## Zero And Doublequantum Operators

Product operators in which both components are in the X-y' plane represent zero-quantum and double-quantum coherences (collectively called "multiple-quantum" coherences). DQC is a superposition of the spin states aIas and ßi ßs, which involves promotion of both nuclei I and S simultaneously from the a state to the ß state or vice versa. ZQC is a superposition of the spin states aI ßs and ßIas, which involves nucleus I flipping from a to ß while nucleus S flips from the ß state to the a state, or the reverse process. Neither of these coherences can be directly observed, but we can convert them into observable (single-quantum) coherence and see the effect of evolution during the time spent as zero- and double-quantum coherences. In product operator notation they look like this

1[2IxSx - 2IySy] = Pure DQC along the x' axis = {DQ}x 2[2IxSy + 2IySx] = Pure DQC along the / axis = {DQ}y

Figure 7.28

1[2IXSx + 2IySy] = Pure ZQC along the X axis = [ZQ}x 1[2IXSy - 2IySx] = Pure ZQC along the y! axis = [ZQ}y

DQC precesses under the influence of chemical shifts at a rate determined by the sum of the two chemical shifts, whereas ZQC precesses at a rate determined by the difference (Fig. 7.28). This can be demonstrated by plugging in [Ix cos(^it) + Iy sin(^it)] for Ix and [Sy cos(^St) — Sx sin(^St)] for Sy, etc., and multiplying the expressions together. The math is rather messy (although very satisfying) so we will not go through it here.

Neither double-quantum nor zero-quantum coherence undergoes /-coupling evolution due to JIS. This is because the energy of interaction of the two nuclear magnets does not change in the transitions aa ^ ftft and aft ^ fta, as they remain either aligned with each other or against each other. Because this energy is not involved, the J coupling has no effect on the evolution. This can also be see by examining the exact energies of the four quantum states: aa, aft, fta, and ftft. For two protons, we have (all energies are divided by Planck's constant h so we can read out frequencies in hertz directly)

aa ftb = —va /2 + vb /2 + J/4 ftaab = < /2 — vb /2 + J/4 aa ab = — < /2 — vb /2 — J/4

Note that the four single-quantum transitions have energy differences corresponding to their exact frequencies in the spectrum:

aft — aa (Hb transition"a") = vb + J/2 ftft — fta (Hb transition"ft") = vb — J/2

fta — aa (Ha transition"a") = + J/2 ftft — aft (Ha transition"ft") = vjJ — J/2

These are the four lines of a1H spectrum with two doublets: one centered on va with splitting J and the other centered on vb with splitting J. In each case, the higher frequency (downfield line) of the pair is the transition in which the other spin remains in the a state—the line we label "a" in the doublet. The two states in which the Ha and Hb spins are aligned (both a or both ft) are slightly lower in energy (by J/4) and the two states in which the Ha and Hb spins are opposed (aft and fta) are slightly higher in energy. This is not universally true, but as we have chosen to always label the downfield component of a doublet the "a" component, we are committed to this relationship.

Now look at the energy differences corresponding to the double-quantum and zero-quantum transitions:

PP - aa (DQ transition) = vao + vb Pa - afi (ZQ transition) = vao - vb

The J terms cancel out in both cases because the interaction between Ha and Hb is the same: it remains opposed (ZQ transition) or it remains aligned (DQ transition).

If there is another spin besides I and S, which is coupled to either I or S, this "passive coupling" will undergo J-coupling evolution. It is only the J coupling between the two nuclei involved in the DQC or ZQC (in this case I and S), which does not lead to any evolution.

As you can see, the bookkeeping for MQCs gets pretty messy, and we will see later (Chapter 10, Section 10.4) that another kind of operator called spherical operators is neater and easier to visualize for MQCs.

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