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For a stationary spherically symmetric state, set $u(r)=r\psi (r)$. The Schrödinger equation becomesA bound state must have $E<0$, while the well's lower bound gives $E>-U$ for a nonzero square-integrable eigenstate. Define $k=\sqrt{2m(U+E)}/\hslash$ and $\kappa =\sqrt{-2mE}/\hslash$. Regularity at the origin requires $u(0)=0$, so the interior cosine solution is excluded; square integrability at infinity excludes the growing exterior exponential. Matching $u$ and $u^{\prime }$ at $r=a$ therefore givesThe limit at the origin is the finite value $Ck$. Continuity of $u,u^{\prime }$ is equivalent here to continuity of $\psi ,{\psi }^{\prime }$ at the finite potential jump. If desired, the normalized state hasThe equation for $k$ implicitly determines the energy, with $k^2+{\kappa }^2=2mU/{\hslash }^2$.

Write $K=a\sqrt{2mU}/\hslash$ and $z=ka$. The matching condition is $z\cot z=-\sqrt{K^2-z^2}$. The first possible solution has $\pi /2<z<\pi$. If $K\le \pi /2$, every permitted interior $z<K$ has nonnegative cotangent and cannot match a negative exterior logarithmic derivative. If $K>\pi /2$, at $z\downarrow \pi /2$ the sum $z\cot z+\sqrt{K^2-z^2}$ is positive; it is negative either at $z\uparrow K$ when $K<\pi$, or at $z\uparrow \pi$ when $K\ge \pi$. Continuity supplies a first bound-state solution. Thus at least one spherically symmetric bound state exists exactly whenAt equality the limiting energy is zero and the exterior wave is proportional to $1/r$, whose norm diverges. Hence the inequality is strict. Deeper wells can support additional radial states; their negative-energy matching solutions have the same displayed form. This is the spherically symmetric bound state of a finite well.