Finite square well: root conditions, simplicity and the bound-state count

[Self-adjointness] This is [F1].

[Energies $E\le -U$ are excluded] Let $H\psi =E\psi$ with $\mathrm{\parallel }\psi {\mathrm{\parallel }}_{L^2}=1$ and $E\le -U$. By [F3], $E=⟨H\psi ,\psi ⟩=\frac{{\mathrm{\hslash }}^2}{2m}\mathrm{\parallel }{\psi }^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2+{\int }_{[-a,a]}V\mathrm{\mid }\psi {\mathrm{\mid }}^2$ and this is at least $-U$. Thus $E=-U$ and equality holds in the lower bound, so $\mathrm{\parallel }{\psi }^{\mathrm{\prime }}{\mathrm{\parallel }}_2=0$. The absolutely continuous representative of an $H^1$ function with derivative zero is constant on every bounded interval by [F2], hence is a single constant on $\mathbb{R}$; the only constant in $L^2(\mathbb{R})$ is $0$. This contradicts $\mathrm{\parallel }\psi {\mathrm{\parallel }}_2=1$, so $E>-U$.

[Parity and matching] Let $E=-\epsilon \in (-U,0)$, $\kappa =\sqrt{2m\epsilon }\mathrm{/}\mathrm{\hslash }$, $k=\sqrt{2m(U-\epsilon )}\mathrm{/}\mathrm{\hslash }$, so $k^2+{\kappa }^2=K^2$. On an exterior interval the classical equation is ${\psi }^{\mathrm{\prime }\mathrm{\prime }}={\kappa }^2\psi$. Put $g={\psi }^{\mathrm{\prime }}-\kappa \psi$; then $g^{\mathrm{\prime }}=-\kappa g$, so $(e^{\kappa x}g{)}^{\mathrm{\prime }}=0$ and $g=C{e}^{-\kappa x}$. Integrating $(e^{-\kappa x}\psi {)}^{\mathrm{\prime }}=C{e}^{-2\kappa x}$ shows that every solution is $A{e}^{\kappa x}+B{e}^{-\kappa x}$. The $L^2$ condition removes the growing term on each exterior half-line, leaving multiples of $e^{-\kappa x}$ on $x>a$ and $e^{\kappa x}$ on $x<-a$. On $(-a,a)$ the equation is ${\psi }^{\mathrm{\prime }\mathrm{\prime }}=-k^2\psi$. The matrix mapping $(A,B)$ to the value and derivative of $A\cos (kx)+B\sin (kx)$ at any $x_0$ has determinant $k\mathrm{\ne }0$, so choose $A,B$ to match the data of any solution there. The difference $w$ solves $w^{\mathrm{\prime }\mathrm{\prime }}=-k^{2}w$ with zero data at $x_0$; differentiating $\mathrm{\mid }{w}^{\mathrm{\prime }}{\mathrm{\mid }}^2+k^2\mathrm{\mid }w{\mathrm{\mid }}^2$ shows this quantity is constant, hence $w=0$. Thus every interior solution is $A\cos (kx)+B\sin (kx)$. The potential is even, so the reflection operator $(\mathrm{\Pi }\psi )(x)=\psi (-x)$ commutes with $H$ on $H^2$ and maps every eigenspace to itself. Eigenspaces are at most one-dimensional: the solutions decaying as $x\to -\mathrm{\infty }$ form a one-dimensional space $c{e}^{\kappa x}$ ($x<-a$), and the transfer through $(-a,a)$ is invertible by the just-proved initial-data representation, so at most one direction matches the decaying solution at $+\mathrm{\infty }$. Hence $\mathrm{\Pi }\psi =\pm \psi$: every eigenfunction is even or odd. For an even eigenfunction, $\psi (x)=A\cos (kx)$ on $(-a,a)$ and $\psi (x)=B{e}^{-\kappa \mathrm{\mid }x\mathrm{\mid }}$ for $\mathrm{\mid }x\mathrm{\mid }>a$; matching $\psi$ and ${\psi }^{\mathrm{\prime }}$ at $a$ gives $A\cos (ka)=B{e}^{-\kappa a},-Ak\sin (ka)=-\kappa B{e}^{-\kappa a},$ which eliminates $A,B$ and gives $k\tan (ka)=\kappa$. For an odd eigenfunction, $\psi (x)=A\sin (kx)$ on $(-a,a)$ and $\psi (x)=B\mathrm{sgn}(x)e^{-\kappa \mathrm{\mid }x\mathrm{\mid }}$ outside; matching gives $A\sin (ka)=B{e}^{-\kappa a}$ and $Ak\cos (ka)=-\kappa B{e}^{-\kappa a}$, i.e. $-k\cot (ka)=\kappa$. Conversely, if either condition holds with $k,\kappa >0$, $k^2+{\kappa }^2=K^2$, then the indicated piecewise function is $C^1$, piecewise $C^2$, satisfies $H\psi =E\psi$ on each constant-potential interval and decays exponentially. Its restrictions and their first derivatives are square-integrable, and its piecewise classical second derivative is square-integrable because the central interval is bounded and the exterior pieces decay exponentially. To verify the global weak derivatives, integrate against a compactly supported smooth test function separately on $(-\mathrm{\infty },-a)$, $(-a,a)$ and $(a,\mathrm{\infty })$: the boundary terms cancel at $\pm a$ for the first derivative because $\psi$ is continuous, and for the second derivative because ${\psi }^{\mathrm{\prime }}$ matches there. Thus the piecewise first and second derivatives are the global weak derivatives and belong to $L^2(\mathbb{R})$, so $\psi \in H^2(\mathbb{R})$ by One-dimensional Sobolev spaces on intervals. It is nonzero and therefore an eigenfunction. This proves (b).

[Monotone branches] For $k\in (0,K)$ put $\kappa (k):=\sqrt{K^2-k^2}>0$, a strictly decreasing continuous function. Define on each interval $B_n:=(\frac{n\pi }{2a},\frac{(n+1)\pi }{2a})$ the function $L_n(k):=\{\begin{array}{ll}k\tan (ka),& n\text{ even},\\ -k\cot (ka),& n\text{ odd}\mathrm{.}\end{array}$ Each $L_n$ is strictly increasing on $B_n$: for $n$ even, $L_n^{\mathrm{\prime }}(k)=\tan (ka)+ka{\sec }^2(ka)>0$ because $ka\in (n\pi \mathrm{/}2,(n+1)\pi \mathrm{/}2)$ with $n$ even lies in a range where $\tan (ka)>0$; for $n$ odd, $L_n(k)=-k\cot (ka)$ with $\cot (ka)<0$ and $L_n^{\mathrm{\prime }}(k)=-\cot (ka)+ka{\csc }^2(ka)>0$. Moreover, ${\lim }_{k\to (\frac{n\pi }{2a}{)}^{+}}{L}_n(k)=0^{+}$ and ${\lim }_{k\to (\frac{(n+1)\pi }{2a}{)}^{-}}{L}_n(k)=+\mathrm{\infty }$ for every $n\ge 0$.

[Energies $E\ge 0$ are excluded] Let $H\psi =E\psi$ with $\psi \mathrm{\ne }0$ and $E\ge 0$. By [F2], on $(a,\mathrm{\infty })$ the classical equation is ${\psi }^{\mathrm{\prime }\mathrm{\prime }}=-k_0^2\psi$ with $k_0=\sqrt{2mE}\mathrm{/}\mathrm{\hslash }\ge 0$ (and ${\psi }^{\mathrm{\prime }\mathrm{\prime }}=0$ when $E=0$). If $E>0$, the classification proved in step 1.3 applies with $k=k_0$ and gives $\psi (x)=A\cos (k_{0}x)+B\sin (k_{0}x)$. The identity $\mathrm{\mid }\psi {\mathrm{\mid }}^2=\mathrm{\mid }A{\mathrm{\mid }}^2{\cos }^2(k_{0}x)+\mathrm{\mid }B{\mathrm{\mid }}^2{\sin }^2(k_{0}x)+2\mathrm{Re}(A\bar{B})\sin (k_{0}x)\cos (k_{0}x)$ and the antiderivatives of ${\cos }^2$, ${\sin }^2$ and $\sin \cos$ give ${\int }_a^X\mathrm{\mid }\psi {\mathrm{\mid }}^2=\frac{\mathrm{\mid }A{\mathrm{\mid }}^2+\mathrm{\mid }B{\mathrm{\mid }}^2}{2}(X-a)+O(1)$ as $X\to \mathrm{\infty }$. Thus $L^2$ membership forces $A=B=0$, so $\psi =0$ on $(a,\mathrm{\infty })$. If $E=0$, every solution of ${\psi }^{\mathrm{\prime }\mathrm{\prime }}=0$ is affine, and an affine $L^2$ function vanishes identically, so again $\psi =0$ there. The same argument on $(-\mathrm{\infty },-a)$ shows $\psi =0$ outside $[-a,a]$. The global $C^1$ representative then has $\psi (a)={\psi }^{\mathrm{\prime }}(a)=0$, and on $(-a,a)$ it solves ${\psi }^{\mathrm{\prime }\mathrm{\prime }}=-\frac{2m(E+U)}{{\mathrm{\hslash }}^2}\psi$ with these terminal data. After translating $(-a,a)$ to a compact interval, uniqueness in Regular interval IVP: existence, a priori bound and entire spectral-parameter dependence makes it vanish identically, so $\psi =0$, a contradiction. Hence $E<0$, and with step 1.2, $E\in (-U,0)$.

[Existence and uniqueness of roots] Fix $n\ge 0$ with $\frac{n\pi }{2a}<K$, and consider $F_n:=L_n-\kappa$ on $B_n\cap (0,K)$, a strictly increasing continuous function. Near the left endpoint, $L_n\to 0^{+}$ while $\kappa >0$ (since the left endpoint is $<K$), so $F_n<0$ there; put $r_n=(n+1)\pi \mathrm{/}(2a)$. If $r_n\le K$, the one-sided limit $L_n(k)\to +\mathrm{\infty }$ as $k\uparrow r_n$ makes $F_n>0$ near that endpoint, including when $r_n=K$; no value of $L_n$ at its pole is taken. If $K<r_n$, then $K$ is interior to the branch, $L_n(K)>0$ is well defined, and $\kappa (k)\to 0$ as $k\uparrow K$, again giving $F_n>0$. Hence by the intermediate value theorem $F_n$ has a zero, and strict monotonicity makes it unique and interior. If instead $\frac{n\pi }{2a}\ge K$, then $k<K$ on $B_n$ never holds, so no root. On each $B_n$ the other parity expression is negative and cannot equal $\kappa >0$. At branch endpoints the matching equations of step 1.3 either require a pole denominator to be nonzero or give a zero left side, again excluding a root with $\kappa >0$. Since $\{B_n:n\ge 0\}$ covers $(0,\mathrm{\infty })$ except the endpoints, these are all roots in $(0,K)$; the number of $n\ge 0$ with $\frac{n\pi }{2a}<K$ is $\lceil 2aK\mathrm{/}\pi \rceil$, which is (d).

[Assembly] By steps 1.2, 2.1 and 1.3 every negative eigenvalue has one of the two root forms, with simple eigenspace, and by step 2.2 the roots are exactly the described ones with the stated count. Steps 1.1, 1.2, 1.3, 2.1 and 2.2 give (a)-(d).