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Question: If the wave function $\psi \to 0$ (pointwise or uniformly in mean), then does $|\psi|^2$ converge in quadratic mean (i.e. in $L^2$) to $0$?

Is this the source of the term "wave function"?

(Since the energy of a wave is proportional to the square of its amplitude, so if the expected energy of a wave goes to zero, then its ensemble amplitude goes to zero in quadratic mean. Likewise, if the expectation of a wave function goes to zero, then its probability density will decay to zero in $L^2$/quadratic mean. And the wave function is supposed to have something to do with energy, hence why it being square-integrable has something to do with it having finite energy, and thus in a certain sense is related to an elementary wave.)

Note: This is a follow-up to my previous question: Do waves and oscillations dissipating energy decrease in amplitude in mean square?

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    $\begingroup$ I'm not sure what you mean by your "technical" question - if the function goes to zero, then yes, it's square and the integral of its square also go to zero. What has this to do with it being called a "wave function"? $\endgroup$
    – ACuriousMind ♦
    Aug 18, 2016 at 11:22
  • $\begingroup$ @ACuriousMind I doubt that there is a good argument for my position, I was just thinking of how in Fourier analysis we are working with convergence in $L^2$ instead of uniformly or pointwise, and this usually makes analysts sad, but if we think about the functions (e.g. sinusoidal waves or wavelets) as graphs of the amplitude of waves, then the fact that they converge in $L^2$ means that their "energy" is converging pointwise to zero, which is actually a very natural condition. So I was wondering if I was correct to draw analogies between Fourier analysis, physical waves, and wave functions. $\endgroup$
    – Chill2Macht
    Aug 18, 2016 at 18:12
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    $\begingroup$ The title question (v3) seems different from the first question in the main body. $\endgroup$
    – Qmechanic ♦
    Aug 4, 2017 at 7:11

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The connection between wavefunctions and waves is much more fundamental than that. The quantum wavefunction $\psi(x)$ has a huge amount in common with the amplitude of a classical wave. For concreteness, let's consider the height $y(x)$ of a wave on a string at some time $t$.

  • The equations of motion for $\psi(x)$ and $y(x)$ are both wave equations, i.e. they both admit traveling wave solutions $e^{i(kx-\omega(k) t)}$. The only difference is the dispersion relation $\omega(k)$.
  • The equations of motion for both types of waves are linear. This implies that wavefunctions, as well as string waves, can display constructive or destructive interference.
  • Given fixed boundary conditions, both $\psi(x)$ and $y(x)$ have standing wave solutions with quantized frequencies/energies $\omega$.
  • The density of 'stuff' in the wave is proportional to $|\psi(x)|^2$ (probability) and to $y(x)^2$ (energy).

This last point is what you found, just stated in simpler words. Physical wave(functions) must be $L^2$ functions, as probability must be normalized and energy must be finite, respectively.

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  • $\begingroup$ I can accept your answer in a minute. So does the decay of a wavefunction or a wave constitute examples of $L^2$ convergence in real life? That's what I am most confused about right now. $\endgroup$
    – Chill2Macht
    Aug 18, 2016 at 5:04
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    $\begingroup$ @William Yes, that's certainly an example. (One possibly confusing thing is that we very often talk about wave(functions) that don't converge in $L^2$, like plane waves. But that's just for convenience; such wave(functions) don't actually exist.) $\endgroup$
    – knzhou
    Aug 18, 2016 at 5:05

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