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I read that:

The smallest wave packet we can build has a size on the order of the de Broglie wavelength $\lambda$ of a free particle moving with the same speed $v$.

I haven't been able to find a clear explanation of why this is. The uncertainty or spread (standard deviation) in position should be something like $$\Delta x \geq \frac{\hbar}{2\Delta p}$$ but I don't see why this necessarily directly relates to the de Broglie wavelength of the momentum (or average of the momentum as it only has a definite value for an infinite plane wave I suppose), and why it means that particles moving more quickly can have more localized wavepackets. Any clear explanation would be much appreciated. (I don't mind a bit of Fourier analysis but I am a bit rusty!)

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As the fourier transform is given by,

$$f(x,t) = \int \mathrm{d}\omega \, \mathrm{d}k \; f(k,\omega) \, e^{i(kx-\omega t)}$$

we see that $x$ and the wavenumber must be related to each other by $x \sim k^{-1}$ as the argument of the exponential must always be dimensionless. The wavenumber is then used to trivially define a wavelength $\lambda \sim k^{-1}$ (which is the same information basically, as we are still in Fourier space). But now as $\lambda$ must have the dimensions of $x$ one often handwavingly equalizes both variables and says that $\Delta x$ is the spread between two wave maxima. This decreases obviously with the uncertainty principle as stated above and increasing velocity or momentum (let's stay classical).

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