Are all bound states normalizeable?

Question

Following Griffiths eq. (2.91) on p. 52 one may define a bound state to be an energy eigenstate $$H|E\rangle=E|E\rangle\tag{1}$$ with an energy being smaller than the potential far away from the origin in the sense $$\lim_{x\rightarrow\pm\infty} V(x)>E.\tag{2}$$ Intuitively we want bound states $|E\rangle$ to be states with a finite spread in position space and to be normalizeable. This means the probability density $|\langle x|E\rangle|^2=|\psi_E(x)|^2$ should vanish at infinity. Is the requirement (2) enough to guarantee this? This Wikipedia article claims this.

Their reasoning seems flawed to me though: Rewriting eq. (1) in position basis $$\frac{\hbar^2}{2m}\psi_E''(x)=\left(V(x)-E\right)\psi_E(x) \tag{3}$$ they directly deduce that $\psi_E(x)$ must decay exponentially for large $x$ as $V(x)-E\gt 0$ for all $x$ bigger than some $X$.

I fail so see why this is true.

So let me expand my question - What really are bound states by definition? Griffiths approach seems to be a too weak condition.

Answer 1

TL;DR: OP has a point. Griffiths' bound state condition $$E~<~ V(\pm\infty) \tag{2.91} $$ on p. 52 is incomplete. The solution $\psi$ to the 1D TISE will be a linear combination of exponentially growing and exponentially damped solutions in the asymptotic regions. One needs to impose an extra condition to get rid of the non-normalizable exponentially growing solution. A heavy-handed (although standard) condition (which Griffiths presumably implicitly has in mind) would be to declare that $\psi$ should be normalizable from the onset. However, it turns out that it is enough to e.g. impose an asymptotic bound (C) on $\psi$, cf. the following Theorem.

Theorem. Let there be given the 1D TISE $$\psi^{\prime\prime}(x)~=~\frac{2m}{\hbar^2}(V(x)-E)\psi(x). \tag{A}$$ If$^1$ $$ \exists k,K>0\forall |x|\geq K:~~ {\rm Re}\left[\frac{2m}{\hbar^2}(V(x)-E)\right] ~\geq~ k^2,\tag{B}$$ and if $\psi$ is asymptotically bounded $$ \exists c,K>0\forall |x|\geq K:~~ |\psi(x)|~\leq ~c,\tag{C} $$ then $\psi$ is exponentially damped: $$\exists C>0 \forall |x|\geq K:~~ |\psi(x)| ~\leq~Ce^{-k|x|}, \tag{D} $$ and hence normalizable/square integrable.

Sketched proof of Theorem:

1. From fairly mild assumptions about the potential $V$, we may assume that the solution $\psi$ is sufficiently many times differentiable, cf. e.g. my Phys.SE answer here.

2. Let us use polar decomposition $$ \psi(x)~=~R(x)e^{i\Theta(x)}. \tag{E} $$ Note that the function $R$ might not be differentiable at its zeros $R(x)=0$, although $R=|\psi|$ is still continuous everywhere.

3. For $|x|\geq K$ away from the zeros $R(x)\neq 0$, we calculate: $$\begin{align} \frac{R^{\prime\prime}(x)}{R(x)}~\geq~&\frac{R^{\prime\prime}(x)-R(x)\Theta^{\prime}(x)^2}{R(x)}\cr ~\stackrel{(E)}{=}~&{\rm Re}\frac{\psi^{\prime\prime}(x)}{\psi(x)}\cr ~\stackrel{(A)}{=}~&{\rm Re}\left[\frac{2m}{\hbar^2}(V(x)-E)\right]\cr ~\stackrel{(B)}{\geq}~& k^2. \end{align} \tag{F} $$ So $R$ is concave upward, $R^{\prime\prime}>0$.

4. It follows that $R$ can not have more than 1 zero in each of the asymptotic regions $]-\infty,-K]$ and $[K,\infty[$. By possibly choosing a bigger $K>0$, we may assume that there are no zeros in the asymptotic regions: $$\forall |x|\geq K:~~ \psi(x)\neq 0. \tag{G}$$

5. Together with the bounds (C), $R^{\prime\prime}>0$ means that $$\begin{align} \forall x\geq K:~~ R^{\prime}(x)~\leq~& 0,\cr \forall x\leq -K:~~ R^{\prime}(x)~\geq~& 0. \end{align}\tag{H}$$

6. Let us define $$S(x)~:=~\ln R(x). \tag{I} $$ Eq. (H) becomes $$\begin{align} \forall x\geq K:~~ S^{\prime}(x)~\leq~& 0,\cr \forall x\leq -K:~~ S^{\prime}(x)~\geq~& 0. \end{align}\tag{J}$$ Eq. (F) transforms into a Riccati inequality: $$\forall |x|\geq K:~~ S^{\prime\prime}(x) +S^{\prime}(x)^2~\geq~ k^2.\tag{K} $$

7. One may now show that $$\begin{align} \forall x\geq K:~~ S^{\prime}(x)~\leq~& -k,\cr \forall x\leq -K:~~ S^{\prime}(x)~\geq~& k. \end{align}\tag{L}$$ Indirect proof of eq. (L): Assume the opposite: $$\begin{align} \exists x_1\geq K:~~ S^{\prime}(x_1)~>~& -k,\cr \exists x_1\leq -K:~~ S^{\prime}(x_1)~<~& k. \end{align}\tag{M}$$ Then we may rewrite the Riccati ineq. (K) as $$ \frac{1}{k}\frac{d}{dx}{\rm artanh}\frac{S^{\prime}(x)}{k} ~=~\frac{S^{\prime\prime}(x)}{k^2-S^{\prime}(x)^2} ~\stackrel{(K)}{\geq}~ 1\tag{N} $$ in a neighborhood of $x_1$. In particular we note that $S^{\prime\prime}(x)>0$, i.e. $S^{\prime}$ is an increasing function. Therefore, in the 2 cases of eq. (M), we may let $x_1\to\pm\infty$, respectively, without violating eq. (M). Integration of ineq. (N) leads to $$ {\rm artanh}\frac{S^{\prime}(x)}{k} ~\left\{\begin{array}{c}\geq\cr\leq\end{array}\right\}~ k(x\!-\!x_0),\tag{O} $$ or equivalently, $$S^{\prime}(x) ~\left\{\begin{array}{c}\geq\cr\leq\end{array}\right\}~ k\tanh k(x\!-\!x_0),\tag{P} $$ in a neighborhood of $x_1$ in the 2 cases, respectively. However, ineq. (P) would violate ineq. (J) in the limit $x\to\pm\infty$, respectively. Contradiction. Hence ineq. (L) holds.

8. Integration of ineq. (L) leads to $$ \forall |x|\geq K\exists C>0:~~ S(x) ~\leq~-k|x|+\ln C. \tag{Q} $$ Now apply the exponential function on each side of the ineq. (Q) to derive the sought-for ineq. (D). $\Box$

--

$^1$ We allow $V(x)-E$ to be complex to include the case of unstable/decaying or resonant states.

Answer 2

I think Griffiths means that we assume that \begin{equation} \lim_{x \rightarrow \infty} V(x) = V_R > E \end{equation} i.e. that the potential tends to a constant value $V_R$ that is larger than $E$. In this case we can prove it as follows:

As OP writes, this means that for some $X_R$ and some $\epsilon > 0$, we have \begin{equation} | V(x) - V_R | < \epsilon \quad \forall x > X_R \end{equation} Let us choose $X_R$ large enough so that \begin{equation} V(x) > E \quad \forall x>X_R \end{equation} and approximate $V(x)$ by $V_R$ in this region. The error will be $\epsilon$, which we can make as small as we want by choosing $X_R$ large enough (but finite).

In this case, the Schrödinger equation reads \begin{equation} -\frac{\hbar^2}{2m}\psi_E''(x) + V_R \psi_E(x) = E\psi_E(x) \quad \forall x>X_R \end{equation} which means

\begin{equation} \psi_E''(x) = \frac{2m}{\hbar^2} (V_R - E) \psi_E(x) \quad \forall x>X_R \end{equation} Since $V_R - E > 0$ by assumption, this is a differential equation with solution \begin{equation} \psi_E (x) = C_1 \exp \left(x \sqrt{\frac{2m(V_R - E)}{\hbar^2}} \right) + C_2 \left(-x \sqrt{\frac{2m(V_R - E)}{\hbar^2}} \right) \end{equation} Unless we want something non-normalizable, we have to set $C_1 = 0$. The value of $C_2$ must be determined by the value of $\psi_E$ and $\psi_E'$ for $x<X_R$. Hence the wave function decays exponentially. The same analysis obviously holds for negative $x$, just replace $V_R$ by $V_L$, $X_R$ by $X_L$, let $x < X_L$ etc. This is just the particle in a (reeeeeeeeally wide) finite box.

That being the case, the whole wave function is normalizable because the interval $[X_L, X_R]$ is finite and the wave function decays exponentially for $x<X_L$ and $x>X_R$.

For potentials that don't have a limit when $x \rightarrow \pm \infty$, e.g. periodic potentials, I don't think we can say much. Periodic potentials are solved by Bloch's theorem and they would give non-normalizable wave functions if we did not use periodic boundary conditions to study the bulk of the lattice.