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.