# Square integrable functions blowing up at infinity

1. Dec 6, 2009

### kof9595995

I've been reading Griffths QM recently, and in the book he mentioned a couple of times that though these pathological functions exist, they're not physically realizable. But what's wrong with these functions? What prevents them to be physically realizable ?

EDIT：Griffths' statement is wave function may not vanish at infinity, not blow up; Thanks to George Jones pointing out.

Last edited: Dec 6, 2009
2. Dec 6, 2009

### George Jones

Staff Emeritus
This is not exactly what Griffiths states.

By blowing up at infinity, do you mean a real-valued function $f$ of a real variable such that

$$\lim_{x \rightarrow \infty} f \left( x \right) = \infty?$$

More precisely, this means that for every $L > 0$, there exists an $x_0 > 0$ with $f \left( x \right) > L$ whenever $x > x_0.$

Using this property, it is fairly easy to show that $\left| f \left( x \right) \right|^2$ is a positive function that blows up at infinity, and that any positive function with this property is not integrable.

Or do you mean for every $L > 0$, there exists an $x_0 > 0$ with $f \left( x \right) > L$ for some $x > x_0$?

There are lots of square-integrable functions with this property. For example, consider the real-valued function of a real variable that is zero except when $x$ is a rational number, and has $f \left( x \right) = x$ when $x$ is a rational number. Then, $\left| f \left( x \right) \right|^2$ integrates to zero. The function in this example is actually (a representative of) the zero vector in Hilbert space, since it differs from the zero function only on a set of measure zero,

3. Dec 6, 2009

### Fredrik

Staff Emeritus
Adding to what George said: The reason why the wavefunction must be square integrable is that the probability interpretation doesn't work if it isn't.

Hurkyl showed me an interesting example of a square integrable function that doesn't even go to zero as x goes to infinty. Link. Those functions aren't physical because a solution of the Schrödinger equation must be differentiable (and all differentiable functions are continuous).

4. Dec 6, 2009

### kof9595995

Yeah you're right, i recheck the book and he only stated that as x goes to infinity f(x) doesn't vanish, and meanwhile it's square integrable.

5. Dec 6, 2009

### kof9595995

Really nice example, and you said it's not valid because it's not differentiable. But isn't it piece-wise differentiable? When we encounter a infinite potential somewhere it seems piece-wise differentiable is enough, like infinite square well or delta function potential.

6. Dec 6, 2009

### Hurkyl

Staff Emeritus
There are continuous, differentiable, and even smooth versions of such functions; you just replace the basic rectangle shape
$$f(x) = \begin{cases} 0 & x < 0 \\ 1 & 0 \leq x < 1 \\ 0 & 1 \leq x \end{cases}$$​
with something smoother.

7. Dec 6, 2009

### Fredrik

Staff Emeritus
And if we just have the width of those shapes decrease more rapidly, we can let their height increase instead of being constant. Now I'm confused again.

8. Dec 6, 2009

### Hurkyl

Staff Emeritus
However, this doesn't contradict George's earlier assertion -- while such a function is square integrable and unbounded, we do not have $\lim_{x \to +\infty} f(x) = +\infty$ -- the limit at infinity simply doesn't exist.

9. Dec 6, 2009

### kof9595995

So Hurkyl, what's your comment on Griffiths' statement, e.g., psi(x) always goes to 0 at infinity in reality?

10. Dec 6, 2009

### Hurkyl

Staff Emeritus
I'm not qualified to comment.

11. Dec 6, 2009

### Count Iblis

Example:

|s> = 6/pi^2 sum over n = 1 to infinity of 1/n |x_n>

Where the |x_n> are position eigenstates that you can replace by smooth normalized states with peak at x_n and with compact support. If the x_n tend to infinity, then the limit of

psi(x) = <x|s> for x to infinity does not exist.