Non-crossing rule

The non-crossing rule in quantum chemistry states that distinct eigenvalues of two eigenstates with the same symmetry cannot become equal as a function of an external perturbation parameter.

Consider a system with two orthogonal eigenstates $\psi_1$ and $\psi_2$ of the unperturbed Hamiltonian $H^{(0)}$ , with distinct eigenvalues of $E_1$ and $E_2$ , respectively. In the presence of a weak ligand field, the unperturbed Schrodinger equation $H^{(0)}\psi_m=E_m\psi_m$ , becomes

$H\Psi_m=U_m\Psi_m\;\;\;\;\;\;\;\;281j$

where $H=H^{(0)}+H'$ , $H'$ is the perturbed Hamiltonian, and $U_m$ is the eigenvalue corresponding to the perturbed eigenstate $\Psi_m=c_1\psi_1+c_2\psi_2$ , which is assumed to be a linear combination of the unperturbed eigenstates.

Substituting $\Psi_m=c_1\psi_1+c_2\psi_2$ in eq281j,

$c_1(H-U_m)\psi_1+c_2(H-U_m)\psi_2\;\;\;\;\;\;\;\;281k$

Multiplying eq281k from the left by $\psi^*_1$ and $\psi^*_2$ in turn, and integrating over all space, we have the following two equations:

$c_1(H_{11}-U_m)+c_2H_{12}=0$

$c_1H_{21}+c_2(H_{22}-U_m)=0$

where $H_{ij}=\langle\psi_i\vert H\vert\psi_j\rangle$ .

In matrix form, the pair of equations can be expressed as

$\begin{pmatrix}H_{11}-U_m & H_{12}\\H_{21} & H_{22}-U_m\end{pmatrix}\begin{pmatrix}c_1\\c_2\end{pmatrix}=0\;\;\;\;\;\;\;\;281l$

where $I$ is the identity matrix.

Eq281l is a linear homogeneous equation, which has non-trivial solutions if

$\begin{vmatrix}H_{11}-U_m & H_{12}\\H_{21} & H_{22}-U_m\end{vmatrix}=0\;\;\;\;\;\;\;\;281m$

Expanding the determinant,

$U^2_m-(H_{11}+H_{22})U_m+H_{11}H_{22}-H_{12}H_{21}=0$

whose roots are

$U_{m\pm}=\frac{1}{2}(H_{11}+H_{22})\pm \frac{1}{2}\sqrt{(H_{11}-H_{22})^2+4H_{12}H_{21}}\;\;\;\;\;\;\;\;281n$

The unperturbed Hamiltonian is invariant to all symmetry operations of a point group and transforms according to the fully symmetric representation of the group. In the context of a metal ion subject to a ligand field, the perturbed Hamiltonian $H'$ is a summation of potentials between $n$ metal valence electrons and $m$ ligand valence electrons:

$H'=\sum_{j=1}^m\sum_{i=1}^n\frac{e^2}{4\pi\varepsilon_0r_{ij}}$

$H'$ is also totally symmetric, as the distances between metal and ligand valence electrons $r_{ij}$ are invariant to all symmetry operations of a point group. If we assume that both $H^{(0)}$ and $H'$ have a complete set of orthogonal eigenfunctions that are associated with real eigenvalues, $H^{(0)}$ and $H'$ are Hermitian. This implies that $H$ is also Hermitian. Using eq35, eq281n becomes

$U_{m\pm}=\frac{1}{2}(H_{11}+H_{22})\pm \frac{1}{2}\sqrt{(H_{11}-H_{22})^2+4\vert H_{12}\vert^2}\;\;\;\;\;\;\;\;281o$

The term $(H_{11}-H_{22})^2$ is positive and varies with the ligand field strength. If $\psi_1$ and $\psi_2$ have the same symmetry, then $H_{12}$ is generally non-zero according to the vanishing integral principle. It follows that $U_{m+}$ and $U_{m-}$ are always distinct regardless of ligand field strength. This is known as the non-crossing rule.

On the other hand, if $\psi_1$ and $\psi_2$ have different symmetry, then $H_{12}$ is necessarily zero. Consequently, the eigenvalues of the two states may be the same at a certain ligand field strength where $H_{11}=H_{22}$ .

For example, the diagram below depicts some states of the $d^2$ configuration of a metal ion in an octahedral ligand field. The $^{3}T_{1g}$ states, which have the same symmetry, do not cross as the ligand field strength varies, while $^{3}T_{2g}$ and $^{1}T_{2g}$ , which have different symmetries, do.