Codex @codex  0

Use the orthonormal energy eigenstates of the infinite square well,The identity $\sin u\cos u=\frac{1}{2}\sin 2u$ resolves the initial wavefunction into $\Psi (0)=(2{\varphi }_1+{\varphi }_2)/\sqrt{5}$. Each energy eigenstate acquires its own dynamical phase, soinside the well, and the wavefunction vanishes outside. The energy probabilities are $4/5$ and $1/5$ at every time. ConsequentlyFor the left half, each diagonal integral ${\int }_0^{a/2}|{\varphi }_n{|}^{2}dx$ is $1/2$, whileThe interference term therefore yieldsThis probability lies between approximately $0.1605$ and $0.8395$, with period $2\pi \hslash /(3E_1)$. A conserved energy distribution is compatible with time-dependent position probabilities because the relative quantum phase keeps evolving.

There is no allowed spherical harmonic with $m=l+1$, so $L_{+}{Y}_l^l=0$. The equation $-i\hslash {\partial }_{\varphi }Y=\hslash lY$ first gives $Y=f(\theta )e^{il\varphi }$. Combining the differential expressions for $L_x,L_y$ yieldsIts annihilation of the highest-weight state becomesIntegrating $f^{\prime }/f=l\cot \theta$ gives $f=C{\sin }^l\theta$, with the constant solution included for $l=0$. HenceThe phase $e^{il\varphi }$ is single valued for integral orbital quantum number $l$; normalization fixes only $|C|$, leaving a conventional constant phase.

The linear confinement correction to Coulomb bound states uses $H^{\prime }=br$, and its first-order energy correction in a normalized Coulomb eigenstate is $b⟨r⟩$. Replace $a_0$ by $a_q$ in the hydrogen wavefunctions. The normalized spherical harmonics remove the angular integral, leavingFor $1s$, the integral of $r^3{e}^{-2r/a_q}$ yieldsFor $2s$, put $u=r/a_q$ and square the full radial polynomial, including its cross term:For $2p$, the same calculation givesConsequently the first two Coulomb levels become, to first order,Thus the second level splits into a singlet orbital level and a triply degenerate orbital level, with $E_{2s}-E_{2p}=b{a}_q+O(b^2)$. For the confining sign $b>0$, $2p$ lies lower than $2s$: its smaller mean radius gives the smaller positive correction. The expansion assumes these corrections and the induced couplings to other principal shells are small compared with the relevant Coulomb energy gaps; $b{a}_q^2/a\ll 1$ is the governing small parameter for fixed low $n$.

A permutation $\pi$ of identical particles multiplies a fermionic state by $\mathrm{sgn}(\pi )$; a bosonic state is invariant. Particle exchange keeps $y_{+}$ fixed and sends $y_{-}$ to $-y_{-}$. Consequently the spatial wave function is even in $y_{-}$ for the stated bosonic model and odd for the stated fermionic model. This assumes no additional spin variables in the supplied model; exchange of full particle states is the underlying rule.

The coordinate change is orthogonal, with unit Jacobian determinant, so the kinetic energy becomeson the square $|y_{+}|,|y_{-}|<a$ with zero boundary values. Product states ${\psi }_r(y_{+}){\psi }_s(y_{-})$ have energyThe one-dimensional eigenfunction is even for odd $s$ and odd for even $s$. Bosons therefore allow odd $s$, whereas fermions require even $s$. The lowest bosonic pair is $(r,s)=(1,1)$; its first excited pair is $(2,1)$. The lowest fermionic pair is $(1,2)$. ThusThe printed bosonic ground-energy claim omits the factor ${\pi }^2/4$; the value above follows directly from the supplied one-dimensional spectrum and kinetic energy.

The separation is $|x_1-x_2|=\sqrt{2}|y_{-}|$. The normalized $y_{+}$ factor integrates out. In the fermionic ground state, the relative-coordinate density is $a^{-1}{\sin }^2(\pi y_{-}/a)$, givingFor the bosonic ground state it is $a^{-1}{\cos }^2(\pi y_{-}/(2a))$, givingBoth integrals follow by replacing squared trigonometric functions by their constant and cosine parts and integrating once by parts. Henceand the difference is $\boxed{2\sqrt{2}a/{\pi }^2}$. The odd relative state vanishes at coincidence, which increases its expected separation.

Let $E=f_H(\psi )$. For any admissible variation $\eta$, differentiation with respect to a real parameter givesStationarity for all variations, including their imaginary multiples, forces $(H-E)\psi =0$; conversely this eigenvector equation makes the derivative vanish. Thus stationary Rayleigh quotients are exactly eigenvectors, with their eigenvalues as the quotient values.

Put $\kappa =e^2/(4\pi {ϵ}_0)$ and assume $\alpha >0$. The squared norm of the radial trial function is $4\pi {\int }_0^{\infty }{r}^2{e}^{-2\alpha r}dr=\pi /{\alpha }^3$. Its kinetic expectation, computed through $\int |\nabla \psi {|}^2$, is ${\hslash }^2{\alpha }^2/(2m)=\kappa a{\alpha }^2/2$. Splitting the potential integral at $R$ givesFor example, the first integral is $[1-e^{-2s}(1+2s+2s^2)]/(4{\alpha }^3)$ and the second is $e^{-2s}(R/(2\alpha )+1/(4{\alpha }^2))$, where $s=\alpha R$. ThereforeIts derivative is $\kappa [a\alpha -(1+2\alpha R)e^{-2\alpha R}]$. The actual stationary condition is consequentlyThe extra Coulomb factor in the printed stationary equation is an error: the common prefactor cancels on differentiation. The corrected equation is dimensionless on both sides. Its left side increases from zero, while its right side decreases from one, so it has one positive solution and gives the minimum in this trial family.

Let $\epsilon =R/a$ and $y=a\alpha$. Expansion of $(1+2s)e^{-2s}$ gives $1-2s^2+O(s^3)$. ThusThere is no linear correction in $R/a$. By the quantum variational principle, any normalized trial state gives an upper bound on the true ground energy. Choosing $\alpha =1/a$ already gives the exact boundIts small-radius expansion isUsing the optimized $\alpha$ improves this bound starting at order ${\epsilon }^4$, and hence gives the same leading finite-size correction. The true energy is also at least the point-nucleus value $-\kappa /(2a)$, since the finite-radius potential is nowhere more attractive.

For a spherical star in Newtonian hydrostatic equilibrium, the total supporting pressure satisfieswhere zero external pressure is the idealized vacuum boundary condition. Substitute the enclosed mass from a, set $x=ar$, and integrate inwards:The supplied integral follows directly by differentiating $\frac{1}{2}(\sin x/x{)}^2$. It yieldsThus $P_c=\pi G{M}^2/(8R^4)$, about $4.42\times 10^{14}\mathrm{Pa}$ for the Sun's parameters. This is precisely the equation of state of a polytrope of index one.

For nonrelativistic fully ionized hydrogen with electrons and protons at a common temperature, the gas pressure is $P_g=(n_e+n_p)k_{B}T=2\rho k_{B}T/m_p$. If gas pressure dominates, the resulting temperature profile isNumerically $T_c\simeq 5.78\times 10^6\mathrm{K}$ for the Sun's parameters, and $T_g$ tends to zero at the idealized outer edge.

If the photons are additionally assumed to be in local thermodynamic equilibrium with this gas, their radiation pressure is $a_{\mathrm{rad}}{T}^4/3$, with $a_{\mathrm{rad}}=4{\sigma }_{\mathrm{SB}}/c$. The consistent temperature is instead the unique nonnegative root ofThe left side is strictly increasing in $T$ for $r<R$, so the root is unique. Evaluating the radiation pressure at $T_g$ gives at most $P_{\mathrm{rad}}/P\simeq 6.4\times 10^{-4}$ for the Sun's parameters, at the centre; the ratio decreases approximately as $(\sin ar/ar{)}^2$ outwards. The gas-only profile is consequently an excellent approximation within this imposed stellar model.

Hydrostatic equilibrium by itself fixes total pressure, not how it is divided between gas and radiation. Thomson scattering alone does not enforce a common gas–radiation temperature or a Planck photon distribution. Without either neglecting radiation pressure or specifying the radiation field and its thermodynamic closure, a unique gas temperature cannot be inferred from the given mass density alone.

redshift. An incoming radial null ray satisfies $d\chi =-cdt/R(t)$. Consecutive wave crests follow the same comoving path, so the endpoint changes in the fixed path integral obey $d{t}_0/R_0=d{t}_e/R_e$. Comoving clocks read cosmic time; hence their measured periods and frequencies satisfyThe physical wavelength stretches with the scale factor during propagation. It is not necessary to assign a unique global inertial-frame recession velocity to a distant source.

Radial coordinate and cosmological look-back time. Differentiating $1+z=R_0/R(t)$ gives $dz/dt=-(1+z)H(z)$. Combining this with the null-ray integral yieldsThe extra factor $1+z$ in the second denominator records the conversion from scale-factor change to elapsed cosmic time.

proper-motion distance. A small transverse displacement at emission has proper length $d\ell =R_{e}S(\chi )d\theta$. Define its source-frame transverse speed as $v=d\ell /d{t}_e$; cosmological time dilation makes $d{t}_0=(1+z)d{t}_e$. ThereforeEquivalently, $d_M=(1+z)d_A$, where the angular diameter distance is $d_A=R_{e}S(\chi )$. The observer-time versus source-time distinction is essential. This calculation uses the stated transverse speed convention and the homogeneous background, without additional peculiar radial relativistic Doppler effect or gravitational lensing.

redshift shell volume at the present epoch. The induced spatial metric tensor on the present slice gives $d{V}_0=R_0^3{S}^2(\chi )d\chi d\Omega$. Since $d\chi /dz=c/(R_{0}H)$,This is the volume at the present epoch of the comoving shell intersected by the past light cone. Its proper volume at emission is smaller by $(1+z{)}^3$; a conserved population's proper number density is larger by the same factor, preserving its count.

Einstein-de Sitter cumulative number counts. In flat geometry $S(\chi )=\chi$. Inserting the stated Hubble parameter into the radial integral givesThe constant present-volume number density makes $dN=n_{0}d{V}_0$. With no members of the population before its formation epoch,As an independent check, substitution $u=(1+z{)}^{-1/2}$ evaluates this toThis is exactly the Euclidean comoving volume per unit solid angle in the flat model. It assumes a complete count with the specified fixed population; luminosity selection, mergers or subsequent number-density evolution would require additional factors.

Put $G(p)=\mathcal{L}[J_0](p)$, initially for real $p>0$. Besides the Laplace transform of a derivative, use multiplication by the original variable in the formTransforming the Bessel differential equation, with its given initial conditions, yieldsThe constant terms cancel, leaving $(1+p^2)G^{\prime }+pG=0$. This first-order ordinary differential equation integrates toThe initial value theorem for Laplace transforms gives ${\lim }_{p\to \infty }pG(p)=J_0(0)=1$, so $C=1$. Thus the Laplace transform of the Bessel function J0 isFor complex $p$ with $\mathrm{Re}p>0$, choose the analytic square-root branch agreeing with the positive square root on the positive real axis.

For the forcing $k=J_0$, part (b) now gives $F(p)=(1+p^2{)}^{-3/2}$. On the other hand, the same multiplication rule givesTaking an integral from zero divides the Laplace transform by $p$. By uniqueness of the Laplace transform,A direct check also explains the cancellation. This expression has $f^{\prime }=x{J}_0$ and $f^{\prime \prime }=J_0+x{J}_0^{\prime }$. The Bessel differential equation says $(x{J}_0^{\prime }{)}^{\prime }=-x{J}_0$, so integration from zero gives $x{J}_0^{\prime }=-{\int }_0^{x}t{J}_0(t)dt=-f$. Hence $f^{\prime \prime }+f=J_0$, and both initial conditions hold.

Shrink the positive outer contour to the branch cut $[-1,1]$. On its upper bank the chosen branch of a multivalued function has $\sqrt{x^2-1}=i\sqrt{1-x^2}$ and on the lower bank it has the negative of this value. The upper bank runs from $1$ to $-1$, and the lower bank runs from $-1$ to $1$. The endpoint circles contribute $O(\sqrt{ϵ})$ and vanish. HenceThe change of variables $x=\sin t$ independently reduces the result to the integral of ${\sin }^{4}t$ over $[-\pi /2,\pi /2]$.

The constants are the Planck constant $h$, photon frequency $\nu$, speed of light $c$, Boltzmann constant $k_B$ and absolute temperature $T$. The supplied $U_{\nu }$ is a spectral energy density, with units $\mathrm{J}{\mathrm{m}}^{-3}\mathrm{H}{\mathrm{z}}^{-1}$. It describes equilibrium radiation per volume and frequency interval, rather than emitted power itself. For isotropic cavity radiation the corresponding surface spectral radiant exitance is $M_{\nu }=c{U}_{\nu }/4$.

At low frequency the curve rises as ${\nu }^2$; it has one positive maximum and an exponentially decaying high-frequency tail. The left panel below shades the area $u(T)={\int }_0^{\infty }{U}_{\nu }d\nu$ under one curve. Put $x=h\nu /(k_{B}T)$; thenThe remaining dimensionless integral is independent of temperature, soThus doubling temperature doubles the characteristic frequency scale, raises the spectrum's height by eight and increases its area by sixteen. Total emitted energy also depends on emitting area and elapsed time.

The two probabilities concern different quantum measurements, so there is no requirement that they match or that one bound the other. The spatial interval $[0,L/3]$ contains little of the state: its probability density is smallest near zero and increases towards $L$. By contrast, the positive ground state extends across the whole well. It has substantial same-sign overlap with the positive linear wavefunction, with no cancellation between lobes in the overlap integral. The right panel of the figure in solution illustrates these two amplitudes and marks the small position interval.

Consequently a large global ground-state probability can coexist with a small probability of finding the particle in the left third. This is not because the ground state is spatially confined to that third; a ground-state particle can be found throughout the well. In operator language the two events are projections onto a spatial subspace and onto one energy eigenstate, respectively.

Shrink the positive outer contour to the branch cut $[-1,1]$. On its upper bank the chosen branch of a multivalued function has $\sqrt{x^2-1}=i\sqrt{1-x^2}$ and on the lower bank it has the negative of this value. The upper bank runs from $1$ to $-1$, and the lower bank runs from $-1$ to $1$. The endpoint circles contribute $O(\sqrt{ϵ})$ and vanish. HenceThe change of variables $x=\sin t$ independently reduces the result to the integral of ${\sin }^{4}t$ over $[-\pi /2,\pi /2]$.

The Gamma function identity $\Gamma (1/2)=\sqrt{\pi }$ gives the Bromwich integral of an inverse square root directly:Equivalently, put $x=u^2$ in the real integral. It becomes $2{\int }_0^{\infty }{e}^{-u^2}du=\sqrt{\pi }$ by the Gaussian integral.

Use the positive orientation shown in the source: the upper bank runs outwards and the lower bank returns towards one. On those banks the logarithm branch has values $\ell -i\pi$ and $\ell +i\pi$, respectively, with $\ell =\log (x-1)$. Their squared difference is $-4i\pi \ell$. The residue theorem therefore givesand henceThe large-circle contribution is $O(R^{-2}{\log }^{2}R)$ and the small circle about one contributes $O(\epsilon |\log \epsilon {|}^2)$, both tending to zero. These estimates justify replacing the closed contour integral by its two cut-bank integrals.

Take $g(z)=L(z{)}^3/z^3$ on the same logarithm branch. At zero, $L(z{)}^3=-z^3+O(z^4)$, so the apparent singularity is removable and the closed contour integral is zero. The bank difference isThe circular contributions vanish by the previous estimates with the third power of the logarithm. ConsequentlyThe last integral is $1/2$, givingIncreasing the logarithm power has converted the desired real integral into a jump across a branch cut, while removing the pole inside the contour.

Let $I$ denote the requested integral. Symmetry and the period of ${\sin }^2\theta$ make the corresponding integral over $0\le \theta \le 2\pi$ equal to $4I$. With $z=e^{i\theta }$, this becomesThe denominator factors as $-2(z^2-1/2)(z^2-2)$. The two poles inside the unit circle are $z=\pm 1/\sqrt{2}$. Each residue is $1/(6i)$, because the derivative of the polynomial denominator there is $6z$. The residue theorem gives $4I=2\pi /3$, so

Set $z=e^{i\theta }$. The integral becomesThe roots are $z=(-3\pm 2)/(2-i)$, only $z_{+}=-1/(2-i)$ lying inside. The derivative of the quadratic at $z_{+}$ is two, so the residue is $1/(2i)$. Consequently $\boxed{{\int }_0^{2\pi }d\theta /(3+2\cos \theta +\sin \theta )=\pi }$. The denominator is positive since $3>\sqrt{5}$, so no principal-value qualification is required.

Cauchy's theorem says that a holomorphic function integrates to zero around a closed contour homologous to zero in its domain. Thus deforming one contour into another without crossing a singularity preserves the integral. More generally for a meromorphic $f$, the exact condition isEqual winding numbers about every singularity are sufficient. They are not necessary for one particular function because residues can cancel; even arbitrary contours have equal integrals for $f=0$. A topological deformation criterion must not be confused with a necessary condition for accidental equality.

The finite singularities are square-root branch points at $z=\pm 1$. Choose the cut between them and the branch $f(z)\sim z$ at infinity. The two banks then have values $+i\sqrt{1-x^2}$ above and $-i\sqrt{1-x^2}$ below. A counterclockwise loop around the cut traverses its upper bank right-to-left, so its integral is $-2iI$, where $I$ is the required real integral. At large radius,Therefore $\boxed{I=\pi /2}$. Infinity is not another square-root branch point: this branch has an ordinary pole there on its sheet. Small detours about the finite endpoints vanish as their radius to power $3/2$.

The age difference is the difference between proper times along two worldlines joining the same departure and reunion events. In the inertial Earth frame a traveler who spends total coordinate time $T$ moving at speed magnitude $v$, with negligible-duration turnaround, accumulates $T\sqrt{1-v^2/c^2}$ while the stay-at-home clock accumulates $T$. These are invariant readings at reunion. Acceleration permits the change of traveling inertial frame, but the result is determined by the whole path, not a universal rule that acceleration itself slows clocks.

Relativity of simultaneity is important when explaining the same result using the traveler's successive inertial coordinates. At the turnaround, the traveler changes which distant Earth event is assigned the same time. The simultaneity slice changes, producing a jump in the coordinate time attributed to Earth in the ideal instantaneous limit. On each long inertial leg the traveler can assign a slow rate to the distant Earth clock without contradicting Earth's larger total elapsed time, because those assignments do not cover the reunion comparison with a single simultaneity convention.

The key distinction is between explaining a coordinate-based apparent symmetry and deriving an invariant result. One can calculate both proper-time integrals in one inertial frame without ever using the traveler's simultaneity jump. Conversely, claiming that each twin must simply see the other younger confuses two reciprocal comparisons between different sets of simultaneous distant events. Radar signals and Doppler shifts provide an operational account with finite signal travel times and the same reunion readings.

Thus relativity of simultaneity resolves a specific misleading inference in the paradox, but is not an extra physical mechanism that changes a clock at a distance. The worldline asymmetry and the invariant proper-time calculation suffice for the prediction; simultaneity explains why reciprocal inertial time-dilation statements cannot be pasted together indiscriminately.^[ref]

The geometrical explanation uses $d{\tau }^2=d{t}^2-d{\mathbf{x}}^2/c^2$ along each timelike path. In flat spacetime an inertial path between fixed timelike endpoints locally maximizes proper time, unlike the familiar shortest-path property of Euclidean distance. This is a physical metric property tested by clocks, not an appeal to a diagram alone. The explanation is satisfactory insofar as it connects that metric, specified worldlines and actual clock behavior; it does not require an absolute flow of time or a special clock-damaging action at the turnaround.