Differentiating summations

Question

Consider the following equation $$ \sum_{m=0}^{M}b_m r^m \left(\frac {d^2}{dr^2}+\frac{2}{r}\frac d {dr}+2E\right)\Psi-2\sum_{k=0}^{K}a_k r^k \Psi=0 \tag{1} $$ Now suppose $$ \Psi(r)=A \exp{[-S(r)]} \tag{2} $$ where $$ S(r)=\sum_{n=1}^N \lambda_n r^n \tag{3} $$ and $$ A=exp[-\lambda_0] $$ If we substitute equation (2) in equation (1), How can we obtain the following equation? $$ \sum_{m=0}^M b_m r^m \left[\sum_{i,j}i j\lambda_i \lambda_j r^{i+j-2}-\sum_j j(j+1) \lambda_j r^{j-2}+2E\right]-2\sum_{k=0}^K a_k r^k=0 \tag{4} $$ and specially I can't understand how $i$ emerges?

Answer 1

Comparison of (1) and (4) shows it is sufficient to focus on the calculation of \begin{align*} \left(\frac {d^2}{dr^2}+\frac{2}{r}\frac d {dr}+2E\right)\Psi\tag{1a} \end{align*}

From (2) we obtain \begin{align*} \frac{d}{dr}\Psi(r)&=\frac{d}{dr}\left(Ae^{-S(r)}\right)\\ &=Ae^{-S(r)}\left(-S^{\prime}(r)\right)\\ &=-S^{\prime}(r)\Psi(r)\\ \frac{d^2}{dr^2}\Psi(r)&=\frac{d}{dr}\left(-S^{\prime}(r)\Psi(r)\right)\\ &=-S^{\prime}(r)\Psi^{\prime}(r)-S^{\prime\prime}(r)\Psi(r)\\ &=\left(\left(S^{\prime}(r)\right)^2-S^{\prime\prime}(r)\right)\Psi(r)\tag{2a} \end{align*}

From (3) we obtain \begin{align*} S(r)&=\sum_{j=1}^N\lambda_jr^j \qquad S^{\prime}(r)=\sum_{j=1}^Nj\lambda_jr^{j-1} \qquad S^{\prime\prime}(r)=\sum_{j=1}^Nj(j-1)\lambda_jr^{j-2}\tag{3a}\\ \left(S^{\prime}(r)\right)^2&=\left(\sum_{i=1}^Ni\lambda_i r^{i-1}\right)\left(\sum_{j=1}^Nj\lambda_j r^{j-1}\right) =\sum_{n=2}^{2N}\sum_{{i+j=n}\atop{i,j\geq 1}}ij\lambda_i \lambda_jr^{i+j-2}\tag{4a} \end{align*}

Putting (1a) - (4a) together we obtain \begin{align*} &\color{blue}{\left(\frac {d^2}{dr^2}+\frac{2}{r}\frac d {dr}+2E\right)\Psi}\\ &\qquad=\left(\left(S^{\prime}(r)\right)^2-S^{\prime\prime}(r)-\frac{2}{r}S^{\prime}(r)+2E\right)\psi\\ &\qquad=\left(\sum_{n=2}^{2N}\sum_{{i+j=n}\atop{i,j\geq 1}}ij\lambda_i \lambda_jr^{i+j-2} -\sum_{j=1}^Nj(j-1)\lambda_jr^{j-2} -2\sum_{j=1}^Nj\lambda_jr^{j-2}+2E\right)\Psi\\ &\qquad\,\,\color{blue}{=\left(\sum_{n=2}^{2N}\sum_{{i+j=n}\atop{i,j\geq 1}}ij\lambda_i \lambda_jr^{i+j-2}-\sum_{j=1}^Nj(j+1)\lambda_jr^{j-2}+2E\right)\Psi}\tag{5a}\\ \end{align*}

We see the index $i$ emerges from the square of the series in (4a).