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Starting from $f_{-1}=\sin \theta e^{-i\varphi }$, part (i) gives $L^2{f}_{-1}=2{\hslash }^2{f}_{-1}$ and $L_3{f}_{-1}=-\hslash f_{-1}$. Apply the angular momentum raising operator explicitly:A second application givesSince $L_{+}$ commutes with $L^2$, both raised functions retain the eigenvalue $2{\hslash }^2$. Their $L_3$ eigenvalues follow either by differentiating their $\varphi$ dependence or from $[L_3,L_{+}]=\hslash L_{+}$. Thus three valid unnormalized functions, in the requested order, areFor later Born rule calculations, normalize with the solid angle measure $d\Omega =\sin \theta d\theta d\varphi$. Their squared norms are $8\pi /3$, $4\pi /3$, $8\pi /3$, respectively. For example ${\int }_0^{\pi }{\sin }^3\theta d\theta =4/3$ and ${\int }_0^{\pi }{\cos }^2\theta \sin \theta d\theta =2/3$. A convenient orthonormal basis is thereforeOrthogonality follows from the distinct Fourier factors in $\varphi$. An overall phase of each basis function is arbitrary; these choices need not impose the usual phase convention for the spherical harmonics.

Put $f_l(\theta ,\varphi )={\sin }^l\theta e^{-il\varphi }$. For the regular, single-valued orbital angular momentum states, $l$ is a nonnegative integer. Differentiation immediately givesFor $0<\theta <\pi$, the derivatives needed for $L^2$ areConsequently the angular differential expression inside the parentheses of $-L^2/{\hslash }^2$ equalsusing ${\csc }^2\theta -{\cot }^2\theta =1$. ThereforeThe expressions extend regularly through the polar coordinate singularities: $f_l$ is the restriction of the polynomial $(x-iy{)}^l$ to the unit sphere. It is proportional to the lowest-projection spherical harmonic of degree $l$; for $l=0$ it is the constant state.

For the generic instrument, magnetic tweezers combine a magnetically responsive bead, a controllable magnetic field gradient, a tether attached to a fixed support, and optical tracking of bead coordinates. For a fixed magnetic dipole moment, the force is $\nabla (\mathbf{m}\cdot \mathbf{B})$; an induced linear isotropic moment instead gives $\mathbf{F}=V\Delta \chi \nabla B^2/(2{\mu }_0)$. A uniform magnetic field can orient an anisotropic bead, but a gradient is needed for translation. Magnet height sets the load, while field rotation can impose torque if bead orientation is coupled to the field. Tracking uses lateral images and a calibrated axial image/defocus response; a fixed reference removes stage drift. A sufficiently slowly varying gradient approximates a force clamp over the bead's displacement. TODO: user-deferred biomolecular implementation; DNA-specific attachment, preparation and response are not treated.

For transverse-fluctuation calibration of a tensioned tether, use an ideal tether of vertical extension $h$ whose endpoint is displaced by small $x,y$. Keeping its length fixed to quadratic order lowers its vertical coordinate by $(x^2+y^2)/(2h)$, increasing load potential by $F_z(x^2+y^2)/(2h)$. The equipartition theorem therefore givesHere fluctuations are about the mean. The height is the effective pendulum length of the measured coordinate, not automatically the polymer contour length. The displayed $h$ formula uses the point-endpoint or fixed-bead-orientation idealization; freely rotating large beads can require a bead-radius correction to that lever arm. The finite-exposure correction to fluctuation force calibration restores variance lost through exposure averaging; ignoring it would overestimate force; localization noise increases it and would underestimate force. Drift, wall coupling, finite observation time and magnetic lateral stiffness must be checked. Vertical fluctuations measure longitudinal stiffness and do not obey the same geometrical formula.

For the stretch and twist coefficients of an elastic filament, take $s$ as reference contour length and $ϵ$ as local axial strain. The three positive quadratic costs represent bending of the tangent, stretching of the material and excess twist about its relaxed orientation. With the paper's coefficient convention,$K_b,K_t$ are bending and torsional rigidities with units of energy times length; $S$ is an axial force modulus. Thus $A,C$ have units of length and $B$ inverse length. Positive quadratic forms describe a stable relaxed filament; odd terms vanish when expansion is about its zero-stress, zero-torque reference configuration.

At weak and intermediate tension, changing thermal bending conformations can extend the filament without appreciably stretching the backbone. Axial strain is $ϵ\simeq F_z/S$, so material stretching must be included when this is no longer negligible. For a long strongly aligned but nearly inextensible three-dimensional worm-like chain, the thermal extension deficit is approximately $\frac{1}{2}\sqrt{k_{B}T/(A{F}_z)}$; stretch adds approximately $F_z/S$ to fractional extension. Twist matters whenever torque or imposed end rotation is present, at any tension. A freely swivelling, torque-free tether need not store excess twist. There is no universal force-only threshold for switching on torsional energy.

The circular bending energy of an elastic filament follows because curvature is $1/R$ over contour length $2\pi R$. Integrating the bending term givesThis is bending energy alone: imposed tension, stretching and twist would add costs. The dimensionless bending action is $(A/2)\int |{\partial }_s\mathbf{t}{|}^{2}ds$. Its tangent propagator in three dimensions obeys rotational diffusion on the unit sphere, ${\partial }_{s}P=(2A{)}^{-1}{\nabla }_{S^2}^{2}P$. The vector spherical harmonic has angular eigenvalue $-2$, givingHence $A$ is the persistence length in this three-dimensional convention. A strictly planar filament instead has tangent-correlation length $2A$; one cannot infer the factor merely by setting a loop's energy equal to $k_{B}T$.

TODO: user-deferred biomolecular applications. Both enzyme-driven unwinding experiments, their experimental sketches, the requested numerical force and plectoneme radius, energy loss per turn and enzyme-specific dissipation mechanisms remain unsolved. None of their biomolecular parameters has been used in a calculation.

Assume a real, central, short-range potential and elastic quantum scattering. At large radius, normalize the stationary wavefunction aswhere $k=\sqrt{2mE}/\hslash$ and $\theta$ is the scattering angle. Equal incoming and outgoing speeds give the differential scattering cross-section $d\sigma /d\Omega =|f(\theta ){|}^2$.

The partial-wave scattering expansion separates the spherical harmonics. The incident plane wave has expansion ${\sum }_{\ell \ge 0}{i}^{\ell }(2\ell +1)j_{\ell }(kr)P_{\ell }(\cos \theta )$, with Spherical Bessel functions $j_{\ell }$ and Legendre polynomials $P_{\ell }$. The regular radial solution outside the potential has asymptotic form $u_{\ell }\propto \sin (kr-\ell \pi /2+{\delta }_{\ell })$. Its real scattering phase shift ${\delta }_{\ell }$ changes the ratio of outgoing to incoming waves to the Partial-wave S-matrix $S_{\ell }=e^{2i{\delta }_{\ell }}$. Matching outgoing coefficients yieldsHere $\ell$ labels the orbital angular momentum, with squared magnitude ${\hslash }^2\ell (\ell +1)$. The assumption of purely elastic scattering is what makes every $S_{\ell }$ have unit modulus.

Let $Y_{\ell }$ be a spherical harmonic of degree $\ell$ on the unit sphere. The Laplacian in spherical coordinates separates its angular part from a radial power, giving the displayed identity for $r>0$. Thus a Poisson equation $\Delta \Phi =r^q{Y}_{\ell }$ has the particular solution $\Phi =r^{q+2}{Y}_{\ell }/[(q+2)(q+3)-\ell (\ell +1)]$ when the denominator is nonzero. Add $A{r}^{\ell }{Y}_{\ell }+B{r}^{-\ell -1}{Y}_{\ell }$ to impose the radial boundary conditions. If the denominator vanishes at $p=q+2$, differentiating the displayed identity with respect to $p$ gives instead $\Delta (r^p\log r{Y}_{\ell })=(2p+1)r^{p-2}{Y}_{\ell }$. Regularity at the origin and decay at infinity must be checked separately; the radial identity alone is local away from zero.

For an isolated solenoidal magnetic field, zero flux through every sphere removes the degree-zero spherical harmonic of $P=\mathbf{B}\cdot \mathbf{x}$. In a poloidal-toroidal decomposition, let ${\mathcal{L}}^2=-{\Delta }_{S^2}$. The spherical poloidal component obeysIf $P={\sum }_{l\ge 1,m}{p}_{lm}(r)Y_{lm}$, put $D_{lm}={\int }_0^{\infty }[r^2|p_{lm}^{\prime }{|}^2+l(l+1)|p_{lm}{|}^2]dr$. The two sides of the bound are respectively $\sum D_{lm}/[l(l+1)]$ and $\frac{1}{2}\sum D_{lm}$. Since $l(l+1)\ge 2$, the bound follows, with equality for degree-one radial scalars. Together with the radial-flow dynamo energy bound it gives $(q/\eta {)}^2\ge 2E_P/E_{\mathrm{total}}$ for a marginal or growing dynamo field.

// Target: fluid-mechanics.bigb

Both $z$ and $p_z$ commute with $L_z$ and are odd under parity. Their angular selection rules therefore require the same magnetic quantum number and opposite orbital parity. Within the nonrelativistic point-Coulomb $n=2$ hydrogen atom eigenspace, the only angularly permitted off-diagonal pair is $|2s⟩$ and $|2p,m=0⟩$.

Those necessary angular conditions do not guarantee a nonzero matrix element. For the Coulomb potential Hamiltonian $H={\mathbf{p}}^2/(2m)+V(r)$,All states within this exact $n=2$ eigenspace have the same energy, so every projected $p_z$ matrix element vanishes. In contrast, the position matrix element is nonzero: with conventional real spherical harmonics and hydrogen radial wavefunction in the first excited shell, $⟨2s|z|2p,0⟩=-3a_0$. Its sign depends on basis phases, while its magnitude does not.

Thus the electric-dipole selection rules for hydrogen allow a linear position perturbation to mix these degenerate states, but a momentum perturbation cannot be assigned the same matrix merely from parity. The conclusion concerns the unperturbed Coulomb eigenspace; degeneracy-breaking interactions require their own energy eigenstates and matrix elements.

The partial-wave scattering channel with orbital angular momentum quantum number one. Its angular dependence is built from degree-one spherical harmonics. For a finite-range potential away from a threshold pole, the finite-range partial-wave threshold law gives phase shift ${\delta }_1=O(k^3)$ and a partial elastic scattering cross-section proportional to $k^4$ as the relative wavenumber $k$ tends to zero.

For radius $a$ and tangential velocity ${\mathbf{u}}_s={\mathbf{I}}_s\mathbf{Ax}$ with symmetric traceless $\mathbf{A}$, the surface divergence is $-3\mathbf{n}\cdot \mathbf{An}$. At small surface Péclet number $a^2|\mathbf{A}|/D_s$, neglecting advection of the concentration perturbation givesThe result follows because this quadratic is a degree-two spherical harmonic with surface Laplacian $-6/a^2$ times itself and zero spherical mean.

For a point-mass companion at separation $a$, subtract the constant and linear terms of its Newtonian gravitational potential in a frame accelerating with the target's center. The leading residual isIt is degree two in spherical harmonics. In a circular orbit with angular frequency ${\omega }_o$, its time-dependent $m=\pm 2$ components force at $2{\omega }_o$ in a frame with no spin; there is also a static $m=0$ component. This is the leading approximation in $r/a$, and the tidal amplitude may be absorbed into a harmonic-dependent constant.

The radial probability density $P(r)$ is defined so that $P(r)dr$ is the probability of finding a particle between radii $r$ and $r+dr$. The spherical volume element gives $P(r)=r^2\int |\psi (r,\Omega ){|}^{2}d\Omega$. For a spherically symmetric wavefunction, $P(r)=4\pi r^2|\psi (r){|}^2$; for $\psi =R(r)Y_l^m(\Omega )$ with normalized spherical harmonics, $P(r)=r^2|R(r){|}^2$. Its maximum need not coincide with the maximum of the density per unit volume: a hydrogenic $1s$ state has its largest spatial density at the origin but its most probable radius at $a_0/Z$.

Separation of variables in a central potential writes a wavefunction as $R_{nl}(r)Y_l^m(\theta ,\varphi )$. With normalized spherical harmonics the radial factor satisfies ${\int }_0^{\infty }{r}^2|R_{nl}{|}^{2}dr=1$. The reduced radial function is $u=rR$, with normalization $\int |u{|}^{2}dr=1$; the two functions have different behavior at the origin. For a regular Coulomb eigenstate, $R_{nl}\propto r^l$ near the origin.

A regular solid harmonic is $r^l{Y}_l^m(\theta ,\varphi )$: it is homogeneous of degree $l$ and solves the Laplace equation. Thus $\mathbf{r}\cdot \nabla F=lF$ and ${\nabla }^{2}F=0$. These identities turn angular spherical harmonic expansions of displacement, density and Newtonian gravitational potential into radial equations.

Rotational Raman spectroscopy observes a frequency change of scattered light associated with molecular rotation. The coupling uses the induced electric dipole moment, ${\mathbf{d}}_{\mathrm{ind}}=\mathbf{\alpha }\mathbf{E}$, where $\mathbf{\alpha }$ is the electric polarizability tensor. For a linear molecule, the useful anisotropy is ${\alpha }_{\parallel }-{\alpha }_{\perp }$: the response differs for an electric field parallel or perpendicular to the molecular axis. A permanent electric dipole moment is not required.^[ref]

For the spin-free linear rigid rotor in one internal quantum state, the anisotropic tensor depends quadratically on the molecular direction. Its angular part has even parity and degree two in spherical harmonics. The triangle inequality for coupled angular momenta permits $|\Delta J|\le 2$, and even parity removes odd $\Delta J$, leaving $\Delta J=0,\pm 2$. The $\Delta J=0$ contribution is unshifted light; shifted rotational lines have $\Delta J=\pm 2$. Consequently the rigid-model Stokes shifts, in ordinary frequency units, areSuccessive allowed shifts are separated by $4B$, rather than the $2B$ absorption spacing of pure rotational electric-dipole selection rules. In spectroscopic wavenumber units replace $B$ by $\tilde{B}=B/c$ and $\Delta \nu$ by $\Delta \tilde{\nu }=\Delta \nu /c$, using the speed of light in $\mathrm{cm}{\mathrm{s}}^{-1}$ for $\mathrm{c}{\mathrm{m}}^{-1}$.

A homonuclear diatomic molecule such as molecular hydrogen has no permanent electric dipole moment in its symmetric internal state, so it lacks the ordinary electric-dipole microwave ladder. Its anisotropic electric polarizability can nevertheless give rotational Raman spectroscopy lines. Nuclear-spin symmetry can change the populations of even and odd $J$ levels, so the spin-free population example above must not be reused unchanged for molecular hydrogen. The method of coupling, as well as the energy eigenvalues, determines what the experiment can see.

For the spin-free linear rigid rotor in thermal equilibrium at temperature $T>0$, assume all integer $J\ge 0$ are allowed and ignore nuclear-spin restrictions. Each $J$ level has $2J+1$ quantum states sharing one energy eigenvalue, an instance of degenerate energy levels, so the Boltzmann distribution givesHere $Q_{\mathrm{rot}}$ is a rotational partition function, $h$ the Planck constant, $k_{\mathrm{B}}$ the Boltzmann constant, and $B$ a frequency in $\mathrm{Hz}$. Thus $hB/(k_{\mathrm{B}}T)$ is dimensionless. The $2J+1$ factor counts magnetic quantum numbers; it does not mean that a whole level is one quantum state. For the stipulated $B=5\mathrm{GHz}$, the most populated levels are $J=2$ at $3\mathrm{K}$ and $J=7$ at $30\mathrm{K}$.

Population alone is insufficient to predict intensity. In weak, unsaturated absorption, the integrated absorption coefficient includes the transition frequency, the squared electric dipole moment matrix element, and the difference of lower and upper substate populations because stimulated emission returns energy to the field. A pulsed FTMW spectroscopy signal additionally depends on excitation, quantum coherence, cavity tuning, and receiver response.^[ref]

For an isotropic sample with polarization of an electromagnetic wave along $z$, write $d_z=d\cos \theta$. Normalized spherical harmonics giveThe last equality follows from ${\sum }_{M=-J}^J{M}^2=J(J+1)(2J+1)/3$. Combining this sum with the per-substate Boltzmann distribution and subtracting stimulated emission gives relative integrated absorption areasThe common factor includes $d^2/3$ and cancels upon normalization. This formula describes integrated weak absorption in a thermal, spin-free linear rigid rotor, not arbitrary peak heights. The example's strongest integrated lines originate at $J=3$ and $J=13$ for $3$ and $30\mathrm{K}$, respectively: they differ from the most populated levels. A missing line can reflect population, coupling, coverage, or sensitivity; it does not prove that the corresponding quantum state is absent.

A scalar spheroidal harmonic is a regular angular solution of $(\sin \theta S^{\prime }{)}^{\prime }/\sin \theta +[c^2{\cos }^2\theta -m^2/{\sin }^2\theta +\Lambda ]S=0$; with $e^{im\varphi }$ it is smooth at both axes. For real $c$, regularity gives discrete eigenvalues ${\Lambda }_{\ell m}(c)$. At $c=0$ it reduces to the angular part of a spherical harmonic, with $\Lambda =\ell (\ell +1)$. The sign of the $c^2$ term distinguishes the convention used here.

A solid harmonic extends a spherical harmonic to a harmonic function in space with either the regular radial power $r^n$ or the irregular radial power $r^{-n-1}$. In three dimensions these are the two separated radial solutions of the Laplace equation at angular degree $n$.

Spectral geometry studies which properties of a Riemannian manifold can be recovered from the spectrum of geometric differential operators, especially the Laplace-Beltrami operator. Explicit spherical harmonics, Fourier series on a flat torus, and isospectral manifolds illustrate three different aspects of the problem.

The laboratory expectation value of the axial electric quadrupole moment in the stretched nuclear quantum state $|I,M=I⟩$; $M\hslash$ is the laboratory $z$ component of total angular momentum. It differs from the intrinsic nuclear quadrupole moment $Q_0$. In an axial nuclear rotational band with $K=0$, projection onto definite angular momentum givesHere $K\hslash$ denotes the component of total angular momentum along the intrinsic symmetry axis. Consequently a positive $Q_0$ for a prolate spheroid corresponds to a negative $Q_s(2)$ in this model. The sign of the laboratory moment alone must be interpreted together with the band assignment.

The $I=0$ result has a short symmetry explanation. A spin-zero state has no preferred spatial direction. Its second charge moments satisfy $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩$, so $⟨2z^2-x^2-y^2⟩=0$. Equivalently, averaging an intrinsic axis uniformly over directions gives $⟨P_2(\cos \theta )⟩=0$, where $P_2(u)=(3u^2-1)/2$ is a Legendre polynomial. This is a statement about a quantum expectation value, not a classical picture of a spinning object sweeping through orientations.

For an independent check of the $I=2$ sign, the $K=0$, $M=2$ rotor orientation probability is proportional to $|Y_2^2{|}^2\propto {\sin }^4\theta$, using a spherical harmonic. With $u=\cos \theta$,A zero static laboratory moment does not make every quadrupole observable zero. With electric multipole operator ${\mathcal{M}}_{2\mu }=\int {\rho }_{\mathrm{ch}}{r}^2{Y}_2^{\mu }{d}^{3}r$, define the reduced transition strength using a reduced matrix element:In the axial $K=0$ rotor, angular projection gives $B(E2;0^{+}\to 2^{+})=5Q_0^2/(16\pi )$ and $B(E2;2^{+}\to 0^{+})=Q_0^2/(16\pi )$. The factor of five is the ratio of the numbers of magnetic substates. Thus an electric quadrupole transition can reveal collective deformation even when $Q_s(0)=0$. These strengths determine $|Q_0|$ within the model; their squares alone cannot distinguish its sign.

A spherical harmonic of degree $\ell$ is an eigenfunction of the angular Laplacian on the unit sphere. In three dimensions, with the sign convention inherited from $\Delta ={\sum }_j{\partial }_j^2$,One chooses an orthonormal basis for each $(2\ell +1)$-dimensional eigenspace. With surface measure $d\Omega$, examples are $Y_0^0=1/\sqrt{4\pi }$ and $Y_1^0=\sqrt{3/(4\pi )}\cos \theta$.

The degree-zero mode is an angular monopole, the degree-one modes form an angular dipole, and higher degrees resolve finer angular structure. Separation of variables in the Laplace equation pairs these angular modes with the solid harmonics $r^{\ell }{Y}_{\ell }^m$ and $r^{-\ell -1}{Y}_{\ell }^m$. This is why the same basis organizes gravitational potential expansions, atomic orbitals and sky-map multipoles.