S-wave phase shift of a declared spherical square well, with its low-energy limit and hard-sphere comparison

For $r<a$ the reduced equation is $u_0^{\mathrm{\prime }\mathrm{\prime }}=-K^2{u}_0$, so its nonzero real regular solution is $u_0(r)=A\sin (Kr)$, $A\mathrm{\ne }0$, with $u_0(0)=0$; the normalization in [F1] corresponds to $A=1\mathrm{/}K$. For $r>a$ a nonzero real solution has the form $B\sin (kr+{\delta }_0)$ with real $B\mathrm{\ne }0$. The regular solution and its derivative are absolutely continuous locally through the bounded step potential; integrating the radial equation across $a$ gives no jump of $u_0^{\mathrm{\prime }}$ or $u_0$. Thus $A\sin (Ka)=B\sin (ka+{\delta }_0)$ and $AK\cos (Ka)=Bk\cos (ka+{\delta }_0)$.

Put $X:=K\cos (Ka)$, $Y:=k\sin (Ka)$ and $z:=X+iY$. Since $k,K>0$ and sine and cosine cannot both vanish, $X^2+Y^2>0$. With $\theta :=ka+{\delta }_0$, the matching equations give $Bk\cos \theta =AX$, $Bk\sin \theta =AY$, hence $e^{2i\theta }=z\mathrm{/}\bar{z}$ regardless of the real sign of $A\mathrm{/}B$. Consequently $S_0(k)=e^{-2ika}(K\cos (Ka)+ik\sin (Ka))\mathrm{/}(K\cos (Ka)-ik\sin (Ka))$ and ${\delta }_0=\arg z-ka$ modulo $\pi$, with a nonzero denominator for every positive energy and depth. This is also the actual Wronskian convention of [F1], since outside the well $W(u_0,e^{ikr})=-Bk{e}^{-i{\delta }_0}$. Where $\cos (Ka)\mathrm{\ne }0$, the equivalent local formula is $\tan (ka+{\delta }_0)=(k\mathrm{/}K)\tan (Ka)$; the complex formula remains valid when either matching value vanishes.

Fix $U>0$ and put $\gamma :=\sqrt{2mU}\mathrm{/}\mathrm{\hslash }$, so $K=\sqrt{{\gamma }^2+k^2}$ and $K-\gamma =k^2\mathrm{/}(K+\gamma )=O(k^2)$. If $\cos (\gamma a)\mathrm{\ne }0$, choose the representative of $\theta$ tending to zero modulo $\pi$. The tangent is smooth near $\gamma a$, so $\tan \theta =(k\mathrm{/}K)\tan (Ka)=k\tan (\gamma a)\mathrm{/}\gamma +O(k^3)$. By [F2], $\tan \theta =\theta +O({\theta }^3)$ near zero; first $\theta =O(k)$ and then $\theta =k\tan (\gamma a)\mathrm{/}\gamma +O(k^3)$. Hence ${\delta }_0=-k{a}_{\mathrm{e}\mathrm{f}\mathrm{f}}+O(k^3)$ modulo $\pi$, where $a_{\mathrm{e}\mathrm{f}\mathrm{f}}:=a-\tan (\gamma a)\mathrm{/}\gamma$, and [F1] gives ${\sigma }_0\to 4\pi a_{\mathrm{e}\mathrm{f}\mathrm{f}}^2$. The expansion uses locally bounded derivatives only where $\cos (\gamma a)\mathrm{\ne }0$.

At an exact threshold $\gamma a={\alpha }_n:=(n+\frac{1}{2})\pi$, $n\ge 0$, write $s_n:=\sin {\alpha }_n=(-1{)}^n$. The identity for $K-\gamma$ and [F2] give $K-\gamma =k^2\mathrm{/}(2\gamma )+O(k^4)$, $X=K\cos (Ka)=-s_{n}a{k}^2\mathrm{/}2+O(k^4)$ and $Y=k\sin (Ka)=s_{n}k+O(k^5)$. Thus $X\mathrm{/}Y=-ak\mathrm{/}2+O(k^3)$ and the angle of $X+iY$ tends to $\pi \mathrm{/}2$ modulo $\pi$. Writing $\theta =\pi \mathrm{/}2+\epsilon$ modulo $\pi$, $\cot \theta =-\epsilon +O({\epsilon }^3)$ gives $\epsilon =ak\mathrm{/}2+O(k^3)$, so ${\delta }_0=\pi \mathrm{/}2-ak\mathrm{/}2+O(k^3)$ modulo $\pi$. Therefore $k^2{\sigma }_0=4\pi {\sin }^2{\delta }_0\to 4\pi$; the finite-scattering-length formula of step 3.1 does not apply at this depth. At zero energy the regular reduced solution is proportional to $\sin (\gamma r)$ inside and to the nonzero constant $s_n$ outside, with matching derivative zero at $a$. Its spatial form is proportional to $1\mathrm{/}r$ outside and is not square integrable by [F3]; this is the zero-energy s-wave resonance meant here.

In the finite-length branch, as $\gamma a\uparrow {\alpha }_n$ from below, $\tan (\gamma a)\to +\mathrm{\infty }$ and $a_{\mathrm{e}\mathrm{f}\mathrm{f}}\to -\mathrm{\infty }$; from above $a_{\mathrm{e}\mathrm{f}\mathrm{f}}\to +\mathrm{\infty }$. Thus the iterated low-energy reading $4\pi a_{\mathrm{e}\mathrm{f}\mathrm{f}}^2$ diverges on either approach, while the exact-threshold law is step 4.1. Along nonresonant depth regimes with $\tan (\gamma a)\mathrm{/}\gamma \to 0$, $a_{\mathrm{e}\mathrm{f}\mathrm{f}}\to a$ and the iterated low-energy reading tends to $4\pi a^2$, four times the geometric area $\pi a^2$. In particular at $\gamma a=n\pi$, $n\ge 1$, $a_{\mathrm{e}\mathrm{f}\mathrm{f}}=a$ and ${\delta }_0=-ka+O(k^3)$; the interior node tends to the boundary. This comparison holds along the stated depth regimes, rather than for every sequence of large depths.

The bound-state interpretation has a separate negative-energy proof. Write $g:=\gamma a$. For ${\alpha }_n<g<(n+1)\pi$, define $f_g(x):=x\cot x+\sqrt{g^2-x^2}$ on $[{\alpha }_n,g]$. It is continuous, $f_g({\alpha }_n)=\sqrt{g^2-{\alpha }_n^2}>0$ and $f_g(g)=g\cot g<0$; [F2] gives a root $x_g\in ({\alpha }_n,g)$. On that open interval $f_g^{\mathrm{\prime }}=\cot x-x{\csc }^{2}x-x\mathrm{/}\sqrt{g^2-x^2}<0$, so the root is unique. Put $q:=x_g\mathrm{/}a$, $\kappa :=\sqrt{g^2-x_g^2}\mathrm{/}a>0$ and $u(r):=\sin (qr)$ for $r\le a$, $u(r):=\sin (qa)e^{-\kappa (r-a)}$ for $r\ge a$. Then $q^2+{\kappa }^2={\gamma }^2$ and $q\cot (qa)=-\kappa$, so both $u$ and $u^{\mathrm{\prime }}$ match at $a$. The spatial function $\psi (x):=u(\mathrm{\mid }x\mathrm{\mid })\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$, with value $q$ at zero, is nonzero and square integrable by [F3]; it is smooth near zero, decays exponentially at infinity and is piecewise smooth with matching first derivatives at $a$. Its weak derivatives through order two are square integrable, with no boundary delta term, hence $\psi \in W^{2,2}$. The radial identity $\mathrm{\Delta }(u(r)\mathrm{/}r)=u^{\mathrm{\prime }\mathrm{\prime }}(r)\mathrm{/}r$ for $r>0$ gives $H\psi =-c{\kappa }^2\psi$, $c={\mathrm{\hslash }}^2\mathrm{/}(2m)$, so normalization produces a genuine s-wave bound eigenvector. As $g\downarrow {\alpha }_n$, $0<\kappa a\le \sqrt{g^2-{\alpha }_n^2}\to 0$, while for $g\le {\alpha }_n$ there is no matching root in this branch $x>{\alpha }_n$ with $x\le g$. Thus a new negative-energy s-wave branch appears just above each stated threshold, whose endpoint is the non-square-integrable zero-energy resonance in step 4.1; no assertion about other angular channels or the full bound-state count is needed.

For the free consistency case $U=0$, $K=k$ and step 2.1 gives $S_0=e^{-2ika}(\cos ka+i\sin ka)\mathrm{/}(\cos ka-i\sin ka)=1$, so ${\delta }_0=0$ modulo $\pi$ at every positive energy, including zeros of sine or cosine. At any positive-depth energy with $Ka=n\pi$, $n\ge 1$, one has $Y=0$ and $X\mathrm{\ne }0$, hence $S_0=e^{-2ika}$ and ${\delta }_0=-ka$ modulo $\pi$: this is exactly the exterior phase for the vanishing radial boundary condition of a hard sphere at that energy. If $\cos (Ka)=0$, the same formula gives $S_0=-e^{-2ika}$ and ${\delta }_0=\pi \mathrm{/}2-ka$ modulo $\pi$. These checks complete the full positive-energy, zero-depth and matching-node cases together with the two low-energy branches above.