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Say you consider spherical waves (momentum eigenstates) propagating outwards from some starting point $r = r_{0} $ (not defined for $r < r_{0}$) with $k \in \mathbb{R} $:

\begin{equation} \psi\left(r, t\right) = \left( A {e^{ikr} \over r} + B {e^{-ikr } \over r } \right) e^{-ikt} \end{equation}

We have boundary conditions at $r = r_{0} $:

\begin{equation} \psi\left(r_{0}, 0\right) = {e^{ikr_{0}} \over r} \end{equation}

This automatically sets $A = 1$ and $ B = 0 $.

Question: Does this wave reach infinity? At first glance it looks like the answer is definitely no:

\begin{equation} \lim_{r \rightarrow \infty } \psi\left(r,t \right) = \lim_{r \rightarrow \infty} {e^{ikr} \over r } e^{-ikt} = {\lim_{r \rightarrow \infty} e^{ikr- ikt} \over \lim_{r \rightarrow \infty} r} = 0 \end{equation}

Since the the $e^{ikr-ikt}$ is just oscillating for any value of $r$ and $t$. However, if we consider the energy of the wave (let $I$ be the intensity):

\begin{equation} I\left(r\right) = \left \vert \psi\left(r,t\right) \right \vert^2 = {1 \over r^2} \end{equation}

And so the energy in a spherical shell at distance $r$:

\begin{equation} 4 \pi r^2 \cdot I\left(r\right) = 4\pi \end{equation}

Notice the energy is constant independent of $r$, so in this sense the energy of the wave reaches infinity. What is happening, is the wave physically reaching infinity or not?

I am looking for an this question that might be generalizable to more complicated situations. Thanks!

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closed as unclear what you're asking by sammy gerbil, Jon Custer, ZeroTheHero, honeste_vivere, JamalS Aug 28 '17 at 11:26

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  • $\begingroup$ If it does not "reach infinity" it must stop at some finite $r$. Can you imagine that? $\endgroup$ – Philip Roe Aug 20 '17 at 23:58
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    $\begingroup$ Your math is all correct, so I think you should start by defining what you mean by "reaching infinity". Maybe then you'll be able to answer the question yourself. $\endgroup$ – Javier Aug 21 '17 at 0:19
  • $\begingroup$ You may be right that a spherical shell at any distance contains a constant amount of energy of $4\pi$, but that constant energy is spread ever more thinly as $r$ increases. Alternatively, in direct answer to the title: Visit infinity and measure it! $\endgroup$ – TripeHound Aug 23 '17 at 8:28
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As stated, your question does not have a well-defined answer. If you had a definition for what you meant by "reaches infinity," then we would have something to work with.

Certainly the wave $\psi$ is not eventually $0$ as the term $e^{ikr}$ keeps it alive for infinitely many $r$. On the other hand, as you note, the limit is $0$, which does not provide any information about the wave itself being nonzero.

Physically, once $r$ is sufficiently large, the amplitude of the wave would be so small that no apparatus we invent could measure it. In that sense, it doesn't reach infinity.

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This is incorrect:

We have boundary conditions at $r = r_{0} $: \begin{equation} \psi\left(r_{0}, 0\right) = {e^{ikr_{0}} \over r} \end{equation} This automatically sets $A = 1$ and $ B = 0 $.

In fact, as written, it's mostly meaningless (you presumably meant to write $\psi\left(r_{0}, 0\right) = {e^{ikr_{0}} / r_0} $ instead), but even that doesn't provide enough information about the system to fix the two coefficients. As a simple consistency check, you had a solution space of dimension $2$ and you've provided one constraint; that's never going to have a unique solution.

If you want to rule out the solution that behaves as $e^{-ik(r+t)}/r$, then you need to do that explicitly: you just say that you're not interested in incoming waves and leave it at that.


Now, as to your actual question, you've mentioned three statements:

  • $\psi(r,t)=e^{ik(r-t)}/r$ obeys $\psi(r,t)\to 0$ as $r\to\infty$
  • $\int |\psi(r,t)|^2\mathrm d\Omega$ is constant and independent of $r$

Those are correct, concrete, meaningful statements about your function

  • $\psi(r,t)$ "reaches" infinity

This one is meaningless unless you provide it with an actual definition. Since you've not done that, then there simply is no answer.

Normally, a physicist would be much more inclined to look at the facts that

  • for any finite $r$, $\psi(r,t)$ is nonzero; and
  • for any fixed surface $S$ and finite $r$, $\int_S |\psi(r,t)|^2\mathrm dS$,

from which we'd conclude that the wave reaches arbitrarily far from the source given a sufficiently precise detector, and leave it at that.

No serious physicist will make statements of what happens "at" infinity: it is not an actual place, it is only a concept in the behaviour of limits. If you want to make statements about that, then you need to provide the precise definitions for them first, and then the question might have an answer. But really, honestly, you shouldn't.

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