On the physical sheet the bound-state amplitude poles lie at $k=i{\kappa }_j$, ${\kappa }_j>0$, and map to real negative $E_j=-{\hslash }^2{\kappa }_j^2/(2m)$. The square-root map has two sheets meeting at the continuum threshold. Analytic continuation can also have virtual-state poles on the negative imaginary $k$ axis and resonance poles in the lower half-plane; these are not normalizable bound states. In the energy plane the resonances belong to the nonphysical sheet, rather than additional negative-axis bound levels.

For the algebraically continued probability, include the mirror poles contributed by $D(-k)$ as explained in part d; the two $k$ signs map to the same numerical energy but need not impose the same boundary condition. A sketch must label the sheet or the amplitude being plotted, rather than treating every singularity as a bound state.

Rewrite the real-energy probability as $T=[1+U_0^2{\sin }^2(\zeta a)/(4E(E+U_0)){]}^{-1}$. Thus $\boxed{T_{\max }=1}$ at $\zeta a=j\pi$, namely $E_j={\hslash }^2{j}^2{\pi }^2/(2m{a}^2)-U_0\ge 0$, with a strictly positive resonance energy in the ordinary scattering problem. Between resonances it is less than one and its oscillations approach one as energy increases. Generically $T(0^{+})=0$; a well tuned to a zero-energy resonance instead has $T(0^{+})=1$. The original transmission plot in S1 illustrates these maxima for a chosen well, not a universal specified depth.

For the amplitude continue to $k=i\kappa$ on the decaying physical sheet. Then $\zeta =q$ and $D=\cos (qa)+({\kappa }^2-q^2)\sin (qa)/(2\kappa q)$. The double-angle identity factors its zero condition into the even and odd matching conditions $\kappa =q\tan (qb)$ and $\kappa =-q\cot (qb)$. Thus physical upper-half-plane amplitude poles are exactly the bound energies.

The literal statement for the probability requires correction. The displayed real-axis expression continued algebraically is $1/[D(k)D(-k)]$, and includes poles of the opposite-boundary-condition amplitude too. A concrete counterexample uses ${\hslash }^2/(2m)=1$, $a=2$, $q=1$, $\kappa =\cot 1$ and $U_0=1+{\cot }^{2}1$. Its continued probability denominator vanishes at $k=i\cot 1$, because $D(-k)=0$. But neither $\kappa =\tan 1$ nor $\kappa =-\cot 1$ holds, so $E=-{\cot }^{2}1$ is not a bound energy. Keeping the amplitude and the decaying boundary condition removes this spurious identification.

Matching $\psi$ and ${\psi }^{\prime }$ at both walls, or propagating the pair $(\psi ,{\psi }^{\prime })$ through the interior by the matrix $\left(\begin{array}{cc}\cos \zeta a& \sin \zeta a/\zeta \\ -\zeta \sin \zeta a& \cos \zeta a\end{array}\right)$, givesHere ${\zeta }^2=k^2+2m{U}_0/{\hslash }^2$. For real positive $k$, taking the modulus givesThe equal exterior velocities make this also the current transmission coefficient.

Continuity has already been built into these forms. At $b$, derivative continuity requires $-q\sin (qb)=-\kappa \cos (qb)$ for even parity and $q\cos (qb)=-\kappa \sin (qb)$ for odd parity. ThusUse $q/\kappa =\sqrt{U_0/|E|-1}$ to obtain the two printed equations. The right side is positive, so only the corresponding allowed tangent/cotangent branches can describe a bound state; squaring without retaining this sign restriction adds spurious roots.

Let $b=a/2$, $q=\sqrt{2m(U_0-|E|)}/\hslash$ and $\kappa =\sqrt{2m|E|}/\hslash$. Bound levels have $-U_0<E<0$. Up to normalization, parity givesThe respective inverse squared normalization constants are $b+\sin (2qb)/(2q)+{\cos }^2(qb)/\kappa$ and $b-\sin (2qb)/(2q)+{\sin }^2(qb)/\kappa$. These include both exponential tails.

The full equation rearranges to $L\psi =-U\psi$. Since a bound state has no incoming free wave, convolution with the decaying inverse gives $\boxed{\psi (x)=-m\int e^{-\kappa |x-x^{\prime }|}U(x^{\prime })\psi (x^{\prime })d{x}^{\prime }/(\kappa {\hslash }^2)}$. Applying $L$ to this identity recovers the differential equation and the decay boundary conditions.

For $E<0$, impose decay and continuity on $G=C{e}^{-\kappa |x-x^{\prime }|}$. Its derivative jump is $-2\kappa C$, so $-{\hslash }^2(G_{+}^{\prime }-G_{-}^{\prime })/(2m)=1$ requiresThere is no nonzero free solution decaying at both infinities that can be added without spoiling the derivative condition.

At the delta potential, the bound equation becomes $\psi (x)=\beta \psi (0)e^{-\kappa |x|}/\kappa$. A nonzero solution therefore requires $\kappa =\beta$, givingThere is one bound level. The transmission amplitude $t(k)$ has its physical-sheet pole at $k=i\beta$, which produces exactly this decaying state. The real-axis probability's rational continuation $T(k)=k^2/(k^2+{\beta }^2)$ has both $+i\beta$ and $-i\beta$ poles; the lower pole is not an additional bound state. It is essential to distinguish the analytic amplitude from an even continuation of its modulus squared.

For $E=-{\hslash }^2{\kappa }^2/(2m)$, the decaying Green's function is $m{e}^{-\kappa |x-x^{\prime }|}/(\kappa {\hslash }^2)$, with the same derivative jump as before. A normalizable negative-energy state has no incident homogeneous part. Applying the inverse to $L\psi =-U\psi$ givesThe condition $\kappa >0$ is necessary for decay on both sides.

The first Born approximation substitutes the incident wave for the exact wave inside the scattering integral. ThusFor the attractive delta this is ${\beta }^2/k^2$, the leading small-$\beta /k$ expansion of the exact result. Physically one neglects the distortion of the incident state by repeated interaction with the potential. At low $k$ the Born result can exceed one, while the exact reflection remains bounded and approaches one; it is not valid in that limit.

Put $\beta =m\alpha /{\hslash }^2$. The delta interaction reduces the integral equation to ${\psi }_s=e^{ikx}+i\beta e^{ik|x|}{\psi }_s(0)/k$. At zero, ${\psi }_s(0)=(1-i\beta /k{)}^{-1}$. HenceTheir sum is one, as required for a real potential. The explicit wave is the indicated incident-plus-reflected form for $x<0$ and $t{e}^{ikx}$ for $x>0$; it is continuous and has derivative jump $-2\beta \psi (0)$.

As $x\to -\infty$, the scattered kernel is $e^{-ikx}{e}^{ik{x}^{\prime }}$; as $x\to +\infty$, it is $e^{ikx}{e}^{-ik{x}^{\prime }}$. Therefore ${\psi }_s\sim e^{ikx}+r{e}^{-ikx}$ on the left and $t{e}^{ikx}$ on the right, withEqual asymptotic wavenumbers give $R=|r{|}^2$, $T=|t{|}^2$. Squaring the first amplitude proves the requested expression. These are current ratios, which would require a velocity factor if the two asymptotic potentials differed.

Write the full equation as $L{\psi }_s=-U{\psi }_s$. Apply the outgoing inverse of $L$ and add the incident homogeneous solution with unit left-incident amplitude. This givesFor sufficiently short-range potentials this is the outgoing Lippmann-Schwinger equation. The minus sign follows from the definition $L=H_0-E$, rather than $E-H_0$.

Away from $x=x^{\prime }$, an outgoing solution has form $G=C{e}^{ik|x-x^{\prime }|}$. Integrate the differential equation across $x^{\prime }$ to get $G_{+}^{\prime }-G_{-}^{\prime }=-2m/{\hslash }^2$. Since the derivative jump is $2ikC$, $C=im/(k{\hslash }^2)$. ThusContinuity excludes a derivative-of-delta contribution. The outgoing condition, equivalently the appropriate limiting resolvent, removes the arbitrary incoming homogeneous solution.

Magnetic pressure is the field energy density $P_B=B_f^2/(2{\mu }_0)$. Using the preceding final field gives $\boxed{P_B\simeq 2.6\times 10^{13}\mathrm{Pa}\simeq 260\mathrm{Mbar}}$. This uses the compressed field, not the initial 36 T field; substituting the latter would omit the square of the already squared convergence amplification.

The final radius is $r_f=0.5\mathrm{mm}/15=33.3\mu$m. For a 3.5 MeV alpha particle, $p\simeq \sqrt{2m_{\alpha }{E}_{\alpha }}$ and the largest perpendicular gyroradius is $p/(2e{B}_f)$. Setting it equal to the core radius givesThis is an order-of-magnitude onset estimate for magnetic deflection, assuming flux freezing and a suitably oriented field. A small gyroradius alone does not ensure complete confinement along open field lines or efficient deposition for every pitch angle.

An inertially confined sphere disassembles on approximately $\tau \sim r/c_s$, while $n_i=\rho /{\bar{m}}_i$. Substitute these in the given number-density confinement product to getFor an equal D-T mix, ${\bar{m}}_i\simeq 2.5m_u$. Taking $c_s\sim \sqrt{2k_{B}T/{\bar{m}}_i}$ at 10 keV gives $A\simeq 0.365$ kg m${}^{-2}=0.0365$ g cm${}^{-2}$. An adiabatic sound-speed convention changes this by an order-one factor. This estimate implements the supplied Lawson product and is not the separate, more stringent alpha-stopping areal-density criterion.

Use a steady planar sonic ablation model with absorbed intensity $I_a$, critical mass density ${\rho }_c=(A_i/Z)m_u{n}_c$, and $n_c={ϵ}_0{m}_e(2\pi c/\lambda {)}^2/e^2$. In the usual isothermal-electron sonic balance, $I_a\simeq 4{\rho }_c{c}_s^3$ and the ablation pressure is $P_A\simeq 2{\rho }_c{c}_s^2$. Eliminating $c_s$ givesFor $A_i/Z\simeq 2$ and complete absorption, the supplied laser parameters give about $1.15\times 10^{13}$ Pa, or $115$ Mbar. Mach one alone fixes neither the material mass per electron nor the enthalpy/absorption coefficient. Different sonic-flow closures change the order-one constant; incomplete absorption multiplies the estimate by its fraction to power $2/3$. The robust result is the two-thirds scaling, rather than a material-independent numerical pressure.

Hydrogen's $n=2$ excitation energy is $3R/4=10.2$ eV. With electronic degeneracy ratio $g_2/g_1=4$, its relative population is $4e^{-10.2/1.60}\simeq 6.8\times 10^{-3}$. Thus the first excited manifold contributes below one percent of the ground population at the inferred temperature, explaining the leading ground-state approximation.

There is a qualification to extending this argument to infinitely many Rydberg states: the isolated-atom partition sum with degeneracy $2n^2$ formally diverges. Plasma screening and level dissolution impose a physical cutoff. The small low-lying populations justify the simple approximation in its intended regime, not an unrestricted infinite partition sum.