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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.