Past exam of the physics A course of the University of Cambridge / 2021

The Born rule gives the required probability by integrating the probability density over the central interval. Put $d=0.1\mathrm{nm}$. For this infinite square well ground state,Substitution of $d/L=0.1$ givesThis exceeds the interval's $20\%$ fraction of the box length because the ground state probability density is largest at the centre; it is not a uniform position distribution.

Inside the rectangular potential barrier, the electron's wavefunction is evanescent, with decay constantFor width $a=1\mathrm{nm}$, the squared exponential suppression in the quantum tunnelling probability isThis is the usual leading-exponential estimate. The rectangular-barrier tunnelling probability, retaining the interface-matching prefactor for equal potentials on the two sides, isHere that prefactor is $16(8)(2)/10^2=2.56$, so the more accurate result is $\boxed{T\simeq 1.30\times 10^{-6}}$. Both estimates show that transmission is of order one part in a million. A classical particle with this kinetic energy cannot cross the barrier at all.

A Stern-Gerlach experiment sends a collimated beam through an inhomogeneous magnetic field and then measures its transverse position after a drift region. The neutron is electrically neutral but has a nonzero magnetic moment proportional to its spin, $\mathbf{\mu }={\gamma }_n\mathbf{S}$. A predominantly $z$-directed field gives the interaction Hamiltonian operator and transverse forceSince a neutron has spin one-half, its spin component along the local field has only two eigenvalues, $S_z=\pm \hslash /2$. Hence there are two opposite force branches, corresponding to ${\mu }_z=\pm {\gamma }_n\hslash /2$, and a beam containing both spin components gives two displaced detector spots. This is neutron Stern-Gerlach splitting, not an electric-charge Lorentz force. A uniform magnetic field can cause Larmor precession but does not provide the required separating magnetic field gradient.

The neutron's gyromagnetic ratio is negative, so its magnetic moment points opposite to its spin; this reverses which spin label goes to which spot without changing the two-way splitting. Equal populations occur for an initially unpolarized beam. A beam prepared entirely in one field-axis spin eigenstate gives only its corresponding branch, explaining the qualification “in general”.

Assume independent background events at a known constant rate, so the background count follows a Poisson distribution. In five months its mean is ${\mu }_b=0.3\times 5=1.5$. The null hypothesis of background alone assigns the observed upward fluctuation the P-valueThusThe tail includes counts above eight because they would be at least as discrepant with the background-only prediction. Even the probability of exactly eight, $e^{-1.5}1.5^8/8!\simeq 1.42\times 10^{-4}$, exceeds the specified threshold, though it is not the appropriate tail test. The Poisson tail discovery probability is a probability of data under a background model; it is not a posterior probability that the background model is true. This interprets the question's discovery rule in its usual statistical significance sense.

For a real signal with highest frequency $f_{\max }=40\mathrm{MHz}$, the Nyquist–Shannon sampling theorem gives the nominal Nyquist sampling rateTo cover an arbitrary component exactly at the band edge, choose $f_s>80\mathrm{MHz}$: a $40\mathrm{MHz}$ sine sampled at exactly $80\mathrm{MHz}$ can have every sample equal to zero for an unfortunate phase. Practical aliasing suppression also calls for a margin above the nominal limiting rate.

With rounding to the nearest digitizer level, quantization with step $q=0.1\mathrm{mV}$ has the uniform-quantization error boundThus the stated digitizer meets the requested nominal amplitude precision per sample, assuming no clipping or other errors. The bandwidth sets the nominal minimum sampling rate; the level spacing sets the rounding precision. Strictly, the sampling theorem reconstructs exactly from exact samples. A uniform $0.05\mathrm{mV}$ error bound on the entire reconstructed continuous waveform is not guaranteed by the level spacing and sampling rate alone; that stronger requirement needs a specified reconstruction and noise model.

The total angular momentum operator is $\mathbf{J}=\mathbf{L}+\mathbf{S}$. Since orbital and spin operators act on different factors of the tensor product, they commute, and expanding $J^2$ givesA coupled angular momentum eigenstate $|l,s;j,m_j⟩$ is a simultaneous eigenstate of these three squared operators, with eigenvalues ${\hslash }^{2}j(j+1)$, ${\hslash }^{2}l(l+1)$ and ${\hslash }^{2}s(s+1)$. Thus the angular factor in the spin-orbit interaction is the numberThe other factor, $r^{-1}dV/dr$, is radial. It commutes with the angular operators, and a central-potential energy eigenstate separates into a radial wavefunction and an angular spinor. Taking its radial expectation value therefore givesThis spin-orbit energy shift in a central potential is independent of $m_j$, so rotational symmetry preserves the $2j+1$ magnetic degeneracy within a fixed $j$ multiplet. It is a first-order fine structure correction; the given coefficient already includes the relativistic factor needed for the electron's spin-orbit coupling.

The attractive Coulomb potential energy is $V(r)=-e^2/(4\pi {ϵ}_{0}r)$, soUse precisely the radial normalization stated in the PDF: the radial probability measure is $4\pi r^2|\psi {|}^{2}dr$. The supplied function hasIts integral is one, since ${\int }_0^{\infty }{r}^4{e}^{-r/a_0}dr=24a_0^5$. In particular,This calculation uses the supplied convention, rather than multiplying a separately normalized standard radial function by an extra $4\pi$.

The addition of angular momentum bracket is $j(j+1)-l(l+1)-s(s+1)=3/4-2-3/4=-2$. Substitution in the spin-orbit energy shift in a central potential therefore givesHere $a_0=5.29\times 10^{-11}\mathrm{m}$ is the supplied Bohr radius. The negative sign means the $j=1/2$ member of the $l=1$ pair is shifted downward by this spin-orbit interaction; it is not the total hydrogen binding energy.

For the isotropic three-dimensional quantum harmonic oscillator, $V(r)=m{\omega }^2{r}^2/2$, hence $r^{-1}dV/dr=m{\omega }^2$ is constant. No radial integral remains, so the spin-orbit energy shift in a central potential reduces toFor $l=0$, the only possibility is $j=1/2$, and the bracket vanishes. For $l=1$, coupling to spin one-half permits $j=1/2$ and $j=3/2$, whose brackets are $-2$ and $+1$. ThusThese spin-orbit shifts of an isotropic harmonic oscillator are a zero line, a downward parabola, and an upward parabola of half the downward magnitude. They are independent of the radial excitation number, and their magnetic multiplicities are respectively two, two and four. Their weighted mean within the $l=1$ manifold is zero: $2\Delta E_{1,1/2}+4\Delta E_{1,3/2}=0$.

Write $a=a_{-}$ and $a^{†}=a_{+}$. Inverting the annihilation operator and creation operator definitions givesThe ladder operators take $|n⟩$ to orthogonal quantum harmonic oscillator energy eigenstates $|n-1⟩$ and $|n+1⟩$. Thus $⟨n|a|n⟩=⟨n|a^{†}|n⟩=0$, andFor $n=0$, the lowering term is simply zero. The conclusion concerns an individual energy eigenstate; interference between different eigenstates can produce nonzero expectation values.

For the harmonic-oscillator number-state second moments, the number operator gives $⟨a^{†}a{⟩}_n=n$, while the canonical commutation relation gives $⟨a{a}^{†}{⟩}_n=n+1$. Terms with $a^2$ or $(a^{†}{)}^2$ have zero diagonal matrix elements, since they change the occupation number by two. ThereforeConsequently the potential energy and kinetic energy means areTheir sum is the oscillator energy eigenvalue $E_n=\hslash \omega (n+1/2)$. Their equality is also the virial theorem result for a quadratic potential, here derived directly from the ladder operators.

The standard deviations defining the position and momentum uncertainties satisfy $(\Delta x{)}^2=⟨x^2⟩-⟨x{⟩}^2$ and $(\Delta p{)}^2=⟨p^2⟩-⟨p{⟩}^2$. Part (a) makes both means zero. Using the second moments from part (b),Multiplication givesThe Heisenberg uncertainty relation requires only $\Delta x\Delta p\ge \hslash /2$. Among the oscillator energy eigenstates, equality occurs solely for the ground state $n=0$; the excited states have larger uncertainty products.

Each energy eigenstate picks up its own phase under unitary time evolution. The relative phase therefore changes at $(E_1-E_0)/\hslash =\omega$. After removing an irrelevant global phase, the state isThe diagonal position matrix elements vanish, but the annihilation operator and creation operator giveThe two off-diagonal cross terms in the expectation value yieldThus the angular frequency is $\omega$, and the magnitude of the oscillation, meaning its amplitude, isThe peak-to-peak displacement is $2A$. The phase $\varphi$ fixes where the oscillation starts but does not change $A$. A pure $|0⟩$ or $|1⟩$ preparation has no mean oscillation, whereas an equal-weight quantum superposition maximizes its amplitude within this two-state family. The motion is produced by interference between energy eigenstates, rather than by a changing occupation of either one.

Take the natural logarithm of the exponential decay model. To keep the logarithm dimensionless, choose the reference rate $R_{\ast }=1\mathrm{count}\mathrm{da}{\mathrm{y}}^{-1}$ and defineThen $y=mx+c$. With time measured in picoseconds, the slope has units ${\mathrm{ps}}^{-1}$ and the intercept is dimensionless. A straight-line linear regression in these variables therefore estimates the inverse mean lifetime and logarithmic normalization.

Use $x=t$ in picoseconds and $y=\ln (R/R_{\ast })$. Since the time uncertainty is negligible, only the vertical errors enter. First-order propagation of uncertainty givesThe transformed measurements areThus the logarithmic transformation converts constant fractional rate uncertainties into approximately equal absolute log-rate uncertainties. The equality is the linear error approximation; finite error bars would become slightly asymmetric under an exact logarithm.

Assume independent approximately Gaussian log-rate errors with known common standard deviation $s=0.15$. The weighted least squares objective isSince all weights agree, the normal equations for linear least squares reduce toFor these data, $\bar{x}=1.5\mathrm{ps}$, $\bar{y}=4.1596533$, $S_{xx}=4.375{\mathrm{ps}}^2$ and $S_{xy}=-10.3782409\mathrm{ps}$. Hence $\hat{m}=-2.3721693{\mathrm{ps}}^{-1}$ and $\hat{c}=7.7179073$.

The covariance matrix of the fitted slope and intercept follows from the inverse weighted design matrix product, or directly by writing both estimates as linear combinations of the independent $y_i$:This givesThe parameter conversion and first-order propagation of uncertainty areConsequently,These are one-standard deviation errors. The estimates are correlated, since $\mathrm{Cov}(\hat{m},\hat{c})=-0.0077143{\mathrm{ps}}^{-1}$. The minimum is ${\chi }_{\min }^2=4.6537$ for four residual degrees of freedom. Because $s$ is supplied as a known measurement uncertainty, it should not be rescaled by the observed reduced ${\chi }^2$. This is a log-linear exponential-decay fit.

For an independent count with a Poisson distribution, collected for exposure $T_i$ at rate $R_i$, the expected count is $N_i=R_i{T}_i$ and its standard deviation is $\sqrt{N_i}$. The fractional rate uncertainty is consequentlyA target of $0.15$ requires $N_i\simeq (0.15{)}^{-2}=44.44$ counts at each proper-time setting. The exposure time for a Poisson rate at fixed fractional precision is therefore $T_i=1/(0.15^2{R}_i)$, givingSince the settings are acquired independently rather than simultaneously,Nearly three quarters of the exposure is spent at the latest, lowest-rate setting. Proper time in picoseconds labels the decay distribution; it is not the laboratory acquisition time in days. This estimate assumes counting statistics dominate, with no additional background-subtraction uncertainty or acquisition overhead.

A chi-squared test compares the sum of squared standardized regression residuals with the appropriate chi-squared distribution. With both line parameters fitted, the data give ${\chi }^2=4.6537$ on $6-2=4$ degrees of freedom, with upper-tail P-value about $0.325$: the fitted exponential decay model describes the data adequately.

First consider fixing the alternative lifetime at its central value ${\tau }_a=0.478\mathrm{ps}$. Its slope is $m_a=-1/{\tau }_a=-2.09205{\mathrm{ps}}^{-1}$. One must refit the normalization rather than retain the earlier intercept. Minimization over the intercept gives $c_a=\bar{y}-m_a\bar{x}=7.29773$, andon $6-1=5$ degrees of freedom, with upper-tail P-value about $0.00130$. Thus that fixed central lifetime gives a substantially worse fit than $\hat{\tau }=0.4216\mathrm{ps}$. Equivalently, the increase after profiling over the intercept isThis fixed-value comparison does not yet include the uncertainty of the alternative measurement.

For the compatibility of independent lifetime measurements, include both errors. Under the approximately Gaussian lifetime-error model, equality of the two true lifetimes is tested byThis has one degree of freedom and P-value approximately $0.032$, equivalent to a $2.15$-standard deviation discrepancy. The two measurements are therefore in mild tension: equality is rejected at a $5\%$ significance threshold, but not at $1\%$. There is no unique yes-or-no compatibility verdict without choosing such a threshold. Treating the alternative central value as exact would overstate the disagreement between the experiments.

The lifetime that provides the best fit to these six rates remains $\boxed{\tau =0.4216\mathrm{ps}}$, not $0.478\mathrm{ps}$. The compatibility calculation above assumes the two experiments have independent uncertainties; a shared systematic error would require its covariance.

The photoelectric effect, black-body radiation and the Davisson–Germer experiment challenge different parts of a fully classical picture. Classical electromagnetic waves successfully describe propagation, interference and diffraction, while classical particles follow trajectories determined by forces. The difficulty is combining that picture with the observed exchange of energy between radiation and matter, the equilibrium distribution of radiation, and electron scattering. The three experiments expose the roles of discrete energy exchange and coherent matter waves.

In a photoelectric effect experiment, monochromatic light illuminates a metal cathode in vacuum and a second electrode collects emitted Electrons. Varying the collecting voltage measures a photocurrent; reversing it until collection ceases determines the stopping potential. The observations distinguish two controls: at fixed frequency above threshold, increasing light intensity increases the emission rate, whereas the largest emitted kinetic energy is determined mainly by frequency. There is a material-dependent photoelectric threshold frequency, and emission above threshold does not require the long charging-up delay predicted by a simple picture of gradual classical energy accumulation at low intensity.

A classical wave carries continuously distributed energy, with intensity proportional to squared field amplitude. On a fully classical absorption picture, increasing intensity would be expected to increase the energy delivered to an electron, and sufficiently long exposure could accumulate the escape energy even below the observed threshold. It does not naturally produce the measured universal straight-line relation between stopping potential and light frequency. In 1905, Albert Einstein instead treated light-energy transfer as localized quanta. His argument distinguished the fastest escaping Electrons from electrons losing energy before escape.^[ref]

In the single-photon regime, conservation of energy gives the Einstein photoelectric equationwhere $\Phi$ is the surface work function. The photon supplies energy $h\nu$; the escape cost is paid first, and the remainder sets the upper electron-energy edge. Increasing intensity at fixed frequency increases the number of available photons, so it raises current without raising this edge. Below ${\nu }_0$, one photon cannot pay the escape cost. In an actual measurement, stable surface preparation, suppression of unwanted higher-frequency light, and the photoelectric contact-potential correction are needed to test the slope $h/e$ reliably. Robert Andrews Millikan's experiment used selected mercury spectral lines, freshly exposed metal surfaces and an electrometer to make such a quantitative test.^[ref] The interpretation is for weak single-photon illumination; multiphoton photoemission at sufficiently intense fields is a different regime.

Black-body radiation tests equilibrium physics rather than the escape of individual electrons. A small aperture in an isothermal absorbing cavity approximates an ideal blackbody: radiation entering the hole undergoes repeated interactions with the walls, and the equilibrium spectrum depends on temperature rather than wall material. A spectrally resolving detector outside the aperture measures how emitted power varies with frequency or wavelength. Experimentally the spectrum has a finite peak, is strongly suppressed at short wavelengths, shifts toward shorter wavelengths when the cavity is heated, and has integrated emission proportional to $T^4$.

For a large cavity, the electromagnetic mode density in a box, including two independent transverse polarizations, is $g(\nu )=8\pi {\nu }^2/c^3$ per volume per unit frequency. The classical equipartition theorem assigns mean energy $k_{B}T$ to each mode, leading to the Rayleigh-Jeans lawThis agrees at low frequencies but diverges as frequency increases; integrating it over all frequencies gives infinite energy density at any nonzero temperature. This is the ultraviolet catastrophe, a failure of classical equipartition applied to the electromagnetic modes.

Max Planck's essential modification was that energy at a frequency is exchanged in units $h\nu$. In a modern oscillator description, the thermal excitation levels are $nh\nu$, $n=0,1,\dots$. Using the Boltzmann distribution and summing a geometric series givesMultiplying by the mode density yields Planck's law for the spectral energy density,When $h\nu \ll k_{B}T$, expansion of the denominator recovers the classical law. When $h\nu \gg k_{B}T$, the large excitation cost suppresses occupancy exponentially, removing the divergence. Its integral gives the Stefan–Boltzmann law, while its peak gives Wien's displacement law. Thus quantization repairs the spectral shape, total emission and temperature dependence together. The displayed thermal energy excludes the oscillator's temperature-independent zero-point energy.

The Davisson–Germer experiment supplies the complementary evidence that matter cannot always be treated as classical particles with definite scattering trajectories. In 1927, Clinton J. Davisson and Lester Halbert Germer sent a narrow, accelerated electron beam normally onto a single nickel crystal in vacuum. A movable collector measured the current of scattered Electrons as direction and accelerating voltage were varied; a retarding potential selected electrons retaining approximately the incident energy. Sharp angular beams depended on crystal orientation and beam energy, with a prominent feature near $54\mathrm{V}$ and detector colatitude $50^{\circ }$. These are scattering directions and currents, not photographs of electron paths.^[ref]

The de Broglie wavelength predictsAt $U=54\mathrm{V}$, this gives $\lambda \simeq 0.167\mathrm{nm}$, comparable to atomic spacings. A crystalline arrangement supplies regularly spaced scattering sites, so scattered probability amplitudes can add constructively in selected directions. For a suitable family of crystal planes, Bragg's law expresses the condition as $2d\sin {\theta }_B=n\lambda$, with ${\theta }_B$ measured relative to those planes. The detector colatitude is not automatically this angle. The measured angular structure and its variation with electron momentum are described by electron diffraction with the de Broglie wavelength. A classical point-particle deflection model has no phase wavelength $h/p$ and does not supply that interference condition.

The common quantum account preserves the successes of wave propagation and particle detection while changing what is quantized. Photons exchange energy in frequency-dependent units; a thermal distribution of those excitations gives the blackbody spectrum. Electrons have a wavefunction whose amplitudes interfere on scattering from a crystal, while detection transfers localized charge. The Born rule turns squared amplitudes into event probabilities. These results invalidate a fully classical description of matter and radiation; the photoelectric energy law alone does not establish every property of a quantized electromagnetic field, but the three phenomena together provide complementary evidence for the quantum description.

For the deep-water gravity wave, the phase velocity and group velocity follow from its dispersion relation:The wave packet travels at $v_g$, while the individual crests travel at $v_p$. Let its effective length be $D\simeq 5\lambda$, where $\lambda =2\pi /k$ is the carrier wavelength. A fixed observer is inside the wave packet for $\Delta t=D/v_g$, and crests arrive with wave period $\lambda /v_p$. ThusCrests enter the rear of the moving wave packet and disappear at its front, so the five visible at one instant are not five permanently identifiable crests. The count uses the intended narrow-band, approximately unchanged wave packet and neglects the precise visibility of its end crests.

A bubble is a thin water layer surrounded by air. At a locally flat patch, let its thickness be $d$, its refractive index be $n$, and the internal angle of refraction be ${\theta }_t$. The two principal reflected light contributions are the reflection at the first surface and the contribution that enters the layer, reflects at its second surface and leaves again. The first reflection, from lower to higher refractive index, introduces a $\pi$ phase shift; the second does not. Their optical path difference is $2nd\cos {\theta }_t$, so their phase difference is, modulo $2\pi$,where ${\lambda }_0$ is the vacuum wavelength. Constructive interference in reflected light therefore occurs approximately whenDestructive interference occurs at $2nd\cos {\theta }_t=m{\lambda }_0$. In the weak-reflection approximation, the reflected light intensity at each wavelength is proportional to ${\sin }^2(2\pi nd\cos {\theta }_t/{\lambda }_0)$. This is a smooth spectral filter, with broad maxima and minima: it does not select three pure wavelengths corresponding to display primaries. Ordinary illumination contains many visible wavelengths, and the eye combines their remaining contributions through color vision. The result is a mixture of colours, often pastel or desaturated.

The layer thickness varies over the bubble because of drainage and flow; the bubble's curvature also changes the local angle of refraction and the viewed optical path difference. The spectral filter therefore changes continuously across the surface and in time. Very thin regions become dark because the two reflected contributions nearly cancel across the visible optical spectrum. Thin-film interference explains both the changing colours and this dark limit.

The ray sketch shows a local patch and is not to scale. The plotted curves use a chosen constant $n=1.33$ at normal incidence and illustrative thicknesses; they show the spectral filter rather than calibrated perceived colours.

Model the room's decaying acoustic waves as a weakly damped resonator. The quality factor from ringdown is related to stored energy $E$ bywhere $A$ is the pressure wave amplitude and $\omega =2\pi f$. The measured frequency $f=440\mathrm{Hz}$ gives $\omega \simeq 2.76\times 10^3{\mathrm{s}}^{-1}$. If the stated $5\mathrm{s}$ is interpreted as an approximate wave amplitude exponential relaxation time,an estimate of order $10^4$. Taking $5\mathrm{s}$ as an energy exponential relaxation time instead gives $Q\simeq 1.4\times 10^4$.

There is a real convention ambiguity: “heard to die away” gives neither an initial sound level nor an audibility threshold, so it does not determine either exponential relaxation time. More generally, if the energy falls by a factor $E(0)/E(t_d)$ in $t_d$,In room acoustics, a reverberation time $T_{60}$ usually denotes a decibel drop of $60\mathrm{dB}$, or an energy factor $10^6$. If $5\mathrm{s}$ means $T_{60}$, the corresponding estimate is instead $\boxed{Q\simeq 2\pi (440)(5)/(6\ln 10)\simeq 1.0\times 10^3}$. The simple decay-time estimate and the conventional reverberation time estimate should not be silently equated.

The three primitive lattice translation vectors join a cube corner to the three adjacent face centres. Thus the Bravais lattice is a face-centered cubic lattice, with conventional cubic side $a$.

Write the given primitive lattice translation vectors as ${\mathbf{a}}_1,{\mathbf{a}}_2,{\mathbf{a}}_3$. Their primitive unit cell volume isThe reciprocal lattice convention ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$ then givesEach reciprocal lattice vector has units ${\mathrm{m}}^{-1}$ when $a$ is in metres. Integer combinations generate a body-centered cubic lattice of conventional cubic side $4\pi /a$; the point $(2\pi /a)(1,1,1)$ is its body centre. Directly taking the nine dot products checks the signs and the $2\pi$ normalization.

Let $N=L/a$ be the integer number of primitive unit cells. A Bloch state can be written as ${\psi }_k(x)=e^{ikx}{u}_k(x)$, with $u_k(x+a)=u_k(x)$. The periodic boundary condition ${\psi }_k(x+L)={\psi }_k(x)$ therefore requiresThe spacing of allowed wavevectors is $\Delta k=2\pi /L$. The first Brillouin zone has width $2\pi /a$, soFor an exact endpoint convention take a half-open Brillouin zone, such as $[-\pi /a,\pi /a)$. Its two endpoints differ by a reciprocal lattice vector and must not both be counted as distinct values of crystal momentum. There are exactly $N$ inequivalent integers $j$ modulo $N$, irrespective of whether $N$ is even or odd. This is state counting in a Bloch band: one allowed wavevector per primitive unit cell per energy band. A separate twofold spin multiplicity gives $2N$ electronic states per energy band, without doubling the number of wavevectors.

Use transverse displacement $y(x,t)$, small slopes and constant tension $T$. For a short string element of length $dx$, Newton's second law gives ${\rho }_{L}dx{y}_{tt}=T[y_x(x+dx)-y_x(x)]$. Hence the wave equation on a string iswhere ${\rho }_L$ is the linear mass density, in $\mathrm{kg}{\mathrm{m}}^{-1}$.

The mechanical impedance of a rightgoing travelling wave is the transverse driving force divided by the transverse velocity at the driven section. Choose the force exerted by the material to the left on the material to the right, $F_y=-T{y}_x$. For $y=f(x-c_{L}t)$, the transverse velocity is $v_y=y_t=-c_L{y}_x$. Thus the positive characteristic impedance of a stretched string isA leftgoing component has $F_y/v_y=-Z_L$ with this fixed orientation of force. Specifying the orientation avoids confusing that propagation sign with a negative material mechanical impedance.

Place the junction at $x=0$ and use the complex amplitude convention $e^{-i\omega t}$. With incident displacement wave amplitude $A_i$, writeHere $r$ and $t$ are displacement wave reflection and transmission coefficients, and $k_j=\omega /c_j$. The junction is assumed massless, with no externally applied transverse force, and both strings have the common equilibrium tension $T$. Continuity of displacement and transverse force givesSince $T{k}_j=\omega Z_j$, these become $1+r=t$ and $Z_L(1-r)=Z_{R}t$. Solving,A larger right-hand mechanical impedance produces a reflected displacement with reversed sign, as expected in the fixed-end limit $Z_R\to \infty$. Equal mechanical impedances give $r=0$ and $t=1$.