Nuclear quadrupole resonance

In the case of NMR, nuclei with spin ≥ 1/2 have a magnetic dipole moment so that their energies are split by a magnetic field, allowing resonance absorption of energy related to the Larmor frequency:

where ${\displaystyle \gamma }$ is the gyromagnetic ratio and ${\displaystyle B}$ is the (normally applied) magnetic field external to the nucleus.

In the case of NQR, nuclei with spin ≥ 1, such as ¹⁴N, ¹⁷O, ³⁵Cl and ⁶³Cu, also have an electric quadrupole moment. The nuclear quadrupole moment is associated with non-spherical nuclear charge distributions. As such it is a measure of the degree to which the nuclear charge distribution deviates from that of a sphere; that is, the prolate or oblate shape of the nucleus. NQR is a direct observation of the interaction of the quadrupole moment with the local electric field gradient (EFG) created by the electronic structure of its environment. The NQR transition frequencies are proportional to the product of the electric quadrupole moment of the nucleus and a measure of the strength of the local EFG:

where q is related to the largest principal component of the EFG tensor at the nucleus. ${\displaystyle C_{q}}$ is referred to as the quadrupole coupling constant.

In principle, the NQR experimenter could apply a specified EFG in order to influence ${\displaystyle \omega _{Q}}$ just as the NMR experimenter is free to choose the Larmor frequency by adjusting the magnetic field. However, in solids, the strength of the EFG is many kV/m^2, making the application of EFG's for NQR in the manner that external magnetic fields are chosen for NMR impractical. Consequently, the NQR spectrum of a substance is specific to the substance - and NQR spectrum is a so called "chemical fingerprint." Because NQR frequencies are not chosen by the experimenter, they can be difficult to find making NQR a technically difficult technique to carry out. Since NQR is done in an environment without a static (or DC) magnetic field, it is sometimes called "zero field NMR". Many NQR transition frequencies depend strongly upon temperature.

Consider a nucleus with a non-zero quadrupole moment ${\textstyle {\textbf {Q}}}$ and charge density ${\textstyle \rho ({\textbf {r}})}$, which is surrounded by a potential ${\textstyle V({\textbf {r}})}$. This potential may be produced by the electrons as stated above, whose probability distribution might be non-isotropic in general. The potential energy in this system equals to the integral over the charge distribution ${\textstyle \rho ({\textbf {r}})}$ and the potential ${\textstyle V({\textbf {r}})}$ within a domain ${\textstyle {\mathcal {D}}}$:

$${\displaystyle U=-\int _{\mathcal {D}}d^{3}r\rho ({\textbf {r}})V({\textbf {r}})}$$

One can write the potential as a Taylor-expansion at the center of the considered nucleus. This method corresponds to the multipole expansion in cartesian coordinates (note that the equations below use the Einstein sum-convention):

$${\displaystyle V({\textbf {r}})=V(0)+\left[\left({\frac {\partial V}{\partial x_{i}}}\right){\Bigg \vert }_{0}\cdot x_{i}\right]+{\frac {1}{2}}\left[\left({\frac {\partial ^{2}V}{\partial x_{i}x_{j}}}\right){\Bigg \vert }_{0}\cdot x_{i}x_{j}\right]+...}$$

The first term involving ${\textstyle V(0)}$ will not be relevant and can therefore be omitted. Since nuclei do not have an electric dipole moment ${\textstyle {\textbf {p}}}$, which would interact with the electric field ${\textstyle {\textbf {E}}=-\mathrm {grad} V({\textbf {r}})}$, the first derivatives can also be neglected. One is therefore left with all nine combinations of second derivatives. However if one deals with a homogeneous oblate or prolate nucleus the matrix ${\textstyle Q_{ij}}$ will be diagonal and elements with ${\textstyle i\neq j}$ vanish. This leads to a simplification because the equation for the potential energy now contains only the second derivatives in respect to the same variable:

$${\displaystyle U=-{\frac {1}{2}}\int _{\mathcal {D}}d^{3}r\rho ({\textbf {r}})\left[\left({\frac {\partial ^{2}V}{\partial x_{i}^{2}}}\right){\Bigg \vert }_{0}\cdot x_{i}^{2}\right]=-{\frac {1}{2}}\int _{\mathcal {D}}d^{3}r\rho ({\textbf {r}})\left[\left({\frac {\partial E_{i}}{\partial x_{i}}}\right){\Bigg \vert }_{0}\cdot x_{i}^{2}\right]=-{\frac {1}{2}}\left({\frac {\partial E_{i}}{\partial x_{i}}}\right){\Bigg \vert }_{0}\cdot \int _{\mathcal {D}}d^{3}r\left[\rho ({\textbf {r}})\cdot x_{i}^{2}\right]}$$

The remaining terms in the integral are related to the charge distribution and hence the quadrupole moment. The formula can be simplified even further by introducing the electric field gradient ${\textstyle V_{ii}={\frac {\partial ^{2}V}{\partial x_{i}^{2}}}=eq}$ , choosing the z-axis as the one with the maximal principal component ${\textstyle Q_{zz}}$ and using the Laplace equation to obtain the proportionality written above. For an ${\textstyle I=3/2}$ nucleus one obtains with the frequency-energy relation ${\textstyle E=h\nu }$:

$${\displaystyle \nu ={\frac {1}{2}}\left({\frac {e^{2}qQ}{h}}\right)}$$