Electric Quadrupole Splitting

A nucleus that has a spin quantum number $ I>\nicefrac{1}{2}$ has a non-spherical charge distribution. The magnitude of the charge deformation, $ Q$, is given by

$\displaystyle eQ=\int \rho r^{2} \left( 3 \cos^{2} \theta- 1 \right) \mathrm{d}\tau$ (2.8)

where $ e$ is the charge on the proton, $ \rho$ is the charge density in a volume element $ \mathrm{d}\tau$ at a distance $ r$ from the center of the nucleus and making an angle $ \theta$ to the nuclear spin quantisation axis. The sign of $ Q$ indicates the shape of the deformation. Negative $ Q$ is due to the nucleus being flattened along the spin axis, an elongated nucleus giving positive $ Q$.[5]

An asymmetric charge distribution around the nucleus causes an asymmetric electric field at the nucleus, characterised by a tensor quantity called the Electric Field Gradient (EFG) $ \nabla{}E$. The electric quadrupole interaction between these two quantities gives rise to a splitting in the nuclear energy levels. The interaction between nuclear moment and EFG is expressed by the Hamiltonian

$\displaystyle \mathcal{H}_{Eq} = -\frac{1}{6}eQ\nabla E$ (2.9)

where $ \nabla E$ may be written as
$\displaystyle \nabla{}E_{ij} = -\frac{\partial^{2}V}{\partial x_{i} \partial x_{j}} = -V_{ij},$     (2.10)
$\displaystyle \{x_{i},x_{j}\} = \{x,y,z\}$      

where $ V$ is the electrostatic potential.

There are two contributions to the EFG i) lattice contributions from charges on distant ions and ii) valence contributions due to incompletely filled electron shells. If a suitable coordinate system is chosen the EFG can be represented by three principal axes, $ V_{xx}$, $ V_{yy}$ and $ V_{zz}$. If an asymmetry parameter is defined using these axes as

$\displaystyle \eta = \left( \frac{V_{xx} - V_{yy}}{V_{zz}} \right)$ (2.11)

where $ \vert V_{zz}\vert \ge \vert V_{yy}\vert \ge \vert V_{xx}\vert$ so that $ 0 \le \eta \le 1$, the EFG can be specified by two parameters: $ V_{zz}$ and $ \eta$.

The Hamiltonian for the quadrupole interaction can be rewritten as

$\displaystyle \mathcal{H}_{Eq} = \frac{e^{2}qQ}{4I(2I-1)} \left[ 3\boldsymbol{I...
...{\eta}{2}\left( \boldsymbol{I^{2}_{+}} + \boldsymbol{I^{2}_{-}} \right) \right]$ (2.12)

where $ \boldsymbol{I_{+}}$ and $ \boldsymbol{I_{-}}$ are shift operators and $ \boldsymbol{I_{z}}$ is a spin operator.[5]

The excited state of $ ^{57}$Fe has a spin $ I=\nicefrac{3}{2}$. The EFG has no effect on the $ I=\nicefrac{1}{2}$ ground state but does remove degeneracy in the excited state, splitting it into two sub-states $ m_{I} = \pm\nicefrac{1}{2}$ and $ m_{I} = \pm\nicefrac{3}{2}$ where the $ m_{I} = \pm\nicefrac{3}{2}$ states are higher in energy for positive $ V_{zz}$. The energy eigenvalues for $ I=\nicefrac{3}{2}$ have exact solutions given by

$\displaystyle E_{Eq} = \frac{e^{2}qQ}{4I(2I - 1)} \left[ 3m^{2}_{I} - I(I+1) \right] \left( 1 + \frac{\eta^{2}}{3} \right)^{\frac{1}{2}}$ (2.13)

whilst the energies for higher spin states require analytical methods to calculate the energies.

The now non-degenerate excited states give rise to a doublet in the Mössbauer spectrum as illustrated in Figure 2.4. The separation between the lines, $ \Delta $, is known as the quadrupole splitting and is given by

$\displaystyle \Delta = \frac{e^{2}qQ}{2} \left( 1 + \frac{\eta^{2}}{3} \right)^{\frac{1}{2}}$ (2.14)

with the line intensities being equal for polycrystalline samples. Texture or orientation effects can lead to asymmetric doublets.

Figure 2.4: The effect on the nuclear energy levels for a $ \nicefrac {3}{2}\to {}\nicefrac {1}{2}$ transition, such as in $ ^{57}$Fe or $ ^{119}$Sn, for an asymmetric charge distribution. The magnitude of quadrupole splitting, $ \Delta $ is shown.

As the nuclear quadrupole moment is fixed the magnitude and sign of $ \Delta $ gives information about the sign of the EFG and magnitude of $ \eta$.

Dr John Bland, 15/03/2003