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The plane waves must be momentum eigenstates $\langle x| k\rangle= e^{ikx}$ normalized so that $$ \langle k|k'\rangle=\int_{-\infty}^{\infty}\langle k| x\rangle \langle x| k'\rangle \,dx = \int_{-\infty}^{\infty} e^{-ikx}e^{ik'x} \,dx = 2\pi \delta(k-k') $$ and with completeness relation $$ {\mathbb I}= \int_{-\infty}^{\infty}\frac {dk}{2\pi} |k\rangle\...


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In the mainstream standard model of particle physics, all matter is made up of point particles with a fixed mass which we measure as best as we can within our experimental errors. There is no width in this masses at the table. Protons and (and neutrons bound in a nucleus) are stable composite particles , made up by a great multitude of quarks antiquarks ...


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You can only absorb energy from the electromagnetic field in units of the photon energy. This is why we call the photon the quantum of electromagnetic energy. Any photodetector that works by absorbing energy from the field it's measuring thus only absorbs that energy in discrete quantities, aka photons. That means that most of the common photodetector ...


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You are correct that having a particle completely at rest is impossible, because of the uncertainty principle: the uncertainties in position and velocity must be related by $\Delta x \Delta v \geq \hbar/2m$. So whenever we say that a proton is at rest, we are doing an approximation: we are neglecting the uncertainty in velocity, compared to other quantities ...


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The theory of special relativity (TSR) doesn't know about quantum mechanics at all. In TSR when you change your frame such that the proton is at rest, it is truly at rest with $v=0$ in the framework of this theory. 'Reality' in physics doesn't exist, only models of it, that describe it at hopefully better and better accuracy. Of course you could ask then, ...


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