Nuclear Overhauser effect/Advanced

{Def|Nuclear Overhauser effect}

The Noe enhancement is quantitatively defined as

$\eta ={\frac {S_{z}-S_{z,equil}}{S_{z,equil}}}Eq.1$

For a pair of nonidentical spins I and S, :

${\frac {d<I_{z}>}{dt}}=-\rho _{I}(<I_{z}>-<I_{z,equil}>)-\sigma (<S_{z}>-<S_{z,equil}>)Eq.2$

${\frac {d<S_{z}>}{dt}}=-\rho _{S}(<S_{z}>-<S_{z,equil}>)-\sigma (<I_{z}>-<I_{z,equil}>)Eq.3$

$\sigma$ is called the cross relaxation rate and is responsible for the Nuclear overhauser effect.

$\rho _{I}={\frac {\gamma _{I}^{2}\gamma _{S}^{2}\hbar ^{2}}{10r^{6}}}(J(w_{I}-w_{S})+3J(w_{I})+6J(w_{I}+w_{S}))Eq.4$

$\sigma ={\frac {\gamma _{I}^{2}\gamma _{S}^{2}\hbar ^{2}}{10r^{6}}}(-J(w_{I}-w_{S})+6J(w_{I}+w_{S})))Eq.5$

${\frac {1}{T_{2}}}={\frac {\gamma ^{2}\gamma _{S}^{2}\hbar ^{2}}{20r^{6}}}(4J(0)+J(w_{I}-w_{S})+3J(w_{I})+6J(w_{I}+w_{S})+6J(w_{S}))Eq.6$

In the steady state ${\frac {d<S_{z}>}{dt}}=0$, when the resonance frequency of spin I is irradiated , <I_z> = 0, therefore:

$(<S_{z}>-<S_{z,equil}>)={\frac {\sigma }{\rho _{S}}}(<I_{z,equil}>)(fromEq.3)$

Therefore,

$\eta ={\frac {<S_{z}>-<S_{z,equil}>}{<S_{z,equil}>}}={\frac {\sigma }{\rho _{S}}}{\frac {\gamma _{I}}{\gamma _{S}}}Eq.7$

This indicates that considerable enhancement in the intensity of the S signal can be obtained by irradiation at the frequency of the I spin, provided that ${\frac {\gamma _{I}}{\gamma _{S}}}>1$, because ${\frac {\sigma }{\rho _{S}}}\rightarrow 1/2$ when $w\tau _{c}<<1$. However, when $w\tau _{c}>>1$, ${\frac {\sigma }{\rho _{S}}}\rightarrow -1$ and negative Noe enhancements are obtained.
The sign of $\eta$ changes from positive to negative when $w\tau _{c}$ is close to one and under such conditions the Noe effect may not be observable. This happens for rigid molecules with relative molecular mass about 500 at room temperature e.g. many hexapeptides.