basis. To transform between them, we additionally need to express the local fields and field gradients in terms of spherical multipoles, which we present using the same transformation matrices as

$$\begin{aligned}
        \begin{split}             \mathbf{E}
            &=\frac{i E_0}{\sqrt{12 \pi}}
            \bar{\bar{\mathrm{O}}}_1
            \mathbf{a}^E_1
        \end{split}&
        \begin{split}             \mathbf{H}
            &=\frac{E_0}{\eta \sqrt{12 \pi}}
            \bar{\bar{\mathrm{O}}}_1
            \mathbf{a}^M_1
        \end{split}\\
        \begin{split}             \frac{1}{k}\Diamond\mathbf{E}
            &=\frac{i E_0}{\sqrt{80 \pi}}
            \bar{\bar{\mathrm{O}}}_{2}
            \mathbf{a}^E_2
        \end{split}&
        \begin{split}             \frac{1}{k}\Diamond\mathbf{H}
            &=\frac{E_0}{\eta \sqrt{80 \pi}}
            \bar{\bar{\mathrm{O}}}_{2}
            \mathbf{a}^M_2
        \end{split}\\
        \begin{split}             \frac{1}{k^2}\Diamond^2\mathbf{E}&=\frac{i E_0}{\sqrt{420 \pi}} \bar{\bar{\mathrm{O}}}_{3}\mathbf{a}^E_3
        \end{split}&
        \begin{split}             \frac{1}{k^2}\Diamond^2\mathbf{H}&=\frac{E_0}{\eta \sqrt{420 \pi}} \bar{\bar{\mathrm{O}}}_{3}\mathbf{a}^M_3
        \end{split}
    \end{aligned}$$

\[eqn:field\]

where $\mathbf{E} = [E_x, E_y, E_z]^\top$, $\Diamond\mathbf{E} = [\partial_xE_x, (\partial_xE_y+\partial_yE_x)/2, (\partial_xE_z+\partial_zE_x)/2, \partial_yE_y, (\partial_yE_z+\partial_zE_y)/2]^\top$, and $$\Diamond^2\mathbf{E}
        =
        \begin{bmatrix}
        \partial_x^2E_x -\frac{1}{5}\nabla^2E_x \\
        \frac{1}{3}(\partial_x^2E_y+2\partial_x\partial_yE_x) -\frac{1}{15}\nabla^2E_y \\
        \frac{1}{3}(\partial_x^2E_z+2\partial_x\partial_zE_x) -\frac{1}{15}\nabla^2E_z \\
        \frac{1}{3}(\partial_y^2E_x+2\partial_x\partial_yE_y) -\frac{1}{15}\nabla^2E_x \\
        \frac{1}{3}(\partial_x\partial_yE_z + \partial_y\partial_zE_x + \partial_z\partial_xE_y) \\
        \partial_y^2E_y -\frac{1}{5}\nabla^2E_y \\
        \frac{1}{3}(\partial_y^2E_z+2\partial_y\partial_zE_y) -\frac{1}{15}\nabla^2E_z
        \end{bmatrix}$$

Now, $\boldsymbol{\alpha}$-tensor and $\mathbf{T}$-matrix can be transformed to each other. However, $\boldsymbol{\alpha}$-tensor is expressed in SI units if Eqns. \[eqn:field\] and \[eqn:multipole\] are directly used. Because $\boldsymbol{\alpha}$-tensor in SI units have different units per components and are difficult to be compared, we use $\boldsymbol{\alpha}$-tensor in units of volume defined as $$\alpha^{pn}_{p'n'} = \frac{6\pi}{ik^3} \bar{\bar{\mathrm{O}}}_{n'} \mathbf{T}^{pn}_{p'n'} \bar{\bar{\mathrm{O}}}_{n}^{-1}
        \label{eqn:T2a}$$ Note that similar expressions for isotropic (scalar) dipolar [@Arango2013; @Liu2016] and quarupolar objects [@Babicheva2019] can be found in literature. $\mathbf{T}$-matrix is inversely obtained from $\boldsymbol{\alpha}$-tensor 