# Proofs ## Polar Fourier expansion and Parseval identity Let $$ D_R=\{(r,\theta):0\le r\le R,\ 0\le\theta<2\pi\} $$ and let $f\in L^2(D_R,r\,dr\,d\theta)$. Define $$ \rho_m(r)=\frac{1}{2\pi}\int_0^{2\pi} f(r,\theta)e^{-im\theta}\,d\theta. $$ For almost every fixed $r$, the function $f(r,\cdot)$ belongs to $L^2(S^1)$. The Fourier basis is complete, so $$ f(r,\theta)=\sum_{m\in\mathbb Z}\rho_m(r)e^{im\theta} $$ in angular $L^2$, and $$ \frac{1}{2\pi}\int_0^{2\pi}|f(r,\theta)|^2\,d\theta = \sum_m|\rho_m(r)|^2. $$ ### Proof Since $$ \int_0^R \left[ \int_0^{2\pi}|f(r,\theta)|^2\,d\theta \right]r\,dr<\infty, $$ Fubini's theorem implies that the inner integral is finite for almost every $r$. Thus $f(r,\cdot)\in L^2(S^1)$ for almost every $r$. The functions $$ \frac{e^{im\theta}}{\sqrt{2\pi}}, \qquad m\in\mathbb Z, $$ form a complete orthonormal basis of $L^2(S^1)$. The coefficients $\rho_m(r)$ are the corresponding Fourier coefficients. Completeness gives angular $L^2$ convergence, and Parseval's identity gives the pointwise-in-$r$ norm relation above. Multiplying by $2\pi r$ and integrating radially gives $$ \int_0^R\int_0^{2\pi}|f|^2r\,d\theta\,dr = 2\pi\sum_m \int_0^R|\rho_m(r)|^2r\,dr. $$ Define $$ P_m=\int_0^R|\rho_m(r)|^2r\,dr. $$ Then the total polar power is $2\pi\sum_mP_m$. ## Exact reconstruction error Let $S\subset\mathbb Z$ be a set of retained harmonics and define $$ f_S(r,\theta)= \sum_{m\in S}\rho_m(r)e^{im\theta}. $$ Then $$ \frac{\|f-f_S\|_{L^2(D_R)}}{\|f\|_{L^2(D_R)}} = \sqrt{ 1- \frac{\sum_{m\in S}P_m} {\sum_mP_m} }. $$ ### Proof The omitted field is $$ f-f_S= \sum_{m\notin S}\rho_m(r)e^{im\theta}. $$ Orthogonality of distinct angular harmonics gives $$ \|f-f_S\|^2 = 2\pi\sum_{m\notin S}P_m, $$ while Parseval gives $$ \|f\|^2 = 2\pi\sum_mP_m. $$ Using $$ \sum_{m\notin S}P_m = \sum_mP_m-\sum_{m\in S}P_m $$ and taking the positive square root gives the result. PETAL2D reports this quantity as a percentage. ## Optimal power-ranked truncation Suppose admissible selection units have powers $$ Q_1\ge Q_2\ge\cdots\ge0. $$ For a complex field, a unit can be one angular channel. For a real field, the admissible units are $m=0$, conjugate pairs $(m,-m)$, and the even-grid Nyquist singleton when present. For any fixed number $K$ of units, retaining the first $K$ units minimizes the reconstruction error. The first $K$ that satisfies the requested error tolerance also uses the minimum number of admissible units. ### Proof The reconstruction-error formula is monotone in retained power. Minimizing the error is equivalent to maximizing $$ \sum_{j\in S}Q_j $$ over all admissible sets $S$ containing $K$ units. If a proposed set contains $Q_a$ and omits a larger $Q_b$, replacing $Q_a$ by $Q_b$ increases retained power. Repeating this exchange produces the set $\{Q_1,\ldots,Q_K\}$. No other $K$-unit set retains more power. If the first $K$ ranked units fail the error target, every other $K$-unit set retains no more power and must also fail. The first successful ranked value of $K$ is minimal. ## Real-valued fields and conjugate pairs If $f$ is real-valued, $$ \rho_{-m}(r)=\rho_m(r)^*. $$ Consequently, $$ P_{-m}=P_m. $$ ### Proof Taking the complex conjugate of the coefficient gives $$ \rho_m(r)^* = \frac{1}{2\pi} \int_0^{2\pi} f(r,\theta)e^{im\theta}\,d\theta, $$ because $f^*=f$. The right-hand side is $\rho_{-m}(r)$. A conjugate pair reconstructs as $$ \rho_me^{im\theta} + \rho_{-m}e^{-im\theta} = 2\operatorname{Re} \left[ \rho_me^{im\theta} \right], $$ which is real. This is why PETAL2D keeps conjugate pairs together for real input. ## Rotation covariance For an active rotation $$ f_\phi(r,\theta)=f(r,\theta-\phi), $$ the angular coefficients transform as $$ \rho_m^{(\phi)}(r) = e^{-im\phi}\rho_m(r). $$ Therefore $$ P_m^{(\phi)}=P_m. $$ ### Proof Insert the rotated field into the coefficient definition: $$ \rho_m^{(\phi)}(r) = \frac{1}{2\pi} \int_0^{2\pi} f(r,\theta-\phi)e^{-im\theta}\,d\theta. $$ Set $u=\theta-\phi$. Periodicity restores the integration interval to $[0,2\pi)$ and $$ e^{-im\theta} = e^{-im\phi}e^{-imu}. $$ Hence $$ \rho_m^{(\phi)}(r) = e^{-im\phi}\rho_m(r). $$ Taking the modulus squared removes the phase factor. ## Rotational selection rules Suppose a field transforms under a rotation by $2\pi/n$ as $$ f(r,\theta+2\pi/n) = e^{i2\pi\ell/n}f(r,\theta), $$ where $\ell$ is an integer modulo $n$. Then any nonzero angular coefficient satisfies $$ m\equiv\ell\pmod n. $$ The invariant scalar case corresponds to $\ell=0$. ### Proof Insert the Fourier expansion: $$ \sum_m \rho_m(r)e^{im\theta}e^{i2\pi m/n} = e^{i2\pi\ell/n} \sum_m\rho_m(r)e^{im\theta}. $$ Uniqueness of Fourier coefficients gives $$ \rho_m(r) \left[ e^{i2\pi m/n} - e^{i2\pi\ell/n} \right] = 0. $$ For $\rho_m(r)\ne0$, $$ e^{i2\pi(m-\ell)/n}=1, $$ so $m-\ell$ is divisible by $n$. ## Discrete angular aliasing For $$ \theta_j=\frac{2\pi j}{N_\theta}, \qquad j=0,\ldots,N_\theta-1, $$ harmonics whose indices differ by an integer multiple of $N_\theta$ are identical on the sampled angular grid. ### Proof For any integer $k$, $$ e^{i(m+kN_\theta)\theta_j} = e^{im\theta_j} e^{i2\pi kj} = e^{im\theta_j}. $$ The samples cannot distinguish $m$ from $m+kN_\theta$. This is the origin of angular aliasing and the Nyquist limit. ## Angular-momentum interpretation for a wavefunction If the analyzed field is a normalized complex wavefunction, $$ \psi(r,\theta) = \sum_m\rho_m(r)e^{im\theta}, $$ the fixed-$m$ subspace is the eigenspace of the planar operator $$ L_z=-i\hbar\frac{\partial}{\partial\theta} $$ with eigenvalue $m\hbar$. The squared norm of the projection onto this subspace is $$ \|\Pi_m\psi\|^2 = 2\pi P_m. $$ Since normalization gives $$ 1=2\pi\sum_nP_n, $$ the angular-momentum probability is $$ \Pr(L_z=m\hbar) = \frac{P_m}{\sum_nP_n}. $$ This interpretation applies to the wavefunction itself. A density such as $|\psi|^2$ does not retain the phase required for the same angular-momentum interpretation. ## Translation of the power centroid Define $$ \mathbf r_c[f] = \frac{ \int \mathbf r\,|f(\mathbf r)|^2\,d^2r }{ \int |f(\mathbf r)|^2\,d^2r }. $$ For $$ g(\mathbf r)=f(\mathbf r-\mathbf a), $$ and an integration domain translated with the field, $$ \mathbf r_c[g] = \mathbf r_c[f]+\mathbf a. $$ ### Proof Set $\mathbf u=\mathbf r-\mathbf a$. The denominator is unchanged. The numerator becomes $$ \int (\mathbf u+\mathbf a) |f(\mathbf u)|^2\,d^2u = \int \mathbf u|f(\mathbf u)|^2\,d^2u + \mathbf a \int |f(\mathbf u)|^2\,d^2u. $$ Dividing by the common denominator gives the result.