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More specifically, I want to understand why a wave is a wave but a wave packet is not considered a wave (as discussed in this question).

I would think that if something have these characteristics: 1. Wavelength; 2. Frequency; 3. Period; 4. Amplitude; and 5. Wave velocity; then it must be a wave. Does a wave packet have these characteristics? Do you have a better set of rules to apply to test if something is a wave?

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A wave packet does not have 1 wavelength, frequency or period. It is built up by the superposition of many waves of single frequency wavelength period.It has a velocity, not the one in a single exponent, but wave packet velocity. An interference phenomenon will tell you of the wave nature of the wave packet. –  anna v Sep 29 '12 at 3:37
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First, I would note that physicists are never strictly particular about their terminology, just about their math. So whatever definition you may get, you might well encounter some context in which someone uses the word "wave" to mean something else.

That being said, the closest thing to a consistent mathematical definition of "wave" in physics is a solution to a wave equation, which in two dimensions takes the general form

$$\frac{\partial^2 f(x,t)}{\partial t^2} - c^2\frac{\partial^2 f(x,t)}{\partial x^2} = -\mu^2c^2 \bigl(f(x,t) - \langle f(x,t)\rangle\bigr)$$

In many cases $\mu = 0$ so the right side disappears.

According to this definition, a wave packet is a wave, since the function that describes the wave packet satisfies the equation.

However, it's often the case that when physicists say "wave," they're talking about a particular kind of solution to the wave equation which is described by the equation

$$f(x,t) = f_+(x + vt) + f_-(x - vt)$$

(still in two dimensions). This consists of a rightward-moving pulse and a leftward-moving pulse, each moving with speed $v$. For a system where $\mu\neq 0$, not all solutions to the wave equation are of this type. In particular, the wave packets you're asking about consist of a combination of pulses with different speeds. So a wave packet does not keep its shape as it travels, and it would not satisfy the definition of a wave in this sense.

In yet other cases, when physicists say "wave" they mean a plane wave, which is something even more specific: it's a sinusoidal function.

$$f(x,t) = A\sin[k(x - vt) + \phi]$$

Naturally, a wave packet does not have to be (and in fact cannot be) sinusoidal, and so it will not be a plane wave.

That may prompt the question, what exactly does the "packet" part of "wave packet" mean? It means a wave that is localized in space, in the sense that outside of some finite region of space, the further away you go from the region, the smaller the amplitude of the wave. Through the "magic" of Fourier analysis, you can prove based on that definition that a wave packet cannot be a plane wave, and also that it cannot be the $f_+ + f_-$ type of wave unless $\mu = 0$ in the wave equation.

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