# Joan Lindsay Orr


# Asymptotic Decay of Bound Solutions to the Schrödinger Equation

In this page we will prove that (sufficiently regular) bound solutions to the Time Independent Schrodinger Equation which do not blow up must have exponential decay and, as result, be normalizable. My argument is inspired by Qmechanic's Physics StackExchange answer. I've fixed a few small points and added in a level of rigour appropriate to a first Analysis course.

## Main Results

The Time Independent Schrodinger Equation (TISE) is $-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + V(x)\psi(x) = E \psi(x)$ where we assume that $$V(x)$$ and $$E$$ are real.

Suppose that there is an $$A\in\RR$$ and a positive $$a$$ such that $$V(x) - E \ge a^2\hbar^2/2m$$ for all $$x\ge A$$. Let $$\psi$$ be a twice continuously differentiable solution to TISE on $$[A, +\infty)$$. Then either $$\lim_{x\rightarrow+\infty}|\psi(x)| = +\infty$$, or else $\psi(x) = O(e^{-ax})$

We shall prove the result for $$\psi$$ assumed to be real-valued, and the general complex-valued result will follow immediately by considering the real and imaginary parts of $$\psi(x)$$ separately.

Assume in everything which follows that $$\psi$$ is a real-valued, twice continuously differentiable solution to TISE. Rewrite TSIE as $\psi''(x) = \frac{2m}{\hbar^2}(V(x) - E)\psi(x) = F(x)\psi(x)$ where $$F(x)\ge a^2$$ for all $$x\ge A$$. In particular, this means that $$\psi(x)$$ and $$\psi''(x)$$ must always have the same sign (or both be zero).

Suppose that for some $$x_0\ge A$$, both $$\psi(x_0)$$ and $$\psi'(x_0)$$ are positive. Then $$\lim_{x\rightarrow+\infty}\psi(x) = +\infty$$.

Proof. The result will follow from the claim that $$\psi'(x) \ge \psi'(x_0)$$ for all $$x\ge x_0$$. Suppose instead that there are $$x > x_0$$ with $$\psi'(x) < \psi'(x_0)$$. Choose an $$0 < \alpha < \psi'(x_0)$$ for which the set $S = \{ x : x > x_0 \text{ and } \psi'(x) < \alpha \}$ is non-empty and let $$x_1 = \inf(S)$$. Clearly by continuity, $$x_1 > x_0$$ and $$\psi'(x_1)\le\alpha < \psi'(x_0)$$. By the Mean Value Theorem there is a $$x_0 < \xi < x_1$$ such that $\psi''(\xi) = \frac{\psi'(x_1) - \psi'(x_0)}{x_1 - x_0} < 0$ However since $$\psi'(x) \ge \alpha > 0$$ on $$[x_0, x_1)$$, it follows that $$\psi(x) > 0$$ on $$[x_0, x_1)$$, contrary the the fact that $$\psi(x)$$ and $$\psi''(x)$$ always have the same sign.

Suppose that $$|\psi(x)|$$ does not converge to $$+\infty$$ as $$x\rightarrow+\infty$$ and also that $$\psi(x)$$ is not eventually zero. Then $$\psi(x)$$ has no zeroes on $$[A, +\infty)$$.

As a consequence, by the Intermediate Value Theorem, $$\psi(x)$$ is always the same sign on $$[A, +\infty)$$.

Proof. Suppose there are $$x\ge A$$ for which $$\psi(x) = 0$$. However since $$\psi$$ is not eventually zero, we can find $$x_1 > A$$ such that $$\psi(x_1) \not= 0$$ but the set $S = \{ x : x < x_1 \text{ and } \psi(x) \le 0 \}$ is non-empty. Let $$x_0 = \sup(S)$$. Replacing $$\psi$$ with $$-\psi$$ if necessary, we may assume that $$\psi(x_1) > 0$$. By continuity, $x_0 < x_1 \text{ and } \psi(x_0) = 0$ By the Mean Value Theorem, there is a $$\xi$$ in $$(x_0, x_1)$$ such that $\psi'(\xi) = \frac{\psi(x_1) - \psi(x_0)}{x_1 - x_0} > 0$ and of course $$\psi(\xi) > 0$$ by the maximality of $$x_0$$. Thus by the previous lemma, $$\lim_{x\rightarrow+\infty}\psi(x) = +\infty$$, contrary to supposition.

Suppose that $$|\psi(x)|$$ does not converge to $$+\infty$$ as $$x\rightarrow+\infty$$. Then $$\psi(x) = O(e^{-ax})$$ as $$x\rightarrow+\infty$$.

Proof. If $$\psi(x)$$ is eventually zero, then it satisfies the asymptotic condition trivially, and so we may as well assume that $$\psi(x)$$ is not eventually zero. Thus by the previous lemma we may assume that $$\psi(x) > 0$$ on $$[A, +\infty)$$. Define $$s(x) = \ln\psi(x)$$. It will be enough to show that $s'(x) \le -a \text{ for all } x \ge A$ Suppose for a contradiction that there is an $$x_0 \ge A$$ with $$s'(x_0) > -a$$.

Observe that $s'(x) = \frac{\psi'(x)}{\psi(x)} \text{ and } s''(x) = \frac{\psi''(x)\psi(x) - \psi'(x)^2}{\psi(x)^2}$ Thus $s''(x) + s'(x)^2 = \frac{\psi''(x)}{\psi(x)} \ge a^2$ Write $$u(x) = s'(x)/a$$ so that $\frac{1}{a}u'(x) + u(x)^2 \ge 1$ and so $u'(x) \ge a(1 - u(x)^2) \tag{A}$ We are assuming that $$u(x_0) > -1$$ and by Lemma 2, $$\psi'(x) < 0$$ for all $$x$$, hence $$s'(x)$$ and $$u(x)$$ are always negative. Thus $$-1 < u(x_0) < 0$$ and so $$u'(x_0) > 0$$.

Now $$u(x)$$ remains increasing on $$x>x_0$$ for as long as $$-1 < u(x) < 0$$. Suppose there is an $$x>x_0$$ with $$u(x) \le -1$$ and let $x_1 = \inf \{ x : x>x_0 \text{ and } u(x) \le -1 \}$ By continuity, $$x_0 < x_1$$ and $$u(x_1) = -1$$. But then by the Mean Value Theorem there is $$x_0 < \xi < x_1$$ with $$u'(\xi) = (u(x_1) - u(x_0))/(x_1 - x_0) < 0$$ while $$-1 < u(\xi) < 0$$ by minimality of $$x_1$$, which is impossible. Thus in fact $$u(x)$$ is increasing on $$[x_0, +\infty)$$, hence $$-1 < u(x) < 0$$ remains true for all $$x\ge x_0$$ and so we can rearrange inequality (A) to read $\frac{d}{dx}\tanh^{-1} u(x) = \frac{u'(x)}{1 - u(x)^2} \ge a$ which is valid for all $$x\ge x_0$$. Integrating this yields $\tanh^{-1} u(x) - \tanh^{-1} u(x_0) \ge a(x - x_0)$ and so $u(x) \ge \tanh(a(x-x_0) + \tanh^{-1} u(x_0))$ Taking the limit as $$x\rightarrow+\infty$$, this increases to $$1$$, which contradicts the fact that $$u(x) < 0$$ for all $$x\ge x_0$$.

From this contraction we conclude that in fact $$s'(x) \le -a$$ for all $$x\ge A$$. Integrating we deduce that $s(x) \le -a(x - A) + s(A)$ and so $0 < \psi(x) \le C e^{-ax}$ as required.

## Corollaries

If $$\psi(x)$$ satisfies TISE on $$\RR$$ with potential $$V(x)$$, then $$\psi(-x)$$ satisfies TSIE on $$\RR$$ with potential $$V(-x)$$. Thus we can immediately conclude:

Suppose that there is an $$A\in\RR$$ and a positive $$a$$ such that $$V(x) - E \ge a^2\hbar^2/2m$$ for all $$x\le A$$. Let $$\psi$$ be a twice continuously differentiable solution to TISE on $$(-\infty, A]$$. Then either $$\lim_{x\rightarrow-\infty}|\psi(x)| = +\infty$$, or else $\psi(x) = O(e^{ax}) \text{ on } (-\infty, A]$

Suppose that there is an $$A\in\RR$$ and a positive $$a$$ such that $$V(x) - E \ge a^2\hbar^2/2m$$ for all $$|x|\ge A$$. Let $$\psi$$ be a twice continuously differentiable solution to TISE on $$|x|\ge A$$. Then one of the following must hold:

1. $$\lim_{x\rightarrow+\infty}|\psi(x)| = +\infty$$,
2. $$\lim_{x\rightarrow-\infty}|\psi(x)| = +\infty$$,
3. $$\psi(x) = O(e^{-a|x|}) \text{ for } |x| \ge A$$

The following result shows that there are no normalizable solutions to TSIE for energies that are entirely lower than the potential $$V$$.

Suppose there is a positive $$a$$ such that $$V(x) - E \ge a^2\hbar^2/2m$$ for all $$x\in\RR$$. Then either $$\psi$$ is identically zero, or else at least one of $$\lim_{x\rightarrow+\infty}|\psi(x)|$$ or $$\lim_{x\rightarrow-\infty}|\psi(x)|$$ must be infinite. In particular, there are no normalizable, or even bounded, solutions to TSIE.

Proof. First suppose that $$\psi$$ is real-valued. Applying Lemma 2 to $$\psi(-x)$$ (with the potential $$V(-x)$$) we can deduce that $$\lim_{x\rightarrow-\infty}\psi(x) = +\infty$$ if there is any $$x$$ at which $$\psi(x) > 0$$ and $$\psi'(x)<0$$. In combination with Lemma 2 in its original form, this implies that unless $$\psi'(x) = 0$$ for all $$x$$, then one of $$\lim_{x\rightarrow+\infty}\psi(x)$$ or $$\lim_{x\rightarrow-\infty}\psi(x)$$ must be infinite. On the other hand if $$\psi'(x) = 0$$ for all $$x$$ then TSIE imples that $$\psi$$ is identically zero.

If $$\psi$$ is complex-valued, apply the previous paragraph to the real and imaginary parts separately. If neither $$\lim_{x\rightarrow+\infty}|\psi(x)|$$ nor $$\lim_{x\rightarrow-\infty}|\psi(x)|$$ is infinite, then both the real and imaginary parts of $$\psi$$ must be identically zero and the result follows.