Why does superposing an infinite number of waves of different wavenumbers eliminate periodicity and may sometimes result in a localised wave?

Question

I am studying how wave packets are defined in quantum mechanics, but I am finding it hard to intuitively understand why superposing an infinite number of waves of different wavenumbers $k$ may sometimes result in a wave packet that is localised.

I understand that constructive and destructive interference take place outside of the wave packet due to many different waves being superposed, resulting in an amplitude of 0. But whey does this not continue through the wave packet? So why does the wave packet exist at all?

In this picture there are many waves being superposed, but there are two points as you can see where the waves interfere constructively, and this results in a localised wave packet, and the more waves you superposed the more spread out these wave packets would be. But my question is: Why, when superposing an infinite number of waves, there could still be one wave packet remaining?

Answer 1

Existing answer is good. I will add some remarks which I hope may further clarify.

1. The question is not essentially about quantum physics. It is about the use of Fourier series and Fourier transforms in any area of physics (or science more generally).

2. The question arises from an intuition that one should not be able to make a localised peak function from a sum of periodic functions. This is indeed true if there is a finite number of terms in the sum. So the question concerns an example of a counter-intuitive fact involving infinity.

3. The physicist will also be concerned with another issue that might or might not bother a mathematician. Which of the functions here is 'physical' in the sense of 'a physical system might be accurately described by this function on its own'? The answer is that strictly periodic functions are unphysical because they involve an infinite number of repetitions, and this does not occur in the natural world. No water wave goes on forever, and neither does any sound wave or light wave or electronic oscillation. Also, in quantum physics no particle ever quite gets into a perfect momentum eigenstate. So when expressing a localised peak function as a sum of sins or cosines, we are doing a somewhat questionable thing: we are writing a physically possible function as a sum of physically impossible ones! We get away with this because there are any number of ways to show it will be ok. One way is to say we are not using strict sin functions that go on forever, we are just using functions that are like sin functions in a region extending well beyond the region of interest, and elsewhere they go smoothly to zero. Another way is to use strict sin function and simply trust the mathematics concerning infinity.

Answer 2

Consider a superposition of the form $$\psi(x)=\sum_{j=1}^n\cos\left(\frac{2\pi j x}{L}\right)$$ Because I chose cosines, this function will automatically form a peak around $x=0$. This function will periodic with period $L$ (convince yourself of this). The period is $L$, because the smallest factor that is multiplied by $x$ in this sum is $2\pi/L$. If I make this factor smaller, by making $L$ larger for example, the wavelength/period of my final function will increase. Now imagine going to the continuum by taking an integral: $$\psi(x)=\int\mathrm dk\, \phi(k)e^{ikx}$$ The value that is multiplied by $x$ is now $k$. Since $k$ can become arbitrarily small in this integral, the period of the resulting function will generally be infinite for an arbitrary function $\phi(k)$. This is why the wavepacket can be non-periodic.

On a related note, consider the identity $$\int\mathrm dk\,e^{ikx}=2\pi\delta(x).$$ Although mathematicians would not like this identity, it tells a lot. It says that for $k\neq0$, integrating a rotating phase factor over all space gives zero$^\dagger$. This might motivate why the wavepacket tends to zero outside of its peak.

$\dagger$ This should be understood in a distributional sense. This means that you should only consider this as true when it appears inside an integral. This is just a disclaimer for the mathematicians