The far-field complex amplitude is the two-dimensional Fourier transform of the aperture amplitude transmittance, evaluated at transverse spatial frequencies $q_x=k{\ell }_x$, $q_y=k{\ell }_y$. The light intensity is proportional to the squared magnitude of this transform.

Use linearity to add the transforms of simpler aperture components, preserving their relative phases. A translated component acquires a complex exponential factor. Repeated identical openings can be expressed as the convolution of one opening with a sum of Dirac delta functions; the convolution theorem then factors the field into a single-opening transform and an array factor. Multiplying aperture functions instead convolves their transforms. These operations simplify complex apertures without repeatedly evaluating the full diffraction integral. It is the complex amplitudes, rather than their wave intensities, that are added for coherent illumination.

For the normally incident plane wave, the incident complex amplitude and phase are uniform across the aperture. Treat the field using scalar diffraction, with the aperture represented by its prescribed amplitude transmittance $h(x,y)$. Let $R$ be the distance from the aperture centre to the observation point and let ${\ell }_x,{\ell }_y$ be the transverse direction cosines of that point.

Across an aperture of characteristic size $a$, the Fraunhofer diffraction assumptions are that $a/R\ll 1$, so $1/r\simeq 1/R$ and the obliquity factor is nearly constant, and that the quadratic contribution to the propagation phase is negligible:The remaining path-length dependence is linear:A coherent monochromatic wave and an aperture large enough relative to wavelength for the scalar aperture model complete the usual physical assumptions. The complex amplitude becomeswhere $C$ includes the common propagation phase, distance factor and nearly constant obliquity factor. For the central part of a planar screen in the paraxial approximation, ${\ell }_x\simeq X/R$ and ${\ell }_y\simeq Y/R$. Small observation angles are useful for that screen-coordinate formula; a general far-field Fourier transform can instead use the exact direction cosines. A lens can also produce this pattern in its focal plane without requiring a physically very distant screen.

Different transverse normal modes have different cutoff frequencies and group velocities. A pulse exciting several of them reaches the output in delayed modal components, causing intermodal dispersion, pulse distortion and overlap of successive signals. Their interference also makes the transverse wave intensity pattern depend on propagation distance; irregularities can transfer energy between modes.

Mitigations are to operate between the first and second cutoff frequencies of a membrane waveguide, excite only the desired mode shape with a suitable source, and use smooth uniform boundaries or gradual transitions to reduce mode conversion. Reducing the signal bandwidth keeps its spectrum inside the single-mode interval and reduces group velocity dispersion within the selected mode. Single-mode operation removes intermodal dispersion but does not remove the dispersion derived in part (b).

The natural interpretation is the fundamental transverse normal mode, $n=1$. Indeedso only $n=1$ propagates over the main part of the given Gaussian distribution of angular frequencies. Its centre travels atDifferentiate the group velocity with respect to angular frequency:At the centre this is $0.010149\mathrm{m}$. The one-standard-deviation group velocity spread is consequentlyOver the central travel time the associated spatial shear isEquivalently, use representative components at ${\omega }_0\pm {\sigma }_{\omega }$, for which $v_g(95)=1.5001\mathrm{m}{\mathrm{s}}^{-1}$ and $v_g(105)=1.6024\mathrm{m}{\mathrm{s}}^{-1}$. Their separation after $t_0$ isAt a fixed detector $10\mathrm{m}$ away, the corresponding arrival-time difference is $10/v_g(95)-10/v_g(105)=0.426\mathrm{s}$.

The paper asks for an estimate of spreading, without specifying a spatial width convention or the initial spectral phase. A Gaussian distribution has no finite full range. Thus $0.326\mathrm{m}$ is a one-standard-deviation dispersive contribution, whereas about $0.65\mathrm{m}$ describes the separation of the $\pm {\sigma }_{\omega }$ representatives. Neither is automatically the increase in the packet's original width. For an initially unchirped Gaussian wave packet in the narrow-band approximation, independent initial and dispersive contributions combine as ${\sigma }_{x,f}^2\simeq {\sigma }_{x,0}^2+{\sigma }_{x,\mathrm{disp}}^2$; a packet with an initial chirp can behave differently. The modal assumption and width convention must accompany the numerical answer.

A nondispersive wave has a linear dispersion relation $\omega =vk$ over the relevant range, so every Fourier transform component has the same phase velocity $v_p=\omega /k=v$ and group velocity $v_g=d\omega /dk=v$. A wave packet then translates without spreading in an ideal homogeneous lossless medium.

For wave dispersion, $v_p$ depends on frequency or wavenumber. Different components accumulate different phases, and in general the group velocity differs from the phase velocity; variation of group velocity across the spectrum produces spreading of a wave packet. The membrane material can be nondispersive while confinement in a membrane waveguide introduces wave dispersion along its length.

Assume independent normally distributed angle errors with known standard deviation ${\sigma }_y=0.002$. The relevant chi-squared test uses the sum of squared standardized residuals,For the model through the origin, the predictions and residuals, expressed in milliradians, areConsequently ${\chi }_0^2=30.0280$. There are $4-1=3$ statistical degrees of freedom. The upper-tail P-value is approximately $1.36\times 10^{-6}$; equivalently the value greatly exceeds the $5\%$ critical value $7.815$. The zero-intercept model is strongly rejected under the stated errors.

For the alternative linear regression, minimize ${\sum }_i(y_i-c^{\prime }{x}_i-d{)}^2$. Setting both partial derivatives to zero gives the normal equations, whose solution isFor these data $\bar{x}=2508.83699{\mathrm{m}}^{-2}$, $\bar{y}=0.0475$, and $S_{xx}=4.830232\times 10^6{\mathrm{m}}^{-4}$. This yieldsThe new predictions are $(66.6548,55.4171,41.4315,26.4966)\mathrm{mrad}$, with residuals $(-0.6548,-0.4171,2.5685,-1.4966)\mathrm{mrad}$. ThereforeThe affine model is compatible with the data. Allowing for its additional fitted parameter, the decrease $\Delta {\chi }^2=27.668$ has a chi-squared distribution with one statistical degree of freedom under the nested zero-intercept hypothesis, corresponding to $p\simeq 1.44\times 10^{-7}$. Thus the model with a fitted intercept provides the better description, rather than merely lowering the residual by adding a parameter.

With the measurement variance treated as known, the coefficient uncertainties areHence $d=(13.11\pm 2.49)\mathrm{mrad}$, about $5.26$ standard deviations above zero. A positive constant systematic error in the angular zero, or an additional nearly constant torque, could account for it. Its presence is evidence against the ideal zero-intercept model; four points alone do not identify the physical cause uniquely. A calibration error in $\kappa$ only rescales the inferred gravitational constant and cannot remove this residual pattern.

Using the fitted slope after allowing for this offset would give $G^{\prime }=\kappa {\hat{c}}^{\prime }/(mML)=(7.27\pm 0.51)\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, with the period and slope errors combined in quadrature. This is conditional on the constant-offset model.

All four angle measurements have the same standard uncertainty, so ordinary least squares minimizesDifferentiation givesThus the unique minimum isThe overbars denote sample means over the four observations. Numerically,soThe fixed $x_i$ have negligible uncertainty. Assuming independent errors in the four $y_i$, propagation of uncertainty in quadrature givesHenceIf the period measurement is independent of the angle measurements, including its calibration uncertainty givesThis last uncertainty includes the stated random measurement errors only. Whether the model describes the data must be checked before treating it as a reliable measurement of the gravitational constant.

Rearrange the given torsion balance equilibrium relation asA useful choice, preserving equal measurement uncertainties on the vertical axis, isWith angle measured in radians, $y$ is dimensionless and $x$ has units ${\mathrm{m}}^{-2}$, so $c$ has units ${\mathrm{m}}^2$, as required. The four $x$ values, in the order of the original table, areThe corresponding $y$ values are $(0.066,0.055,0.044,0.025)$, each with standard uncertainty ${\sigma }_y=0.002$. The uncertainty of $x$ is negligible by assumption, making linear regression through the origin appropriate for the proposed model.

The original apparatus has two small masses at distance $L/2$ from the suspension axis. Neglecting the rod's moment of inertia, their total moment of inertia is $I=2m(L/2{)}^2=m{L}^2/2$, consistent with the supplied torsion pendulum period. Treat $m$ and $L$ as exact because no uncertainties are specified. Rearranging the period relation givesThe torsion constant is a restoring torque per unit angle. In dimensional calculations the radian is dimensionless.

Using first-order propagation of uncertainty, with the quoted period uncertainty interpreted as a standard uncertainty,Consequently

Identical spin one-half particles are fermions. By fermionic exchange symmetry, the complete spatial-and-spin wavefunction must change sign when the two particle labels are exchanged. Define the normalized antisymmetric spin-one-half singlet stateand the three symmetric spin-one-half triplet statesThe Hamiltonian is independent of spin, so these spin functions carry no additional energy.

Both particles can occupy $u_0$ if their spin factor is antisymmetric. The unique ground state wavefunction isTwo equal-spin particles could not share this orbital, by the Pauli exclusion principle.

At the first excited energy, $E=ϵ={\hslash }^2/(2m{R}^2)$, one orbital is $u_0$ and the other is $u_s$ with $s=+1$ or $-1$. Define symmetric and antisymmetric normalized spatial functionsFor each $s$, all allowed wavefunctions are spanned byThere are $2(1+3)$ orthonormal states, hence first-excited-state degeneracy $g=8$. In each product, exactly one factor is antisymmetric, ensuring the required total fermionic exchange symmetry.

Put $ϵ={\hslash }^2/(2m{R}^2)$. The two-particle product energy eigenstates areBecause the particles are distinguishable, $(n_1,n_2)$ and $(n_2,n_1)$ are distinct when $n_1\ne n_2$. Also, the signs of nonzero integers distinguish opposite angular momenta. The quantum degeneracy is therefore the number of ordered integer pairs at each sum of squares.

The first distinct energy levels are $0,ϵ,2ϵ,4ϵ,5ϵ,\dots$. This ordering matters: $4ϵ$ is the third excited level, while $5ϵ$ is the fourth excited level.

These counts concern energy eigenspaces; arbitrary quantum superpositions within each one do not create additional independent states.

Solving the time-independent Schrödinger equation with $\psi \propto e^{ik\varphi }$ gives $E={\hslash }^2{k}^2/(2m{R}^2)$. The periodic boundary condition requires $e^{2\pi ik}=1$, hence $k=n\in \mathbb{Z}$. With the angular integration measure $d\varphi$, wavefunction normalization givesso a convenient orthonormal eigenbasis isThe $n=0$ ground state is nondegenerate. For each $|n|>0$, $u_n$ and $u_{-n}$ span a two-dimensional energy eigenspace; real sine and cosine combinations are an equally valid basis. These opposite-circulation eigenstates have angular momentum $n\hslash$ and $-n\hslash$.

If instead the wavefunction is normalized with respect to physical arc length $ds=Rd\varphi$, its expression is $e^{in\varphi }/\sqrt{2\pi R}$. The two conventions describe the same probability, since $|u_n{|}^{2}d\varphi =|u_n/\sqrt{R}{|}^{2}ds$. The remaining parts use $d\varphi$.

For a particle in a ring, the arc-length coordinate is $s=R\varphi$. The tangential kinetic energy isThe constraint removes radial motion; the constant potential energy along the ring can be chosen as zero. The tangential momentum operator is ${\hat{p}}_s=-i\hslash {\partial }_s$, or equivalently the angular momentum operator is ${\hat{L}}_z=-i\hslash {\partial }_{\varphi }$. ThusIts physical domain has periodic boundary conditions, $\psi (\varphi +2\pi )=\psi (\varphi )$ and ${\psi }^{\prime }(\varphi +2\pi )={\psi }^{\prime }(\varphi )$. These make the Hamiltonian self-adjoint and reflect that $\varphi$ and $\varphi +2\pi$ describe the same point. No magnetic flux or additional angle-dependent potential energy is present.

The specified energy requires $n_x+n_y=2$, so the relevant energy eigenspace has the orthonormal basisApplying the creation and annihilation operators givesThus its matrix in this ordered basis isWriting $q=\hslash \lambda$, the characteristic polynomial gives $\lambda ({\lambda }^2-4)=0$. The possible angular momentum eigenvalues are thereforeFor completeness, normalized simultaneous energy and angular momentum eigenstates areEach value occurs once in this energy eigenspace. Equivalently, circular ladder operators of a two-dimensional oscillator give $\hat{H}=\hslash \omega ({\hat{N}}_{+}+{\hat{N}}_{-}+1)$ and ${\hat{L}}_z=\hslash ({\hat{N}}_{+}-{\hat{N}}_{-})$. With $N_{+}+N_{-}=2$, the difference can only be $2,0,-2$.

Set $\hat{B}={\hat{a}}_x^{†}{\hat{a}}_y-{\hat{a}}_y^{†}{\hat{a}}_x$. Taking the Hermitian adjoint reverses the order of products, so ${\hat{B}}^{†}=-\hat{B}$. Complex conjugation also reverses the sign of $-i$, and thereforeThus $\hat{Q}$ is a self-adjoint operator on the usual oscillator states.

Using the coefficients in part (a), products containing two position operators or two momentum operators cancel in the difference. Cross-coordinate operators commute, leavingHenceIt is the component of orbital angular momentum perpendicular to the plane, with units $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ (equivalently $\mathrm{J}\mathrm{s}$). The rotational symmetry of the two-dimensional isotropic harmonic oscillator makes this a conserved observable; directly, each term in $\hat{B}$ preserves $N_x+N_y$, so $[\hat{H},\hat{Q}]=0$.

The number operator commutators established above imply that ${\hat{a}}_x^{†}|n⟩$ is proportional to $|n+1⟩$ and ${\hat{a}}_x|n⟩$ to $|n-1⟩$. Their norms determine the magnitudes of the proportionality constants:Choose the phase of the normalized ground eigenstate once, then define recursivelyThis removes the otherwise arbitrary relative phases of the eigenstates. The Hermitian adjoint relation then fixes the lowering coefficient with the same positive sign, givingAt $n=0$, the second expression means the zero vector, ${\hat{a}}_x|0⟩=0$; it does not introduce a physical state $|-1⟩$.

For either number operator $\hat{N}={\hat{a}}^{†}\hat{a}$, positivity follows fromHence an eigenvalue $\nu$ cannot be negative. The commutatorsshow that $\hat{a}|\nu ⟩$, if nonzero, is an eigenstate with eigenvalue $\nu -1$. For a normalized eigenstate, $\Vert \hat{a}|\nu ⟩{\Vert }^2=\nu$.

If $\nu$ were not an integer, successive nonzero lowerings would eventually reach an eigenstate with $0<{\nu }^{\prime }<1$. Its next lowering would have nonzero norm but a negative eigenvalue, contradicting positivity. The ladder must therefore terminate at $\nu =0$, with $\hat{a}|0⟩=0$. Conversely, successive applications of the creation operator never vanish, becauseThe normalizable bottom state exists explicitly: solving $\hat{a}{\psi }_0=0$ gives ${\psi }_0(x)\propto \exp [-m\omega x^2/(2\hslash )]$. Thus the number operator has exactly the non-negative integer eigenvalues.

Since the two coordinate number operators commute, use the product eigenstates $|n_x,n_y⟩$. Their energy eigenvalues areAt fixed $N=n_x+n_y$, the allowed pairs are $(0,N),(1,N-1),\dots ,(N,0)$. Thus the energy level $E_N=(N+1)\hslash \omega$ has quantum degeneracy $N+1$, and the ground-state energy is $\hslash \omega$.