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The uncertainty principle holds for pairs of certain observables, such as position and momentum. All these observables have a relation to spacetime. Other particle properties, by contrast, such as mass or electric charge can be measured at arbitrary precision. Quantum theories do not even model them as observables, but as parameters to the field.

So, what is the philosophical argument for why there is a uncertainty principle for some quantities, but not for others?

Qmechanic
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    "Quantum theories do not even model them as observables, but as parameters to the field." What does this mean? Since 4-momentum is an operator, mass is clearly the eigenvalue of the operator $p^2$, isn't it? On the Hilbert space of all particles, there also evidently is an operator that assigns to every state of definite charge its charge. I don't see any reason why you'd think mass or charge are any different from any other observable. – ACuriousMind Jun 26 '23 at 19:10
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    @ACuriousMind for example, mass is a parameter to a Klein Gordon field. I have never seen it appear as an operator, multiplication by it commutes with everything – Nicolas Malebranche Jun 26 '23 at 19:25
  • While it is not in the same context, charge and magnetic flux are conjugate variables in an LC circuit. – LPZ Jun 26 '23 at 19:34
  • I don't see how such a question is answerable. It's like asking why the laws of physics are what they are. There's no reason why, it's just how this universe works. – CPlus Jun 27 '23 at 01:04
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    Related: https://physics.stackexchange.com/a/130074/150174 – More Anonymous Jun 27 '23 at 07:19

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An uncertainty relation for mass occurs with clock variables (i.e., proper time $\tau$ as an observable) in the form

$$c^{2}\,\Delta m\,\Delta\tau\geq\frac{\hbar}{2}.$$

The problem is, with clock variables, you end up sacrificing positive-definitedness of mass, or some other desired property.

The classic paper with no-go theorems on quantum clock variables:

  • William G. Unruh and Robert M. Wald, "Time and the interpretation of canonical quantum gravity". Physics Review D 40 (1989) 2598 doi:10.1103/PhysRevD.40.2598

For the mass uncertainty relation, see, e.g.,

  • Shoju Kudaka, Shuichi Matsumoto, "Uncertainty principle for proper time and mass". Part I and Part II
Alex Nelson
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  • That's just the energy/time uncertainty recasted – LolloBoldo Jun 26 '23 at 20:42
  • In Einstein's light box, mass is measured by stretching a spring. So the answer is that any practical measurement involves some quantity related to spacetime, introducing thereby one of the standard observables? And having constant mass is nothing but a nice property of the theory? – Nicolas Malebranche Jun 26 '23 at 21:32
  • @NicolasMalebranche Unless you have a good reason for the violation of the conservation of mass, we would greatly desire it in these simple models of a quantum clock using free particles. Unruh and Wald's paper demonstrates some severe problems with this whole scheme, which you should seriously examine more carefully. – Alex Nelson Jun 26 '23 at 21:46
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You are simply wrong on every point you are trying to make.

  1. If you really scrutinise things in modern physics, then there is no such thing as mass. What you really have, is the invariant or rest energy. When $c=1$, there is not even a unit difference between them.

Because of that, if you think a bit about it, the uncertainty relationship $$\tag1\Delta E\Delta t\geqslant\frac\hslash2\qquad\implies\qquad c^2\Delta m\Delta\tau\geqslant\frac\hslash2$$ which has a very pleasing interpretation: The proper time lifetime of a particle limits the possible determination of the invariant rest energy of the particle. So decays and resonant states are all energy-uncertain in a well-defined manner.

N.B. However, that the energy-time uncertainty relationship is controversial even within QM.

  1. The QFT charge operator is given by $$\tag2\hat Q=\int\frac{\mathrm d^3p}{(2\pi)^3}\hat b^\dagger_p\hat b_p-\hat c^\dagger_p\hat c_p$$ and its uncertainty is related to the number-phase uncertainty relation.

  2. You cannot ask some of these questions in basic QM because basic QM is a broken approximation working in fixed number representation, where some of these relationships are hidden because, by definition of fixed number, the uncertainty in there is zero, leading to complete indetermination of its conjugate variable. As is now obvious from QFT, once you let go of the basic QM restrictions, these apparent conjugate variables reappear.

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First, we are not philosophers. We compute math calculations to model esperimental observations.

In QFT charge and mass are different entities. Charge is an observable, the charge you refer to is the eigenvalue of the observable (operator). Mass is also observable but it's complicated.

You also don't understand the uncertainty principle.

You can measure a position as precise as you want. You can't measure position AND momentum together with arbitrary precision. If you don't care about momentum you can measure position as precise as you want.

The uncertainty principle then is used when you have a very special couple of observable to measure together. Those are like energy and time, momentum and position. Those are called conjugate variables and is the simultaneous measure of such a couple which has a restricted precision.

LolloBoldo
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    Physicist are philosophers, the Copenhagen interpretation is essentially philosophy. The charge operator of an electron is multiplication by a scalar, and has no spectrum of eigenvalues. – Nicolas Malebranche Jun 26 '23 at 19:54
  • We don't deal with it in fact – LolloBoldo Jun 26 '23 at 19:56
  • Philosophers do, we calculate transition amplitudes. – LolloBoldo Jun 26 '23 at 19:56
  • Also, an electron is an eigenstate of the U(1) charge operator. Use the same identical operator on a left handed lepton and you will see that charge is not a scalar multiplication – LolloBoldo Jun 26 '23 at 19:57
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    Aren’t you a little harsh in suggesting the OP doesn’t understand the uncertainty principle? After all, uncertainty relations do exist even for non-conjugate variables, like components of angular momentum. – ZeroTheHero Jun 26 '23 at 21:10
  • Angular momentum ones still come from the conjugate pair x,p if i recall it – LolloBoldo Jun 26 '23 at 22:41
  • but the commutator is fundamentally different: angular momenta are conjugate to angles (at least in class. mech.), not to other angular momenta. Pick spin if you insist on the $q-p$ analogy. – ZeroTheHero Jun 26 '23 at 22:48