Why are multiple-quantum coherences (coherences with changes in magnetic quantum numbers, $|\Delta m| \neq 1$) unobservable in regular NMR experiments?


2 Answers 2


I'm not sure what you mean by 'regular' NMR experiments. NMR experiments which observe multi quantum signal are still 'regular' experiments, so I'll assume you asking why you wouldn't see DQ signal in a basic 1D 1H spectrum, for instance. I will attempt to explain without the use of product operator formalism, and scary looking maths.

In its simplest terms, the frequency of a double quantum coherence is about twice the normal observation frequency, and so cannot be observed without transformation back to single quantum coherence. Similarly, zero quantum coherence is typically less than a few kilohertz, and so is also not observable. If you set your receiver to the correct frequency, this transition would be observable, although significantly lower in intensity due to the lower probability of this transition path.

Energy transitions

In the case of an isolated, uncoupled spin, there are only two energy states, and the transition between these two states (the single quantum transition) is the only possible transition. For the case of coupled nuclei, the state of spin I is affected by the spins state of nucleus S. The diagram above shows the typical spin transitions for a spin system, IS. The most common transition is a single quantum transition, where only a single spin changes state, and this generates magnetisation in the transverse plane. When both spins change state simultaneously, you can get either zero quantum (the spins flip to opposite states) or double quantum (spins flip the same way) transitions. The frequency of the double quantum transition is roughly twice that of a single quantum transition. In the vector model, double quantum transitions have no magnetisation in the transverse plane (only transverse magnetisation can be observed), but can be transferred to single quantum transition through the process of generation of SQC(first pulse), evolution (linked to couplings) and generation of MQC (second pulse) in a multi quantum experiment. This is typical of a DQF/TDF COSY type experiment, but equally utilised in studies of dipolar couplings of small organics in liquid crystal media (where quantum orders of >5 are typically observed)

  • $\begingroup$ This makes sense. I guess what I don't understand now is why within the product operator formalism the DQ/ZQ coherence operators represent transitions and not states/polarizations. $\endgroup$
    – J. LS
    Commented May 6, 2015 at 8:21
  • 1
    $\begingroup$ What do you think about @levineds answer? $\endgroup$
    – Mithoron
    Commented Jun 16, 2017 at 16:03

long's answer (above the figure) is fundamentally incorrect. The most basic thing to recall is that both your detector and amplifier produce dipolar fields and so only dipole-allowed transitions are observable and multiple-quantum transitions are higher order multipoles.

In a liquid state sample, tuning your detector to twice the frequency will not produce a detectable signal. While it is sometimes possible to see such a weak transition in some systems, this is due to an overtone transition (in which mixing between dipole-allowed states makes the transition weakly allowed) not a multiquantum transition. Similarly, zero quantum coherences are also higher order. They cannot be observed because dipoles and higher order multipoles have zero overlap, not because the energy scale is small.

If you had a quadrupolar detector, you could detect multiquantum coherences.

Another way to look at it is that magnetization (like dipoles, a rank-1, vector-like operator) is always 0 for all transitions of higher order at all times and so there is never any signal to pick up in the coil no matter what frequency you are looking.


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