In PNMR, one does not directly observe the behavior of individual nuclei (protons in our case). What is measured are the macroscopic bulk magnetization, $\bf M$, arising from the alignment of large numbers of nuclei.
First consider consider the case of a large number (order $10^{19}$) of protons when no external magnetic field is present. In this case, there is no preferred direction in space, no quantization axis, and no magnetic torque exerted on the protons. Thus, individual protons will be randomly oriented in space, there will be no precession, and the two energy states shown in Fig. 2 will be degenerate such that $\Delta E =0$. Therefore, if one were to do the vector sum of all of the proton dipole moments along any axis, the average would be zero; there would be no bulk magnetization in the sample.
If we now turn on an external magnetic field,
| ${\bf B} = B_0 \bf{\hat{z}}$, | (11) |
two important changes will occur to the ensemble of protons. One is that the protons will precess about the z-axis, which is now defined by the direction of the external magnetic field, at a frequency given by Eq. (8). The other is the spin energy states are no longer degenerate. At any given moment, some number of protons will be aligned parallel ($m_l = +1/2$) to ${\bf B}$ and the remainder anti-parallel ($m_l = - 1/2$) to ${\bf B}$. If the sample is in thermal equilibrium, the average number of protons in each of the two possible states will be given by a Boltzmann distribution. If $N_1$ and $N_2$ are the number of spins per unit volume in the higher and lower energy states respectively, then the population ratio $(N_1/N_2)$ in thermal equilibrium is,
| $\dfrac{N_1}{N_2} = e^{\frac{-\Delta E}{k_BT}} = e^{{-\hbar\omega_0/k_BT}}$, | (11) |
where $T$ is the absolute temperature, $k_B$ is the Boltzmann constant, and $\Delta E = \hbar\omega_0$ is the energy difference between the two states. At room temperature, $k_BT \approx 1/40 \,eV$ and the ratio between $N_1$ and $N_2$ is nearly equal to, but not quite, 1; there is a slightly larger number of protons in the low energy state (i.e. aligned along +$z$) than in the high energy state. Now, when we sum up all of the proton dipole moments,
| $\bf{M} =\displaystyle \sum_i \boldsymbol{\mu}_i$, | (12) |
we get a non-zero net magnetization along the $z$-axis,
| ${\bf{M}} = \left( N_1 - N_2\right) \mu = M_Z \hat{\bf{z}}$. | (13) |
Because the angular momentum is quantized only along the $z$-axis, there is no preferred direction in the $xy$-plane. Therefore, when the precessing components of the proton dipole moments in the $xy$-plane are summed, ($\mu_{xy}$, see Fig. 1), they average out to zero. Figure 3 depicts this equilibrium bulk magnetization.
Note that individual protons in the sample are continually flipping back and forth between the two possible orientations (i.e. energy states) with respect to the $z$-axis. A proton in the low energy state can be excited to the upper state by absorbing a quanta of energy,$\Delta E = \hbar \omega_0$, from the local environment. Similarly, a proton in the excited state can de-excite to the low energy state by giving up a quanta of energy. It is the physical processes by which the protons in the sample exchange energy with their environment (referred to as the lattice) that determine how frequently proton spins flip between energy states. How long it takes an ensemble of protons to reach equilibrium depends on how rapidly they are changing states which in turn depends on the details of their environment. In liquids, the protons and other molecules are loosely bound and undergo Brownian motion which results in random fluctuations in the local electric and magnetic fields the protons see. These fluctuating fields are an important factor in driving the rate of spin flips in liquids.
In solids, the protons are locked into a rigid lattice structure where quantized lattice vibrations and the exchange of phonons come into play. The more rapidly protons flip spin states, the faster the ensemble reaches equilibrium.
PNMR is typically used to make measurements of the characteristic time constants associated with how an ensemble of protons reaches equilibrium in an external magnetic field. These characteristic times are $T_1$ and $T_2$. PNMR techniques – as we will see in more detail in the next section – operate roughly as follows:
The time constant $T_1$ is a measure of how fast the magnetization on the z-axis ($M_Z$) relaxes to equilibrium. Recall that $M_Z$ arises from the fact that each proton has to be in one of two possible states, either parallel or anti-parallel to $\bf B$. If there are more protons in one of the two states, the difference will add up to a net magnetization along the z-axis. Now imagine that our sample of protons is at thermal equilibrium; this yields the situation depicted in Fig. 4(a). We then do the following:
The rate at which the system approaches equilibrium is proportional to how far away from equilibrium it is ($M_0$ - $M_Z$) and inversely proportional to the characteristic relaxation time ($T_1$):
| $\dfrac{dM_Z}{dt} = \dfrac{M_0 - M_Z}{T_1}$. | (14) |
Integrating this with the initial condition $M_Z (t=0) = -M_0$, we get
| $M_Z(t) = M_0(1-2e^{-t/T_1})$. | (15) |
The other time constant of interest is $T_2$, known as the spin-spin relaxation time. $T_2$ is a measure of how long it takes for a non-equilibrium bulk magnetization in the $xy$-plane to relax back to the equilibrium value of zero. Recall that when our ensemble of protons is in the presence of an external magnetic field and at equilibrium, all of the proton dipole moments will be precessing about the $z$-axis at the same frequency. However, since angular momentum is not quantized in the $xy$-plane, the precessing component of the proton dipoles, $\mu_{xy}$, are randomly oriented and consequently sum to zero. Therefore, there is no bulk magnetization in the $xy$-plane at equilibrium. If, however, some or all of the dipoles are precessing in phase, summing over all $\mu_{xy}$ would result in a bulk magnetization in the $xy$-plane, $M_{xy}$. Furthermore, $M_{xy}$ would itself precess about the $z$-axis at frequency $\omega_0$. We normally assume that each proton dipole sees exactly the same magnetic field and therefore precesses at exactly the same frequency. If this were the case, the net magnetization$M_{xy}$ would keep the same magnitude and would maintain the constant precession indefinitely. That is not the case, though. Even if the applied magnetic field is perfectly uniform, each proton will still experience a slightly different net field because of nearest-neighbor effects. For example, each proton sees not just the external magnetic field, but also the magnetic dipole fields of the other protons (or other magnetic nuclei) in the sample which warp the local value. In addition, in a liquid sample, the nuclei are all moving about randomly, meaning that the local magnetic field fluctuations also change in time as particles come and go.
The net result is that there is a small spread of precession frequencies among the proton magnetic dipoles in the sample. The protons which initially are precessing in phase will gradually become more and more out of phase with each other until the entire ensemble is back to being randomly oriented. How quickly this happens is related to the magnitude of the spread of precession frequencies which is related to the magnitude of the local magnetic field fluctuations. This process is called spin-spin relaxation because it is driven by the interaction of the spin (equivalently, the dipole moment) of the proton with the spin of its neighboring protons.
As we did above when discussing $T_1$, let's start by imagining that we have a sample of protons at thermal equilibrium such that the net magnetization lies along the $z$-axis as in Fig. 4(a) above. We then do the following:
The differential equations which describe this relaxation are
| $\dfrac{d\vec{M}_{x*}}{dt} = -\dfrac{\vec{M}_{x*}}{T_2}$ | (16a) |
and
| $\dfrac{d\vec{M}_{y*}}{dt} = -\dfrac{\vec{M}_{y*}}{T_2}$, | (16b) |
the solutions of which are
| $M_{x*}(t) = M_0 e^{-t/T_2}$ | (17a) |
and
| $M_{y*}(t) = M_0 e^{-t/T_2}$. | (17b) |