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

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

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

where $\nabla E$ may be written as

$$\nabla{}E_{ij} = -\frac{\partial^{2}V}{\partial x_{i} \partial x_{j}} = -V_{ij},$$ (2.10)

$$\{x_{i},x_{j}\} = \{x,y,z\}$$

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

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

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

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

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