The expansion of $\frac{1}{r_{ij}}$ (the reciprocal of the inter-electron distance)

The expansion of $\frac{1}{r_{ij}}$ enables the Hartree equations to be solved analytically. Consider two electrons $q_i$ and $q_j$ as depicted in the diagram below:

According to the law of cosines $\vert\boldsymbol{\mathit{r}}_i-\boldsymbol{\mathit{r}}_j\vert^2=r_i^2+r_j^2-2r_ir_jcos\gamma$ , which is equal to $r_i^2\left[1+\left(\frac{r_j}{r_i}\right)^2-2\left(\frac{r_j}{r_i}\right)cos\gamma\right]$ or

$\frac{1}{\vert\boldsymbol{\mathit{r}}_i-\boldsymbol{\mathit{r}}_j\vert}=\frac{1}{r_{ij}}=\frac{1}{r_i}(1+\lambda)^{-1/2}\;\;\;\;\;\;\;\;(35)$

where $\lambda=\left(\frac{r_j}{r_i}\right)\left(\frac{r_j}{r_i}-2cos\gamma\right)$ .

Due to inter-electronic repulsion, $r_i\gg r_j$ if $\gamma$ is acute and hence $\lambda\ll 1$ , which implies that we can express eq35 as a binomial series, where $\frac{1}{r_{ij}}=\frac{1}{r_i}\left(1-\frac{1}{2}\lambda+\frac{3}{8}\lambda^2-\frac{5}{16}\lambda^3+\cdots\right)$ . Substituting $\lambda=\left(\frac{r_j}{r_i}\right)\left(\frac{r_j}{r_i}-2cos\gamma\right)$ in the series, rearranging and displaying the terms of up to $\left(\frac{r_j}{r_i}\right)^3$ , we have

$\frac{1}{r_{ij}}=\frac{1}{r_i}\left[1+cos\gamma\frac{r_j}{r_i}+\frac{1}{2}(3cos^2\gamma-1)\left(\frac{r_j}{r_i}\right)^2+\frac{1}{2}(5cos^3\gamma-3cos\gamma)\left(\frac{r_j}{r_i}\right)^3+\cdots\right]$

Since the coefficients of the powers of $\frac{r_j}{r_i}$ are the Legendre polynomials $P_n(cos\gamma)$ and that $r_i\gg r_j$

$\frac{1}{r_{ij}}=\frac{1}{r_>}\sum_{n=0}^{\infty}\left(\frac{r_<}{r_>}\right)^nP_n(cos\gamma)\;\;\;\;\;\;\;\;(36)$

As mentioned in the previous article, $cos\gamma$ is dependent on $\theta_i$ , $\theta_j$ , $\phi_i$ and $\phi_j$ (see diagram below). This implies that a function of $cos\gamma$ like $P_n(cos\gamma)$ is a function of four angles. If we rotate the z-axis such that it lies on $\boldsymbol{\mathit{r}}_j$ , the vectors are expressed in a new coordinate system, where the new polar angle and azimuthal angle are $\gamma$ and $\beta$ respectively. Consequently, we have $P_n(cos\gamma)=f(\gamma,\beta)$ , which is a simpler function to work with. We can also simplify the function in the original coordinate system by holding $\theta_j$ and $\phi_j$ as constants, resulting in a function of the variables $\theta_i$ and $\phi_i$ , or $P_n(cos\gamma)=f(\theta_i,\phi_i)$ .

With reference to the new coordinate system, $P_n(cos\gamma)=f(\gamma,\beta)$ and its complex conjugate are eigenfunctions of $\hat{L}^2=-\hbar^2\left[\frac{1}{sin^2\gamma}\frac{\partial^2}{\partial\beta^2}+\frac{1}{sin\gamma}\frac{\partial}{\partial\gamma}\left(sin\gamma\frac{\partial}{\partial\gamma}\right)\right]$ with eigenvalue $l(l+1)\hbar^2$ . For the original coordinate system, $P_n(cos\gamma)=f(\theta_i,\phi_i)$ and its complex conjugate are eigenfunctions of $\hat{L}^2=-\hbar^2\left[\frac{1}{sin^2\theta_i}\frac{\partial^2}{\partial\phi_i^2}+\frac{1}{sin\theta_i}\frac{\partial}{\partial\theta_i}\left(sin\theta_i\frac{\partial}{\partial\theta_i}\right)\right]$ with the same eigenvalue $l(l+1)\hbar^2$ .

Question

Show that the eigenvalue of $\hat{L}^2$ is $l(l+1)\hbar^2$ in any three-dimensional coordinate system.

Answer

From eq108 and eq109, the quantum orbital angular momentum ladder operators for the original and new coordinate systems are $\hat{L}_{\pm}=\hat{L}_x\pm i\hat{L}_y$ and $\hat{L}_{\pm}^{'}=\hat{L}_{x'}\pm i\hat{L}_{y'}$ respectively. The corresponding eigenfunctions of $\hat{L}_{\pm}$ and $\hat{L}_{\pm}^{'}$ are $Y$ and $Y'$ respectively. Following the steps of determining the eigenvalues of $\hat{L}^2$ and $\hat{L}^{2'}$ , we have $\hat{L}^2Y=l(l+1)\hbar^2Y$ and $\hat{L}^{2'}Y'=l(l+1)\hbar^2Y'$ respectively.

Furthermore, the general solution of the eigenfunction of $\hat{L}^2=-\hbar^2\left[\frac{1}{sin^2\theta_i}\frac{\partial^2}{\partial\phi_i^2}+\frac{1}{sin\theta_i}\frac{\partial}{\partial\theta_i}\left(sin\theta_i\frac{\partial}{\partial\theta_i}\right)\right]$ is $\sum_{m=-n}^nA_{nm}P_n^{\vert m\vert}(cos\theta_i)e^{i\vert m\vert\phi_i}$ , where $P_n^{\vert m\vert}(cos\theta_i)$ are the associated Legendre polynomials and $A_{nm}$ are the coefficients of the basis polynomials in the linear combination. Therefore,

$P_n(cos\gamma)=\sum_{m=-n}^nA_{nm}P_n^{\vert m\vert}(cos\theta_i)e^{i\vert m\vert\phi_i}\;\;\;\;\;\;\;\;(37)$

Question

Why is $P_n(cos\gamma)=\sum_{m=-n}^nA_{nm}P_n^{\vert m\vert}(cos\theta_i)e^{i\vert m\vert\phi_i}$ ?

Answer

For a particular value of $n$ , the linear combination is a sum of basis functions of a particular quantum angular momentum value $l$ . For example, when $n=1$ , we have a linear combination of the hydrogenic $p$ -orbital wavefunctions. Therefore, the eigenvalues of each term of the sum is always degenerate, which ensures that $P_n(cos\gamma)$ is an eigenfunction of $\hat{L}^2$ .

To solve for $A_{nm}$ , we multiply both sides of eq37 by the eigenfunction $P_n^{\vert m'\vert}(cos\theta_i)e^{-i\vert m'\vert\phi_i}$ and integrate over all space:

$\int P_n^{\vert m'\vert}(cos\theta_i)e^{-i\vert m'\vert\phi_i}P_n(cos\gamma)d\tau$ $=\int\sum_{m=-n}^nA_{nm} P_n^{\vert m'\vert}(cos\theta_i)e^{-i\vert m'\vert\phi_i}P_n^{\vert m\vert}(cos\theta_i)e^{i\vert m\vert\phi_i}d\tau$

Due to the orthogonal property of associated Legendre polynomials, only the $m=m'$ term in the RHS expansion of the above equation survives,

$\int P_n^{\vert m'\vert}(cos\theta_i)e^{-i\vert m'\vert\phi_i}P_n(cos\gamma)d\tau$ $=2\pi A_{nm}\int_{-1}^1 P_n^{\vert m\vert}(cos\theta_i)P_n^{\vert m\vert}(cos\theta_i)d(cos\theta_i)$

From eq98 of notes on Legendre functions, $\int_{-1}^1 \vert P_n^{\vert m\vert}(x)\vert^2dx=\frac{(n+\vert m\vert)!}{(n-\vert m\vert)!}\frac{2}{2n+1}$ . Substitute this equation in the above equation and rearranging,

$A_{nm}=\frac{2n+1}{4\pi}\frac{(n-\vert m\vert)!}{(n+\vert m\vert)!}\int P_n^{\vert m\vert}(cos\theta_i)e^{-i\vert m\vert\phi_i}P_n(cos\gamma)d\tau\;\;\;\;\;\;\;\;(38)$

To evaluate the integral in eq38, we expand the eigenfunction $P_n^{\vert m\vert}(cos\theta_i)e^{-i\vert m\vert\phi_i}$ as a linear combination of basis functions in the new coordinate system:

$P_n^{\vert m\vert}(cos\theta_i)e^{-i\vert m\vert\phi_i}=\sum_{k=-n}^nB_{nk}P_n^{\vert k\vert}(cos\gamma)e^{i\vert k\vert\beta}\;\;\;\;\;\;\;\;(39)$

Question

Why can we express $P_n^{\vert m\vert}(cos\theta_i)e^{-i\vert m\vert\phi_i}$ in the original coordinate system as a linear combination of basis functions in the new coordinate system?

Answer

As mentioned earlier in the article, $\hat{L}^2$ and $\hat{L}^{2'}$ share the same set of eigenvalues. For a particular value of $n$ , $P_n^{\vert m\vert}(cos\theta_i)e^{-i\vert m\vert\phi_i}$ is an eigenfunction of $\hat{L}^2$ with a specific eigenvalue of $l(l+1)\hbar^2$ , which is the same eigenvalue associated with $\sum_{k=-n}^nB_{nk}P_n^{\vert k\vert}(cos\gamma)e^{i\vert k\vert\beta}$ for the same value of $n$ . In other words, $\hat{L}^2Y'=\hat{L}^2\sum_iY_i=l(l+1)\hbar^2\sum_iY_i=\hat{L}^{2'}Y'$ , which implies that $\hat{L}^2$ is invariant with respect to rotation of the coordinate system.

To solve for $B_{nk}$ , multiply both sides of eq39 by $P_n^{\vert k'\vert}(cos\gamma)e^{-i\vert k'\vert\beta}$ , integrate over all space and repeat the steps in evaluating $A_{nm}$ . We have

$B_{nk}=\frac{2n+1}{4\pi}\frac{(n-\vert k\vert)!}{(n+\vert k\vert)!}\int P_n^{\vert k\vert}(cos\gamma)e^{-i\vert k\vert\beta}P_n^{\vert m\vert}(cos\theta_i)e^{-i\vert m\vert\phi_i}d\tau\;\;\;\;\;\;\;\;(40)$

Next, eq39 must hold for the case of $\gamma=0$ , with $\theta_i=\theta_j$ and $\phi_i=\phi_j$ . Expanding eq39,

$P_n^{\vert m\vert}(cos\theta_j)e^{-i\vert m\vert\phi_j}=B_{n(-n)}P_n^{\vert -n\vert}(1)e^{i\vert -n\vert\beta}$ $+B_{n(-n+1)}P_n^{\vert -n+1\vert}(1)e^{i\vert -n+1\vert\beta}+\cdots+B_{n(n)}P_n^{\vert n\vert}(1)e^{i\vert n\vert\beta}\;\;\;\;\;\;\;\;(41)$

From eq33b of notes on Legendre functions, $P_n^{\vert k\vert}(cos0)=(1-cos^20)^{\frac{\vert k\vert}{2}}\frac{d^{\vert k\vert}}{dcos\gamma^{\vert k\vert}}p_n(cos0)$ , which is equal to zero if $k\neq 0$ and is equal to one if $k=0$ (because $P_n^0(1)=P_n(1)=1$ ), i.e.

$P_n^{\vert k\vert}(1)=\left\{\begin{matrix} 0 &k\neq 0 \\ 1 &k=0 \end{matrix}\right.\;\;\;\;\;\;\;\;(42)$

So the only term that survives in eq41 is $B_{n(0)}P_n^{\vert 0\vert}(1)e^{i\vert 0\vert\beta}=B_{n(0)}P_n^{\vert 0\vert}(1)$ and eq41 becomes

$P_n^{\vert m\vert}(cos\theta_j)e^{-i\vert m\vert\phi_j}=B_{n(0)}\;\;\;\;\;\;\;\;(43)$

Substitute eq40, where $k=0$ and noting that $P_n^{\vert 0\vert}(cos\gamma)=P_n(cos\gamma)$ , in eq43,

$P_n^{\vert m\vert}(cos\theta_j)e^{-i\vert m\vert\phi_j}=\frac{2n+1}{4\pi}\int P_n(cos\gamma)P_n^{\vert m\vert}(cos\theta_i)e^{-i\vert m\vert\phi_i}d\tau\;\;\;\;\;\;\;\;(44)$

Substitute eq44 in eq38

$A_{nm}=\frac{(n-\vert m\vert)!}{(n+\vert m\vert)!}P_n^{\vert m\vert}(cos\theta_j)e^{-i\vert m\vert\phi_j}\;\;\;\;\;\;\;\;(45)$

Combining eq45, eq37 and eq36

$\frac{1}{r_{ij}}=\sum_{n=0}^{\infty}\sum_{m=-n}^n\frac{(n-\vert m\vert)!}{(n+\vert m\vert)!}\frac{r_<\, ^n}{r_>^{\,n+1}}P_n^{\vert m\vert}(cos\theta_i)P_n^{\vert m\vert}(cos\theta_j)e^{i\vert m\vert(\phi_i-\phi_j)}\;\;\;\;\;\;\;\;(46)$

Substitute $Y_n^{\vert m\vert}(\theta_i,\phi_i)=\sqrt{\frac{2n+1}{4\pi}\frac{(n-\vert m\vert)!}{(n+\vert m\vert)!}}P_n^{\vert m\vert}(cos\theta_i)e^{i\vert m\vert\phi_i}$ in eq46, we have

$\frac{1}{r_{ij}}=\sum_{n=0}^{\infty}\sum_{m=-n}^n\frac{4\pi}{2n+1}\frac{r_<\, ^n}{r_>^{\,n+1}}Y_n^{\vert m\vert}(\theta_i,\phi_i)\left[Y_n^{\vert m\vert}(\theta_j,\phi_j)\right]^*\;\;\;\;\;\;\;\;(47)$

The expansion of \(\frac{1}{r_{ij}}\) (the reciprocal of the inter-electron distance)

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