-MHz spectrometer va + vb is around 1200 MHz! So we can assume that these diagonal terms are the correct energies and that the aa and 33 states are true stationary states. We only need to look at the central 2 x 2 matrix in the Schrodinger equation:

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