Derivations

$L^2$ norm integrations

The $L^2$ norm in the radially symmetric case is

\[\left\lVert f(r) \right\rVert = 2\pi \int_0^\infty \left\vert f(r)\right\vert^2 r\,\mathrm{d}r\,.\]

In the case where $f(r)$ is an electric field, this can be used to calculate the total power $P$:

\[P(t) = \frac{\epsilon_0 c}{2} \left\lVert E(t, r) \right\rVert\,.\]

When using a QDHT, this integral can be approximated by a simple dot product, which is implemented in integrateR and integrateK.

In the following, we use the $0^{\mathrm{th}}$-order QDHT, but the same derivation holds for other orders: $J_0$ becomes $J_p$ and $J_1$ becomes $J_{p+1}$ Following eqs. (21)-(24) in Yu et al. we take the integral, expand $f(r)$ in a Fourier-Bessel series, and use the Hankel transform to obtain the series coefficients:

\[f(r) = \frac{1}{\pi R^2}\sum_{m=1}^M \tilde{f}\left(\frac{j_m}{2\pi R}\right) J_1^{-2}(j_m)J_0\left(\frac{j_m r}{R}\right)\,,\]

which turns the integral into

\[\int_0^R \left\vert f(r)\right\vert^2 r\,\mathrm{d}r = \frac{1}{\pi^2 R^4} \sum_{nm}\tilde{f}\left(\frac{j_n}{2\pi R}\right) \tilde{f}^*\left(\frac{j_m}{2\pi R}\right) J_1^{-2}(j_n) J_1^{-2}(j_m) \int_0^R J_0\left(\frac{j_n r}{R}\right) J_0\left(\frac{j_m r}{R}\right)r\,\mathrm{d}r\,.\]

The integral expression here is

\[\int_0^R J_0\left(\frac{j_n r}{R}\right) J_0\left(\frac{j_m r}{R}\right)r\,\mathrm{d}r = \frac{1}{2}R^2 J_1^2(j_m) \delta_{mn}\,,\]

where $\delta_{mn}$ is the Kronecker delta, which reduces the sum back to one variable:

\[\int_0^R \left\vert f(r)\right\vert^2 r\,\mathrm{d}r = \frac{1}{2\pi^2 R^2} \sum_{m=1}^M\left\vert\tilde{f}\left(\frac{j_m}{2\pi R}\right)\right\vert^2 J_1^{-2}(j_m)\,.\]

If we now define $\tilde{F}$ (this is $F_2$ in Yu et al.):

\[\tilde{F}(m) = \tilde{f}\left(\frac{j_m}{2\pi R}\right) J_1^{-1}(j_m) \frac{K}{2\pi}\,,\]

we arrive at

\[\int_0^R \left\vert f(r)\right\vert^2 r\,\mathrm{d}r = \frac{2}{K^2R^2}\sum_{m=1}^M \vert\tilde{F}(m)\vert^2\]

From Parseval's theorem we know that

\[\int_0^R \left\vert f(r)\right\vert^2 r\,\mathrm{d}r = \int_0^K \vert \tilde{f}(k)\vert^2 k\,\mathrm{d}k\,.\]

Following the same procedure as above for the $k$ integral we find

\[\int_0^K \left\vert \tilde{f}(k)\right\vert^2 k\,\mathrm{d}k = \frac{2}{K^2R^2}\sum_{n=1}^N \vert F(n)\vert^2\,,\]

and hence

\[\int_0^R \left\vert f(r)\right\vert^2 r\,\mathrm{d}r = \frac{2}{K^2R^2}\sum_{n=1}^N \vert F(n)\vert^2 = \frac{2}{K^2} \sum_{n=1}^N \left\vert f\left(\frac{j_n}{K}\right)\right\vert^2 J_1^{-2}(j_n)\,.\]

This is just the dot product between $\vert f(r_n) \vert^2$ and a scaling vector $S_R$ (called scaleR in QDHT):

\[\int_0^R \left\vert f(r)\right\vert^2 r\,\mathrm{d}r = S_R \cdot \vert f(r_n) \vert^2\]

with

\[S_R(n) = 2K^{-2} J_1^{-2}(j_n)\]

The same derivation holds for the $k$ integral and results in

\[S_K(n) = 2R^{-2} J_1^{-2}(j_n)\,.\]

Integration of functions

For integration of the form

\[\int_0^R f(r) r\,\mathrm{d}r\,,\]

we cannot use Parseval's theorem. Instead, following a similar approach we arrive at

\[\int_0^R f(r) r\,\mathrm{d}r = \frac{1}{\pi R^2} \sum_{m=1}^M \tilde{f}\left(\frac{j_m}{2\pi R}\right)J_1^{-2}(j_m) \int_0^R J_0\left(\frac{j_m r}{R}\right)r\,\mathrm{d}r\,.\]

Using

\[\int_0^x x' J_0(x')\,\mathrm{d}x' = xJ_1(x)\,,\]

the integral evaluates to

\[\int_0^R J_0\left(\frac{j_m r}{R}\right)r\,\mathrm{d}r = \frac{R^2}{j_m} J_1(j_m)\,,\]

and so

\[\int_0^R f(r) r\,\mathrm{d}r = \frac{1}{\pi} \sum_{m=1}^M \frac{1}{j_m}\tilde{f}\left(\frac{j_m}{2\pi R}\right) J_1^{-1}(j_m) = \frac{2}{K} \sum_{m=1}^M \frac{1}{j_m} \tilde{F}(m)\,.\]

If we now use the QDHT as defined in Yu et al.,

\[\tilde{F}(m) = \sum_{n=1}^N C_{mn}F(n)\]

with

\[C_{mn} = \frac{2}{S} J_0\left(\frac{j_n j_m}{S}\right)J_1^{-1}(j_n)J_1^{-1}(j_m)\,,\]

after some algebra we arrive at

\[\int_0^R f(r) r\,\mathrm{d}r = \frac{4}{K^2} \sum_{n=1}^N J_1^{-2}(j_n)f(r_n) \sum_{m=1}^M \frac{1}{j_m}J_0\left(\frac{j_n j_m}{S}\right)J_1^{-1}(j_m)\,.\]

When taking into account that $S = j_{N+1}$, the sum over $m$ is approximately $1/2$ for all values of $n$ (this was verified numerically). This leaves us with

\[\int_0^R f(r) r\,\mathrm{d}r \approx \frac{2}{K^2} \sum_{n=1}^N J_1^{-2}(j_n)f(r_n)\]

as in the case of integrals over $\vert f(r)\vert^2$.