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An acoustically rigid boundary has zero perturbation normal velocity. For a reflected acoustic plane wave, pressure amplitudes add in phase and the reflection coefficient is $+1$. It is the infinite-surface acoustic impedance limit. A pressure-release boundary instead fixes the pressure perturbation to zero and reflects with coefficient $-1$.

For a progressive acoustic plane wave in a homogeneous fluid, the pressure-to-particle-speed ratio is $Z=\rho c$, where $\rho$ is mass density and $c$ is the speed of sound. Its units are $\mathrm{Pa}s/m$. This specific impedance differs from pressure divided by a volume flux, whose value also depends on the port area.

At angle $\theta$ from an interface normal, the normal particle-velocity amplitude is $U_n=U\cos \theta$. Therefore the normal acoustic impedance isFor two lossless fluids with propagating incident and transmitted waves, continuity of pressure and normal velocity givesand hence the pressure-amplitude reflection coefficientZero reflection requires equal normal acoustic impedances; equal bulk values $\rho c$ alone suffice only at normal incidence. Using particle displacement rather than velocity multiplies both normal impedances by the same frequency factor and does not replace $Z/\cos \theta$ by $Z\cos \theta$.^[ref]

For a film of refractive index $n$ and thickness $d$ in air, its round-trip propagation phase is $\delta =(4\pi d/{\lambda }_0)\sqrt{n^2-{\sin }^2{\theta }_i}$. Two neighbouring same-type interference fringes differ by $2\pi$. At small angular spacing in radians,The relation assumes slowly varying interface reflection coefficients and no rapidly varying reflection phase. At normal incidence the first derivative vanishes, so a local linear angular-spacing formula must be replaced by a quadratic expansion.

At a lossless vascular junction, pressure continuity and volumetric flow rate balance give $1+r=t$ and $Y_1(1-r)=Y_{d}t$, where $Y_d$ is the sum of daughter characteristic admittances of an arterial wave. Eliminating transmission amplitude $t$ gives the displayed reflection coefficient. A decrease in total area yields positive reflection if parent and daughter wave speeds are equal; differing stiffnesses require comparison of the admittances rather than areas alone.

The characteristic impedance is the voltage-to-current ratio of a single forward travelling wave on a uniform transmission line. For real $Z_0$ and a real load resistance $R_L$, the voltage-wave reflection coefficient is $(R_L-Z_0)/(R_L+Z_0)$ and the power accepted by the load is the incident power times $4R_L{Z}_0/(R_L+Z_0{)}^2$.

An etalon coating group delay is the frequency derivative of the reflection phase contributed by the two coatings during one internal round trip. At fixed incidence and polarization, write the internal reflection coefficients as $r_j(\omega )=|r_j(\omega )|e^{i{\theta }_j(\omega )}$. With the phase convention of etalon round-trip phase,The phases must refer to consistently chosen surfaces: moving a reference plane transfers a propagation contribution between the terms. Only the consistently defined total is directly comparable with the resonance spacing.

A constant coating phase changes the positions of resonances but contributes zero group delay. A linear phase produces constant group delay. Curvature produces group delay dispersion:A highly reflecting coating therefore cannot be characterized by reflectance alone. The manufacturer's ultrafast-mirror documentation separately specifies reflection dispersion and shows multiple-bounce arrangements for compensation.^[ref]

For an original constant-slope example, suppose propagation contributes $10\mathrm{ps}$ and the coatings together contribute $0.20\mathrm{ps}$. The local free spectral range is approximately $1/(10.20\mathrm{ps})=98.04\mathrm{GHz}$, rather than $100\mathrm{GHz}$. Ignoring coating delay would overassign that $0.20\mathrm{ps}$ to propagation when inferring a spacer thickness or group index. The values are stipulated, not a coating specification.

For an independent curvature example, let the summed coating phase contain $D(\omega -{\omega }_0{)}^2/2$, with $D=100\mathrm{f}{\mathrm{s}}^2$. Across a frequency interval $10\mathrm{THz}$, its group delay changes by $D2\pi \Delta f\approx 6.28\mathrm{fs}$. The derivative is with respect to angular frequency; omitting $2\pi$ changes the result.

Experimentally, one may measure adjacent resonance frequencies and compare their spacing with an independently characterized spacer and group index. A discrepancy alone cannot uniquely identify coating delay: spacer dispersion, incidence angle and geometry also affect the result. Etalon delay inferred from free spectral range gives an interval-averaged phase slope; identifying it with a single-frequency derivative requires sufficiently slow variation. The simple manufacturer's spacer relation explicitly neglects spacer dispersion.^[ref]

Etalon reference-plane dependence is the change in the assigned propagation and reflection phases when the mathematical planes defining an etalon's field amplitudes are moved. A consistently computed etalon round-trip phase and its resonance frequencies remain unchanged. The decomposition into spacer delay and etalon coating group delay depends on those planes.

Use time dependence $e^{-i\omega t}$ and a lossless uniform spacer with wavenumber $k(\omega )=n(\omega )\omega /c$. Let the physical interfaces be separated by $L$, with internal reflection coefficients $r_1,r_2$ referred to their surfaces. The round-trip multiplier isThe multilayer formalism explicitly assigns forward and backward amplitudes at specified interfaces and propagates their phases through each layer. Its sign conventions must remain consistent when comparing reflection phases.^[ref]

Move the left mathematical reference plane a distance $\delta >0$ into the spacer, leaving the actual coating in place. A wave launched from that plane travels to the coating and back, so the effective left reflection coefficient becomes $r_1^{\prime }=r_1{e}^{2ik\delta }$. The explicit remaining distance is $L^{\prime }=L-\delta$. HenceThis original derivation changes the representation, not the physical cavity. Moving a real mirror would instead change the system.

Let ${\tau }_c=d\arg (r_1{r}_2)/d\omega$, using continuous wave phase. Since $dk/d\omega =n_g/c$, the new assigned terms obeyTheir sum equals ${\tau }_c+2n_{g}L/c$. If the spacer is dispersive, the corresponding second-derivative contributions likewise transfer between the two terms; the total group delay dispersion is unchanged.

For an original example, take $n_g=1.5$ and $\delta =15\mu \mathrm{m}$. The assigned reflection delay increases by about $150.1\mathrm{fs}$ while the assigned propagation delay decreases by the same amount. The free spectral range cannot reveal a change of bookkeeping. An apparent discrepancy of that size between two coating calculations can arise solely from different reference planes.

Report the incident medium, polarization, reference surfaces and phase convention with a coating-delay result. A measured resonance spectrum constrains the total phase; independently identifying a coating contribution requires a compatible spacer model. Etalon delay inferred from free spectral range therefore should not be compared with an isolated coating derivative until both calculations use the same reference planes.

Equality of the relevant wave impedances across an interface. In the elementary normal-incidence model this makes the reflection coefficient vanish, even when other material properties differ.

For a rapidly decaying KdV potential, the scattering data consist of the continuous reflection coefficient, the negative discrete eigenvalues $-{\kappa }_n^2$, and a positive norming constant for each bound state.

For a slower surface layer supporting propagating SH-waves with vertical wavenumber $q>0$ above an evanescent half-space with decay exponent $p>0$, continuity of displacement field and traction gives the displayed reflection coefficient for a downward incident displacement field wave. Its modulus is one and its wave phase is $-2\arctan (\mu p/({\mu }_{0}q))$. A traction-free top reflects displacement field with coefficient $+1$. The round-trip condition $R_h{e}^{2iqh}=1$ then gives $\tan (qh)=\mu p/({\mu }_{0}q)$, equivalent to direct Love wave mode matching.

An ideal open circuit permits no conduction electric current through a terminal pair, corresponding to infinite load electrical impedance. At the end of a lossless transmission line, its voltage-wave reflection coefficient is $+1$ and the terminal current is zero. Real open terminals may retain parasitic capacitance and radiation loss.

Write the complex displacements as $y_1=(A{e}^{i{k}_{1}x}+B{e}^{-i{k}_{1}x})e^{-i\omega t}$ and $y_2=C{e}^{i{k}_{2}x-i\omega t}$. At a massless join, continuity of displacement and transverse force givesEliminating $B$ gives the displacement transmission coefficient and reflection coefficient,The angular frequency is shared across a stationary interface, whereas the wavelength changes. The equal-tension assumption is used in the slope condition.

For the reflection of a string wave from a freely moving endpoint mass, let $c=\sqrt{T/\mu }$, $k=\omega /c>0$ and use the complex time-harmonic wave conventionThe string pulls its endpoint toward negative $x$, so its transverse force on the attached mass is $-T{y}_x(0,t)$. Newton's second law gives the boundary condition $m{y}_{tt}(0,t)=-T{y}_x(0,t)$. Substitution yieldsHere $Z$ is the characteristic impedance of a stretched string. Since the incident and reflected waves share the same string, their mean power ratio is the squared modulus of this displacement reflection coefficient:There is no remaining fraction of the mean incident energy: the freely moving mass provides no dissipation and no onward transmission channel. It exchanges kinetic energy with the string during each cycle but has zero mean absorbed power. Indeed the energy flux of a transverse string wave satisfieswhose cycle average is zero. As checks, $m\to 0$ gives $R=+1$, the free-end reflection of a transverse string wave, while $m\to \infty$ at fixed $\omega >0$ gives $R=-1$, the fixed-end limit.

Use a linear, lossless arterial pressure wave model with an incompressible fluid of mass density $\rho$, small wall displacement and negligible mean-flow advection compared with the wave speed. Let the parent vessel have equilibrium area $A_1$ and speed $c_1$. Linearizing mass conservation and axial momentum gives $a_t+A_1{u}_x=0$ and $\rho u_t=-p_x$. With wall compliance $p=(dp/dA)a$, these imply $c_1^2=(A_1/\rho )dp/dA$. A stiffer wall therefore has larger wave speed. A downstream travelling pressure wave carries flux $Q_{+}=Y_1{p}_{+}$, where $Y_1=A_1/(\rho c_1)$ is the characteristic admittance of an arterial wave; a returning wave has $Q_{-}=-Y_1{p}_{-}$.

Place the reflecting junction at $x=L$. Treat the daughters as outgoing-wave loads, with combined admittance $Y_d={\sum }_j{A}_j/(\rho c_j)$ and no further returning waves. At the junction, pressure continuity and flux conservation give $p_I+p_R=p_T$ and $Y_1(p_I-p_R)=Y_d{p}_T$. Thus the bifurcation reflection of an arterial pressure wave hasA decrease in total area gives $0<r<1$ if the parent and daughter wave speeds are equal. More generally this requires $Y_d<Y_1$, an assumption about both area and stiffness. To isolate the effect of ageing, assume all relevant wave speeds increase by the same factor, so their ratios and hence $r$ stay fixed. Unchanged geometry alone does not ensure unchanged $r$ if the vessels stiffen by different factors.

At a station $x<L$, put $\varphi =\omega (t-x/c_1)$ and $\vartheta =2\omega (L-x)/c_1$, the round-trip phase to the reflector. The incident pressure harmonic and its reflection giveFor time convention $e^{i\varphi }$, the two phasors are $P=1+r{e}^{-i\vartheta }$ and $V=1-r{e}^{-i\vartheta }$. Their maximum times are shifted by $-\arg P/\omega$ and $-\arg V/\omega$, respectively. Hence the pressure-flow phase lag from one arterial reflection isIndeed $V/P$ has the argument of $1-r^2+2ir\sin \vartheta$. On the branch $0<\vartheta <\pi$, this phase lag is positive: the pressure maximum follows the flow maximum. There is no cycle-wrap ambiguity on this branch because $1-r^2>0$.

The age trend need not have the same sign at every station. Differentiating the phase givesIncreasing $c_1$ decreases $\vartheta$. At a distal station with $0<\vartheta <\pi /2$, the lag therefore decreases. At a more proximal, ascending-aortic station with $\pi /2<\vartheta <\pi$, it increases. Thus one sufficient set of assumptions reproducing both trends isAs long as ageing keeps the stations in these ranges, the distal lag falls while the proximal lag rises. The effect is interference between pressure waves that add and flow waves that subtract, not merely the shortening of one travel time. The lag is largest at a round-trip phase of $\pi /2$.

This is a conditional explanation, with fixed representative harmonic frequency, fixed reflector location, negligible attenuation, negligible additional reflections and common stiffness scaling. The given stiffness and geometry facts alone do not specify the phase ranges, so they do not force either age trend without these extra assumptions. Here the two stations lie on the incident-plus-reflected segment. A station beyond the sole junction in an ideal outgoing daughter has only a transmitted wave and zero pressure-flow lag in this model. Actual peripheral measurements beyond that junction require returning waves from further junctions or terminal loads; a single-reflector calculation cannot establish their lag. Non-sinusoidal waveforms, distributed viscosity, mean flow and differently evolving reflection coefficients can also shift peak times. These limitations matter when applying the harmonic explanation to a real arterial network.

Choose depth $z$ positive downwards and harmonic time dependence $e^{-i\omega t}$. At normal incidence, P waves and the two polarizations of S waves are independent scalar problems: a P wave uses the longitudinal modulus $\rho {\alpha }^2$, and an S wave uses the shear modulus $\rho {\beta }^2$. In a uniform layer with wave speed $c$ and seismic impedance $q=\rho c$, write velocity and compressive traction asAt a perfectly bonded interface the velocity and traction are continuous. For incident velocity wave amplitude one from medium 1 into medium 2, these two conditions are $1+r_v=t_v$ and $q_1(1-r_v)=q_2{t}_v$. ThusReflection is determined by seismic impedance, not by speed alone. A step to larger seismic impedance reverses the reflected velocity/displacement field waveform; the pressure reflection coefficient has the opposite sign. Its transmitted pressure coefficient is $t_p=2q_2/(q_1+q_2)$. Equal seismic impedance eliminates the reflection even when speed and wavelength change. At a free surface traction vanishes and velocity reflects with coefficient $+1$; at a rigid surface velocity reflects with coefficient $-1$.

The mean elastic-wave energy flux is $\frac{1}{2}\mathrm{Re}(P{V}^{\ast })=q(|A{|}^2-|B{|}^2)/2$ for positive real $q$. Therefore the reflected and transmitted energy fractions areA displacement field transmission coefficient greater than one is compatible with conservation of energy; its flux must be weighted by the receiving seismic impedance.

For many layers, use the seismic layer transfer matrix. A layer of thickness $h_j$ has wave phase thickness ${\delta }_j=\omega h_j/c_j$ andThis follows either by eliminating $A,B$ between the two faces or by integrating $V^{\prime }=i\omega P/(\rho c^2)$ and $P^{\prime }=i\omega \rho V$. For the complete stack put $M=M_N\cdots M_1=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. With upper and lower half-space impedances $q_a,q_b$, an outgoing lower wave satisfies $P_b=q_b{V}_b$. Solving this condition gives the input seismic impedanceHere $T$ is lower-face velocity divided by upper-face incident velocity, including the propagation wave phase. For lossless layers and real frequency, $|R{|}^2+(q_b/q_a)|T{|}^2=1$. This supplies a sensitive check on any spreadsheet or numerical implementation; it is not correct simply to multiply the individual interface transmissions and discard the elastic waves.

The physical reason is repeated wave interference. In one layer between media 0 and 2, let $r_{ij},t_{ij}$ denote the velocity coefficients above and $\delta =\omega h/c_1$. The first transmitted wave has wave amplitude $t_{01}{t}_{12}{e}^{i\delta }$. Every extra round trip supplies $r_{10}{r}_{12}{e}^{2i\delta }$. Summing the geometric series givesIn the time domain the successive arrivals are delayed by $2h/c_1$, producing reverberation. In the frequency domain the same delayed terms produce resonant maxima and minima. For a layer between identical half-spaces of impedance $q_0$,At $\delta =n\pi$ the layer transmits all the incident energy: the internally elastic waves cancel the direct reflection. Halfway between these resonances, a large impedance contrast gives strong reflection. For unequal exterior impedances, a quarter-wave layer with $q_1^2=q_a{q}_b$ transforms its lower load to $Z_{\mathrm{in}}=q_1^2/q_b=q_a$, again giving zero reflection at its design frequency. This matching is narrow-band.

Periodic alternation produces a periodic seismic multilayer stop band. For a cell of two layers with total thickness $d$, the cell matrix has reciprocal eigenvalues $e^{\pm iKd}$, soWhen the absolute value of this expression exceeds one, the infinite periodic medium has no propagating cell mode; a long finite stack has exponentially small transmission. The energy is reflected, not absorbed. Pass bands instead contain oscillatory transmission peaks, becoming more numerous as the stack length increases. Disorder removes the exact cell periodicity but not multiple scattering or frequency-selective interference.

For a concrete reproducible investigation, take $q_H/q_L=4$, equal one-way travel times $\tau$ in both layer types, exterior impedance $q_L$, and frequency coordinate $\nu =2\omega \tau /\pi$. At $\nu =1$ each layer is a quarter wavelength; one cell has diagonal matrix entries $-4,-1/4$. For $N$ cells,It is about $0.22145$ for one cell, $6.10\times 10^{-5}$ for four cells and $9.31\times 10^{-10}$ for eight cells. The infinite-cell stop band is $2\arcsin (0.8)/\pi <\nu <2-2\arcsin (0.8)/\pi$. The accompanying calculation also shows the first transmitted arrivals for a single high-impedance slab: their wave amplitudes are $0.64(0.36{)}^n$ at times $(2n+1)\tau$. Their sum is one, consistent with unit zero-frequency transmission. These calculations illustrate both the spectrum and the pulse train without assuming access to the course's spreadsheet.

Finally, as $\omega \to 0$ for a fixed finite stack, every $M_j\to I$, so the response approaches that of a direct interface between the exterior half-spaces. Equal exterior impedances give $R\to 0,T\to 1$. For wavelengths much longer than a repeating cell, expansion of $M_j$ to first order instead gives an effective medium with averaged mass density ${\rho }_{\mathrm{eff}}=\sum h_j{\rho }_j/d$ and harmonic-mean modulus $M_{\mathrm{eff}}^{-1}=\sum h_j/M_j^{\mathrm{mod}}/d$, where $M_j^{\mathrm{mod}}={\rho }_j{c}_j^2$. Its speed is $\sqrt{M_{\mathrm{eff}}/{\rho }_{\mathrm{eff}}}$, not generally the arithmetic mean of the layer speeds. Attenuation makes the energy balance an inequality, while for a fixed oblique horizontal slowness the scalar SH calculation uses $q_j={\mu }_j{p}_{\beta ,j}$ and ${\delta }_j=\omega p_{\beta ,j}{h}_j$. Oblique P-SV waves require coupled displacement field and traction matrices because of mode conversion at a planar elastic interface. Thick evanescent layers require scattering or impedance recursion rather than exponentially ill-conditioned transfer-matrix products.

At a sharp interface, continuity of velocity and traction gives, for unit incident velocity from the left,Thus $r_v=(Z_a-Z_b)/(Z_a+Z_b)$ and $t_v=2Z_a/(Z_a+Z_b)$. Normalizing the transmitted velocity by its elastic-wave energy flux gives $T_{ba}=\sqrt{Z_b/Z_a}{t}_v$, exactly the result above. Right incidence gives the reversed reflection sign and the same flux-normalized transmission. A stress reflection coefficient has the opposite sign to the velocity reflection coefficient; the convention matters.

Put $r=R_{aa}$ and $t=T_{ab}$. The incoming-to-outgoing matrix is $S_0=\left(\begin{array}{cc}r& t\\ t& -r\end{array}\right)$, with $r^2+t^2=1$. Therefore $S_0^{†}{S}_0=I$, proving equality of incoming and outgoing time-averaged flux for arbitrary coherent incident wave amplitudes, not just incidence from one side.

For all real frequencies the harmonic equations instead giveTheir right sides are equal; the diagonal terms $\pm i\omega /\beta$ have zero real contribution. ConsequentlyThe mean flux for real-part phasors is $(|{\varphi }_{+}{|}^2-|{\varphi }_{-}{|}^2)/2$. Evaluating it at the two endpoints proves exactly that total outgoing flux equals total incoming flux at every real frequency. This relies on real lossless moduli and mass density; absorption would change the identity.

Refer all layer wave amplitudes to the lower interface. Let a downward incident SH displacement field have unit wave amplitude there, an upward reflected wave have wave amplitude $R_h$, and the decaying half-space wave wave amplitude $T_h$. Interface continuity givesSolving these equations proves the Love-wave interface reflection phase and transmission wave amplitude:Thus $|R_h|=1$: the evanescent substrate field has nonzero wave amplitude but no net downward mean flux. At the traction-free top, an upward incident wave and its downward reflection must have canceling derivatives, so their displacement field reflection coefficient is $R_0=+1$.

A layer wave acquires wave phase $2qh$ during a complete round trip. Constructive interference requiresTaking the tangent gives again $\tan (qh)=\mu p/({\mu }_{0}q)$. The reflection phases, rather than only geometric travel distance, are essential to the mode condition.

The one-dimensional probability current isAt a finite jump of the potential, both $\psi$ and ${\psi }^{\prime }$ are continuous. A jump in $\psi$ would create a delta derivative in the stationary Schrödinger equation, and integrating that equation across an increasingly small interval then shows continuity of ${\psi }^{\prime }$. Accordingly impose continuity of both quantities at zero and at $a$.

For an incident wave of positive energy, define $k=\sqrt{2mE}/\hslash$. With no wave incident from the right, the left region hasThe interference terms cancel from the imaginary part in the current formula. If $E>V_0$, let $k_0=\sqrt{2m(E-V_0)}/\hslash$. In the right region,For $0<E<V_0$, let ${\kappa }_0=\sqrt{2m(V_0-E)}/\hslash$. Boundedness at positive infinity excludes the growing exponential, soThe decaying solution has zero current even if its amplitude is nonzero. These exterior forms do not require $E$ to lie on a particular side of $V_1$: in the finite middle layer the solution is oscillatory for $E>V_1$, exponential for $E<V_1$, and linear at $E=V_1$; matching determines the amplitudes.

Define the reflection coefficient as reflected current magnitude divided by incident current, and the transmission coefficient as outward right current divided by incident current. ThenThe denominator is $j_{\mathrm{incident}}=\hslash k|A{|}^2/m$. For a real potential, differentiation givesby the stationary Schrödinger equation. Continuity at each interface preserves this conserved current, so $j_{\mathrm{left}}=j_{\mathrm{right}}$. Dividing by incident current yieldsThere can be penetration into the barrier without transmission to infinity, because the entire right half-line remains classically forbidden.

Translation invariance of the stationary surface statistics implies that a single incident tangential wavenumber retains that wavenumber in its coherent mean. Hence, away from the surface, writeThe unknown effective reflection coefficient $r_b$ is referenced to the mean plane $z=0$. It may be the rigid or pressure-release coefficient just derived, or a more general rough-boundary response. Lossless microscopic reflection does not require $|r_b|=1$: diffuse scattering removes power from the coherent specular field.

For the added planar upper interface, keep the same tangential wavenumber $\alpha$. Let ${\beta }_2=\sqrt{k_2^2-{\alpha }^2}$ on its outgoing sheet. The normal acoustic impedances in the layer and upper medium areContinuity of pressure and normal velocity gives the reflection coefficient for upward incidence from the layer and transmission coefficient for downward incidence from above:If a separate impedance sheet rather than an ordinary fluid interface is intended, its specified effective impedance replaces the upper termination; a sheet impedance has not otherwise been given. These formulas correspond to the stated two-fluid interface.

Let the upper incident pressure amplitude at $z=d$ be $I$. If $D$ and $U$ denote downward and upward layer amplitudes at that plane, their interface and lower-reflection relations areSolving, or summing the geometric series of repeated round trips, gives the coherent acoustic layer above a rough reflector:The expression describes the field at heights where the common homogeneous layer exists; perturbative mean fields near the moving rough boundary need their physical trace evaluated separately. For an upper incident wave normalized as $e^{i\alpha x-i{\beta }_{2}z}$, set $I=e^{-i{\beta }_{2}d}$. With a matched upper medium, $r_{12}=0$ and $t_{21}=1$, the formula reduces to the single lower-reflection field. Zeros of the denominator give ideal layer resonances, interpreted with outgoing continuation or small loss.

This is a coherent specular description: a single effective $r_b$ closes the downward/upward mean amplitudes. Diffuse paths reflected back from the upper interface can modify the mean response; they must either be neglected in this approximation or included in the effective lower reflection coefficient for the actual layered environment. Without a full roughness spectrum and boundary type, a unique numerical mean field cannot be specified beyond this general coefficient form.

Put $p=\sqrt{2mE}/\hslash$ and $\gamma =mk/{\hslash }^2$. Integrating the Schrödinger equation across the delta potential gives continuity of $\psi$ andFor incidence from the left, write $\psi =e^{ipx}+r{e}^{-ipx}$ on $x<0$ and $\psi =t{e}^{ipx}$ on $x>0$. Matching gives $t=1+r$ and $ip(t-1+r)=2\gamma t$, henceThe incident and transmitted probability currents have the same velocity factor, so the reflection coefficient and transmission coefficient areFor $k<0$, an even bound state has $\psi =A{e}^{-\kappa |x|}$, $\kappa =\sqrt{-2mE}/\hslash$. The derivative jump now gives $-2\kappa A=2\gamma A$, so $\kappa =-\gamma >0$ andThere is exactly one positive value of $\kappa$, hence exactly one even bound state. Its normalized amplitude is $A=\sqrt{\kappa }$. An odd decaying solution would have to vanish at zero and is therefore trivial.

In the shrinking finite square well, its integrated potential is $k=-2Ul$, not $-Ul$: the full width is $2l$. With $Ul$ fixed and $l\to 0$, $0\le (\alpha l{)}^2\le 2mU{l}^2/{\hslash }^2\to 0$. The matching condition then bounds $\beta$ by $(2mUl/{\hslash }^2)(1+o(1))$, so $E$ remains bounded. Consequently $El\to 0$, and the previous matching condition givesUsing $E=-{\hslash }^2{\beta }^2/(2m)$ recovers the same bound-state energy. All other even levels disappear as the width shrinks.