Proof of an equation with the aid of Laguerre functions

Question

The generating function for the Laguerre functions {$\phi_m(x_3;\alpha)$} is:
\begin{align}\frac{\alpha ^{1/2}e^{-(1/2)\alpha x_3 (1+s)/(1-s)}}{1-s}= \sum_{m=0}^{\infty} s^m\phi_m(x_3;\alpha)\end{align}
and
\begin{align}
\phi_m(x_3;\alpha)=\alpha^{1/2}e^{-(1/2)\alpha x_3} L_n(\alpha x_3)= |n\rangle
\end{align}
here $L_n(x)$ is the $n$th Laguerre polynomial.

Thus
\begin{align}
&\int_{0}^{\infty}dx_3\frac{\alpha ^{1/2}e^{-(1/2)\alpha x_3 (1+s)/(1-s)}}{1-s}e^{-\beta x_3}\frac{\alpha ^{1/2}e^{-(1/2)\alpha x_3 (1+t)/(1-t)}}{1-t} \tag{1}\\
&\quad=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}s^m t^n\left \langle m|e^{-\beta x_3}|n \right \rangle \tag{2}\\
&\quad=\frac{\alpha}{(\alpha+\beta-\beta t)-(\beta+(\alpha-\beta) t) s} \tag{3}\\
&\quad=\alpha \sum_{m=0}^{\infty} s^{m} \frac{(\beta+(\alpha-\beta) t)^{m}}{(\alpha+\beta-\beta t)^{m+1}} \tag{4}\\
&\quad=\alpha \sum_{m=0}^{\infty} s^{m} \sum_{n=0}^{\infty} t^{n} \sum_{p=0}^{n} \theta(m-p) \frac{(m+n-p) !}{(m-p) !(n-p) ! p !} \frac{(\alpha-\beta)^{p} \beta^{m+n-2 p}}{(\alpha+\beta)^{m+n+1-p}} \tag{5}
\end{align}

where $\theta(n)=1$ for $n=0,1,2, \ldots$ and $\theta(n)=0$ for $n=-1,-2,-3, \ldots$

We can see that (3) is the result of (1) and can be rewritten to (5). Comparing (2) and (5), and extracting the different parts, we can obtain the conclusion:

\begin{equation}
\left \langle m|e^{-\beta x_3}|n \right \rangle = \sum_{p=0}^{min(m,n)} \frac{(m+n-p) !}{(m-p) !(n-p) ! p !} \frac{\alpha(\alpha-\beta)^{p} \beta^{m+n-2 p}}{(\alpha+\beta)^{m+n+1-p}}
\end{equation}

However, I don't know how to proof the following equation using the similar method mentioned above. Can you show me how to proof the following equation? Thanks.
\begin{align}
&\langle m|e^{-\beta z}\frac{d^2}{dz^2}|n \rangle = a_{mn} + 4\sum_{p=0}^{n-1}a_{mp}(n-p) \label{eq:new4}
\end{align}
where

\begin{equation}
a_{mn}=\frac{1}{4}\alpha^3 \sum_{p=0}^{min(m,n)} \frac{(m+n-p) !}{(m-p) !(n-p) ! p !} \frac{(\alpha-\beta)^{p} \beta^{m+n-2 p}}{(\alpha+\beta)^{m+n+1-p}}
\end{equation}

Answer 1

We want to show \begin{align*} \color{blue}{\bigg \langle m}&\color{blue}{\left|e^{-\beta z}\frac{d^2}{dz^2}\right|n \bigg \rangle=a_{m,n}+4\sum_{l=0}^{n-1}(n-l)a_{m,l}}\tag{1.0} \end{align*}

We have according to (1) - (5) and the definition of $a_{m,n}$: \begin{align*} &\frac{\alpha}{(1-s)(1-t)}\int_{0}^{\infty}e^{-\frac{\alpha(1+s)}{2(1-s)}z} e^{-\beta z} e^{-\frac{\alpha(1+t)}{2(1-t)}z}\,dz\tag{1.1}\\ &\qquad=\frac{\alpha}{(1-s)(1-t)}\int_{0}^{\infty} e^{-\left(\frac{\alpha}{2}\left(\frac{1+s}{1-s}+\frac{1+t}{1-t}\right)+\beta\right)z}\,dz \\ &\qquad=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}s^m t^n\left \langle m|e^{-\beta z}|n \right \rangle\tag{1.2}\\ &\qquad=\frac{\alpha}{(\alpha+\beta-\beta t)-(\beta+(\alpha-\beta) t) s}\tag{1.3}\\ &\qquad=\frac{4}{\alpha^2}\sum_{m=0}^\infty\sum_{n=0}^\infty a_{m,n}s^mt^n\tag{1.4} \end{align*}

Since we want to calculate $\color{blue}{\langle m|e^{-\beta z}\frac{d^2}{dz^2}|n \rangle} $ we consider according to (1.1) - (1.4) \begin{align*} &\color{blue}{\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}s^m t^n\left \langle m\left|e^{-\beta z}\frac{d^2}{dz^2}\right|n \right \rangle}\tag{2.1}\\ &\qquad=\frac{\alpha}{(1-s)(1-t)}\int_{0}^{\infty}e^{-\frac{\alpha(1+s)}{2(1-s)}z} e^{-\beta z} \left(\frac{d^2}{dz^2} e^{-\frac{\alpha(1+t)}{2(1-t)}z}\right)\,dz\\ &\qquad=\frac{\alpha}{(1-s)(1-t)}\int_{0}^{\infty}e^{-\frac{\alpha(1+s)}{2(1-s)}z} e^{-\beta z}\left(\frac{\alpha^2}{4} \frac{(1+t)^2}{(1-t)^2}\right)e^{-\frac{\alpha(1+t)}{2(1-t)}z}\,dz\\ &\qquad=\frac{\alpha^2}{4}\, \frac{(1+t)^2}{(1-t)^2}\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}s^m t^n\left \langle m|e^{-\beta z}|n \right \rangle\\ &\qquad=\frac{\alpha^2}{4}\, \frac{(1+t)^2}{(1-t)^2}\,\frac{\alpha}{(\alpha+\beta-\beta t)-(\beta+(\alpha-\beta) t) s}\\ &\qquad\,\,\color{blue}{=\frac{(1+t)^2}{(1-t)^2}\,\sum_{k=0}^\infty\sum_{l=0}^\infty a_{k,l}s^kt^l}\tag{2.2} \end{align*}

We use the coefficient of operator $[s^m]$ to denote the coefficient of $s^m$ of a series.

We obtain from (2.1) and (2.2): \begin{align*} \color{blue}{\bigg \langle m}&\color{blue}{\left|e^{-\beta z}\frac{d^2}{dz^2}\right|n \bigg \rangle}\\ &=[s^mt^n]\frac{(1+t)^2}{(1-t)^2}\,\sum_{k=0}^\infty\sum_{l=0}^\infty a_{k,l}s^kt^l\\ &=[t^n](1+t)^2\sum_{j=0}^\infty\binom{-2}{j}(-t)^j\sum_{l=0}^\infty a_{m,l}t^l\tag{3.1}\\ &=[t^n](1+t)^2\sum_{j=0}^\infty(j+1)t^j\sum_{l=0}^\infty a_{m,l}t^l\tag{3.2}\\ &=[t^n](1+t)^2\sum_{q=0}^\infty\left(\sum_{{j+l=q}\atop{j,l\geq 0}}^\infty(j+1)a_{m,l}\right)t^q\tag{3.3}\\ &=\left([t^n]+2[t^{n-1}]+[t^{n-2}]\right)\sum_{q=0}^\infty \left(\sum_{l=0}^q(q-l+1)a_{m,l}\right)t^q\tag{3.4}\\ &=\sum_{l=0}^n(n-l+1)a_{m,l}+2\sum_{l=0}^{n-1}(n-l)a_{m,l}+\sum_{l=0}^{n-2}(n-l-1)a_{m,l}\tag{3.5}\\ &=\left(\sum_{l=0}^{n-1}(n-l)a_{m,l}+\sum_{l=0}^na_{m,l}\right)+2\sum_{l=0}^{n-1}(n-l)a_{m,l}\\ &\qquad+\left(\sum_{l=0}^{n-1}(n-l)a_{m,l}-a_{m,n-1}-\sum_{l=0}^{n-2}a_{m,l}\right)\tag{3.6}\\ &\,\,\color{blue}{=a_{m,n}+4\sum_{l=0}^{n-1}(n-l)a_{m,l}} \end{align*} and the claim (1.0) follows.

Comment:

• In (3.1) we select the cofficient of $s^m$ and make a binomial series expansion of $\frac{1}{(1-t)^2}$.

• In (3.2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (3.3) we calculate the Cauchy-product of the two series.

• In (3.4) we expand $(1+t)^2$ and apply the rule $[t^p]t^qA(t)=[t^{p-q}]A(t)$. We also eliminate $j=q-l$ in the inner sum of the Cauchy-product.

• In (3.5) we select the coefficients of $[t^n],[t^{n-1}]$ and $[t^{n-2}]$.

• In (3.6) we split the sums to collect relevant parts conveniently in the last step.