Use the principal complex square root, $\sqrt{z}=\sqrt{r}{e}^{i\theta /2}$ with $-\pi <\theta <\pi$. For $F(z)=e^{z-\sqrt{z}}/z$, the local expansion on this holomorphic branch isThe fractional power prevents a single-valued Laurent series in a punctured disc. In particular, zero is a branch point, not an isolated singularity classified as a pole. Merely observing that $zF(z)\to 1$ on the slit plane would not establish a simple pole: $zF(z)$ does not extend as a holomorphic function through a whole disc around zero.

The negative real half-line is a chosen branch cut. Its upper and lower boundary values differ because $\sqrt{-x+i0}=i\sqrt{x}$ and $\sqrt{-x-i0}=-i\sqrt{x}$ for $x>0$. Continuation once around zero changes the sign of the complex square root, and hence changes $F$ in general. Other branch cuts are possible; this one defines the principal holomorphic branch.

On the upper semicircular contour, $|z|=R$ and the reverse triangle inequality gives $|z^4+1|\ge R^4-1>0$. Its arc length is $\pi R$, so the estimation lemma givesThe arc is open, so an ordinary integral sign is appropriate even though the source uses a closed-integral symbol there. Subtracting the vanishing arc contribution from the closed contour integral gives the real-line improper integral $\pi /\sqrt{2}$. Its integrand is an even function, and its tail is $O(x^{-2})$, ensuring convergence. Thus

The poles are the four roots of $z^4=-1$. The upper half-plane contains $a=e^{i\pi /4}$ and $b=e^{3i\pi /4}$, both enclosed when $R>1$. At a simple pole $q$,The sum of the two residues isThe residue theorem, with the real segment traversed from $-R$ to $R$ and the upper arc from $R$ to $-R$, yields

Let $f$ be holomorphic on a neighbourhood of a positively oriented, piecewise smooth simple closed curve and its interior except at finitely many isolated singularities $a_j$ strictly inside. The residue theorem statesThe residue is the coefficient of $(z-a_j{)}^{-1}$ in the local Laurent series. No singularity may lie on the contour. For more general closed contours, each residue is multiplied by the corresponding winding number.

Parametrize the positively oriented unit circle by $z=e^{i\theta }$. Then $dz=izd\theta$ andConsequently the contour integral already evaluated satisfies

Assume positive, counterclockwise orientation of a contour. Factor the polynomial denominator asOnly $z=\pm (\sqrt{3}-\sqrt{2})$ lie inside the unit circle. These are simple poles. Writing the denominator as $P(z)$, their residues areThe residue theorem therefore givesReversing orientation of a contour changes the sign.

The geometric series for $(2+z{)}^{-1}$ converges for $|z|<2$, while the cancelled sine terms have an everywhere convergent Taylor series. Hence the requested Laurent series has no negative powers:For example,The extension through zero also makes this a Taylor series. The pole at $-2$ fixes its radius of convergence at two.

The apparent singularity at zero cancels: the Taylor series of the sine givesThus zero is a removable singularity, with extension $f(0)=1/2$. At $z=-2$, the other terms are holomorphic, whereas $(z+2{)}^{-1}$ has a simple pole. ThereforeThere are no other finite singularities. If “all” includes the point at infinity on the Riemann sphere, it is an essential singularity: $f(1/w)=w/(1+2w)+w^2\sin (1/w)-w$ has infinitely many negative powers in its Laurent series at $w=0$.

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.

An energy measurement selects the ground state with probability equal to the squared inner product with ${\psi }_1$. Using $N=\sqrt{3}/L^{3/2}$ and integration by parts,ThusThe full Fourier sine series has $c_n=(-1{)}^{n+1}\sqrt{6}/(n\pi )$, so ${\sum }_n|c_n{|}^2=(6/{\pi }^2){\sum }_n{n}^{-2}=1$, an independent wavefunction normalization check. However $E_n|c_n{|}^2=3{\hslash }^2/(m{L}^2)$ for every $n$, hence ${\sum }_n{E}_n|c_n{|}^2$ diverges. This explicitly verifies the infinite-energy expectation qualification in part (b), without invalidating the individual projection probabilities.

Apply the position version of the Born rule to the normalized wavefunction:The result is independent of $L$ because the state and the measurement interval scale together. Most position probability lies towards the right wall, since the probability density grows as $x^2$.

Assume the given wavefunction is zero outside $[0,L]$. Wavefunction normalization requiresAn arbitrary overall quantum phase is unobservable, so choose the positive real representative $N=\sqrt{3}/L^{3/2}$.

The linear ramp state in an infinite square well has a boundary subtlety: the limiting value at $L$ is $\sqrt{3/L}$, rather than zero. The function is still a normalized element of the well's Hilbert space $L^2(0,L)$, so position measurement and energy-projection probabilities are meaningful. It is not in the domain of the Dirichlet Hamiltonian, nor is it a finite-energy box state. Altering the value at the single endpoint does not cure its nonzero boundary trace; a physical finite-energy preparation must smooth the function to zero near the wall. The following probabilities refer to the supplied idealized state.

Inside the infinite square well, $V=0$, and the time-independent Schrödinger equation isFor $E>0$, write $k^2=2mE/{\hslash }^2$ and $\psi =A\sin kx+B\cos kx$. The first Dirichlet boundary condition gives $B=0$, and the second requires $kL=n\pi$ for a nonzero state. The case $n=0$ is identically zero; negative $n$ only changes an overall quantum phase. For $E=0$ a linear solution satisfying both walls is zero, and for $E<0$ the hyperbolic solutions likewise cannot satisfy both walls nontrivially.

Wavefunction normalization, ${\int }_0^L|\psi {|}^{2}dx=1$, fixes $|A|=\sqrt{2/L}$. The orthonormal energy eigenstates and energy eigenvalues are thereforeEach state vanishes outside the well. Its stationary time-dependent form is ${\Psi }_n(x,t)={\psi }_n(x)e^{-i{E}_{n}t/\hslash }$. General states are normalized superpositions of these modes, rather than additional single energy eigenstates.

Retain the backscattering geometry from part (c). With $\lambda =10\mathrm{nm}$, ${\lambda }^{\prime }=\lambda +2{\lambda }_C$ differs from $\lambda$ by only $4.85\times 10^{-4}$ of its value. Momentum conservation givesThe Lorentz-factor correction is of order $v^2/c^2\simeq 2.4\times 10^{-7}$, so the electron recoil is safely nonrelativistic. Its kinetic energy is only about $0.060\mathrm{eV}$, compared with incoming photon energy about $124\mathrm{eV}$. This validates the free-electron recoil approximation's relativistic aspect; actual binding or thermal motion in matter can require separate corrections. No larger recoil occurs at another scattering angle for the same initial photon.

The ordinary electron Compton wavelength isA photon of that wavelength has energy equal to the electron rest energy, $hc/{\lambda }_C=m_e{c}^2$. It therefore sets the characteristic scale at which photon recoil cannot generally be treated nonrelativistically, and it sets the magnitude of the Compton wavelength shift. The reduced Compton wavelength, ${\bar{\lambda }}_C=\hslash /(m_{e}c)={\lambda }_C/(2\pi )$, is a related but different convention. Localization on this reduced scale needs momenta of order $m_{e}c$, so a nonrelativistic single-particle description becomes inadequate. Neither wavelength is a measured geometrical radius of the electron.

For complete photon backscatter $\theta =\pi$, the electron recoils forward and momentum conservation givesCombining this with conservation of energy and writing $\Delta \lambda ={\lambda }^{\prime }-\lambda$ givesFor $\Delta \lambda \ll \lambda ,{\lambda }^{\prime }$, the final ratio is approximately four, soWithin the stated nonrelativistic equations the unsimplified relation is implicit and has higher-order recoil corrections; the displayed final approximation is all that the small-shift assumption supports. Exact relativistic Compton recoil kinematics gives $\Delta \lambda =(h/m_{e}c)(1-\cos \theta )$, and therefore exactly $2{\lambda }_C$ at backscatter. Agreement at leading order does not make the nonrelativistic derivation exact at arbitrary photon energy.

Take a stationary initial electron of mass $m_e$, a photon incident along $x$, and a nonrelativistic recoil velocity of magnitude $v$. Choose $\theta$ above $x$ for the scattered photon and $\varphi$ below $x$ for the electron. Conservation of momentum givesThe electron's unchanged rest energy cancels in conservation of energy, leavingEquivalently use $h\nu /c=h/\lambda$ and $h\nu =hc/\lambda$. The nonrelativistic limit requires $v/c\ll 1$, and treating the electron as free requires its binding energy to be negligible on the scale of the collision.

Compton scattering transfers energy and momentum from a photon to an electron. For an initially stationary, approximately free electron, a photon scattered through a nonzero angle generally has lower frequency and longer wavelength; the recoiling electron carries the difference. The effect demonstrates the photon momentum $p_{\gamma }=h\nu /c=h/\lambda$, beyond a purely elastic classical scattering description.

The middle panel of the figure in solution shows the incident photon along $+x$, the outgoing photon at angle $\theta$ above that direction, and the recoiling electron at deflection angle $\varphi$ below it. Each arrow labels its particle and momentum; the initial electron is at rest at the interaction point. The scattered photon and electron have opposite transverse momentum components, while their forward components sum to the incident photon momentum.

Using the stipulated $570\mathrm{nm}$ as a wavelength-density peak, Wien's displacement law givesThis is a rough estimate from the supplied characteristic wavelength, not an accurate measurement of the Sun's effective temperature. The Sun emits a broad, approximately thermal visible spectrum and appears white outside the atmosphere; its familiar yellow appearance does not identify a monochromatic thermal peak. A photospheric effective temperature near $5770\mathrm{K}$ instead corresponds to a wavelength-density maximum near $502\mathrm{nm}$.^[ref][ref]

The core must be much hotter to provide the pressure required for hydrostatic equilibrium and to sustain nuclear fusion. Outward energy transport through an opaque interior proceeds down a temperature gradient, by radiative diffusion and then convection through much of the outer envelope. The observable photosphere radiates to space at the temperature needed to carry the solar luminosity, $L_{\odot }=4\pi R_{\odot }^2\sigma T_{\mathrm{eff}}^4$; it is not an unobstructed view of the hot core. The printed core estimate above $20$ million degrees is also too high for the present Sun: the standard central temperature is about $1.56\times 10^7\mathrm{K}$. This correction leaves the large core-to-surface contrast and its explanation intact.^[ref]

First convert from a spectral energy density per unit frequency to one per unit wavelength, including the Jacobian matrix and determinant:Set $d{U}_{\lambda }/d\lambda =0$ and define $x=hc/(\lambda k_{B}T)$. The nonzero maximum satisfies $5(1-e^{-x})=x$, with numerical root $x=4.96511$. ConsequentlyThis is Wien's displacement law. One must maximize the wavelength density, rather than find the frequency maximum and merely put $\lambda =c/{\nu }_{\max }$: the frequency and wavelength peaks of Planck's law are different because a spectral density changes under a change of coordinate.