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!)