The NMR spectrum of a system of nuclear spins in thermal equilibrium is directly proportional to $\frac{\mu H_0}{kT}$, where $\mu H_0$ is the difference in Zeeman energy between adjacent nuclear magnetic states and $T$ is the temperature. The metastable system, solid ortho-hydrogen, is far from thermal equilibrium because of the large rotational energy of the ($J=1\mathrm{}$) molecules. Thus the populations of the three magnetic states of the ($I=1\mathrm{}$) total-nuclear-spin wave functions are affected not only by the magnetic field and the temperature but also by the rate of ortho-para conversion from each of the three states. In this paper we calculate the difference $D$ between the ortho-para conversion rates from the $m_I=0\mathrm{}$ and the $m_I=\pm 1$ states for a crystal of ortho-hydrogen in the ordered state. It is found that $D$ depends on ($3{cos}^2\beta -1\mathrm{}$), where $\beta$ is the angle between the magnetic field and the symmetry axis of the molecular wave function. We then compute the steady-state populations of the nuclear-spin states as a function of $\frac{\mu H_0}{kT}$, $D$, and the nuclear spin-lattice relaxation time $T_1$. These are used to calculate the shape of the NMR spectrum of a powder sample for values of $T_1$ which are appropriate to the ordered state. The result is that the usual Pake line shape is distorted by an enhancement which is linear in frequency shift and proportional to $T_{1}D$. An expression is also derived for the average ortho-para conversion rate as a function of molar volume and the Debye energy which shows that the conversion rate, which we have calculated for the two-phonon process, is negligible below 20 ${\mathrm{cm}}^3$/mole. By contrast, experiments show that at this molar volume the rate increases sharply with $\frac{1}{V}$. Our conclusion is that the increasing rate is due to a one-phonon process which is only effective for $V$ less than about 22 ${\mathrm{cm}}^3$/mole.

• Received 8 June 1973

DOI:https://doi.org/10.1103/PhysRevB.8.5013