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The quadrupole moment can be written as

${\mathit{Q}}_{\mathbf{i}}\mathit{j}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{鈭�}\mathbf{\left[}\mathbf{3}{{\mathbf{r}}_{\mathbf{i}}}^{\mathbf{\text{'}}}{{\mathbf{r}}_{\mathbf{j}}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{(}{\mathbf{r}}^{\mathbf{\text{'}}}{\mathbf{)}}^{\mathbf{2}}{\mathbf{未}}_{\mathbf{i}}\mathbf{j}\mathbf{\right]}\mathit{蚁}\mathbf{\left(}\mathit{r}{\mathbf{鈨�}}^{\mathbf{\text{'}}}\mathbf{\right)}\mathit{d}{\mathit{蟿}}^{\mathbf{\text{'}}}$

The Kronecker Delta function can be defined as

$\mathit{未}\mathit{\_}\mathit{i}\mathit{j}\mathbf{=}\mathbf{\{}\begin{array}{lll}\mathbf{1}& \mathbf{if}& \mathbf{i}\mathbf{=}\mathbf{j}\\ \mathbf{0}& \mathbf{if}& \mathbf{i}\mathbf{鈮�}\mathbf{j}\end{array}\mathbf{鈥�}\mathbf{鈥�}$

The quadrupole term in the potential multipole expansion can be written as

$V_{"}quad(r鈨�)=\frac{1}{8蟺{蔚}_0}\frac{1}{r^3}鈭�蚁\left(r{鈨�}^{\text{'}}\right)d{蟿}^{\text{'}}(3co{s}^2{胃}^{\text{'}}-1){r^{\text{'}}}^2鈥�鈥�鈥�鈥�鈥�.....\left(4\right)$

Let us start with with the right hand side of equation (1) and substitute equation (2) in it,

$$\begin{gathered} \frac{1}{4{\mathrm{蟺蔚}}_0}\frac{1}{{\mathrm{r}}^3}\underset{\mathrm{i},\mathrm{j}=1}{\overset{3}{鈭�}}{\hat{\mathrm{r}}}_{\mathrm{i}}{\hat{\mathrm{r}}}_{\mathrm{j}}{Q}_{\mathrm{ij}}=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{1}{{\mathrm{r}}^3}\underset{\mathrm{i},\mathrm{j}=1}{\overset{3}{鈭�}}{\hat{\mathrm{r}}}_{\mathrm{i}}{\hat{\mathrm{r}}}_{\mathrm{j}}\frac{1}{2}鈭�\left[3{{\mathrm{r}}_{\mathrm{i}}}^{\text{'}}{{\mathrm{r}}_{\mathrm{j}}}^{\text{'}}-({\mathrm{r}}^{\text{'}}{)}^2{\mathrm{未}}_{\mathrm{i}}\mathrm{j}\right]蚁\left(r{鈨�}^{\text{'}}\right)d{蟿}^{\text{'}}鈥� \\ =\frac{1}{8{\mathrm{蟺蔚}}_0}\frac{1}{{\mathrm{r}}^3}鈭�蚁\left(\hat{r}\right)d{蟿}^{\text{'}}3\underset{i}{鈭�}{\hat{\mathrm{r}}}_{\mathrm{i}}{\hat{\mathrm{r}}}_i\text{'}\underset{j}{鈭�}{\hat{\mathrm{r}}}_j{\hat{\mathrm{r}}}_j\text{'}-\underset{i,j=1}{\overset{3}{鈭�}}{\hat{\mathrm{r}}}_{\mathrm{i}}{\hat{\mathrm{r}}}_{\mathrm{j}}{\left(r\text{'}\right)}^2 \end{gathered}$$

But

$\underset{i}{鈭�}{\hat{r}}_i\widehat{r_i}=\underset{j}{鈭�{\hat{r}}_j\widehat{r_j}}=r{鈨�}^{\text{'}}鈰�\stackrel{鈬赌}{r}路r^{\text{'}}cos{胃}^{\text{'}}$

and

$\underset{i,j=1}{\overset{3}{鈭�}}{\hat{\mathrm{r}}}_{\mathrm{i}}{\hat{\mathrm{r}}}_{\mathrm{j}}{未}_{ij=}\underset{i}{鈭�{\hat{r}}_i\widehat{r_i}}=\hat{r}路\hat{r}=1$

We substitute these results in the right hand side of equation (1) and get

$\frac{1}{4蟺{蔚}_0}\frac{1}{r^3}\underset{i,j=1}{\overset{3}{鈭�}}{\hat{r}}_i\widehat{r_j}{Q}_{ij}=\frac{1}{8蟺{蔚}_0}\frac{1}{r^3}鈭�蚁\left(r{鈨�}^{\text{'}}\right)d{蟿}^{\text{'}}(3co{s}^2{胃}^{\text{'}}-1){r^{\text{'}}}^2$

This is exactly the quadrupole term in the multipole expansion as defined in equation (4).

There are four charges in the configuration

$$\begin{gathered} +qat\left(\frac{a}{2},\frac{a}{2},0\right)-qat\left(-\frac{a}{2},\frac{a}{2},0\right) \\ +qat\left(-\frac{a}{2},-\frac{a}{2},0\right)-q\left(\frac{a}{2},\frac{a}{2},0\right) \end{gathered}$$

With these we will start calculating the components of the quadrupole moment

$$\begin{gathered} Q_{xx}=\frac{1}{2}\left[q\left(3\frac{a^2}{4}-\frac{a^2}{4}\right)-q\left(3\frac{a^2}{4}-\frac{a^2}{4}\right)+q\left(3\frac{a^2}{4}-\frac{a^2}{4}\right)-q\left(3\frac{a^2}{4}-\frac{a^2}{4}\right)\right] \\ {Q}_{yy}=\frac{1}{2}\left[q\left(3\frac{a^2}{4}-\frac{a^2}{2}\right)-q\left(3\frac{a^2}{4}-\frac{a^2}{2}\right)+q\left(3\frac{a^2}{4}-\frac{a^2}{2}\right)-q\left(3\frac{a^2}{4}-\frac{a^2}{2}\right)\right] \\ {Q}_{zz}=\frac{1}{2}\left[q\left(-\frac{a^2}{4}\right)-q\left(-\frac{a^2}{4}\right)+q\left(-\frac{a^2}{4}\right)-q\left(-\frac{a^2}{4}\right)\right] \\ {Q}_{xy}=\frac{1}{2}\left[q\left(3\frac{a^2}{4}\right)-q\left(3\left(-\frac{a^2}{4}\right)\right)+q\left(3\frac{a^2}{4}\right)-q\left(3\left(-\frac{a^2}{4}\right)\right)\right] \\ =\frac{3q{a}^2}{2}=Q_{yx} \\ {Q}_{xz}=\frac{1}{2}\left[q脳0-q脳0+q脳0-q脳0\right] \\ =0=Q_{zx}=Q_{yz}=Q_{zy} \end{gathered}$$

Thus, the only two non-zero components are $Q_{xy}=Q_{yx}=\frac{3q{a}^2}{2}$.

Let the origin be shifted by . Then the new quadrupole moment becomes

$$\begin{gathered} Q{\text{'}}_{ij}=\frac{1}{2}鈭�\left[3\left(r_i\text{'}-d_i\right)\left(r_j\text{'}-d_j\right)-{\left(\stackrel{鈬赌}{r}\text{'}-\stackrel{鈬赌}{d}\right)}^2{未}_{ij}\right]蚁d蟿\text{'} \\ =Q_{ij}-\frac{3d_i}{2}鈭�r_j\text{'}蚁d蟿\text{'}-\frac{3d_j}{2}鈭�r_i\text{'}蚁d蟿\text{'}+\frac{3d_i{d}_j}{2}鈭�蚁d蟿\text{'}+\stackrel{鈬赌}{d}路鈭�\stackrel{鈬赌}{r}{未}_{ij}蚁d蟿\text{'}-d^2鈭�{未}_{ij}蚁d蟿\text{'} \\ =Q_{ij}-\frac{3}{2}{d}_i{p}_j-\frac{3}{2}{d}_j{p}_i+\frac{3}{2}{d}_i{d}_{i}Q+\stackrel{鈬赌}{d}路\stackrel{鈬赌}{p}{未}_{ij}-d^2{未}_{ij}Q \end{gathered}$$

Thus, if the the monopole moment $Q=0$ and the dipole moment $p鈨�=0$ then

$Q_{ij}\text{'}=Q_{ij}$

The quadrupole moment is independent of the origin.

The fourth term in the multipole expansion can be written as

$V_{oct}=\frac{1}{4蟺{蔚}_0{r}^4}\underset{i,j,k}{鈭�}{\hat{r}}_i{\hat{r}}_j{\hat{r}}_k{Q}_{ijk}$

From the multipole expansion series of the potential, the octopole term can be written as

$V_{oct}=\frac{1}{4蟺{蔚}_0{r}^4}鈭�r{\text{'}}^3\frac{1}{2}(5co{s}^3胃-3cos胃)蚁d{蟿}^{\text{'}}$

We can compare these two and write the octopole moment as

$V_{oct}=\frac{1}{4蟺{蔚}_0{r}^4}鈭�r{\text{'}}^3\frac{1}{2}(5co{s}^3胃-3cos胃)蚁d{蟿}^{\text{'}}$

This is the expression for the octopole moment.