Since the momentum operator P commutes with the Hamiltonian H, it does not change with time. I.e., it is conserved. Accordingly, a basis for stationary particle states can be labelled by their momentum eigenvalues.

It is unclear what you intend to say with your two follow-up sentences. In LHC the particles do experience external forces, since they complete a full circle of 26,659 m in about 18 microseconds (about 11 250 revolutions per second --- crossing the Swiss-French border 22 500 times per second without a passport!). Measured on a human scale, this corresponds to a very large acceleration.

However, the relevant unit of time for hadronic processes is 10^(-24) s (or a few orders of magnitude smaller), and the relevant unit of length is 10^(-15) m (or a few orders of magnitude smaller). Measured in such units the acceleration is essentially zero, and the conservation of momentum in LHC is one of the few things in life you can really rely on.

Momentum conservation is a consequence of the invariance of physical laws with respect to spatial translations, while the conservation of energy is a consequence of the invariance of physical laws w.r.t. the time translations. The invariance of physical laws w.r.t. spatial and time translations holds both, in the classical and in the quantum physics. In particular, in the quantum field theory it manifests itself via Heisenberg equation for the field operators Phi(x): 

                  [Pmu ,Phi(x)]=idmuPhi(x)   (here dmu is a partial derivative),

where Pmu is an operator of a 4-momentum of the field, and [a ,b ] means commutator of operators a and b, ab-ba.

Note that in quantum mechanics p is an operator. When it acts on a state, the resulting eigenvalue gives us the momentum of that state.

According to Ehrenfest's theorem, expectation value of quantum mechanichal observable p  obey the following:

d/dt

= - < del V /del x>

here p is an quantum mechanical operator

is the expectation value of the operator p. Also V is a classical potential which is a polynomial in x. If you put V=0 then

d/dt

=0

This means that the expectation value of the momentum operator is unchanged if the particle state in not in a force field.

The answer, in quantum, as well as classical mechanics, is, of course, Yes, since, in the absence of a potential, that's a function of the position(s), the Hamiltonian is a function of the norm of the momentum and thus commutes with the momentum operator, thus momentum is conserved. This is true in non-relativistic, as well as relativistic, mechanics. However the partition of the momentum among the particles that participate in the process is much more complicated and interesting and its study can provide insight in the dynamics, beyond the global conservation laws, deduced from Galilean or Poincaré invariance. 

The cross section of any process *is* the probability, for the process to occur,  appropriately normalized. In particle collisions, such as those that take place in accelerators, total momentum is conserved, due to global Poincaré invariance, that's a symmetry of the interactions (were the processes non-relativistic, total momentum would be conserved from global Galilean invariance)-and what's interesting is its distribution among the particles that participate in the process.

What's important to note here is that it's impossible to write a potential term, i.e. a function of the coordinates, that does not commute with the momentum operator, for a finite number of particles, in a way consistent with Poincaré invariance-the so called ``no interaction theorem'', first proved by Leutwyler in 1964. Indeed, one needs a field, that creates and annihilates an indefinite number of particles, in order to write relativistically invariant interactions. The intuitive reason is simple: if more than one particles are subject to forces, then they can attain an enegy greater than their rest mass and create more particles-already classically-unless the interaction can be shown to define an integrable system, i.e. a non-interacting system in disguise. One particle can't do this, because the conservation laws don't allow it, even in the presence of an external field. 

Just a small comment to add to what has already been said. The condition for the conservation of the total momentum for a real manybody Hamiltonian with interacting particles, i.e. H*=H is that the many electron wave function also is real or can be chosen so in the case of a degeneracy. The real character of the wavefunction automatically imparts total momentum conservation.

This is a rather general theorem and it also follows, in the case of bound states, that using the linear variational principle one can further show that the real or imaginary part of the wavefunction optimizes its expectation value. So the theorem also holds for trial or approximate bound state wave functions as well. However, in a non-linear variational situation, like the Hartree Fock etc. the theorem may not hold.

Finally even if the total momentum is conserved, the individual particles do not, in general, have to uphold this conservation law.

In fact, conservation of the total momentum means not that the wavefunction is real, but that a space-like translation of the coordinates of all the particles implies that it gets multiplied by a globally defined phase, that, therefore,  does not contribute to any physical quantity. This statement holds in non-relativistic as well as relativistic, quantum mechanics. 

Just for the sake of completeness: neither momentum nor energy of a free particle is conserved in General Relativity due to the universe expansion, which destroys the translational invariance of physical laws.

     A trivial example: relic radiation.

Stam,

Since pioneering QM imparts that the multiplication of a complex phase factor does not  violate the eigenvalue relation it is in agreement with what you just said. Furthermore since the momentum, being the conjugate quantity of operator x, conservation of momentum is of course a consequence of Noether's theorem. Nevertheless one can prove that under quite general conditions it imparts that the wavefunction must be real, see appendix of the IJQC paper enclosed.

I should perhaps have said that the wave function is real mod a complex factor. Hence a complex wavefunction must have real and imaginary parts that are not linearly dependent. Then a proof of the theorem I referred to are a consequence of Ehrenfest's relations.

Actually I did some work on this long time ago, see enclosed J. Mol. Spectry paper also enclosed.

Best erkki

In CERN when high energy quarks collide they themselves carry color charge, as well as electroweak charges (e.g hypercharge). Therefore they are not free particles at all. According to Noether's theorem color charge and electroweak charges are conserved at each vertex of interaction. But then again if the vertex is fixed then there is a large uncertainty in momentum. While calculating the cross section one has to integrate over all possible momenta because the interaction vertex is fixed.

In the question you mention quantum mechanics of possibility. You are right. If you fix the point of interaction, you must take into account the spread over momentum, and consequently integrate the scattering cross section over the allowed range of momentum.

The fact that the particles at the CERN accelerators are or  aren't free, indeed,  isn't relevant-what's relevant is that their interactions are invariant under global Poincaré transformations, which means that the total momentum is conserved (from Noether's theorem, for on-shell states, from the identities satisfied by the correlation functions, that express the fact that the partition function is invariant under Lorentz transformations of the fields for off-shell states). 

Indeed, by the Coleman-Mandula theorem, the conservation of the charges of internal symmetries is independent of the conservation of the charges under spacetime transformations. 

The question was: In case of no external field of force, is the total momentum conserved in quantum mechanics and if so why?

Without external interaction the quantum mechanical many-body system can be represented by a transformation from the laboratory system (assuming the potential being invariant under translation and rotation of the latter) to the center of mass system, with the translational motion of the total center of mass rigorously separated off. Hence the internal motion will be represented by a coordinate system with origo at the center of total mass.

The price to be paid is that electrons, nuclei etc. contain reduced masses and the kinetic energy will contain mass polarization terms (however no coupling between electrons and nuclei). Nevertheless the kinetic energy of the total system is conserved, the center of mass moves as a free particle (continuous energy spectrum) while the internal motion gives rise to discrete energies.

The center of mass can therefore be chosen at rest for the description of the internal motion, and in this representation the total momentum is zero.

Binding energies will be heavily dependent on the correct treatment using reduced masses, and furthermore accurate calculations by Kolos and Wolniewicz brought the QM analysis to such an accuracy (hydrogen molecule) that Herzberg and Monfils had to redo the experiment finally finding excellent agreement with QM!

My previous posting regarding the proof of the wavefunction being real, thus concerns the character of a stationary state

=0 (from Ehrenfest's relations).

Spiro,

If somebody not yet participated in this thread is obstructing a serious RG conversation by anonymous downvoting general information from an interdisciplinary perspective it is against the idea of free communication and detrimental to scientific exchange!

Maybe you should intervene or protest as being the one asking the question and requesting supplementary information.

I can possibly understand downvoting when there are conflicting views and some hard words are exchanged, but not here!

Poincaré transformations include 4 more translations (one for time) over the set of Lorentz transformations. Time translation invariance of a system give conservation of energy and space translation invariance gives conservation of momentum of the center of mass. The center of mass moves with a uniform velocity in 3D, then there is a constant kinetic energy term. This is a constant piece of Lagrangian which disappears in the dynamics i.e  equations of motion. However such constant energy term in principle can have observable effects elsewhere, which are unrelated to the equations of motion. A similar case involves  the constant term in  ground state energy of the harmonic oscillator.  This cannot be determined from spectroscopic analysis which focus on differences between energy levels of different states of a particle. Sum of all ground state energies of oscillators, however, leads to vacuum energy, which is physically significant.

Yes, you can find why it is so on any quantum mechanics book. See, for instance, the one by Cohen

https://books.google.com.br/books?id=e_Ec0IlSbMMC&dq=quantum+mechanics+cohen&hl=pt-BR&source=gbs_navlinks_s