# What does energy, momentum, etc mean in quantum mechanics?

I see all those nice operators that are used to give the expectation values of dynamical varaibeles, but what does it actually mean to measure kinetic energy, momentum, etc of a particle? How is it done? For instance, is the kinetic energy just $$(1/2)mv^2$$, so I havd to measure the velocity of a particle and put it into the formula to know the kinetic energy at that moment, or is this not the way it's done?

Elementary particles are quantum mechanical entities. The quantum field theory model of describing their interactions developed in order to fit experimentally found values of crossections and decays. Experiments are designed to measure energy and momentum of the participating particles in the interaction, i.e. the four vectors of the observed particles.

For clarity let us take an event from a bubble chamber experiment, a famous one because it is a first observation of the $$Ω^-$$ particle.

It is interpreted as particles:

$${\newcommand{Subreaction}[2]{{\rlap{\hspace{0.38em} \lower{25px}{{\rlap{\rule{1px}{20px}}} {\lower{0.5ex}{\hspace{-1px} \longrightarrow {#2}}}}}} {#1} }} {K}^{-} ~~ p ~~ {\longrightarrow} ~~ {\Subreaction{{\Omega}^{-}}{ {\Subreaction{{\Lambda}^{0}}{p ~~ {\pi}^{-}}} ~~ {K}^{-}}} ~~ {K}^{+} ~~ {K}^{+} ~~ {\pi}^{-}$$

and shows the generation and decay of an $${\Omega}^{-},$$ the particle that fills up the prediction in the decuplet of hadrons.

The four momentum of the classically behaving tracks out of the interaction vertex are found by using the classical equations of charged particles in a magnetic field,

$$Bqv=mv^2/r$$

and using the ionization curves to get at the different mass identifications.

The quantum mechanical interaction, the creation of an $${\Omega}^{-}$$ is modeled with quantum field theory (QFT) and the crossection computed , i.e the probability of this interaction happening. Also the decay of an $${\Omega}^{-}$$ can be calculated with QFT.