Is there a uncertainty relation for charge $q$ of the form $\Delta q \Delta? \geq \hbar$ in quantum mechanics? From checking the units ($[q] = A\cdot s$) I guess that $?$ would have to be the magnetic flux $[\Phi] = V \cdot s$, so we would have $$ \Delta q \Delta \Phi \geq \hbar $$ If I play the same game with mass $m$ the units would suggest a uncertainty relation $$ \Delta m \Delta \Phi_g \geq \hbar $$ where $[\Phi_g] = \frac{m^2}{s}$ is the gravitomagnetic flux (which happens to have the same units as the kinematic viscosity, the specific angular momentum, the mass diffusivity and the thermal diffusivity).

Do these two uncertainty relations exist in quantum mechanics?


Regarding electrical charge the answer is definitely negative: In Quantum Mechanics there exists a so-called superselection rule of the charge which requires that the charge is always definite in every quantum state of any quantum physical system carrying electrical charge. So $\Delta q_\psi =0$ in every state $\psi$ and no Heisenberg relations are possible for whatever choice of a conjugated variable of $q$.

Regarding mass, in non relativistic QM the situation is identical: There is an analogous superselection rule (Bargmann's superselection rule).

In the quantum relativistic realm, for elementary systems the mass observable is actually a Casimir operator and thus, again it is always defined.

For composed systems or non-elementary relativistic systems, the situation is more delicate, but I think the only possibility to define some notion of mass is just the Hamiltonian observable defined with respect to a preferred reference frame at rest with the system. This way one immediately faces the problem of time-energy uncertainty relations which do not have a well-defined status in QM, since there is no self-adjoint operator representing the time observable (this is the so called Pauli's theorem).

  • $\begingroup$ Does this imply that there must also be a superselection rule for the conjugate variable (magnetic or gravitomagnetic flux) in non relativistic QM? $\endgroup$ – asmaier Aug 20 '16 at 17:24
  • $\begingroup$ No, exactly the contrary: as $\Delta q=0$, for the conjugate variable we would have $\Delta X = +\infty$ and thus no state at definite $X$ may exist in contradiction with a superselection rule for $X$. $\endgroup$ – Valter Moretti Aug 20 '16 at 17:26

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