# How to tell if spherical waves reach infinity? [closed]

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

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

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

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

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:

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

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):

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

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

$$4 \pi r^2 \cdot I\left(r\right) = 4\pi$$

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!

• If it does not "reach infinity" it must stop at some finite $r$. Can you imagine that? Aug 20, 2017 at 23:58
• 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. Aug 21, 2017 at 0:19
• 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! Aug 23, 2017 at 8:28

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.

This is incorrect:

We have boundary conditions at $r = r_{0}$: $$\psi\left(r_{0}, 0\right) = {e^{ikr_{0}} \over r}$$ 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 $r$

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