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Inside walls at $\pm a$, a central delta potential of strength $\kappa$ leaves odd parity operator eigenstates unchanged because their wavefunctions vanish at zero. For even states, continuity and the derivative jump ${\psi }^{\prime }(0+)-{\psi }^{\prime }(0-)=2m\kappa \psi (0)/{\hslash }^2$ give the displayed quantization equation. For $\kappa >0$, one root lies between $(n+1/2)\pi /a$ and $(n+1)\pi /a$ for every $n\ge 0$.

For two identical fermions of mass $m$ in the same one-dimensional quantum harmonic oscillator, use center of mass $X=(x_1+x_2)/2$ and separation $r=x_1-x_2$. The corresponding oscillator masses are $2m$ and reduced mass $m/2$, with the same frequency $\omega$. If $N,n$ are their occupation numbers, the unperturbed energy is $\hslash \omega (N+n+1)$. The fermionic exchange symmetry requires even $n$ with a spin-one-half singlet state and odd $n$ with a spin-one-half triplet state.

For a delta potential $\lambda \delta (r)$ the first-order shift is $\lambda |{\phi }_n^{\mathrm{rel}}(0){|}^2$. Every odd relative wavefunction vanishes at contact, so these states remain exact unaffected eigenstates of the spatial problem. For the even relative ground state, the oscillator length is $\sqrt{2}\ell$, where $\ell =\sqrt{\hslash /(m\omega )}$, and the shift is $\lambda /(\sqrt{2\pi }\ell )$, independent of $N$. This separates interaction-sensitive singlets from contact-insensitive triplets.

Take a nonrelativistic particle of constant mass $m$ with energy $E>0$, 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 $k=\sqrt{2mE}/\hslash$ and makes $T=|t{|}^2$. Here $\hslash =h/(2\pi )$ is the reduced Planck constant. This model has a finite jump in potential energy, no delta potential, and no absorbing term.

For $0<E<V_0$, the time-independent Schrödinger equation gives the spatial wavefunctionThe full stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. 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 $\kappa$ 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 $A{e}^{\kappa a}$ and $B{e}^{-\kappa a}$ in terms of $t$, then substitute into the left-face equations. Taking the squared modulus gives the rectangular barrier transmission probability.

A hard wall at the origin and an attractive delta potential at distance $a$ form a leaky quantum cavity. For unit incident amplitude from outside, the interior standing wave has coefficient $A=-i\lambda x/[1-e^{2ix}+i\lambda x]$, where $x=ka$ and $\lambda ={\hslash }^2/(mqa)>0$. Its modulus squared is the title expression. A resonance pole $p_n=x_n-i{\gamma }_n$ produces an enhancement with a locally Lorentzian line shape, proportional to $[(x-x_n{)}^2+{\gamma }_n^2{]}^{-1}$. At fixed low $n$ and small $\lambda$, $x_n\simeq n\pi (1+\lambda /2+{\lambda }^2/4)$ and ${\gamma }_n\simeq (\lambda n\pi /2{)}^2$.

The coefficient can exceed one because it measures a standing-wave amplitude, while the net interior probability current is zero. Conservation of probability current gives unit reflected flux: no current passes through the hard wall. Near zero, $|A{|}^2\to {\lambda }^2/(2-\lambda {)}^2$ when $\lambda \ne 2$; the threshold case $\lambda =2$ requires a separate limit.

Write the trap angular frequency as $\omega$ and the single-particle oscillator length as $\ell =\sqrt{\hslash /(m\omega )}$. The center of mass coordinate $X=(x_1+x_2)/2$ and relative coordinate $r=x_1-x_2$ give independent quantum harmonic oscillators, both of frequency $\omega$, with masses $2m$ and reduced mass $m/2$. Label their excitation numbers by $N$ and $n$. The unperturbed energy eigenvalue is $\hslash \omega (N+n+1)$.

Particle exchange leaves $X$ unchanged and sends $r$ to $-r$. The exchange symmetry of two identical spin-one-half fermions therefore requires even $n$ to accompany the spin-one-half singlet state, and odd $n$ to accompany a spin-one-half triplet state. For the contact shifts of two spin-one-half fermions in a harmonic trap, the delta potential depends only on $r$. For odd $n$, the relative wavefunction vanishes at $r=0$, so its contact shift is zero. Within the first-excited manifold, both perturbations are diagonal in the labels $(N,n,S,M)$: this is a perturbation diagonal within a degenerate eigenspace. For $n=0$, the relative oscillator length is $\sqrt{2}\ell$, giving the first-order nondegenerate perturbation theory shiftThe Zeeman splitting is $2{\mu }_B{B}_{z}M$ for total spin projection $\hslash M$. Thus the five lowest states, counted with multiplicity, haveunder the assumed conditions $|U|,2{\mu }_B|B_z|\ll \hslash \omega$. The three triplet states have slopes $-2{\mu }_B,0,2{\mu }_B$, while both singlet states are field independent at this order. The three triplet state branches coincide at zero magnetic field. Their crossing with the excited singlet state is allowed because the total-spin sectors do not mix. The odd relative wavefunctions are in fact unaffected by this delta potential at any coupling; the displayed singlet state shifts retain only first order.

For a real potential energy and a stationary state, the probability current isThe time-independent Schrödinger equation gives ${\psi }^{\prime \prime }=2m(V-E)\psi /{\hslash }^2$, so $j^{\prime }=\hslash \mathrm{Im}({\psi }^{\ast }{\psi }^{\prime \prime })/m=0$. In an exterior plane wave superposition, interference terms contribute no current andConsequently $|A{|}^2+|D{|}^2=|B{|}^2+|C{|}^2$ for every choice of incoming amplitudes. If $u=(A,D{)}^T$, this says $u^{†}(I-S^{†}S)u=0$ for every complex vector $u$. Since $I-S^{†}S$ is Hermitian, vanishing of its quadratic form implies that the matrix itself vanishes. ThereforeThe second identity follows by choosing $u=(1,0{)}^T$. Since $S$ is a square matrix, the first also implies $S{S}^{†}=I$, proving unitarity. A real delta potential preserves this proof: continuity of $\psi$ and the real derivative jump make the current continuous across the interaction point.

Assume $a>0$ and put $\mu =mg/{\hslash }^2$, $q=k/\hslash$, and $\gamma =\mu /q$. At a delta potential located at $x_0$, integration of the Schrödinger equation gives the matching conditionsWriting the adjacent plane wave coefficients as $(A_L,B_L)$ and $(A_R,B_R)$ givesSolving these two linear equations yields the transfer matrix of a delta potential:For the symmetric double-delta potential, use separate plane wave pairs in $x<-a$, $-a<x<a$, and $x>a$. Successive matching gives $M=D(a)D(-a)$ between the two exterior pairs. For real $q$, multiplication givesThe determinant identity $\det M=1$ reads $|d{|}^2-F^2=1$. Left incidence means $(t,0{)}^T=M(1,r{)}^T$, whenceHere the first relation also follows from $t=M_{11}-M_{12}{M}_{21}/M_{22}=\det M/M_{22}$. Therefore the requested transmission probability and quantum reflection probability, as functions of the printed momentum $k$, areThey obey $R+T=1$. As checks, $g=0$ gives $T=1$, and the coincident limit $a\to 0$ gives $T=q^2/(q^2+4{\mu }^2)$, the answer for a single delta potential of strength $2g$.

Take fixed $E>0$ and define $k=\sqrt{2mE}/\hslash$. For sufficiently large $V_0$, the barrier region has $\kappa =\sqrt{2m(V_0-E)}/\hslash$ and solves ${\psi }^{\prime \prime }={\kappa }^2\psi$. Solving this constant-coefficient equation gives the exact transfer relationIn the specified limit, $a=U/V_0$ and $\kappa a\to 0$, whileThe transfer matrix therefore tends to $\left(\begin{array}{cc}1& 0\\ \eta & 1\end{array}\right)$. As the two boundaries collapse to the origin, this proves continuity and the derivative jump for a delta potential:The vanishing exterior phase shift $e^{ika}\to 1$ introduces no further factor.

Write the incoming and reflected waves as $e^{ikx}+r{e}^{-ikx}$ on the left and the transmitted wave as $t{e}^{ikx}$ on the right. Matching givesThe denominator is nonzero for real $\eta$ and $k>0$, so the limiting matching system is well defined. The probability currents on either side have the same factor $\hslash k/m$, making the transmission probability $|t{|}^2$. Thus scattering by a delta potential givesThe limit retains a finite reflection probability despite the barrier's vanishing width because its integrated strength remains finite.

For a normalized trial state in the quadratic-form domain of a self-adjoint Hamiltonian operator bounded below, its expectation of $H$ is at least the ground energy $E_0$. Minimizing this Rayleigh quotient over a tractable trial family therefore supplies an upper bound.

For the two states, set $a_{\pm }=(a_1\pm a_2)/\sqrt{2}$. Their quotient isThe strict assumption $ϵ<sE$ excludes $|s|=1$, since equality in the overlap Cauchy-Schwarz bound would then force the two states to be proportional and $ϵ=sE$. Thus $|s|<1$. The two generalized eigenvalues are $(E\pm ϵ)/(1\pm s)$, and their difference has the sign of $ϵ-sE$. The symmetric trial is the lower one, so

For the single attractive delta well, the supplied wavefunction has norm one because $\int \lambda e^{-2\lambda |x|}dx=1$. Its kinetic energy in quadratic-form notation is ${\hslash }^2\int |{\psi }_1^{\prime }{|}^2/(2m)={\hslash }^2{\lambda }^2/(2m)$, and its delta potential energy is $-K|{\psi }_1(0){|}^2=-K\lambda$. With $\lambda =mK/{\hslash }^2$ this givesIts derivative jump is ${\psi }_1^{\prime }(0+)-{\psi }_1^{\prime }(0-)=-2\lambda {\psi }_1(0)$, exactly the delta boundary condition, so it is an eigenstate. A decaying bound state on the two half-lines must have a common decay rate fixed by this same jump; there is only this one negative-energy state, confirming it is the ground state.

For $R>0$, integrate the overlap on the three regions $x<0$, $0<x<R$, $x>R$. Put $a=\lambda R$. The result isThe second equality is the single-well energy plus the extra delta expectation; the third follows by applying either single-well eigenvalue equation in the cross matrix element. Moreover $ϵ-sE=K\lambda e^{-a}[(1+a)e^{-2a}-1]<0$, so the preceding bound applies. Substitution givesFor widely separated wells, each localized orbital is close to an isolated-well ground state and the low-energy subspace is accurately their span; the symmetric combination captures the leading tunneling energy lowering. In the limit both exact and trial energies approach $E_B$. This illustrates electronic molecular binding: sharing an orbital between attractive centers can lower its energy below the separated-center value. A full equilibrium bond length would also require the repulsion of the positive centers.

Insert the position and momentum resolutions of the identity to obtain the unitary Fourier transformThe kinetic term is multiplication by $p^2/(2m)$, and the potential has momentum kernel $⟨p|V|p^{\prime }⟩=(2\pi \hslash {)}^{-1}\int V(x)e^{-i(p-p^{\prime })x/\hslash }dx$. Thus the Schrödinger equation becomesFor the attractive delta potential, $V(x)=-({\hslash }^2\lambda /m)\delta (x)$, the kernel is the constant $-\hslash \lambda /(2\pi m)$, giving the stated positive right-hand side. For $\tilde{\psi }=N/(p^2+{\alpha }^2)$ with $\alpha >0$, constancy of the left side requires $E=-{\alpha }^2/(2m)$. Since $\int dp/(p^2+{\alpha }^2)=\pi /\alpha$, its coefficient equation givesFinally, transforming $\sqrt{\lambda }{e}^{-\lambda |x|}$ gives $\tilde{\psi }=2{\lambda }^{3/2}{\hslash }^2/[\sqrt{2\pi \hslash }(p^2+{\hslash }^2{\lambda }^2)]$, verifying the same momentum-space form and energy.

The parity operator acts by $(P\psi )(x)=\psi (-x)$. For an even potential, differentiating twice and using $V(-x)=V(x)$ proves $HP=PH$ on the parity-invariant domain of the Hamiltonian operator. Therefore each energy eigenspace is invariant under $P$. The projectors $(I+P)/2$ and $(I-P)/2$ split it into even and odd subspaces, giving a basis of simultaneous eigenstates. An odd state obeys $\psi (0)=-\psi (0)$, hence ${\psi }_{\mathrm{odd}}(0)=0$.

The infinite walls impose $\psi (-a)=\psi (a)=0$, with the wavefunction zero outside. It is continuous at the central delta potential. Integrate the stationary Schrödinger equation over $[-\epsilon ,\epsilon ]$; the energy integral tends to zero and the Dirac delta function contributes $\kappa \psi (0)$. This givesFor the given piecewise positive-energy form, $E={\hslash }^2{\lambda }^2/(2m)$. Replacing $x$ by $-x$ swaps the two pieces and leaves their values unchanged, proving $P\psi =\psi$. The value at zero is $A$, and the one-sided derivatives are $-B\lambda$ and $B\lambda$. The jump condition gives $B=-m\kappa A/({\hslash }^2\lambda )$. The wall at $a$ gives $A\cos (\lambda a)-B\sin (\lambda a)=0$, henceFor $\kappa =0$ the undivided wall equation instead gives the usual free even roots. For $\kappa >0$, set $c=m\kappa /{\hslash }^2>0$ and $k_n=(n+1/2)\pi /a$. In every interval $k_n<\lambda <(n+1)\pi /a$ there is exactly one root: writing $\delta =a(\lambda -k_n)\in (0,\pi /2)$ gives $\cot \delta =\lambda /c$, whose two sides are respectively strictly decreasing and increasing. These are all positive even roots; the quadratic form ${\hslash }^2\int |{\psi }^{\prime }{|}^2/(2m)+\kappa |\psi (0){|}^2$ is nonnegative, so there are no negative eigenvalues. Thus ${\lambda }_n>k_n$ and ${\eta }_n={\hslash }^2({\lambda }_n^2-k_n^2)/(2m)>0$.

The printed limit of this absolute energy difference is false. The exact quantization equation gives ${\delta }_n=\arctan (c/{\lambda }_n)$, so ${\delta }_n\to 0$ but ${\lambda }_n{\delta }_n\to c$. ConsequentlyThus the high-energy shift from a central delta barrier is a finite positive constant, not zero. The corrected vanishing statement is the relative shift ${\eta }_n/E_n^{(0)}\to 0$, or the wave-number difference ${\lambda }_n-k_n\to 0$. Physically, highly energetic particles are weakly affected in relative terms by a fixed point barrier. The free even wavefunction normalized in $[-a,a]$ has density $1/a$ at the origin, so the barrier's first-order expectation is $\kappa /a$, matching the exact limit.

Odd states vanish at zero, so the delta term and the derivative jump both vanish. Their normalized wavefunctions and energies arewith zero wavefunction outside. They do not depend on $\kappa$. These even and odd sectors describe the central delta barrier in a symmetric infinite well.

With outgoing-channel index first, the scattering matrix convention is ${\psi }_b\sim I_b+{\sum }_{a=\pm }{S}_{ab}{O}_a$ as $|x|\to \infty$. An even potential commutes with parity, so its incoming even and odd channels remain separate: $\boxed{S_{+-}=S_{-+}=0}$.

For the odd channel, on $0<x<a$ write ${\psi }_{-}(x)=C\sin (kx)$, while on $x>a$ its prescribed normalization is $e^{-ikx}-S_{--}{e}^{ikx}$. A delta potential keeps the wavefunction continuous and imposes ${\psi }^{\prime }(a+)-{\psi }^{\prime }(a-)=U_0\psi (a)$, with $U_0=2m{V}_0/{\hslash }^2$. Thus the exterior logarithmic derivative is $k\cot (ka)+U_0$. Matching givesMultiplying numerator and denominator by $2\sin (ka)$ and using exponential forms yieldsFor real $k,U_0$, the numerator of the fraction is the complex conjugate of its denominator, and the prefactor has modulus one. Hence $\boxed{|S_{--}{|}^2=1}$, conservation of probability flux in this single elastic parity channel. In particular the minus sign built into $O_{-}$ makes the free odd-channel matrix element equal to one.

For an odd bound state, continue to $k=i\kappa$, $\kappa >0$. A pole means the outgoing solution is exponentially decaying on both sides. Its denominator vanishes whenThe function $2\kappa /(1-e^{-2\kappa a})$ rises strictly from $1/a$ to infinity: differentiation reduces positivity to $e^{2\kappa a}>1+2\kappa a$. Thus there is exactly one positive solution if $-U_0>1/a$, and none otherwise. The numerator does not vanish at that solution, so the pole is genuine. ConsequentlyAt equality the zero-energy threshold solution is not square integrable.

For repulsive strong barriers put $u=U_{0}a\gg 1$ and $z=ka=\pi +\delta$. The pole equation is $z\cot z+u-iz=0$. Since $\cot \delta ={\delta }^{-1}-\delta /3+O({\delta }^3)$, multiplication by $\delta$ gives $\pi +(u+1-i\pi )\delta +O({\delta }^2)=0$. Iterating in $u^{-1}$ yieldsThis lower-half-plane resonance pole is an odd standing wave trapped between the barriers and slowly leaking by tunnelling. Its complex energy is $E_R-i\Gamma /2$, with $\Gamma \simeq 2{\hslash }^2\pi \gamma /(m{a}^2)>0$. The factor $e^{-iEt/\hslash }$ then decays rather than grows; $\gamma >0$ encodes the positive escape rate.

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.

A bound state has $E<0$ and $\kappa =\sqrt{-2mE}/\hslash >0$, since outside the delta potentials a square-integrable solution must decay exponentially. The wavefunction is continuous at each delta: a jump would produce a delta derivative in ${\psi }^{\prime \prime }$ with no matching term in the equation. Integrating the time-independent Schrödinger equation across either $x_0=\pm a$ givesThe symmetry of the symmetric double-delta potential permits even and odd solutions. Up to a nonzero normalization factor, the even candidate is $\cosh (\kappa x)$ for $|x|\le a$, and $\cosh (\kappa a)e^{-\kappa (|x|-a)}$ outside. Its derivative jump at $a$ givesThe odd candidate is $\sinh (\kappa x)$ inside and $\mathrm{sgn}(x)\sinh (\kappa a)e^{-\kappa (|x|-a)}$ outside. Its jump givesParity ensures the jump at $-a$ as well. Put $t=\kappa a$ and $L=2m\lambda a/{\hslash }^2$. The continuous strictly increasing function $t(1+\tanh t)$ ranges from $0$ to infinity, so there is exactly one even bound state for every $L>0$. The continuous strictly increasing function $t(1+\coth t)$ ranges from $1$ to infinity, so the odd bound state of a symmetric double-delta potential exists if and only if $L>1$. Strict increase follows from the allowed monotonicity facts plus the strictly increasing term $t$. ThereforeAt equality the odd limiting solution has $\kappa =0$ and is not square integrable, so it is not a bound state. No nonnegative energy can supply another full-line square-integrable state, since its exterior solutions are oscillatory or affine.

In the Parity basis of a one-dimensional S-matrix, parity acts as $\mathcal{P}=\mathrm{diag}(1,-1)$. Parity invariance requiresand thereforeIndependently, unitarity givesCombining parity and unitarity leaves two phases,

For the symmetric double-delta potential, consider the odd channel on $x\ge 0$. Choose the asymptotic conventionand write $psi(x)=C\sin (kx)$ for $0<x<a$; the sine enforces odd parity at the origin. Continuity and the delta potential derivative jump at $x=a$ giveEliminating $C$ yields the Odd-parity S-matrix of a symmetric double-delta potentialUsing $2i\sin (ka)=e^{ika}-e^{-ika}$ rewrites this as the requested expressionFor real $k$, the two factors in the ratio are complex conjugates, directly confirming $|S_{--}|=1$.

A bound state occurs at a pole $k=i\kappa$, $kappa>0$. Setting the denominator to zero givesorFor $U_0<0$, write $g=-U_0>0$. The right-hand side $g(1-e^{-2\kappa a})$ is increasing and concave, starts at zero with slope $2ga$, and approaches $g$, whereas the left-hand side is the straight line $2\kappa$. A positive intersection therefore exists exactly whenThis is the odd bound state of a symmetric double-delta potential; equality is its zero-energy threshold.

For $U_0>0$, analytically continue the same pole equation into the lower half of the complex $k$-plane. PutIt becomesWhen $G\gg 1$, the lowest odd mode lies near the hard-wall value $z=\pi$. Writing $z=\pi +\eta$ and expanding givessoIn particular, the odd resonance of a strongly repulsive symmetric double-delta potential hasThe negative imaginary part identifies a decaying scattering resonance. Ifthen, to leading nonzero order,Its time-dependent amplitude is proportional toso its survival probability decays as $e^{-\Gamma t/\hslash }$ and its lifetime isPhysically, this is the lowest odd standing wave trapped for a long time between two strong repulsive barriers and leaking out by quantum tunnelling.

The required dipole matrix element iswhere differentiating the elementary Laplace integral for $\sin (kx)$ evaluates the last integral. Hence

Put $E_f=E_0+\hslash \omega >0$ and $k_f=\sqrt{2m{E}_f}/\hslash$. The continuum transition probability at long times usesSincethe total escape probability becomesFor the attractive delta potential,Substitution gives the requested leading-order result: