When I studied a spin, the textbook said spin is an intrinsic quantity like mass. However, while we can calculate just expectation values $ \langle \textrm{S}^2\rangle $ or $ \langle S_z\rangle $, the mass $m$ seems to be fixed everywhere in quantum mechanics. Is it right? If so, why isn't there the uncertainty about the mass of particles?
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The rest mass is fixed because it's a law of physics. For all states that are allowed in Nature, the rest mass of an electron is always 511 keV. So the mass $m$ is an operator on the Hilbert space that is proportional to the unit operator/matrix. We say that it is a $c$-number, a classical number. The equally big value is the only eigenvalue of the operator. Every state of the Hilbert space is an eigenstate of this operator with the same eigenvalue. Almost just like $\pi$, is a fixed parameter of the laws of physics, not a dynamical variable. – Luboš Motl Aug 28 '15 at 09:22
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Apologies, @AcidJazz, I probably don't understand the question. The constancy of electron's mass and the attractiveness of gravity are 2 laws of physics. They're different laws but their being laws means the "same thing", like for all other laws of physics. I don't understand what the other thing that a law of physics could mean, "set at that for experimental work", could mean. All laws of physics are reversely extracted from the experiments. We're never sure that they're the 100% final answer. They rarely are. But they're still laws of physics. – Luboš Motl Aug 28 '15 at 14:36
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The numerical value of the electron mass depends on the units, obviously. But once you fix your units by a well-defined definition, the numerical value is completely universal. Nature guarantees that. One may derive that assuming that our Standard Model etc. is right. And one may also observe the constancy of the electron masses from observations. It's in principle plausible that the mass will be observed to be non-constant in the future, and a more accurate theory will have to be modified, but this is a pure speculation unbacked by anything. – Luboš Motl Aug 28 '15 at 14:37
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When we describe the observations - e.g. a quantum system by a Hilbert space and operators - the whole formalism implicitly assumes that all such things we haven't observed yet and we have no reason to believe to be observed in the future (like the variability of the electron mass) - are indeed absent. So the formalism doesn't allow $m_e$ to be anything else than a fixed parameter whose value is simply inserted. It's a universal constant like $\pi$ except that the value is less unique and it's not mathematically calculable out of nothing (so far). – Luboš Motl Aug 28 '15 at 14:40
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@LubošMotl I am very sorry about that, I have the flu, and my question is worded dreadfully, it does not make sense to me at the moment either. I will delete it, thanks for your time – Aug 28 '15 at 15:40
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Please get well and don't erase questions. Maybe someone will understand both of us. ;-) – Luboš Motl Aug 28 '15 at 16:48
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Possible duplicate: https://physics.stackexchange.com/q/19424/2451 – Qmechanic Feb 15 '23 at 02:43
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The mass is given by the Casimir operator $\hat{P^2}$, the square of the momentum, generated by translation invariance. This will give you the actual mass of the particle, as opposed to the bare mass that you will find in the theory, as the actual mass can be affected by the particle's interactions.
Slereah
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1Well, yes and no. The operator $P_\mu P^\mu$ is an operator but relativistic laws mean that in a free theory, it's equal to the $c$-number $m^2$ for each particle. And when we do include the interactions, $P_\mu P^\mu$ becomes a rather hard to interpret operator because in the interacting theory, single- and multi-particle states gets mixed into superpositions. There are different values of masses we insert or measure - bare, pole mass, one in this or that renormalization scheme - but the difference between them has nothing to do with their being $c$-numbers. – Luboš Motl Aug 28 '15 at 14:32