Infinite momentum is impossible, boundary term in integration by parts

[George Keeling]

TL;DR Summary

Does boundary term containing momentum space wave function with ±∞ limits vanish?

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?

 

[DrClaude]

$$\left.e^\frac{ipx}{\hbar}\Phi\left(p,t\right)\right|_{-\infty}^\infty=0$$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.

 

[martinbn]

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.

 

[Demystifier]

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).

 

[martinbn]

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).

 

[George Keeling]

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!

 

[Demystifier]

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?

 

[martinbn]

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?

 

[gentzen]

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.

 

[martinbn]

Yes, feels artificial to me.

As function or as a wave function?

 

[Demystifier]

What do you mean by artificial?

Not appearing in physics applications.

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.

 

[martinbn]

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.

 

[gentzen]

As function or as a wave function?

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.

 

[martinbn]

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.

What practice? Do you not read what I comment on?

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

So? How does this related to the original post that I commented on?

 

[gentzen]

So? How does this related to the original post that I commented on?

That your example was essentially no better than the example Demystifier already gave. If the problem is point evaluation, all another example can do is expose whether a certain technique to resolve that problem will work or not. The problem in Demystifier's example can be fixed by modifying the function on a set of measure zero. This no longer works for your example. For your example, for any $\epsilon >0$, one can modify the function on a set of measure $<\epsilon$. Fine, in this sense, your example is perhaps better. But this stuff with "modify functions on sets of measure ..." is still a bit mathematical and theoretical.

What practice?

For square integrable functions, looking at results for test functions might be closer to what is relevant in practice. That is why I suggested that as criterion. Not sure whether it is a good one.

Do you not read what I comment on?

Yes, I always read stuff that I comment on. I get irritated when I get the feeling that some established members (not you or Demystifier) don't seem to try to follow that practice.

 

[martinbn]

That your example was essentially no better than the example Demystifier already gave. If the problem is point evaluation, all another example can do is expose whether a certain technique to resolve that problem will work or not. The problem in Demystifier's example can be fixed by modifying the function on a set of measure zero. This no longer works for your example. For your example, for any $\epsilon >0$, one can modify the function on a set of measure $<\epsilon$. Fine, in this sense, your example is perhaps better. But this stuff with "modify functions on sets of measure ..." is still a bit mathematical and theoretical.

Yes, that is why I gave the other example. Because his gives the zero element in the $L^2$ space. I thought he would then object to that.

For square integrable functions, looking at results for test functions might be closer to what is relevant in practice. That is why I suggested that as criterion. Not sure whether it is a good one.

Yes, distributions might be more natural than functions.

Yes, I always read stuff that I comment on. I get irritated when I get the feeling that some established members (not you or Demystifier) don't seem to try to follow that practice.

 

[DrClaude]

A good mathematician can supply you with pathological counterexamples, but they do not arise in physics; for us the wave function always goes to zero at infinity.

D. J. Griffiths, Introduction to Quantum Mechanics (2nd ed.).

 

[George Keeling]

A good mathematician can supply you with pathological counterexamples, but they do not arise in physics; for us the wave function always goes to zero at infinity.
D. J. Griffiths, Introduction to Quantum Mechanics (2nd ed.).

I have the third edition. Did I miss that? Where does Griffiths write it?

 

[DrClaude]

I have the third edition. Did I miss that? Where does Griffiths write it?

In the 2nd ed., it is footnote 12 on page 14, in relation to eq. (1.26).

 

[George Keeling]

Thanks! I did miss that. In the 3rd edition it's on page 14, note 15.

Slightly new question: is a function square integrable if and only if it is normalisable?

 

[martinbn]

Thanks! I did miss that. In the 3rd edition it's on page 14, note 15.

Slightly new question: is a function square integrable if and only if it is normalisable?

Except for the identically zero function.

 

[Paul Colby]

Okay, isn’t the operator $i\frac{\partial}{\partial p}$ required to be hermitian? This would force the boundary terms to be zero.

 

[Demystifier]

Except for the identically zero function.

There are other exceptions as well.

 

[martinbn]

There are other exceptions as well.

They are all the same element in the $L^2$ space, the zero.

 

[martinbn]

Apparently similar questions are asked often even here. After some searching I found this

https://www.physicsforums.com/threads/wave-function-zero-at-infinity.242250/