Qm Three Dimensional Scattering — Examples

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.

Since $E_r>0$, the theorem can be evaluated at the physical energy $E_r$, giving ${\sin }^2{\delta }_0(E_r)=L(E_r)=1$ and ${\sigma }_0(E_r)=4\pi \mathrm{/}{k}_r^2=4\pi {\mathrm{\hslash }}^2\mathrm{/}(2m{E}_r)$ in ${\mathrm{m}}^2$. The real-axis factor has $L(E_r\pm \mathrm{\Gamma }\mathrm{/}2)=1\mathrm{/}2$. The upper point is always positive; the lower point is positive exactly if $\mathrm{\Gamma }<2E_r$, equals the excluded boundary $0$ if $\mathrm{\Gamma }=2E_r$, and is negative if $\mathrm{\Gamma }>2E_r$. Only positive points have physical phase and cross-section readings.

For every physical $E>0$, with $x:=E-E_r$ and $y:=\mathrm{\Gamma }\mathrm{/}2>0$, dividing by the positive resonance-centre value yields ${\sigma }_0(E)\mathrm{/}{\sigma }_0(E_r)=(k_r\mathrm{/}k(E){)}^2{y}^2\mathrm{/}(x^2+y^2)=(E_r\mathrm{/}E)L(E)$. At the always-positive upper point the ratio is $E_r\mathrm{/}(2(E_r+y))$; at the lower point it is $E_r\mathrm{/}(2(E_r-y))$ only when $E_r-y>0$. If that lower point is zero or negative, the real-axis identity $L(E_r-y)=1\mathrm{/}2$ is still valid but no physical ratio is evaluated there. The factor $E_r\mathrm{/}E$ is energy dependent and need not vary slowly for arbitrary $\mathrm{\Gamma }$.

Take the declared parameters $E_r=4\mathrm{\Gamma }$ (both in joules). Then at $E=E_r+\mathrm{\Gamma }\mathrm{/}2=(9\mathrm{/}2)\mathrm{\Gamma }$ the Lorentzian factor is $1\mathrm{/}2$ and the ratio of step 2.1 is $(4\mathrm{/}(9\mathrm{/}2))(1\mathrm{/}2)=4\mathrm{/}9$; at $E=(7\mathrm{/}2)\mathrm{\Gamma }$ the factor is again $1\mathrm{/}2$ and the ratio is $(4\mathrm{/}(7\mathrm{/}2))(1\mathrm{/}2)=4\mathrm{/}7$. The first value is strictly below $1\mathrm{/}2$ and the second strictly above, so this is the explicit asymmetry produced by the $1\mathrm{/}E$ factor, so these factor half-height locations are not the locations where ${\sigma }_0(E)={\sigma }_0(E_r)\mathrm{/}2$; no maximum of ${\sigma }_0$ is used in this comparison.

On the full real axis, $L$ reaches $1$ at $E_r$ and $1\mathrm{/}2$ exactly at $E_r\pm \mathrm{\Gamma }\mathrm{/}2$, so its full width at half maximum is $\mathrm{\Gamma }$. On the positive physical axis both half-height locations are present exactly when $\mathrm{\Gamma }<2E_r$; for $\mathrm{\Gamma }\ge 2E_r$ the lower location is excluded. This qualifies the physical interpretation without narrowing the allowed $E_r>0$, $\mathrm{\Gamma }>0$ parameters or changing the real-axis width identity. The example of step 3.1 keeps both half-height points positive and shows that the cross-section ratios there differ from $1\mathrm{/}2$.

By the Born theorem, $f_B(\mathrm{\Omega })=-\frac{1}{4\pi }{\int }_{B_R}{e}^{i{\mathbf{q}}_{\mathrm{t}\mathrm{r}}\cdot y}W(y)dy$ with $W=V_{0}2m\mathrm{/}{\mathrm{\hslash }}^2$ constant on $B_R$; in polar coordinates the ball integral is ${\int }_{B_R}{e}^{iq\cdot y}dy=2\pi {\int }_0^R{\int }_{-1}^1{e}^{iqrs}{r}^{2}dsdr=4\pi (\sin (qR)-qR\cos (qR))\mathrm{/}{q}^3$ for $q=q_{\mathrm{t}\mathrm{r}}>0$, and equals $4\pi R^3\mathrm{/}3$ at $q=0$ by continuity.

The kernel $G_k(x-y)W(y)$ on $B_R\times B_R$ has squared Hilbert-Schmidt norm ${\int }_{B_R}{\int }_{B_R}\mathrm{\mid }{G}_k(x-y)W(y){\mathrm{\mid }}^{2}dxdy=\frac{\mathrm{\mid }{V}_0{\mathrm{\mid }}^2(2m\mathrm{/}{\mathrm{\hslash }}^2{)}^2}{16{\pi }^2}{\int }_{B_R}{\int }_{B_R}\mathrm{\mid }x-y{\mathrm{\mid }}^{-2}dxdy$; for each $y$ the inner integral over $x\in B_R$ is at most the integral over the ball of radius $2R$ around $y$, which is $8\pi R$. Hence $\mathrm{\parallel }{K}_k\mathrm{\parallel }\le \mathrm{\parallel }{K}_k{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}\le \frac{2m\mathrm{\mid }{V}_0\mathrm{\mid }}{{\mathrm{\hslash }}^2}{R}^2\sqrt{2\mathrm{/}3}=:q_{\mathrm{H}\mathrm{S}}$.

Therefore $f_B(\mathrm{\Omega })=-\frac{2m{V}_0}{{\mathrm{\hslash }}^2}\cdot \frac{\sin (q_{\mathrm{t}\mathrm{r}}R)-q_{\mathrm{t}\mathrm{r}}R\cos (q_{\mathrm{t}\mathrm{r}}R)}{q_{\mathrm{t}\mathrm{r}}^3}$ for $q_{\mathrm{t}\mathrm{r}}>0$, with the forward value $f_B(\theta =0)=-\frac{2m{V}_0}{{\mathrm{\hslash }}^2}\frac{R^3}{3}$; the amplitude has the dimension of length, since $2m{V}_0\mathrm{/}{\mathrm{\hslash }}^2$ is in ${\mathrm{m}}^{-2}$ and the bracket in ${\mathrm{m}}^3$.

Consequently, if $\frac{2m\mathrm{\mid }{V}_0\mathrm{\mid }}{{\mathrm{\hslash }}^2}{R}^2\sqrt{2\mathrm{/}3}<1$, then $q=\mathrm{\parallel }{K}_k\mathrm{\parallel }\le q_{\mathrm{H}\mathrm{S}}<1$ and the controlled Born estimate of the theorem applies; since $t\mapsto t\mathrm{/}(1-t)$ is increasing on $[0,1)$ and $\mathrm{\parallel }{e}_{\mathrm{i}\mathrm{n}\mathrm{c}}{\mathrm{\parallel }}_{L^2(B_R)}=\mathrm{\mid }{B}_R{\mathrm{\mid }}^{1\mathrm{/}2}$, the uniform error bound is at most $\mathrm{\mid }{f}_k(\mathrm{\Omega })-f_B(\mathrm{\Omega })\mathrm{\mid }\le \frac{\mathrm{\parallel }W{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }{B}_R\mathrm{\mid }}{4\pi }\frac{q_{\mathrm{H}\mathrm{S}}}{1-q_{\mathrm{H}\mathrm{S}}}=\frac{2m\mathrm{\mid }{V}_0\mathrm{\mid }}{{\mathrm{\hslash }}^2}\frac{R^3}{3}\frac{q_{\mathrm{H}\mathrm{S}}}{1-q_{\mathrm{H}\mathrm{S}}}$ in metres; this is an explicit condition on the declared strength, radius, mass and action constant, verified here without any numerical value of the constants.

Under the same condition the beam model reads $d\sigma \mathrm{/}d\mathrm{\Omega }(0)=\mathrm{\mid }{f}_k(0){\mathrm{\mid }}^2$, and the theorem controls the difference from $\mathrm{\mid }{f}_B(0){\mathrm{\mid }}^2$ through the bound of 2.2; the pair asserts the display $\mathrm{\mid }{f}_B(0){\mathrm{\mid }}^2=(\frac{2m{V}_0}{{\mathrm{\hslash }}^2}{)}^2\frac{R^6}{9}$ as the Born forward reading, read in ${\mathrm{m}}^2\mathrm{s}{\mathrm{r}}^{-1}$, and no measured value.

Put $V_{\mathrm{l}\mathrm{o}\mathrm{c}}:=V_{g,\mu }{\mathbf{1}}_{\{\mathrm{\mid }x\mathrm{\mid }<L_{\ast }\}}$ and $V_{\mathrm{t}\mathrm{a}\mathrm{i}\mathrm{l}}:=V_{g,\mu }{\mathbf{1}}_{\{\mathrm{\mid }x\mathrm{\mid }\ge L_{\ast }\}}$. The local part is supported in the closed radius-$L_{\ast }$ ball, and polar integration gives $\int \mathrm{\mid }{V}_{\mathrm{l}\mathrm{o}\mathrm{c}}{\mathrm{\mid }}^{2}dx=4\pi g^2{\int }_0^{L_{\ast }}{e}^{-2\mu r}dr\le 4\pi g^2{L}_{\ast }<\mathrm{\infty }$. Thus it is compactly supported $L^2$, while the tail is bounded by $g\mathrm{/}{L}_{\ast }$, in joules.

For $s=\mathrm{\mid }x\mathrm{\mid }\mathrm{/}{L}_{\ast }\ge 1$, $\mathrm{\mid }{V}_{\mathrm{t}\mathrm{a}\mathrm{i}\mathrm{l}}(x)\mathrm{\mid }\le (g\mathrm{/}{L}_{\ast })e^{-{\mu }_{\ast }s}$. For any dimensionless $\delta >0$, choose an integer $N>1+\delta$; for $s\ge 1$, $(1+s{)}^{1+\delta }\le (2s{)}^N$, so The exponential dominates every fixed nonnegative integer power at $+\mathrm{\infty }$ at $a={\mu }_{\ast }$ and $m=N$ gives $e^{-{\mu }_{\ast }s}(1+s{)}^{1+\delta }\to 0$. This and continuity on compact intervals give the finite dimensionless constant $C_{\delta }:={\sup }_{s\ge 0}{e}^{-{\mu }_{\ast }s}(1+s{)}^{1+\delta }$. Hence $\mathrm{\mid }{V}_{\mathrm{t}\mathrm{a}\mathrm{i}\mathrm{l}}(x)\mathrm{\mid }\le (g\mathrm{/}{L}_{\ast })C_{\delta }(1+\mathrm{\mid }x\mathrm{\mid }\mathrm{/}{L}_{\ast }{)}^{-1-\delta }$ on that region; on $\mathrm{\mid }x\mathrm{\mid }<L_{\ast }$ the tail is zero, so the bound holds everywhere.

By 1.1 and 2.1 the declared Yukawa potential has the decaying-tail form of Asymptotic completeness for the resolvent-tail short-range class, with the decaying-tail example and the explicit exclusion of the Coulomb tail clause (c) with any $\delta >0$; hence, by that theorem, the model's interacting operator is self-adjoint on $W^{2,2}$, its wave operators exist and are asymptotically complete with vanishing singular continuous spectrum, and its scattering operator is unitary.

For each $x\mathrm{\ne }0$, as $\mu \to 0^{+}$ the potential $V_{g,\mu }(x)$ tends to $-g\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$; assigning any value to the limiting Coulomb potential at the origin gives the same multiplication operator. For that limiting potential, and for every split into a compactly supported part and a remainder, the quotient $(1+\mathrm{\mid }x\mathrm{\mid }\mathrm{/}{L}_{\ast }{)}^{1+\delta }\mathrm{\mid }{V}_{\mathrm{t}\mathrm{a}\mathrm{i}\mathrm{l}}(x)\mathrm{\mid }=g(1+\mathrm{\mid }x\mathrm{\mid }\mathrm{/}{L}_{\ast }{)}^{1+\delta }\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ tends to $+\mathrm{\infty }$ as $\mathrm{\mid }x\mathrm{\mid }\to \mathrm{\infty }$ for every $\delta >0$, so the sufficient decay hypothesis of that theorem fails and the Coulomb case is not covered by the short-range class; its treatment is the separate declared long-range model The declared long-range Coulomb model and its Dollard-modified free comparison, for which the theorem supplies no ordinary-comparison statement.