Definition of single particle basis

In this section, we define some single particle basis used in the JASMINE component to write down the atomic Hamiltonian $\hat{H}_{\text{atom}}$. We set $\hbar=1$ in this note.

Complex spherical harmonics basis

The complex spherical harmonics $Y_{l}^{m}(\theta,\phi)$ is the eigenstate of $l^{2},l_{z}$,

$$\begin{equation} l^{2}Y_{l}^{m}=l(l+1)Y_{l}^{m}, \end{equation}$$ $$\begin{equation} l_{z}Y_{l}^{m}=mY_{l}^{m},\ m=-l,-l+1,\cdots,l. \end{equation}$$

Real spherical harmonics basis

The real spherical harmonics $Y_{lm}$ is defined as

$$\begin{gather} Y_{lm}=\begin{cases} \frac{i}{\sqrt{2}}\left(Y_{l}^{-|m|}-(-1)^{m}Y_{l}^{|m|}\right) & \text{if}\ m<0,\\ Y_{l}^{0} & \text{if}\ m=0,\\ \frac{1}{\sqrt{2}}\left(Y_{l}^{-|m|}+(-1)^{m}Y_{l}^{|m|}\right) & \text{if}\ m>0. \end{cases} \end{gather}$$

For $p$ system, $$\begin{eqnarray} p_{x} & = & Y_{11}=\frac{1}{\sqrt{2}}\left(Y_{1}^{-1}-Y_{1}^{1}\right),\nonumber \\ p_{y} & = & Y_{1,-1}=\frac{i}{\sqrt{2}}\left(Y_{1}^{-1}+Y_{1}^{1}\right),\\ p_{z} & = & Y_{10}=Y_{1}^{0}.\nonumber \end{eqnarray}$$

For $d$ system, $$\begin{eqnarray} d_{z^{2}} & = & Y_{20}=Y_{2}^{0},\nonumber \\ d_{zx} & = & Y_{21}=\frac{1}{\sqrt{2}}\left(Y_{2}^{-1}-Y_{2}^{1}\right),\nonumber \\ d_{zy} & = & Y_{2,-1}=\frac{i}{\sqrt{2}}\left(Y_{2}^{-1}+Y_{2}^{1}\right),\\ d_{x^{2}-y^{2}} & = & Y_{22}=\frac{1}{\sqrt{2}}\left(Y_{2}^{-2}+Y_{2}^{2}\right),\nonumber \\ d_{xy} & = & Y_{2,-2}=\frac{i}{\sqrt{2}}\left(Y_{2}^{-2}-Y_{2}^{2}\right).\nonumber \end{eqnarray}$$

For $t_{2g}$ system $(l\approx-1)$, we have a $T-P$ equivalence, $$\begin{eqnarray} d_{zx} & \rightarrow & p_{y}=\frac{i}{\sqrt{2}}\left(Y_{1}^{-1}+Y_{1}^{1}\right),\nonumber \\ d_{zy} & \rightarrow & p_{x}=\frac{1}{\sqrt{2}}\left(Y_{1}^{-1}-Y_{1}^{1}\right),\\ d_{xy} & \rightarrow & p_{z}=Y_{1}^{0}.\nonumber \end{eqnarray}$$

For $f$ system, $$\begin{eqnarray} f_{z^{3}} & = & Y_{30}=Y_{3}^{0},\nonumber \\ f_{xz^{2}} & = & Y_{31}=\frac{1}{\sqrt{2}}\left(Y_{3}^{-1}-Y_{3}^{1}\right),\nonumber \\ f_{yz^{2}} & = & Y_{3,-1}=\frac{i}{\sqrt{2}}\left(Y_{3}^{-1}+Y_{3}^{1}\right),\nonumber \\ f_{z(x^{2}-y^{2})} & = & Y_{32}=\frac{1}{\sqrt{2}}\left(Y_{3}^{-2}+Y_{3}^{2}\right),\\ f_{xyz} & = & Y_{3,-2}=\frac{i}{\sqrt{2}}\left(Y_{3}^{-2}-Y_{3}^{2}\right),\nonumber \\ f_{x(x^{2}-3y^{2})} & = & Y_{33}=\frac{1}{\sqrt{2}}\left(Y_{3}^{-3}-Y_{3}^{3}\right),\nonumber \\ f_{y(3x^{2}-y^{2})} & = & Y_{3,-3}=\frac{i}{\sqrt{2}}\left(Y_{3}^{-3}+Y_{3}^{3}\right).\nonumber \end{eqnarray}$$

Cubic spherical harmonics basis

The cubic spherical harmonics is defined as the basis of the irreducible representation of cubic point group $O_{h}$.

For $p$ orbitals, $$\begin{equation} T_{1u}:p_{x},p_{y},p_{z}. \end{equation}$$

For $d$ orbitals, $$\begin{gather} \begin{cases} E_{g}: & d_{z^{2}},d_{x^{2}-y^{2}},\\ T_{2g}: & d_{zx},d_{zy},d_{xy}. \end{cases} \end{gather}$$

For $f$ orbitals, $$\begin{eqnarray} f_{x^{3}} & = & -\frac{\sqrt{6}}{4}f_{xz^{2}}+\frac{\sqrt{10}}{4}f_{x(x^{2}-3y^{2})},\nonumber \\ f_{y^{3}} & = & -\frac{\sqrt{6}}{4}f_{yz^{2}}-\frac{\sqrt{10}}{4}f_{y(3x^{2}-y^{2})},\nonumber \\ f_{z^{3}} & = & f_{z^{3}},\nonumber \\ f_{x(y^{2}-z^{2})} & = & -\frac{\sqrt{10}}{4}f_{xz^{2}}-\frac{\sqrt{6}}{4}f_{x(x^{2}-3y^{2})},\nonumber \\ f_{y(z^{2}-x^{2})} & = & \frac{\sqrt{10}}{4}f_{yz^{2}}-\frac{\sqrt{6}}{4}f_{y(3x^{2}-y^{2})},\nonumber \\ f_{z(x^{2}-y^{2})} & = & f_{z(x^{2}-y^{2})},\nonumber \\ f_{xyz} & = & f_{xyz}.\nonumber \end{eqnarray}$$

NOTE: $$\begin{gather} \begin{cases} T_{1u}: & f_{x^{3}},f_{y^{3}},f_{z^{3}}\\ T_{2u}: & f_{x(y^{2}-z^{2})},f_{y(z^{2}-x^{2})},f_{z(x^{2}-y^{2})}\\ A_{2u}: & f_{xyz} \end{cases} \end{gather}$$

$j^{2},j_{z}$ diagonal basis

Define $\phi_{ljm_{j}}$ as the eigenstate of $j^{2},j_{z}$,

$$\begin{equation} j^{2}\phi_{ljm_{j}}=j(j+1)\phi_{ljm_{j}}, \end{equation}$$

$$\begin{equation} j_{z}\phi_{ljm_{j}}=m_{j}\phi_{ljm_{j}}. \end{equation}$$

For $j=l+\frac{1}{2},m_{j}=m+\frac{1}{2}$, $$\begin{equation} \phi_{ljm_{j}}=\sqrt{\frac{l+m+1}{2l+1}}Y_{l}^{m}\uparrow+\sqrt{\frac{l-m}{2l+1}}Y_{l}^{m+1}\downarrow. \end{equation}$$

For $j=l-\frac{1}{2},m_{j}=m+\frac{1}{2}$, $$\begin{equation} \phi_{ljm_{j}}=-\sqrt{\frac{l-m}{2l+1}}Y_{l}^{m}\uparrow+\sqrt{\frac{l+m+1}{2l+1}}Y_{l}^{m+1}\downarrow. \end{equation}$$

Natural basis

The natural basis is defined as the diagonal basis of on-site term $E_{\alpha\beta}$.