# Normalizable wavefunction that does not vanish at infinity

I was recently reading Griffiths' Introduction to Quantum Mechanics, and I stuck upon a following sentence:

but $\Psi$ must go to zero as $x$ goes to $\pm\infty$ - otherwise the wave function would not be normalizable.

The author also added a footer: "A good mathematician can supply you with pathological counterexamples, but they do not arise in physics (...)".

Can anybody give such a counterexample?

• Go back to the definition of an integral as the area under a curve. What does it look like if $f(x)$ tends to a nonzero limit, keeping in mind that the integrand is $|f(x)|^2>0$? – Michael Brown Aug 28 '13 at 13:50
• If you allow very badly behaved functions (things which don't even possess limits) you can have, for example, $f(x)=\sin(|x|)^{|x|}$ which you think might work since the width of the peaks decrease as $x\to\infty$, but in fact the integral of $f(x)^2$ still diverges, although slowly. Needless to say this is a horrible function that never comes up in physics. :) – Michael Brown Aug 28 '13 at 14:14
• Would math.stackexchange.com be a better home for this question? – Qmechanic Aug 30 '13 at 3:20
• Possible duplicates: physics.stackexchange.com/q/92517/2451 and links therein. – Qmechanic May 9 '17 at 12:56
• If you are interested in these questions, I highly recommend this. It's an easy read that taught me a lot. – Steven Mathey May 9 '17 at 13:43

Take a gaussian (or any function that decays sufficiently quickly), chop it up every unit, and turn all the pieces sideways.

• Minor clarification: Since the wave function $\psi$ should be square integrable, (the proof becomes more neat if) the above picture shows the probability distribution $|\psi|^2$ (rather than the wave function $\psi$ itself). – Qmechanic May 10 '17 at 9:18
• @Qmechanic It works for either $\psi$ or $|\psi|^2$. Squaring a function less than 1 makes it strictly smaller. – Nick Alger May 10 '17 at 12:29
• @Nick Alger: Yeah, I know. I just find it neater to have equal integrals in a picture proof :) – Qmechanic May 10 '17 at 12:39
• @svavil It's a nice example, I find it a lot more pathological than Qmechanic's smooth one. – leftaroundabout May 10 '17 at 14:59
• Wonderful answer! – Diracology May 10 '17 at 16:56

Let $$\psi(x) = \begin{cases} 1 & \exists\, n \in \mathbb N: x \in [n, n+\frac 1 {n^2}]\\ 0 & \text{otherwise.} \end{cases} = \sum_{n \in \mathbb N} \mathbf 1_{[n,n+\frac 1 {n^2}]}(x) ,$$ where $\mathbf 1_A$ is the characteristic function of the set $A$. Then $$\int_{-\infty}^\infty |\psi(x)|^2 dx = \sum_{n=1}^\infty \frac 1 {n^2} < \infty,$$ but $\psi(x)$ doesn't converge to zero as $x \to +\infty$.

Note that $\psi \in L^2(\mathbb R)$, but it is not twice (weakly) differentiable and can therefore not be the solution to Schrödinger's equation with $H = -\Delta + V$. However, that problem can easily be solved by replacing the rectangle function with a smooth pulse with compact support. Alternatively, use $$\psi(x) = x^2 \mathrm e^{-x^8 \sin^2 x} ,$$ as discussed in Example 2 of §2.1 in arXiv:quant-ph/9907069 -- this is even analytical.

• Shouldn't a wave function be twice differentiable w.r.t space to satisfy Shroedinger's equation? Yours has infinite number of discontinuities! – xletmjm May 9 '17 at 12:53
• @xletmjm you could just smear out all the edges and the expression would still hold. – noah May 9 '17 at 12:54
• Yes, you can smoothen it without changing the integral. Also, when we talk about wave functions, we usually just require $L^2$. Not every such wave function will be in the domain of the Hamilton operator, but what that domain exactly is depends on the Hamilton, of course. (It was will usually be some Sobolev space of weakly differentiable functions.) – Noiralef May 9 '17 at 13:57
• I would also mention that by tweaking the example (i.e., making the intervals even thinner), one can get $\psi$ to be unbounded. – Martin Argerami May 9 '17 at 16:29

Emilio Pisanty and Eckhard Giere have already given discontinuous, piecewise constant counterexamples in their answers. Here we provide for-the-fun-of-it a smooth infinitely-many-times-differentiable counterexample $f\in C^{\infty}(\mathbb{R})$ of a square integrable function $f:\mathbb{R} \to [0,1]$ that does not satisfy $\lim_{|x|\to \infty}f(x)=0$. Our counterexample is

$$\tag{1} f(x)~:=~ e^{- g(x)} ~\in ~]0,1], \qquad g(x)~:=~x^4 \sin^2 x~\in ~[0,\infty[.$$

Intuitive idea: If we imagine $x$ as a time variable, then the function $f$ returns periodically to its maximum value

$$\tag{2} f(x) =1 \quad\Leftrightarrow\quad g(x) =0 \quad\Leftrightarrow\quad \frac{x}{\pi}\in \mathbb{Z} ,$$

but spends most if its time close to the $x$-axis in order to be square integrable.

Proof: We leave a detailed rigorous epsilon-delta mathematical proof to the reader, but a sketched heuristic proof goes like this. For each very large integer $|n|\gg 1$, define a shifted variable

$$\tag{3} y~:=~x-\pi n.$$

For the fixed integer $n\in\mathbb{Z}$, always assume from now on that the $y$-variable belongs to the interval

$$\tag{4} |y|~\leq~ \frac{\pi}{2}.$$

For $|y|\ll\frac{\pi}{2}$ very small, we may approximate $g(x) \approx (\pi n)^4y^2$, so that in the interval (4), we have

$$\tag{5} g(x)~\lesssim~ \pi^4 |n| \quad \Leftrightarrow\quad |y| ~\lesssim~ |n|^{-\frac{3}{2}}.$$

Thus we may form a square integrable majorant function $h\geq f$ (outside a compact region on the $x$-axis) by defining

$$\tag{6} h(x)~:=~\left\{\begin{array}{lcl} 1 &{\rm for}& |y| ~\lesssim~ |n|^{-\frac{3}{2}}, \cr e^{-\pi^4 |n|}&{\rm for}& |n|^{-\frac{3}{2}}~\lesssim~ |y| ~\leq~ \frac{\pi}{2}, \end{array} \right. \qquad |n|\gg 1.$$

The function $h\in {\cal L}^2(\mathbb{R})$ is square integrable on the whole $x$-axis, since

$$\tag{7} \sum_{n\neq 0} |n|^{-\frac{3}{2}} ~<~ \infty$$

and

$$\tag{8} \pi \sum_{n\in\mathbb{Z}}e^{-2\pi^4 |n|}~<~\infty$$

are convergent series.

Apart from not being sufficient to prove convergence of the integral $$\int |f(x)|^2\text dx<\infty,$$ having the vanishing limig $\lim_{x\rightarrow\infty}f(x)=0$ is only necessary for the convergence within a suitable class of "nice" functions.

Consider, for example, the function $$f(x)=\sum_{n=1}^\infty\chi_{\left[n,n+\frac1{n^2}\right]}(x)=\left\{\begin{array}&1\text{ if } n\leq x\leq n+1/n^2 \text{ for some }n=1,2,3,\ldots,\\0\text{ otherwise.}\end{array}\right.$$ (Here $\chi_A$ is the characteristic function of a set $A\subseteq\mathbb R$.) This function has a convergent $L^2$ integral but does not have a well-defined limit at infinity. While this function is not continuous, but using suitable bump functions you can make a similar $C^\infty$ function with the same properties. This is the kind of function you're allowing when you do not impose vanishing limits at infinity - i.e., pretty ugly.

More to the point, say your wavefunction obeys a stationary Schrödinger equation with energy $E$ for some potential $V$ such that $\lim_{x\rightarrow} V(x)>E$ (i.e. a bound state). Then you know that, at infinity, $f''(x)$ has the same sign as $f$, which we can assume to be positive. If $f'(x)$ is ever zero in that region, then you know that it will be positive for all $x$ after that and $f(x)$ will monotonically increase, in which case the $L^2$ integral has no chance of convergence. In this particular setting, then you can restrict yourself to monotonically decreasing functions, and those are nice enough that the vanishing limit at infinity is necessary for $L^2$ convergence.

(To be followed by a more rigorous argument if I find the time.)

• Your ugly function is the type of thing I was aiming at in my comment above, but didn't quite get. :) Good example. I think you mean the inequality to go the other way, i.e. $E<V(x)$, but nice argument. – Michael Brown Aug 30 '13 at 4:21

Some simple example that illustrates that the condition $$\lim_{|x|\to \infty} f(x) = 0 \quad (1)$$ is not necessary. If the condition were necessary $f\in L^2$ would imply that the limit in (1) holds.

Take in dimension 1 the function $$f(x) = \sum_{n=2}^{\infty} \chi_{I_n}(x)$$ where $\chi_{I_n}$ is the characteristic function of the interval $I_n = [n-\frac{1}{n^2}, n+\frac{1}{n^2}]$ then the integral evaluates to $$\int |f(x)|^2 dx = \sum_{n=2}^{\infty} |I_n| = \sum_{n=2}^{\infty} \frac{2}{n^2} < \infty\ .$$ But the function does not converge to zero for $|x|\to \infty$.

Sorry: Forgot to center the intervals around n. Now corrected.