- Saturation

Saturation

Previously, we considered the interaction between the spins and lattice and found the characteristic relaxation time $T_{1}$. That is, when the spin temperature $T_{_{S}}$ is different than the lattice temperature $T$ how long does it take for the spin temperature to return to the lattice temperature? Answer: $T_{1}$.

We then discussed spin-spin interactions which led to the notion of inhomogeneous broadening by a spatially-varying local magnetic field. This led to the concept of a phase-memory time constant $T_{2}$. That is, when two nearly-resonant spins are excited then allowed to freely decay (RF field off) how long before the two spins are out of phase? Answer: $T_{2}$.

We now investigate the steady-state behavior of the spin-lattice system with an applied transverse RF field.

In the absence of the transverse RF field, the differential equation describing the time variation of $n$, the excess number of nuclei in the lower state, is ${\tfrac {dn}{dt}}={\tfrac {n_{_{0}}-n}{T_{_{1}}}}$. This makes sense because if $T_{S}>T$ then ${\tfrac {dn}{dt}}$ is greater than zero ($n<n_{_{0}}$) and there's a net flow of nuclei into the lower state with a characteristic time constant $T_{_{1}}$.

When the transverse RF field is present, another term must be added to account for the induced upward transitions corresponding to a net absorption of energy so that we get ${\tfrac {dn}{dt}}={\tfrac {n_{_{0}}-n}{T_{_{1}}}}-2nP$, where $P$ is the probability per unit time that a nuclei makes an upward transition. A steady state is reached when ${\tfrac {dn}{dt}}=0$ so that a steady-state value of $n_{_{ss}}$ is given by ${\tfrac {n_{_{ss}}}{n_{_{0}}}}={\tfrac {1}{1+2PT_{_{1}}}}$. It can be seen that $n_{_{ss}}$ decreases as $P$ or $T_{_{1}}$ increases.

We must now evaluate the transition probability $P$. If a transverse RF field is applied with the correct sense of rotation at the resonant frequency the probability of a transition in unit time between the states is given (with the help of Fermi's golden rule) by

$P_{_{m\rightarrow m'}}={\tfrac {1}{2}}\gamma ^{2}H_{_{1}}^{1}|<m|I|m'>|^{2}g(\nu )$, where $<m|I|m'>$ is the appropriate matrix element of the nuclear spin operator.

For $m'=m-1$ it can be shown that $|<m|I|m'>|^{2}={\tfrac {1}{2}}(I+m)(I-m+1)$. For the case of $I={\tfrac {1}{2}}$ this reduces to

$P={\tfrac {1}{4}}\gamma ^{2}H_{_{1}}^{2}g(\nu )$ so that we have

${\tfrac {n_{_{ss}}}{n_{_{0}}}}=[1+{\tfrac {1}{2}}\gamma ^{2}H_{_{1}}^{2}T_{_{1}}g(\nu )]^{-1}\equiv Z$.

If an RF field is applied, whose amplitude $H_{_{1}}$ is large, ${\tfrac {n_{_{ss}}}{n_{_{0}}}}$ becomes quite small; the spin temperature $T_{_{S}}$ becomes very high and the spin system is said to be saturated. Remembering that $T_{_{2}}={\tfrac {1}{2}}g(\nu )_{_{max}}$ we can write

${\tfrac {n_{_{ss}}}{n_{_{0}}}}=[1+\gamma ^{2}H_{_{1}}^{2}T_{_{1}}T_{_{2}}]^{-1}=Z_{_{0}}$, where $Z_{_{0}}$ is the value of the saturation factor at the maximum of the lineshape function $g(\nu )$.

Recall that, in thermal equilibrium, the lattice temperature $T$ is defined in terms of $n_{_{0}}$ as $T={\tfrac {N\mu H_{_{0}}}{n_{_{0}}k}}$. Similarly, when the system is not in thermal equilibrium, the spin temperature $T_{_{S}}$ is related to $n$ as $T_{_{S}}={\tfrac {N\mu H_{_{0}}}{kn}}$. These two results yield

$T_{_{S}}=T{\tfrac {n_{_{0}}}{n_{_{ss}}}}={\frac {T}{Z}}$.

The spins can easily be heated up to extremely high temperatures. For example, at just below room temperature for $^{1}H$, a transverse field of 0.1 Gauss, $T_{_{1}}=1{\text{sec}}$ (cold water) and $T_{_{2}}=10^{-4}$ seconds, $Z\approx 0.05$ giving a spin temperature $T_{_{S}}=T/Z\approx 5400K$.