Limit of expression containing unknown function.

[D_Tr]

I would like to get an answer or pointers to suitable material, on the following question:

I know that ∫|f(x)|²dx is finite. Can we say that lim x→±∞ x*(d/dx)|f(x)|² is zero? Are there any theorems about such limits with unknown functions that have some known properties? Basically this question arose from Griffith's introductory QM textbook which I started studying today (self study). f(x) is Ψ(x) and the author at page 16 takes the above limits to be zero "on the ground that Ψ goes to zero at ±∞". But it may go to zero asymptotically, and x goes to infinity, so we have an indeterminate form and we know only that the function's absolute value squared has a finite integral. I posted here, however, since I would like to get a more general answer on these types of limits and to be at least aware of the theory that deals with them.

Thanks in advance for your precious time people!

 

[Simon Bridge]

Given: $$I=\int_{-\infty}^\infty |f(x)|^2\;\text{d}x$$ ... if $I$ converges, then; $$\lim_{x\rightarrow\pm\infty} x\frac{d}{dx}|f(x)|^2 = 0$$

(We also know from context that f(x) is a solution to the Schrödinger equation.)

I suspect there are other implied conditions... but I think it pretty much follows from the definition of an integral and the conditions for convergence.

What was the author trying to show by that particular limit?

 

[D_Tr]

Thanks a lot :) The author calculated the time derivative of <x>, just before introducing the notion of the operator. In one of the few steps he eliminated the following term which appeared as a result of integration by parts:
[x$\frac{d}{dx}$|Ψ|$^{2}$]$^{-∞}_{∞}$, "on the grounds that Ψ goes to zero at plus and minus infinity". It seems obvious now but I was a bit confused by the fact that the expression was multiplied by x and that the derivative was written in its expanded form containing he complex Ψ and Ψ* functions.

 

[Simon Bridge]

Thanks a lot :) The author calculated the time derivative of <x>, just before introducing the notion of the operator. In one of the few steps he eliminated the following term which appeared as a result of integration by parts:
[x$\frac{d}{dx}$|Ψ|$^{2}$]$^{-∞}_{∞}$, "on the grounds that Ψ goes to zero at plus and minus infinity". It seems obvious now but I was a bit confused by the fact that the expression was multiplied by x and that the derivative was written in its expanded form containing he complex Ψ and Ψ* functions.

No worries - it can take a while to learn to read these things ;)

 

[Office_Shredder]

Given: $$I=\int_{-\infty}^\infty |f(x)|^2\;\text{d}x$$ ... if $I$ converges, then; $$\lim_{x\rightarrow\pm\infty} x\frac{d}{dx}|f(x)|^2 = 0$$

(We also know from context that f(x) is a solution to the Schrödinger equation.)

This isn't actually true...for example f(x)² can be the function which is 0 everywhere except for in the interval [n,n+1/n] where it bumps up to a value of 1/n and then back down to zero. Then the integral is equal to the sum
$$\sum_{n=1}^{\infty} 1/n^2$$
which converges, but when it bumps up the derivative could be arbitrarily large; so that limit could fail to be zero even before you include multiplying by x.

Probably the condition that f(x) is supposed to be satisfying here is something like |f(x)|² is convex if f(x) is large enough (equivalently |f(x)|² is decreasing for large enough values of x - if |f(x)|² is decreasing down to zero and you don't have those bumps, then it has to die faster than 1/x to be integrable, so the derivative will look like 1/x² or something smaller (this is a handwavy argument but can probably be made rigorous).

 

[Simon Bridge]

I had thought of that ...
Does it matter that the derivative (per your example) in the interval that f is non-zero could be arbitrarily large?
The limit is taken arbitrarily far away from that interval - where the derivative is zero.