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For $l=0$, put $u(r)=r{R}_0(r)$. The radial Schrödinger equation reduces to $u^{\prime \prime }+k^{\prime 2}u=0$ inside and $u^{\prime \prime }+k^{2}u=0$ outside, where $k^{\prime 2}=2m(E+V_0)/{\hslash }^2$ and $k^2=2mE/{\hslash }^2$. Regularity at the origin requires $u(0)=0$, excluding the inner cosine. A real outer sine-cosine combination can be written with an amplitude and a phase shift:The inner expression has the finite limiting value $A{k}^{\prime }$ at the origin. An overall complex normalization can carry the asymptotic phase convention of part a.

Write $\kappa =\sqrt{2m{V}_0}/\hslash$ and $q=\kappa a$. At low energy, the $S$ wave normally dominates. Under the additional stated near-resonance assumption $ka\ll |{\delta }_0|$, neglect $ka$ inside the cotangent of the matching condition and set $k^{\prime }\simeq \kappa$:Substitution in ${\sigma }_0=4\pi {\sin }^2{\delta }_0/k^2$ givesThis is a near-resonance square-well approximation, not a uniform formula for every well radius. A phase shift is defined modulo $\pi$; its representative must be chosen consistently when applying the comparison with $ka$.

Away from the shell, the reduced radial equation is free. Regularity at the origin and the definition of the S-wave phase shift giveThe reduced radial wave function is continuous at $R$; otherwise its second derivative would contain a delta derivative not present in the equation. Integrating the equation across the shell gives the jumpDivide by the common shell value and substitute the two free solutions. This yieldsIf the shell value vanishes, use the undivided matching equations or a continuous limit rather than this logarithmic-derivative division. For an explicit phase shift of a delta-shell potential, write $x=kR$ and $g=2m{V}_{0}R/{\hslash }^2$. The cotangent addition formula givesThe phase is defined modulo $\pi$; choose its branch continuously as energy or coupling varies. At zero coupling the physical phase shift is zero.

The scattering length is $a_0=-{\lim }_{k\to 0}\tan {\delta }_0/k$. Since $\sin (kR)\sim kR$ and $\cos (kR)\sim 1$, the exact phase shift gives, away from the pole,Thus plot $a_0/R$ against $g$, so that both axes are dimensionless.

For repulsion, $g>0$, the curve rises from zero to the horizontal asymptote $a_0/R=1$, staying between zero and one. For weak attraction, $-1<g<0$, the scattering length is negative and tends to $-\infty$ as $g\to -1^{+}$. For stronger attraction, $g<-1$, it is positive and greater than $R$, tending to $+\infty$ as $g\to -1^{-}$ and to $R$ as $g\to -\infty$. Its only zero is $g=0$.

The pole at $g=-1$ is the zero-energy threshold of an S-wave bound state. To see this directly, use $u\propto \sinh (\kappa r)$ inside and $u\propto e^{-\kappa r}$ outside for energy $-{\hslash }^2{\kappa }^2/(2m)$. The same derivative jump givesThis approaches $-1$ as $\kappa \downarrow 0$ and admits a positive $\kappa$ for $g<-1$. At the threshold the scattering length diverges; this is a zero-energy scattering resonance, not a divergence of the finite-energy cross-section beyond its partial-wave unitarity bound. These are the main features of the scattering length of a delta-shell potential.

A phase shift is a change in an oscillation's wave phase, specified modulo $2\pi$. It can arise from propagation, reflection or an imposed response. With the time convention $e^{-i\omega t}$, a delay $\tau$ multiplies the complex amplitude by $e^{i\omega \tau }$. A sign reversal is a $\pi$ phase shift.

For $V(r)=V_0\delta (r-R)$, put $g=2m{V}_{0}R/{\hslash }^2$. The regular reduced radial wave function is proportional to $\sin kr$ inside the shell and to $\sin (kr+{\delta }_0)$ outside. It is continuous at $R$, while its radial derivative jumps by $(g/R)u(R)$. Dividing the jump condition by the common value gives $k\cot (kR+{\delta }_0)-k\cot (kR)=g/R$, which implies the displayed phase shift modulo $\pi$. At nodes, the undivided matching equations give the continuous limiting interpretation. For $g\ne 0$, an equivalent formula is $k\cot {\delta }_0=-k\cot (kR)-k(kR)/[g{\sin }^2(kR)]$. The exact S-wave scattering amplitude is $[k\cot {\delta }_0-ik{]}^{-1}$.

The wave phase specifies position within an oscillation cycle. For a sinusoidal travelling wave,where $k$ is the wavenumber, $\omega$ the angular frequency and ${\varphi }_0$ an initial phase shift. Adding $2\pi$ to $\Phi$ leaves the field unchanged; one complete cycle advances the wave phase by $2\pi$ radians.

A surface of equal wave phase is a wavefront. Tracking one such surface gives the phase velocity, whereas comparing two fields at the same point gives their phase difference. Two equal-frequency disturbances add constructively when their wave phases agree and can cancel when their phase difference is $\pi$. The sign assigned to a phase lag depends on the chosen time convention. In a complex amplitude, the complex argument carries this information while the modulus sets the wave amplitude.

A reactive-load phase shift in a finite-output-impedance amplifier occurs because the load and output impedance form a frequency-dependent divider. In the model of loaded voltage gain, assume a positive real output resistance $R_o$, an ideal capacitive load $C_L$ and open-circuit gain $A_0$ approximately constant over the band of interest. With $\omega =2\pi f$,The divider follows from the terminal equivalent described by the university circuit notes.^[ref]

Putting $\tau =R_o{C}_L$ givesThus even an ideal capacitor can change both amplitude and phase when the source driving it has finite resistance. Its reactive current and energy exchange are described by complex electrical power; zero average dissipation in the ideal capacitor does not mean zero output-current demand.

For an original example, choose $R_o=100\Omega$, $C_L=10\mathrm{nF}$ and $A_0=10$. Then $\tau =1\mu \mathrm{s}$ and the load pole is $f_p=1/(2\pi \tau )\approx 159.2\mathrm{kHz}$. At that frequency the gain magnitude is $10/\sqrt{2}\approx 7.071$ and the extra phase lag is $45^{\circ }$. A measured input of $10\mathrm{mV}$ root mean square gives about $70.7\mathrm{mV}$ root mean square at the capacitor, requiring approximately $0.707\mathrm{mA}$ root-mean-square capacitive current.

A bench test holds the actual amplifier input fixed while sweeping frequency and compares loaded and unloaded output amplitudes and phases. Include probe capacitance in $C_L$. The group delay introduced by a capacitive amplifier load also matters when comparing pulse waveforms.

This single-pole calculation assumes the stated constant open-circuit gain and resistive output impedance. The complete amplifier may have additional frequency dependence and feedback effects. The manufacturer's ideal feedback-gain examples are defined within their operating assumptions.^[ref] Check the actual output-current and linear operating range before interpreting a departure from this model as a capacitor-only effect.

An experimental physicist who coauthored a practical reconstruction of the Fizeau experiment. Comparing counterpropagating beams in moving water makes the small motion-dependent phase shift observable through interference. The setup links historical questions about light propagation to a reproducible optical measurement.^[ref]

The sinusoidal steady state of a stable linear time-invariant system driven by a sinusoid is its persistent response at the driving frequency, after the relevant transient response has decayed. A phasor calculation finds that response; it does not generally satisfy the circuit's switch-on initial conditions. The late-time assumption is explicit in Berggren's treatment of sinusoidal circuit analysis.^[ref]

Using peak phasors, write the source as $v_s(t)=\mathrm{Re}({\hat{V}}_s{e}^{i\omega t})$. If $H(i\omega )$ is the relevant transfer function, its particular solution isThe complete response adds a homogeneous solution determined by the initial conditions. Approaching the displayed expression requires the modes affecting the measured output to decay; an unchanged amplitude and a fixed phase shift then characterize its frequency response.

For an original turn-on example, drive a series RC circuit with $v_s(t)=V_0\cos \omega t$ for $t\ge 0$, measure the capacitor's electric potential difference, and take $v_C(0)=0$. With time constant $\tau =RC>0$, Kirchhoff's voltage law gives the first-order linear differential equationPut $x=\omega \tau$. Substitution verifies the sinusoidal particular solution $V_0(\cos \omega t+x\sin \omega t)/(1+x^2)$. Adding $A{e}^{-t/\tau }$ and imposing the initial condition fixes $A=-V_0/(1+x^2)$, givingIts sinusoidal part has amplitude $V_0/\sqrt{1+x^2}$ and lags by $\arctan x$. The exponential settling bound for RC sinusoidal turn-on, relative to that amplitude, isWith chosen $R=1\mathrm{k\Omega }$, $C=1\mu \mathrm{F}$, $V_0=1\mathrm{V}$ and $\omega =1000rad/s$, we have $\tau =1\mathrm{ms}$ and $x=1$. At turn-on the phasor approximation predicts 0.5 volt while the capacitor is initially at zero. After 5 milliseconds, their difference is about 3.37 millivolts, or 0.476 percent of the sinusoidal amplitude. A 1 percent settling time threshold for this specific switch-on is about $4.259\tau$.

A useful measurement compares a long, low-frequency square-wave response with sinusoidal responses at several frequencies. An oscilloscope can identify the exponential decay and then measure amplitude ratios and phase shifts. MIT's laboratory uses that sequence to connect the time constant to the sinusoidal response; probe loading belongs in the measured circuit model.^[ref]

Undamped resonant forcing has no bounded sinusoidal steady state. For an ideal series LC circuit, let $q$ be electric charge and ${\omega }_0=1/\sqrt{LC}$. At resonance, the equationhas the particular solution $q_p(t)=V_{0}t\sin ({\omega }_{0}t)/(2L{\omega }_0)$. Its growing envelope prevents a bounded attracting sinusoidal response. This ideal counterexample explains why decay of the relevant homogeneous solutions is part of the physical steady-state assumption.

The square-root collapse of capacitive bridging bandwidth near the DC error floor describes how the usable frequency interval disappears when a resistive divider has almost exhausted a complex-error allowance. Use the model and exact formula in capacitive bridging complex-error bandwidth: $g=R_s/R_{\mathrm{in}}>0$, ${\tau }_s=R_s{C}_{\mathrm{in}}$, and the DC floor ${\eta }_0=g/(1+g)$. Write the available extra tolerance as $\Delta =\eta -{\eta }_0>0$. The exact bandwidth becomesAt fixed positive $g$, the leading numerator is linear in $\Delta$, so $f_{\eta }$ is proportional to $\sqrt{\Delta }$. Thus demanding four times less extra tolerance approximately halves the bandwidth; it does not reduce it by a factor of four.

For an original example, choose $R_s=20\mathrm{k\Omega }$, $R_{\mathrm{in}}=1\mathrm{M\Omega }$, and $C_{\mathrm{in}}=50\mathrm{pF}$. Then $g=0.02$, ${\tau }_s=1\mu \mathrm{s}$, and ${\eta }_0\approx 0.019608$. Taking $\Delta =0.00020$ gives about $456\mathrm{Hz}$ from the exact formula. Taking $\Delta =0.00005$ gives about $228\mathrm{Hz}$. The small deviation from an exact two-to-one ratio comes from the retained quadratic term and denominator.

The qualification fixed positive $g$ matters. If the receiving resistance is infinite, $g={\eta }_0=0$ andwhich is linear in small $\eta$. For a small but nonzero floor, the square-root approximation needs $\Delta$ small compared with ${\eta }_0$ as well as with the remaining distance to unit error. The crossover from square-root to linear capacitive bridging bandwidth scaling prevents using the fixed-floor asymptotic expression uniformly as $g$ also tends to zero.

This is a limit of the chosen passive input impedance model, not an empirical limit for every amplifier. The manufacturer's shunt-capacitance model motivates the resistive/capacitive paths; an actual bench comparison must include both amplitude and phase shift and the probe's added capacitance.^[ref]

Thin-film interference combines coherent light contributions reflected or transmitted at the two surfaces of a layer. For a film of refractive index $n$, thickness $d$ and internal angle ${\theta }_t$, the round-trip optical path difference is $2nd\cos {\theta }_t$. The relative reflection phase shift must also be included. For a nonabsorbing higher-index film surrounded by air, one of the two leading reflected contributions has a $\pi$ phase shift and the other does not. Reflection maxima therefore satisfy $2nd\cos {\theta }_t=(m+1/2){\lambda }_0$. A smooth reflection spectrum filters many visible wavelengths at once; varying thickness and angle produce spatially varying mixed colours.