By xletmjm

2017-05-09 12:31:32 8 Comments

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?


@Noiralef 2017-05-09 12:46:42

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.

@xletmjm 2017-05-09 12:53:15

Shouldn't a wave function be twice differentiable w.r.t space to satisfy Shroedinger's equation? Yours has infinite number of discontinuities!

@noah 2017-05-09 12:54:19

@xletmjm you could just smear out all the edges and the expression would still hold.

@Noiralef 2017-05-09 13:57:21

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.)

@Martin Argerami 2017-05-09 16:29:34

I would also mention that by tweaking the example (i.e., making the intervals even thinner), one can get $\psi$ to be unbounded.

@Nick Alger 2017-05-09 17:22:53

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

enter image description here

@Qmechanic 2017-05-10 09:18:44

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).

@Nick Alger 2017-05-10 12:29:43

@Qmechanic It works for either $\psi$ or $|\psi|^2$. Squaring a function less than 1 makes it strictly smaller.

@Qmechanic 2017-05-10 12:39:23

@Nick Alger: Yeah, I know. I just find it neater to have equal integrals in a picture proof :)

@leftaroundabout 2017-05-10 14:59:26

@svavil It's a nice example, I find it a lot more pathological than Qmechanic's smooth one.

@Diracology 2017-05-10 16:56:40

Wonderful answer!

@user2357112 2017-05-10 20:46:26

@NickAlger: But if you do this with the function instead of its square, doesn't the process change the asymptotics of the integral of the function's square? The result might not be normalizable. I don't know how it works out for a gaussian, but it seems like this would matter for some functions.

@user2357112 2017-05-10 20:53:27

For example, the even function that takes value $1/n$ from $n-1$ to $n$ for positive $n$ (and symmetric values for negative input) is square-integrable, but it's not square-integrable if you chop it up and rotate the pieces like this.

@Nick Alger 2017-05-10 21:20:11

@user2357112 You would need the original function to decay like $n^{-1-\epsilon}$ or faster, asymptotically. $n^{-1}$ barely doesn't work. A gaussian decays exponentially so it will definitely work.

@M.Herzkamp 2017-05-11 09:41:17

Only nitpicking here: The chosen function should be strictly monotonic and its maximum should not be greater than the chosen bin width ;-)

@Qmechanic 2013-08-30 10:37:15

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

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


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

are convergent series.

@Emilio Pisanty 2013-08-29 22:26:00

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.)

@Michael Brown 2013-08-30 04:21:25

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.

@Eckhard Giere 2013-08-29 20:09:43

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.

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