Editing Nuclear quadrupole resonance (section)

==Principle==
Any nucleus with more than one unpaired nuclear particle (protons or neutrons) will have a charge distribution which results in an electric quadrupole moment. Allowed nuclear energy levels are shifted unequally due to the interaction of the nuclear charge with an electric field gradient supplied by the non-uniform distribution of electron density (e.g. from bonding electrons) and/or surrounding ions.  As in the case of NMR, irradiation of the nucleus with a burst of RF electromagnetic radiation may result in absorption of some energy by the nucleus which can be viewed as a [[perturbation theory|perturbation]] of the quadrupole energy level. Unlike the NMR case, NQR absorption takes place in the absence of an external magnetic field.  Application of an external static field to a quadrupolar nucleus splits the quadrupole levels by the energy predicted from the [[Zeeman effect|Zeeman interaction]]. The technique is very sensitive to the nature and symmetry of the bonding around the nucleus.  It can characterize [[phase transition]]s in solids when performed at varying temperature. Due to symmetry, the shifts become averaged to zero in the liquid phase, so NQR spectra can only be measured for solids.

===Analogy with NMR===
In the case of NMR, nuclei with [[Spin (physics)|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 precession#Larmor frequency|Larmor frequency]]:
{{center|
<math>\omega_L = \gamma B</math> 
}}
where <math>\gamma</math> is the [[gyromagnetic ratio]] and <math>B</math> is the (normally applied) magnetic field external to the nucleus.

In the case of NQR, nuclei with spin ≥ 1, such as <sup>14</sup>[[nitrogen|N]], [[Oxygen-17|<sup>17</sup>O]], <sup>35</sup>[[chlorine|Cl]] and <sup>63</sup>[[copper|Cu]], also have an [[Quadrupole#Electric quadrupole|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 [[spheroid|prolate]] or [[spheroid|oblate]] shape of the nucleus.  NQR is a direct observation of the interaction of the quadrupole moment with the local [[Electric field gradient|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: 
{{center|
<math> \omega_Q \sim \frac{e^2 Q q}{\hbar} = C_q</math>
}}
where q is related to the largest principal component of the EFG tensor at the nucleus. <math>C_q</math> is referred to as the quadrupole coupling constant.

In principle, the NQR experimenter could apply a specified EFG in order to influence <math>\omega_Q</math> 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.

=== Derivation of resonance frequency ===
Source:<ref name=":0">{{Cite journal|last=Smith|first=J. A. S.|date=January 1971|title=Nuclear Quadrupole Resonance Spectroscopy|url=https://pubs.acs.org/doi/pdf/10.1021/ed048p39|journal=Journal of Chemical Education|volume=48|pages=39–41|doi=10.1021/ed048p39 |url-access=subscription}}</ref>

Consider a nucleus with a non-zero quadrupole moment <math display="inline">\textbf{Q}</math> and charge density <math display="inline">\rho(\textbf{r})</math>, which is surrounded by a potential <math display="inline">V(\textbf{r})</math>. 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 <math display="inline">\rho(\textbf{r})</math> and the potential <math display="inline">V(\textbf{r})</math> within a domain <math display="inline">\mathcal{D}</math>:

<math display="block">U = - \int_{\mathcal{D}}d^3r \rho(\textbf{r})V(\textbf{r})</math>One can write the potential as a [[Taylor series|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):

<math display="block">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] + ...</math>

The first term involving <math display="inline">V(0)</math> will not be relevant and can therefore be omitted. Since nuclei do not have an [[electric dipole moment]] <math display="inline">\textbf{p}</math>, which would interact with the electric field <math display="inline">\textbf{E} = - \mathrm{grad} V(\textbf{r})</math>, 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 <math display="inline">Q_{ij}</math> will be diagonal and elements with <math display="inline">i \neq j</math> 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:

<math display="block">U = - \frac{1}{2} \int_{\mathcal{D}}d^3r \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^3r \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^3r \left[\rho(\textbf{r}) \cdot x_i^2 \right]</math>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]] <math display="inline">V_{ii} = \frac{\partial^2 V}{\partial x_i^2} = eq </math> , choosing the z-axis as the one with the maximal principal component <math display="inline">Q_{zz} </math> and using the [[Laplace's equation|Laplace equation]] to obtain the proportionality written above. For an <math display="inline">I = 3/2</math> nucleus one obtains with the [[Planck–Einstein relation|frequency-energy relation]] <math display="inline">E = h\nu</math>:

<math display="block">\nu = \frac{1}{2}\left(\frac{e^2qQ}{h}\right)</math>