Infinite momentum is impossible, boundary term in integration by parts

I am meeting the momentum space wave function ##\Phi\left(p,t\right)## in chapter 3 of Griffiths & Schroeter. I have an integral which I can integrate by parts:

$$\int_{-\infty}^{\infty}{\frac{\partial}{\partial p}\left(e^\frac{ipx}{\hbar}\right)\Phi d p}=\left.e^\frac{ipx}{\hbar}\Phi\left(p,t\right)\right|_{-\infty}^\infty-\int{e^\frac{ipx}{\hbar}\frac{\partial\Phi}{\partial p}dp}$$which I got from my cheat sheet and it says that the boundary term vanishes
$$\left.e^\frac{ipx}{\hbar}\Phi\left(p,t\right)\right|_{-\infty}^\infty=0$$which puzzled me for a while but then I thought that momentum can't be infinite so the probability of that is zero, that is ##\Phi\left(\pm\infty,t\right)=0##.

Is that the reason the boundary term vanishes?

George Keeling said:

$$\left.e^\frac{ipx}{\hbar}\Phi\left(p,t\right)\right|_{-\infty}^\infty=0$$which puzzled me for a while but then I thought that momentum can't be infinite so the probability of that is zero, that is ##\Phi\left(\pm\infty,t\right)=0##.

Is that the reason the boundary term vanishes?

No. This is non-relativistic, so there is no upper limit to the momentum. But the wave function must be square integrable (makes no difference if it is in position or momentum space), which requires that it goes to zero at infinity.

DrClaude said:

No. This is non-relativistic, so there is no upper limit to the momentum. But the wave function must be square integrable (makes no difference if it is in position or momentum space), which requires that it goes to zero at infinity.

Square integrable is not enough to imply that it goes to zero at infinity.

martinbn said:

Square integrable is not enough to imply that it goes to zero at infinity.

Counterexamples are too artificial and don't appear in practice (in theoretical physics).

Demystifier said:

Counterexamples are too artificial and don't appear in practice (in theoretical physics).

I don't think they are artificial, but in any case my comment was about the statement that square integrable implies decay at infinity. May be it is reasonable to consider smooth functions (in position and momentum space), then they will be going to zero at infinity (they will be in the Schwartz space).

DrClaude said:

No. This is non-relativistic, so there is no upper limit to the momentum. But the wave function must be square integrable (makes no difference if it is in position or momentum space), which requires that it goes to zero at infinity.

Thanks!

martinbn said:

I don't think they are artificial

Consider the function ##f(x)## defined as ##f(x)=1## if ##x## is integer, and ##f(x)=0## otherwise. It does not vanish at infinity but its integral is zero. Would you call such function artificial?

Demystifier said:

Consider the function ##f(x)## defined as ##f(x)=1## if ##x## is integer, and ##f(x)=0## otherwise. It does not vanish at infinity but its integral is zero. Would you call such function artificial?

What do you mean by artificial?

Consider a function that looks like little bumps above the x axis with height one. And the width of the bumps get smaller and smaller. Say 1/2, 1/4, 1/8, ... It will be integrable and it will not go to zero at infinity. Would you call this one artificial?

martinbn said:

What do you mean by artificial?

Consider a function that looks like little bumps above the x axis with height one. And the width of the bumps get smaller and smaller. Say 1/2, 1/4, 1/8, ... It will be integrable and it will not go to zero at infinity. Would you call this one artificial?

Yes, feels artificial to me.

gentzen said:

Yes, feels artificial to me.

As function or as a wave function?

martinbn said:

What do you mean by artificial?

Not appearing in physics applications.

martinbn said:

Consider a function that looks like little bumps above the x axis with height one. And the width of the bumps get smaller and smaller. Say 1/2, 1/4, 1/8, ... It will be integrable and it will not go to zero at infinity. Would you call this one artificial?

Yes.

Demystifier said:

Not appearing in physics applications.


Yes.

But the whole ##L^2## space appears in applications of physics. All the functions are needed. By the same logic you can say that only rational numbers and a few real numbers are needed and all the rest of the real numbers are artificial.

All that misses the point! My comment was simply about the implication " it is square integrable therefore it goes to zero at infinity". If you replace it by "all functions needed for physics applications which are square integrable go to zero at infinity" I wouldn't have a problem with it.

martinbn said:

As function or as a wave function?

martinbn said:

But the whole ##L^2## space appears in applications of physics. All the functions are needed. By the same logic you can say that only rational numbers and a few real numbers are needed and all the rest of the real numbers are artificial.

All that misses the point! My comment was simply about the implication " it is square integrable therefore it goes to zero at infinity". If you replace it by "all functions needed for physics applications which are square integrable go to zero at infinity" I wouldn't have a problem with it.

Your example is artificial, because point evaluation is not the important criteria in practice. And your example just artificially tries to hide this point, which was already exposed by Demystifiers example why such counterexamples are artificial.

If you look at the convolution of some arbitrary test function with your example function, the result goes to zero at infinity.