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We want to calculate the expectation value $\langle q^2\rangle$ in polar coordinates which gives us the following integral, for integer values of $p$: \begin{equation}\int_0^\infty dq~q^3 \left(\int_0^\infty dr~ J_{0}(q r) e^{-r^2/2} L_{p}^{1}(r^2) r^2\right)^2.\end{equation} With Bessel function of the first kind $J_n(x)$ and generalized Laguerre polynomial $L_{p}^{l}(x)$.

We can't solve it because we don't find a nice form of the Fourier-Bessel transform of our waves: (with $l\geq1$) \begin{equation}\psi_p^l(q)=\int_0^\infty dr~ J_{l-1}(q r) e^{-r^2/2} L_{p}^{l}(r^2)r^{l+1}.\end{equation} In the literature (http://hdl.handle.net/1794/3778, Eq. 3.26) we find a form which is similar but not quite the same. \begin{equation}\int_0^\infty dr~ J_{l}(q r) e^{-r^2/2} L_{p}^{l}(r^2)r^{l+1}=(-1)^p q^l L_p^l(q^2) e^{-q^2/2}.\end{equation}

Can anyone help us to solve the first or second integral?

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  • $\begingroup$ I'm not sure, but integrating out the $q$ variable using Hankel transform en.wikipedia.org/wiki/Hankel_transform might be possible. I did some calculations and it seems $\int_0^\infty q^3J_0(qr_1)J_0(qr_2)dq=-\frac{\delta'(r_1-r_2)}{r_1^2}-\frac{\delta''(r_1-r_2)}{r_1}$. Then one of the $r$ variables can be integrated from the delta function, and the resulting integral over the second $r$ would be equal to a finite sum. $\endgroup$
    – Negan
    Sep 10 at 11:24
  • $\begingroup$ The majestic, royal, use of the plural of the first person has not passed unnoticed... $\endgroup$
    – Alex M.
    Sep 10 at 12:23
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    $\begingroup$ @Nemo Thanks, this helped a lot, I can now solve it! By the way, there should be a "+" instead of a "-" in front of the second delta function. $\endgroup$
    – Corne Koks
    Sep 11 at 12:43
  • $\begingroup$ I will post the full solution tomorrow. $\endgroup$
    – Corne Koks
    Sep 11 at 12:44
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This is my solution. Might still contain mistakes. Hope this helps others too!

The bessel function

Using the following properties \begin{equation} \int_0^\infty dq q J_v(q \rho_1) J_v(q \rho_2)= \frac{\delta(r_1-r_2)}{r_1} \end{equation} \begin{equation} J'_v(z)=- J_{v+1}(z)+\frac{v}{z}J_{v}(z) \end{equation} \begin{equation} J_0'(z)=-J_1(z) \end{equation} \begin{equation} J_{-n}=(-1)^n J_{n}. \end{equation}

Therefore; \begin{equation} q^3 J'_{-1}(q \rho_1) J'_{-1}(q \rho_2)= q^3 J_0(q \rho_1) J_0(q \rho_2)+ q \frac{1}{\rho_1 \rho_2} J_{-1}(q\rho_1)J_{-1}(q\rho_2)+q^2\left( \frac{1}{\rho_2}J_0(q \rho_1) J'_0(q \rho_2)+\frac{1}{\rho_1}J'_0(q \rho_1) J_0(q \rho_2) \right) \end{equation} This gives \begin{equation} \int_0^\infty dq q^3 J_0(q \rho_1) J_0(q \rho_2)=\left(\partial_{\rho_1}\partial_{\rho_2}-\frac{1}{\rho_1 \rho_2}-\frac{\partial_{\rho_1}}{\rho_1}-\frac{\partial_{\rho_2}}{\rho_2}\right)\frac{\delta(\rho_1-\rho_2)}{\rho_1} \end{equation}

Solving the integral

\begin{equation} I=\int_0^\infty d \rho f(\rho) \left(-\frac{\partial^2_{\rho}}{\rho}+\frac{3\partial_{\rho}}{\rho^2}-\frac{3}{\rho^3} \right)f(\rho). \end{equation} define $x\equiv \rho^2$. \begin{equation} I=-\int_0^\infty dx f(x)\left[ 2\partial^2_x-2\frac{\partial_x}{x}+\frac{3}{2x^2}\right] f(x). \end{equation}

Using the identities \begin{equation} \partial_x(x L^1_p(x))=(p+1) L_{p}^0 \end{equation} \begin{equation} \partial_x(L_p^0)=-L_{p-1} (\text{for}~ p\geq1) ~(=0~ \text{otherwise}) \end{equation}

the derivatives to $f(x)=x L_p^1(x) e^{-x/2}$ are \begin{equation} \partial_x f(x)=(- \frac{x}{2} L_p^1(x)+(p+1)L_p^0(x) ) e^{x/2} \end{equation} \begin{equation} \partial^2_x f(x)=(\frac{x}{4} L_p^1(x)-(p+1)L_p^0(x)-L^1_{p-1}(!)) e^{x/2} \end{equation} where the $(!)$ points out that this term is only there is $p\geq1$.

This gives the following integral \begin{equation} I=-\int_0^\infty dx e^{-x} \left[L_p^1(x)L_p^1(x)(\frac{3}{2}+x+\frac{x^2}{2}) -x (p+1) L_p^0(x) L_p^1(x)- 2 x L_{p-1}^1(x) L_{p}^1(x)\right] \end{equation}

Orthogonality conditions

Making us of the orthogonality conditions

\begin{equation} \int_0^\infty dx x^a e^{-x} L_n^a L_m^a =\frac{(n+a)!}{n!} \delta_{n,m} \end{equation} \begin{equation} \int_0^\infty dx x^{a+1} e^{-x} (L_n^a(x))^2 =\frac{(n+a)!}{n!} (2n+a+1) \end{equation} and \begin{equation} L_p^1=\sum_{i=0}^p L_{i}^0(x) \end{equation} \begin{equation} L_n^a=L_n^{a+1}-L_{n-1}^{a+1}. \end{equation} then \begin{equation} I=-(3p/2+(p+1)+(p+1)(2p+1+1)/2-(p+1)(p+1)-0)=-\frac{5p}{2}-1 \end{equation}

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