# For energy eigenstates $\psi(x)$, does $\psi(x)\to0$ as $|x|\to\infty$ imply $E<V(\pm\infty)$ and vice-versa? If so, how can it be proved?

Let $$\psi(x)$$ be a solution of the time-independent Schrodinger equation (TISE) in one-dimension $$\psi''(x)+\gamma\left[E-V(x)\right]\psi(x)=0$$ where $$\psi(x)$$ is called an energy eigenfunction, $$\gamma=2m/\hbar^2$$ and $$E$$ is a fixed parameter, called the energy.

Claim 1

If an energy eigenfunction $$\psi(x)$$ vanishes as $$|x|\to\infty$$, then examples point out that the energy $$E$$ of that state satisfies $$E.

Claim 2

If the energy $$E$$ of an energy eigenfunction $$\psi(x)$$ satisfies $$E, then examples point out that $$\psi(x)$$ vanishes as $$|x|\to\infty$$.

Can we prove the above claims mathematically, at least under reasonable assumptions, from the time-independent Schrodinger equation?

• Related: physics.stackexchange.com/a/68060/2451 and links therein. Jan 10, 2021 at 19:14
• If I recall correctly there are several relevant prior discussions, if not straight duplicates. It's worth having a thorough search of the site archives. Jan 10, 2021 at 19:19
• @EmilioPisanty I saw a few posts but could not conclude whether normalizability of energy eigenfunctions imposes the criterion $E<V(\pm\infty)$. However, in the mean time, I am able to find a problem in Griffiths which asks to prove that for any energy eigenfunction $\psi(x)$ to be normalizable, it must have $E\geq V_{\rm min}$ i.e. the total energy must exceed the minimum of the potential energy $V(x)$. Jan 11, 2021 at 17:01
• If nothing else, note that normalizability is not equivalent with to tending to zero at infinity (see in particular Qmechanic's answer there). If you want the question to be about normalizability, then you should edit the text. (Also, see the Linked questions there; there are several useful threads there.) Jan 11, 2021 at 18:41
• @EmilioPisanty Thanks. QMechanic's example demonstrates that it is possible to cook up functions which are square-integrable, yet do not vanish at infinity. Also I can think of delta functions which vanish at $\pm\infty$ but not square-integrable. Therefore, vanishing at infinity is neither sufficient nor necessary condition for normalizability. Please ignore my previous comment. However, the question remains the same: Can we show $E<V(\pm \infty)$ from the condition of localized energy eigenfunction i.e. $\psi\to 0$ as $|x|\to 0$ and vice-versa. Jan 11, 2021 at 19:07

The condition that the wave function vanishes at $$\pm\infty$$ can be formally stated as $$\int dx |\psi(x)|^2 =const,$$ where the constant is by convention taken to be one, i.e., the wave function is normalized. The integration here and everywhere below is implied to be from $$-\infty$$ to $$+\infty$$.
We can now calculate $$\int dx \psi^*(x)\left[-\frac{\hbar^2}{2m}\partial_x^2 + V(x)\right]\psi(x) = E\int dx |\psi(x)|^2=E$$ Furthermore $$\int dx \psi^*(x)\partial_x^2\psi(x) = \psi^*(x)\partial_x\psi(x)|_{x=-\infty}^{+\infty} - \int dx |\partial_x\psi(x)|^2 = - \int dx |\partial_x\psi(x)|^2,$$ where I used the fact that the wave function vanishes in $$+\infty$$. Thus, we can write $$E = \frac{\hbar^2}{2m}\int dx |\partial\psi(x)|^2 + \int dx V(x)|\psi(x)|^2 \leq \int dx V(x)|\psi(x)|^2 \leq \max_{x\in\mathbb{R}}V(x)=V_{max}.$$ This answers claim 1.
Claim 2 is actually not totally true. In the region where $$E the solution of the Schrödinger equation $$\psi''(x) + \gamma(E-V_{max})\psi(x)=0$$ has an approximately exponentially decaying and exponentially growing solutions. Thus, we may have a finite energy solution with a wave function diverging in the infinity. Scattering theory deals with the solutions that do not decay in the infinity, i.e., which are not normalizable (but correspond to a finite particle flux). The solutions growing are mostly discarded on th grounds of physical interpretation, as a mathematical artefact.
• You proved that $E\leq V_{\rm max}$ not $E\leq V({\pm \infty})$ Jan 15, 2021 at 10:28
• Can you tell what is the definition of a bound state? Griffiths says, bound states are those stationary states that satisfy $E<V(\pm\infty)$. But my hunch is that $\psi(x)\to 0$ as $|x|\to\infty$ can also be taken as the definition of a bound state. Jan 15, 2021 at 14:44