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Coherence order classifies an operator component of NMR coherence by its response to rotation about the chosen total $z$ angular momentum. With dimensionless $F_z$ and rotation convention $U(\varphi )=e^{-i\varphi F_z}$, order $p$ meansThe order describes a phase transformation, rather than the amount of magnetization or the duration for which coherence persists.^[ref]

An original two-spin example shows why order zero is not synonymous with a diagonal population. Let $F_z=I_z+S_z$ for two spin one-half nuclei. For raising and lowering components with $[I_z,I_{+}]=I_{+}$ and $[S_z,S_{\pm }]=\pm S_{\pm }$,The first is zero-quantum coherence; the second is a double-quantum coherence component. In a product basis, the first connects the states with projections $(+1/2,-1/2)$ and $(-1/2,+1/2)$: different states with the same total projection. The second connects total projections−1 and+1. These operator components are not individually normalized density matrices.

At $\varphi =\pi /2$, the order-zero term remains unchanged while the order-two term changes sign. At $\varphi =\pi$, both remain unchanged. A phase comparison must therefore be chosen to discriminate the orders of interest; one phase change cannot distinguish every possible component.

Order zero also does not require zero time evolution. Under the Quantum Hamiltonian $H/\hslash ={\Omega }_I{I}_z+{\Omega }_S{S}_z$, the first component acquires $e^{-i({\Omega }_I-{\Omega }_S)t}$ when the two offsets differ. Total order and the separate offsets answer different questions.^[ref]

NMR phase cycling uses such phase responses to select contributions to an experiment. Selecting an order does not ensure that the spin system or pulse sequence actually generated it, nor that it is directly observable before further conversion.^[ref]

Zero-quantum coherence is a component of NMR coherence with total coherence order $p=0$, invariant under a common $z$ rotation. In a multispin system it can connect distinct basis states having the same total spin projection. This makes it different from merely specifying diagonal populations.^[ref]

For two spin one-half nuclei, consider $A=I_{+}{S}_{-}$. With $F_z=I_z+S_z$,It transfers one projection upward and the other downward, keeping their total fixed. The operator is proportional to $|\uparrow \downarrow ⟩⟨\downarrow \uparrow |$, an off-diagonal matrix element in the product basis.

An original state comparison usesThe pure-state density matrix has a zero-quantum off-diagonal pair. An equal mixture of the two basis states has the same populations and no such pair. Both have total projection zero, but their coherent properties differ.

Invariance under a common rotation does not imply that the component never evolves. For $H/\hslash ={\Omega }_I{I}_z+{\Omega }_S{S}_z$, its evolution isFor invented offsets ${\Omega }_I/(2\pi )=100\mathrm{Hz}$ and ${\Omega }_S/(2\pi )=70\mathrm{Hz}$, its coefficient completes a phase cycle in $1/30\mathrm{s}$. Adding the same offset to both nuclei leaves that difference unchanged. Unequal perturbations can still change it; the example is not a universal protection against noise.^[ref]

The zero order helps distinguish contributions in NMR phase cycling and gradient-based coherence selection. Longitudinal magnetization also has order zero, so an order label alone does not identify a unique physical component or guarantee direct detection.^[ref]