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For $U_{0}a\gg 1$, the lowest odd scattering resonance pole lies near the first odd mode trapped between the barriers:in its real part and to leading nonzero order in its imaginary part. Writing $E={\hslash }^2{k}^2/(2m)=E_R-i\Gamma /2$, the survival probability decays as $e^{-\Gamma t/\hslash }$.

For fixed $0<E<V_0$ and $\kappa a\gg 1$, the exact rectangular barrier transmission probability givesThis is the leading term of an asymptotic expansion: the wavefunction amplitude scale is $e^{-\kappa a}$, and the transmission probability scale is $e^{-2\kappa a}$, with an interface prefactor. Increasing width by $\Delta a$ multiplies this approximation by $e^{-2\kappa \Delta a}$. At fixed incident energy, increasing barrier height increases the tunnelling decay constant as $\sqrt{V_0-E}$; heavier nonrelativistic particles also have stronger suppression. For the numerical example in rectangular barrier transmission probability, the approximation gives $0.01397$, compared with the exact $0.01388$.

The regime condition matters. Near $E=V_0$ at fixed width, $\kappa a$ tends to zero and the opaque barrier approximation fails; use the exact threshold value. For $\kappa a\ll 1$, expansion of the hyperbolic sine givesThis tends to one as $a\to 0$ at fixed $E,V_0$, but the condition $\kappa a\ll 1$ alone does not imply nearly perfect transmission if the remaining factor is large. At $E>V_0$, use the oscillatory formula and its scattering resonances, rather than an exponential decay law.

The incoming and outgoing coefficients give the scattering matrixIt obeys $S(-k)=S(k{)}^{-1}$ and $S(k^{\ast }{)}^{\ast }=S(k{)}^{-1}$, so $|S(k)|=1$ for real $k$, as required by elastic unitarity. With $S=e^{2i\delta }$, the scattering phase shift satisfies $\tan \delta =k(\kappa +\alpha )/(k^2-\kappa \alpha )$. The inverse tangent needs a continuous branch: taking $\delta (\infty )=0$ gives $\delta (0)=\pi$. Equivalently $\delta =\arctan (\kappa /k)+\arctan (\alpha /k)$ for $k>0$.

The scattering length follows from $k\cot \delta =-1/a_s+O(k^2)$:For a putative bound state put $k=i\eta$, $\eta >0$. The incoming term grows as $e^{\eta r}$, so its coefficient must vanish. The incoming Jost function is $F(k)=g(-k)=(k-i\kappa )/(k+i\alpha )$; its only upper-half-plane zero is $i\kappa$. Under the regular-potential assumption this gives one bound state, with $\boxed{E=-{\hslash }^2{\kappa }^2/(2m).}$ The pole at $i\alpha$ in $S$ instead comes from the outgoing numerator: it is a redundant pole of a scattering matrix, not a second normalizable state. The S-matrix alone, without its incoming/outgoing analytic structure, would tempt an incorrect count of two states.

For completeness, the nonsingular Eckart realization of this Jost function requires $\alpha >\kappa$, as discussed in dipot.ulb.ac.be/dspace/bitstream/2013/373712/4/2306.12216.pdf. The supplied positive constants alone do not establish such a realization in every ordering; a coincident pole is likewise not evidence for two states.

There is no positive-energy scattering resonance. Indeed ${\delta }^{\prime }(k)=-\kappa /(k^2+{\kappa }^2)-\alpha /(k^2+{\alpha }^2)<0$, and the s-wave total partial-wave total scattering cross-section is $4\pi (\kappa +\alpha {)}^2/[(k^2+{\kappa }^2)(k^2+{\alpha }^2)]$, strictly decreasing for $k>0$. All poles are on the imaginary axis, with no oscillatory decaying resonance pole.

Let $g=a{U}_0\gg 1$ and $z=ka$. A denominator zero away from the removable zero at $z=0$ obeysNear $z=\pi$, write $z=\pi +\delta$ and choose the logarithm branch with $\delta \to 0$. Expanding givesConsequentlyThe lower-half-plane pole is a scattering resonance: an odd standing wave is nearly confined between the two high repulsive barriers but leaks out slowly. If $E=E_R-i\Gamma /2={\hslash }^2{k}^2/(2m)$, then $\Gamma \simeq 2{\hslash }^2\pi \gamma /(m{a}^2)$ and its probability lifetime is $\hslash /\Gamma$. It is not a genuine bound state of the repulsive potential.

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.

Since $\tan {\delta }_0=\gamma /(K_0-k)$,The exact nearby stationary point satisfies $K_0-k={\gamma }^2/K_0+O({\gamma }^4/K_0^3)$, so to leading order in $\gamma /K_0$ the local maximum is at $k=K_0$. This is a scattering resonance withwhere $\Gamma$ is the full energy width of the Breit-Wigner peak.

For the resonant contribution only, subtract nonresonant Bhabha scattering and its interference before interpreting a fitted area. The integrated narrow vector resonance area follows from the Breit-Wigner resonance cross section, with $J=1$, $s_{e^{-}}=s_{e^{+}}=1/2$, and $p\simeq M/2$, becomesAcross a narrow scattering resonance, treat the prefactor and partial decay widths as constant and extend the integral over $E-M$ to the real line. Since ${\int }_{-\infty }^{\infty }du/(u^2+a^2)=\pi /a$,A normalized beam-energy response convolves the line shape without changing this full area. It can completely change the observed peak height and apparent width. If a uniform acceptance $ϵ$ is not already unfolded, use ${\mathcal{I}}_{ee}={\mathcal{I}}_{ee}^{\mathrm{obs}}/ϵ$. The distinction between $B_{ee}^2$ for this exclusive final state and $B_{ee}$ for an inclusive final state is essential.

TODO: the numerical resonance-area reading is unavailable from the retained source. Page 6 retains the upper data panels but loses the lower part of the scanned data image, including the numerical horizontal energy calibration. A vertical cross section scale and an apparent peak cannot supply $\int \sigma dE$ without that calibration. The decoded image contains 2,834,898 of 4,437,216 required RGB bytes; the archived bundle has the same truncation. No horizontal scale or natural decay width is inferred from the display width.

The recoverable numerical conversion is nevertheless definite. With $M\simeq 3.1\mathrm{GeV}$, $B_{ee}=0.060$, $ϵ=0.50$, and $1{\mathrm{GeV}}^{-2}=389400\mathrm{nb}$, b givesThis coefficient assumes the plotted exclusive electronic sample has not already been corrected for acceptance. A restored area, with continuum subtraction and acceptance convention checked, would complete the estimate. A tabulated modern J/psi meson width would be external information rather than a reading of these data.

The narrow scattering resonance indicates suppressed strong interaction decay. In the quark model, $c\bar{c}$ is below the open-charm quark $D\bar{D}$ threshold, so its constituents cannot simply emerge in separate charmed mesons. The OZI rule suppresses annihilation into light hadrons; for a $1^{--}$ color-singlet state the leading purely gluonic channel needs three gluons. The observed leptonic channels remain competitive. Its mass and $I=0$ fit charmonium; the small width alone would not uniquely identify the new quark flavour without the wider spectrum and threshold evidence.

Interpret the repeated species label in the Yukawa interaction as summed over $i=1,2$: each Feynman vertex is diagonal in species and equals $-ig$. Define the invariant scattering amplitude by $S-1=i(2\pi {)}^4{\delta }^{(4)}(P_f-P_i)\mathcal{M}$. There is one tree-level Feynman diagram, an $s$-channel scalar exchange. In the left panel, $j=2$ for this part; the right panel only occurs in d. Arrows show fermion-number flow, so an antifermion arrow opposes its physical momentum.

Writing $k=p_1+p_2=p_3+p_4$ and $s=k^2$, the momentum-space Feynman rules giveThe scalar carries no spin label. Average over the $2\times 2$ incoming spin states and sum the four outgoing combinations. For the real scalar Yukawa interaction, the conjugate of ${\bar{v}}_2{u}_1$ is ${\bar{u}}_1{v}_2$. The fermion spin sums then yieldThe bar denotes the initial spin average, not a Dirac adjoint. The gamma matrix trace identities giveand the second trace has the same value because $p_3+p_4=p_1+p_2$ and all external masses are $m$. ThereforeThis is the fixed-order expression away from $s=M^2$. If that pole is accessible, a physical scattering resonance requires a scalar self-energy and decay width; the tree formula alone is not finite there.

For $g=0$, the invariant scattering amplitude from the tree-level Feynman diagram in i has a pole at $s=M^2$. The intermediate heavy antiparticle can go on shell and live for a time of order its mean lifetime. Near this scattering resonance, inserting a single self-energy is no longer a uniformly small correction. Dyson resummation gives, to leading accuracy in a narrow decay width,Here $M$ is understood as the renormalized resonance mass; the real self-energy shifts the original pole position. The imaginary term removes the unphysical divergence of the fixed-order invariant scattering amplitude.

For equal incoming and outgoing light masses, the two-body momentum ratio cancels in the cross section. The angle-independent invariant scattering amplitude, integrated over the full solid angle with the identical final-state symmetry factor $1/2!$, gives $\sigma =|\mathcal{M}{|}^2/(32\pi s)$. Thus the Breit-Wigner resonance cross section isAt $s\gg M^2$ it reproduces $h^4/(32\pi s^3)$. At the resonance peak,This is also $8\pi /k_{\ast }^2$ for $k_{\ast }=M\beta /2$, the elastic identical-boson S wave partial-wave unitarity limit when this is the only open decay channel. The approximation assumes ${\Gamma }_{\mathrm{phys}}\ll M$ and a resonance sufficiently far from threshold for the width to vary slowly across it.

If one mechanically uses the unsymmetrized ${\Gamma }_{\mathrm{printed}}=2{\Gamma }_{\mathrm{phys}}$ from d, the same displayed Breit-Wigner resonance cross section would have that doubled width and predict a peak $8\pi /(M^2{\beta }^2)$. This is the literal-width substitution, rather than the consistent physical result: the given high-energy cross section already includes the identical-final-particle factor that the printed decay width integral omits.

For equal exterior potential energies and constant mass, the exact transmission probability isBelow the barrier, the hyperbolic sine is positive for $a>0$, so $0<T<1$. At $E=V_0$, the interior wavefunction is linear in $x$; either neighbouring formula has the displayed finite limit, since $\sinh z/z$ and $\sin z/z$ tend to one. Above the barrier, the interior wavefunction oscillates: the quantum reflection probability generally remains nonzero, but $qa=n\pi$ for a positive integer $n$ gives perfect transmission probability $T=1$ through constructive wave interference. These are scattering resonances, not below-barrier quantum tunnelling.

At half height, $E=V_0/2$, we have $k=\kappa$ and $T={\mathrm{sech}}^2(\kappa a)$, consistent with finite square barrier transmission at half barrier height after replacing this width $a$ by that example's width $2a$. In the illustration, the width is four times $\ell =\hslash /\sqrt{2m{V}_0}$ and $E/V_0=1/2$, giving $T\simeq 0.01388$. The plotted wavefunction comes from solving the four interface equations; the transmission probability curve uses the closed formula.