Free Complex scalar field and separate conservation of particle and antiparticle number For a free complex scalar field, the difference between the number of particles and antiparticles is conserved by Noether's theorem. This constraint can be satisfied in two ways:
(i) Number of particles and antiparticle remains separately conserved in time, and therefore, their difference too.
(ii) By simultaneous creation of equal number of particles and antiparticles so that the number of Particles and antiparticles need not be conserved separately.  To illustrate, suppose one started with a system of $N$ particles and $M$ antiparticles at $t=0$, such that $N-M=$conserved. The time evolution of the system might be such that it can change $N$ to $N_1=N+n$, and $M$ to $M_1=M+n$ at a later time $t>0$, keeping $N_1-M_1=N-M=$conserved. 
As far as Noether's theorem is concerned, it doesn't forbid any of the above two possibilities. But dynamics might forbid the second one.
If the possibility (ii) is forbidden by the dynamics in a free theory, why is that? How do I convince myself that the dynamics separately conserves particle and antiparticle number?
 A: In an interacting theory, violation of individual particle and antiparticle number happens all the time via pair production and annihilation.  In a free theory, each is conserved individually.  The conceptual reason is that is the dynamics of a free theory is trivial - the initial condition can be decomposed via Fourier transform into a sum of plane waves, and the plane waves just pass right through each other without "talking," so no possible process could create or destroy a particle or antiparticle.  To see this formally, just consider the Hamiltonian for a free complex scalar field in equation (3.26) of Srednicki's QFT textbook adapted to the complex case:
$$ H = \text{const.} + \frac{1}{2} \int \widetilde{dk}\ \omega(k) \left( a^\dagger(k) a(k) + b^\dagger(k) b(k) \right) := \text{const.} + \frac{1}{2} \int \widetilde{dk}\ \omega(k) \left( n^a(k) + n^b (k)\right),$$
where $a(k)$ annihilates particles and $b(k)$ antiparticles.  It's obvious that this Hamiltonian commutes with both of the total number operators $N^a := \int \widetilde{dk}\ n^a(k)$ and $N^b := \int \widetilde{dk}\ n^b(k)$, so both particle and antiparticle number are independently conserved.
A: The possibility (ii) is forbidden by the dynamics only if the dynamics is to be covariant under symmetries from the Poincare group. This is explained in detail in Weinberg's book on quantum field theory (Vol.1).
In nonrelativistic theories with Galilei covariance, particle and antiparticle number are separately conserved. 
