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For a zero-potential interval of width $a$, an infinite wall at its left end, and constant exterior potential $U_0$ to its right, the $j$th bound state branch exists precisely when $U_0>{\mathcal{U}}_j$, for $j=0,1,\dots$. Matching the wavefunction and its derivative gives $z\cot z=-\sqrt{z_0^2-z^2}$ with $z_0=a\sqrt{2m{U}_0}/\hslash$. At equality the new threshold solution is not normalizable, so exactly one bound state corresponds to $U_0\in ({\mathcal{U}}_0,{\mathcal{U}}_1]$.

A physical-strip pole of a diagonal S-matrix can come from exchange of an existing particle in the crossed channel. For external masses $m_a,m_b$, use the momentum difference, rather than the sum, to identify the exchanged mass through the displayed invariant. Crossing symmetry turns the pole at $i\alpha$ into a direct pole at $i(\pi -\alpha )$ of the particle-antiparticle channel. For $m_B=2m\cos (u/2)$, the pole of $S_{BA}$ at $iu/2$ has $t=m^2$ and exchanges the existing charge-one particle; it is not a new charge-three bound state.

An electron hole is the absence of an electron in an otherwise filled band, described as a mobile positive-charge excitation. It can form a bound state with an electron, giving an exciton.

An exciton is a bound state of an electron and an electron hole. Coupling this material excitation strongly to a photon can form a polariton.

For $V=-A{e}^{-\mu r}/r$, the normalized exponential with scale $\alpha$ gives $E(\alpha )={\hslash }^2{\alpha }^2/(2m)-4A{\alpha }^3/(\mu +2\alpha {)}^2$. Choosing $\alpha =\mu /2$ proves binding if $\mu <Am/{\hslash }^2$. Failure to obtain a negative trial expectation outside this range does not prove absence of a bound state. At equality a small positive admixture of the scale-$\mu$ exponential has a negative cross matrix element and proves binding too.

Compare the same total number of finite-box modes in the soliton and vacuum sectors before removing the ultraviolet cutoff. A localized bound state replaces a continuum mode, and its contribution must be retained. For the Sine-Gordon kink fluctuation operator and the odd phase branch $\delta (k)=2\arctan (1/k)$, the bound translation mode replaces the free $k=0$ oscillator. Thus the complete oscillator correction is $-m/2-(m/2)\int dk\delta (k)k/[2\pi \sqrt{k^2+1}]$ in this convention. The continuum integral alone misses the finite $-m/2$ contribution.

Put $p=\sqrt{2mE}/\hslash$ and $\gamma =mk/{\hslash }^2$. Integrating the Schrodinger 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.