Square integrable functions blowing up at infinity

[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.

 

[George Jones]

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 ?

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,

 

[Fredrik]

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).

 

[kof9595995]

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,

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.

 

[kof9595995]

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).

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.

 

[Hurkyl]

There are continuous, differentiable, and even smooth versions of such functions; you just replace the basic rectangle shape

$$f(x) = \begin{cases} 0 & x < 0 \\<br /> 1 & 0 \leq x < 1 \\<br /> 0 & 1 \leq x \end{cases}$$​

with something smoother.

 

[Fredrik]

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.

 

[Hurkyl]

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.

 

[kof9595995]

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

 

[Hurkyl]

I'm not qualified to comment.

 

[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.