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5

Phenomena in quantum mechanics may be expressed using any basis. It doesn't mean that all bases are equally useful for a given situation. In particular, a fundamental postulate of quantum mechanics says that right after every measurement, the system is found in one of the eigenstates of the observable that was just measured. That's why the basis of the ...


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This is just very sloppy language on the paper's part. As you say, gauge bosons are very real and their existence has physically measurable consequences (otherwise, why would we ever waste time talking about them?). (By the way "photons and electrons" are not good examples of non-gauge particles, because photons are also gauge bosons :) .) The paper just ...


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Heisenberg's uncertainty principle is in fact not a principle but a consequence of the operator formalism of QM. If we associate to the operator $X$ the standard deviation $$\Delta_X = \sqrt{ \langle{X^2}\rangle -\langle X \rangle^2}$$ it can be then shown that, given two operators $A,B$ $$\Delta_A \Delta_B \geq \frac{1}{2} \left| \langle ...


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The point is that the domain $D(P)$ of $P$ must be such that $P$ is (essentially) self-adjoint thereon. Otherwise it does not represent an observable. I am assuming that $D(X)= L^2([0,L],dx)$ instead, where $X$ is automatically self-adjoint. The vector $\psi$ you use to prove Heisenberg inequality has to belong to $D(PX) \cap D(XP)$ as you see by direct ...


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I think you are probably misinterpreting the context here. If you read the previous line carefully it says "there is always an undetermined interaction between observer and observed; there is nothing we can do to avoid the interaction or to allow for it ahead of time. And later he just says due to the fact that photon can be scattered within the 2θ' angle ...



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