Finite square well

A finite square well is a bounded interval on which the potential is lower than outside, supporting matched oscillatory and exponential states.

For the finite square well $V=-V_0$ on $|x|<a$ and $V=0$ outside, with $a,V_0>0$, there is always an even bound state, however small $V_0$ is. Define $\rho =a\sqrt{2m{V}_0}/\hslash$, the dimensionless decay constant $\alpha =a\sqrt{-2mE}/\hslash$, and interior oscillation constant $\beta =a\sqrt{2m(E+V_0)}/\hslash$. Wavefunction matching at a finite potential step gives $\alpha =\beta \tan \beta$ and ${\alpha }^2+{\beta }^2={\rho }^2$.

On $0<\beta <\pi /2$, these equations reduce to $\beta /\cos \beta =\rho$. The left side is continuous, strictly increasing from zero to infinity, with derivative $\sec \beta (1+\beta \tan \beta )>0$. The intermediate value theorem gives exactly one root on this first branch for every $\rho >0$. It has $0<\beta <\rho$, hence $\alpha >0$ and $-V_0<E<0$. The matched cosine interior and exponential tails therefore give a nonzero even square-integrable function with continuous first derivative, which can be normalized. As the well becomes shallow the binding energy approaches zero but remains negative; see shallow finite-square-well energy.

If a normalized wavefunction has exterior tails $C{e}^{-q|x|}$ outside $|x|<a/2$, integration over both tails gives the title probability. For an even cosine interior $A\cos (kx)$, the remaining normalization contribution is $A^2[a/2+\sin (ka)/(2k)]$. Tail amplitudes defined at the edge rather than at the origin absorb the factor $e^{-qa/2}$, so their probability formula looks different despite describing the same state.

For a symmetric attractive well of width $a$ and depth $V_0$, the first odd bound state obeys $-k\cot (ka/2)=q>0$, where $q$ is the inverse tail length. Its lowest branch has $\pi <ka<2\pi$. Since $k^2<2m{V}_0/{\hslash }^2$, existence requires $V_0>V_{\mathrm{odd}}$; it is also sufficient by continuity of the matching equation on that branch. At equality the exterior solution has zero decay rate and is not square-integrable. Thus a well at or below this threshold has only its even ground state.

For a symmetric one-dimensional finite square well, nondegenerate bound states alternate even and odd parity, starting with an even ground state. The $n$th state counted from zero has $n$ interior nodes. Outside the well, it decays with inverse length $\sqrt{2m(U-E_n)}/\hslash$. Matching both the wavefunction and its first derivative at a finite step distinguishes it from the infinite-well approximation.

For $V(x)=0$ on $0<x<a$ and $V(x)=V_w>0$ outside, the lowest bound state is even about $a/2$, has no nodes, and has $0<E<V_w$. Write $k=\sqrt{2mE}/\hslash$ and $\kappa =\sqrt{2m(V_w-E)}/\hslash$. The interior wavefunction is proportional to $\cos [k(x-a/2)]$; the exterior tails decay exponentially. Wavefunction matching at a finite potential step gives the displayed condition. Its lowest root has $0<ka<\pi$, so $E<{\pi }^2{\hslash }^2/(2m{a}^2)$, the infinite square well ground state energy. As $V_w$ increases, the penetration length $1/\kappa$ tends to zero and that limiting energy is approached from below.

For a well of depth $V_0$ and full width $2a$, measured from zero exterior potential energy, even-state matching gives $q=k\tan ka$. If $a\sqrt{m{V}_0}/\hslash \ll 1$, then $ka\ll 1$ and $q\simeq 2m{V}_{0}a/{\hslash }^2$. Using $E_0=-{\hslash }^2{q}^2/(2m)$ gives the displayed leading binding energy. It agrees with the attractive delta-potential result for integrated strength $2a{V}_0$.

For a square well of width $L$, exterior wavenumber $k$ and interior wavenumber $q>k$, matching the wavefunction and its derivative gives the reflection probability $R=C{\sin }^2(qL)/(1+C{\sin }^2(qL))$, where $C=(q^2-k^2{)}^2/(4k^2{q}^2)$. For example $q=3k$ and $kL=\pi /2$ give $R=16/25$, whereas zero potential gives $R=0$. Moreover deeper wells at $qL\in \pi \mathbb{Z}$ have zero reflection by resonant transmission through a square well. Reflection therefore cannot be ordered solely by pointwise ordering of attractive potentials.

For an attractive finite square well of width $a$, outside wave number $k$ and inside wave number $q$, matching the wavefunction and its first derivative at both endpoints gives the displayed transmission probability. It equals one when $qa$ is an integer multiple of $\pi$: the internal reflections cancel. For $q=2k$ the coefficient of ${\sin }^2(qa)$ is $9/16$.

A spherical square well is a central potential equal to a negative constant inside a sphere and zero outside it. For an S-wave scattering state with exterior wave number $k$, the interior wave number obeys ${\kappa }^2=k^2+2m{V}_0/{\hslash }^2$. Continuity of the reduced radial wave and its derivative gives $\kappa \cot (\kappa a)=k\cot (ka+{\delta }_0)$, relating the depth to the scattering phase shift.

For binding energy $\epsilon >0$, match the inner reduced radial wave $\sin (Kr)$ to an outer decaying exponential. The resulting equation is $K\cot (Ka)=-\chi$, with $K^2=2m(V_0-\epsilon )/{\hslash }^2$ and ${\chi }^2=2m\epsilon /{\hslash }^2$. A bound level reaches zero energy when $\kappa a=(n+1/2)\pi$. These are divergences of the spherical square-well scattering length. The exact zero-energy threshold wave is not square-integrable; a normalizable bound state appears on the more attractive side.

The exact spherical square well matching equation is $k^{\prime }\cot (k^{\prime }a)=k\cot (ka+{\delta }_0)$. For $E\ll V_0$ near a threshold peak, neglecting $ka$ relative to the chosen scattering phase shift and replacing $k^{\prime }$ by $\kappa$ gives ${\sigma }_{\mathrm{res}}\simeq (4\pi /k^2)E/[E+V_0{\cot }^2(\kappa a)]$. Its maxima approach the S-wave bound-state threshold in a spherical square well. Its formal zeros at $\kappa a=n\pi$ are outside that approximation's validity: the exact zero-energy scattering length is then $a$, whereas genuine leading-order zeros satisfy $\tan (\kappa a)=\kappa a$.

For an attractive spherical square well, let $\kappa =\sqrt{2m{V}_0}/\hslash$. At zero energy, the outer reduced radial wave is proportional to $r-a_s$ and the regular inner wave to $\sin (\kappa r)$. Matching their logarithmic derivatives gives $a_s=a-\tan (\kappa a)/\kappa$. Away from a threshold and at sufficiently low energy, the total S-wave scattering scattering cross-section tends to $4\pi a_s^2$. The length can be negative, zero or divergent.

For a spherical square well of radius $a$, depth $V_0$ and reduced mass $\mu$, the regular zero-orbital angular momentum solution has reduced radial wavefunction $U=A\sin kr$ inside and $U=B{e}^{-qr}$ outside. Wavefunction matching at a finite potential step gives $k\cot (ka)=-q$, where $k^2=2\mu (V_0+E)/{\hslash }^2$ and $q^2=-2\mu E/{\hslash }^2$. The $n$th radial branch first reaches $E=0$ at $ka=(n+1/2)\pi$, for $n=0,1,\dots$. A normalizable negative-energy state on that branch requires $V_0>V_{s,n}$. Thus exactly one such state exists for $V_{s,0}<V_0\le V_{s,1}$; the new zero-energy $s$-wave solution at the upper equality is not normalizable.

For the first $\ell =1$ branch, the zero-energy exterior solution has $U\propto r^{-1}$. The regular interior solution is $U\propto r{j}_1(kr)$, where $j_1$ is a Spherical Bessel function. Matching logarithmic derivatives gives $1+z{j}_1^{\prime }(z)/j_1(z)=-1$, or $j_0(z)=0$, with $z=a\sqrt{2\mu V_0}/\hslash$. Its first threshold is $z=\pi$, so $V_{p,0}={\pi }^2{\hslash }^2/(2\mu a^2)$. Higher values of orbital angular momentum increase the positive centrifugal potential, so none can bind below this first $p$-wave threshold. Spin-dependent nuclear forces are absent from this central-well model.

For well depth $U$ and radius $a$, let $u=r\psi$ for a regular spherically symmetric wavefunction. At energy $-U<E<0$, the interior is $u=C\sin (kr)$ and the exterior is proportional to $e^{-\kappa r}$, with $k^2=2m(U+E)/{\hslash }^2$ and ${\kappa }^2=-2mE/{\hslash }^2$. Matching gives $k\cot (ka)=-\kappa$. The first negative-energy solution exists strictly above the displayed threshold. At equality the exterior zero-energy wave behaves as $1/r$ and is not square-integrable, so the threshold is not itself a bound state.

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 $U_0\in ({\mathcal{U}}_0,{\mathcal{U}}_1]$.

For radii $0<a<b$, a hard-core spherical square-well potential has $V(r)=\infty$ for $r\le a$, $V(r)=-V_0$ for $a<r\le b$, and $V(r)=0$ for $r>b$. Its radial wavefunction vanishes at the hard-core radius and is matched continuously with its radial derivative at $r=b$.

For outer radius $b=2a$ and zero-energy well wavenumber ${\kappa }_0=\sqrt{2m{V}_0}/\hslash$, matching the S-wave scattering logarithmic derivative givesIt diverges when ${\kappa }_{0}a=(n+\frac{1}{2})\pi$, where the well has a zero-energy scattering resonance.