Measurement of Mass and Momentum of a particle simultaneously In quantum mechanics can the mass and the linear momentum of a particle be measured precisely or do they commute ?
 A: I'll provide an answer in non-relativistic quantum mechanics.
The short answer is that momentum and mass commute, so a particle can have a well-defined momentum and mass simultaneously.
But really, mass isn't considered an operator in quantum mechanics; it's a parameter, a number. So for some system, it is presumed that the mass is known always. There's no "mass measuring operator" that could give different results. Mass is input into quantum mechanics, not a measurement in the traditional QM sense.
A: Well, the two are constrained by
$$ p_{\mu}p^{\mu} = -m^2 $$
or
$$ -\mathcal{E}_{\vec{p}}^2 + \vec{p}^2 = -m^2 $$ so I would say that yes they can.
Let us consider the types of interactions people make in the LHC at CERN.
I am not an experimentalist, but from my undergrad particle physics course, I seem to remember that linear-momentum can be measured (at least for electromagnetically charged particles) by observing the bending of the particle's path in a magnetic field.
Then the energy of the particle can be measured by adding up the energy of all the 'stuff' that comes out of the interaction, using all sorts of fancy detectors.
If you measure the linear-momentum, and you measure the energy, then you can calculate (which is, to be fair, the way that the linear-momentum is calculated from the measured bend in the path) the particle mass.
(See here for a description of one such type, namely a Calorimeter.)
Of course this will be constrained by the systematic errors of how good your measurement device is.
You refer to the uncertainty principle, that is
$$[ \vec{x}, \vec{p} ]  \neq 0 $$
for example, such that we can't measure both the position and linear-momentum precisely at the same time.
However the mass is just a scalar real-valued number (at least for all Standard Model particles). Thus it will have trivial commutator relations.
In fact, as a proof we could write
$$ [m^2,\vec{p}] = [\mathcal{E}_{\vec{p}}^2 - \vec{p}^2,\vec{p}] = [\mathcal{E}_{\vec{p}}^2, \vec{p} ] - [\vec{p}^2,\vec{p}] = (0) - (0) = 0 $$
This feels a little silly though, since the third line and the first line are esentially saying the same thing, that is, a scalar will have vanishing commutator.
