Finite rectangular barrier matching Created 2026-10-14 Updated 2026-10-20
Take a nonrelativistic particle of constant mass with energy , and choose the same zero of potential energy on both sides of a rectangular potential barrier:The equality of the exterior potential energies matters: it gives the same exterior wavenumber and makes . Here is the reduced Planck constant. This model has a finite jump in potential energy, no delta potential, and no absorbing term.
For , the time-independent Schrödinger equation gives the spatial wavefunctionThe full stationary state is . We set the incident probability amplitude to one and impose no incoming wave from the right; this scattering wavefunction is not a normalizable wavefunction. The tunnelling decay constant describes exponential decay, rather than an oscillatory wavenumber in the barrier. Both exponential functions are needed on the finite interval: the growing term is not excluded by a condition at infinity.
The wavefunction and its derivative are continuous at each face, giving four linear matching conditions:Their solution supplies the transmission amplitudeA short derivation is to use the two right-face equations to express and in terms of , then substitute into the left-face equations. Taking the squared modulus gives the rectangular barrier transmission probability.
Past exam of the mathematics course of the University of Cambridge 2010 ii Paper 2 34B Solution Created 2026-09-24 Updated 2026-10-14
The asymptotic scattering wavefunction is . For a real central potential, the scattering phase shift multiplies the outgoing component of each partial wave relative to the incoming one by . Matching the incident-wave normalization givesSubtract the given incident plane wave expansion. Since , the scattering amplitude isOrthogonality eliminates cross terms in , giving .
For the spherical square well, the regular reduced radial wave is inside and outside, with . Continuity of the wave and its derivative gives the logarithmic-derivative matching relation , orAt zeros of the denominators the original matching relation supplies the limiting interpretation.
Under the stated deep-well approximation, put and . The condition is . If , its first positive solution lies in ; if , it is . For , a solution lies in because there rises continuously from zero to infinity. If , the first positive solution lies in , where rises from one to infinity. The limiting pole case gives by the cotangent relation. Thus , without calculating the integer phase index. The use of presumes the located wave numbers remain small compared with .
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 50 2 Solution Created 2026-10-03 Updated 2026-10-14
Choose boundary states with nonzero overlap with the lowest-energy state in the vacuum sector and in the one-kink topological sector . In a finite spatial box, a fixed scalar field configuration eigenstate may be used formally, with temporal endpoints equal to the chosen configuration. More generally, smear the endpoints with wavefunctionals . Their unitary time evolution kernels areThe inner scalar field path integral remains in the chosen topological sector, with . A zero-total-momentum projection can be included to select the rest state; alternatively, the translational prefactor does not change the large-time exponential. After a Wick rotation to physical Euclidean time , the energy eigenstate expansion gives . Thus the exact vacuum-subtracted soliton mass isEquivalently it is at large time with a damping prescription. The ratio subtracts the vacuum energy; the Hamiltonian operator and action here are the fully regulated and renormalized ones, not merely their classical approximations.
With the dimensionless coordinates of this paper, the correctly normalized classical action isFor the static Sine-Gordon kink, and , so . Write . Expanding and integrating by parts givesThe first variation is , plus boundary terms. It vanishes because the kink satisfies the Euler-Lagrange field equation and the fluctuations have the prescribed temporal endpoints and admissible spatial boundary behavior. This is the principle of stationary action, not a symmetry assumption about .
The printed expansion omits despite the stated Lagrangian density. Its displayed form is the expansion of ; it is not the physical at arbitrary coupling. Alternatively, writing puts the quadratic term in canonical normalization, while leaving the classical term unchanged. This normalization repair does not change or the physical fluctuation frequencies.
The Sine-Gordon kink fluctuation operator has the useful factorizationIt is nonnegative, and gives the normalized translational zero mode of a sine-Gordon kink:A displacement changes by . Thus the zero mode in field theory is the position collective coordinate of the kink, reflecting translation invariance. It has no restoring force and no oscillator zero-point energy. Integrate that collective coordinate separately rather than inserting a zero factor into the Gaussian functional determinant.
Sine-Gordon kink fluctuation potential and normalized translational zero mode
. Details. In the Gaussian fluctuation approximation, the formal oscillator contribution to the one-loop soliton mass correction, before adding any counterterm, isUse a common regulator for the two sums, include all discrete modes, and treat the translation mode as above. The vacuum sum is essential: subtracting only classical vacuum energy would leave an extensive oscillator energy. A periodic fluctuation and its derivative are matched at the two ends of the large box; the one-kink background lies in the twisted topological sector, and its infinite-line profile is accurate up to exponentially small boundary corrections.
For a continuum scattering wavefunction, equality of its two asymptotic values gives the periodic-box phase-shift quantizationAway from the threshold, expand at a matched mode number:Since consecutive free wave numbers are separated by , replacing the continuum-mode sum by an integral proves the displayed continuum contribution:The ultraviolet cutoff is retained until the counterterm is added.
There is a finite threshold issue if this expression is identified with the complete oscillator correction. It can be settled directly using the factorization: a continuum eigenfunction is . Its transmission phase obeyswith the odd phase branch that tends to zero at large . For , the continuum roots have labels ; there is no periodic continuum root at , since the limiting eigenfunction has opposite signs at the two ends. The bound state at replaces the free oscillator with . Consequently, in mode-number regularization of soliton masses,The PDF's continuum-only formula misses this finite under this standard phase and mode-counting convention. It has the correct logarithmic ultraviolet divergence, but the missing term is not suppressed by large . Changing the phase branch requires changing the mode labels and endpoint terms consistently; it cannot erase a physical mode from the formal spectrum sum.
At high momentum, , so both expressions have divergent part . The canonical field has a quartic interaction with coupling . Its vacuum tadpole diagram shifts the squared mass by , whereThe Sine-Gordon vacuum tadpole counterterm has and adds potential energy density . Since , its vacuum-subtracted kink energy iswhich cancels the logarithmic ultraviolet divergence. Finite parts require a specified renormalization condition. As a consistency check, with this tadpole subtraction and matched mode-number cutoff, integration by parts yieldsThe complete semiclassical soliton mass is then in that convention. This last finite result uses the bound-mode term and is additional to the requested ultraviolet cancellation.
Past exam of the mathematics course of the University of Cambridge 2018 ii Paper 3 35A Solution Created 2026-09-24 Updated 2026-10-14
The asymptotic scattering wavefunction isIn each partial wave, the Partial-wave S-matrix is the ratio of its outgoing coefficient to its incoming coefficient. With the conventions in the question,For elastic scattering by a real central potential, unitarity gives , so for a real scattering phase shift . HenceUsing the Orthogonality of Legendre polynomials in yields the partial-wave total scattering cross-section
For the S wave, write the reduced radial wavefunction aswhere regularity at selected the hyperbolic sine. Continuity of and at , equivalently logarithmic-derivative matching, givesorFor , let the scattering length be . Expanding the matching relation givesOnly the S wave contributes at leading order, soIn the hard-sphere limit , penetration is suppressed, , andThis is four times the geometric area , a wave effect associated with diffraction from an impenetrable sphere.
Past exam of the theoretical physics II course of the University of Cambridge 2018 2 a Solution 2026-10-18
For a short-range potential, an incident plane wave of unit amplitude and wave vector has the large-distance scattering wavefunctionThe coefficient is the scattering amplitude. For a central potential it depends only on the scattering angle . The incident probability current is , while the scattered radial current is to leading order. Their ratio givesThese are the differential scattering cross-section and total cross section for elastic scattering with the same incident and outgoing wave number.
The expression is asymptotic: it specifies the outgoing wave's leading term, not its finite-distance radial corrections or its form inside the interaction region. In particular, for angle-dependent , the spherical term by itself has an angular-Laplacian contribution of order to the free wave equation, which is cancelled by subleading terms in the exact solution.
Periodic-box phase-shift quantization Created 2026-10-06 Updated 2026-10-20
For a reflectionless scattering wavefunction whose two asymptotic phases differ by , a large periodic box gives . Compared with the free value , away from thresholds. Discrete bound states and phase-branch changes alter mode indexing and must be counted when comparing complete spectra.
