Why we neglect the $\hbar ω/2$ in the Hamiltonian of the the Electromagnetic Field? After the quantization of the electric and the magnetic field, we get the Hamiltonian of the electromagnetic field: 
$$H= \hbar ω(a^{\dagger}a +1/2) .$$
with $\hbar$ the planck constant and $a^{\dagger}$ the creation operator.
Why can we neglect the term $\hbar ω/2$ in many cases, e.g. when we want to describe the Rabi Hamiltonian, where we just take $H= \hbar ωa^{\dagger}a$ .
 A: If the number of oscillators is finite, the term is finite and usually constant (if frequencies do not change). Thus it is just an additive constant in the Hamiltonian, which never changes anything important, so it can be dropped to make analysis less cumbersome.
If the term changes (such as due to changing boundary conditions) or if it is infinite (if we consider full Hamiltonian of EM field with no cutoffs), then the term is important. However, infinite term brings a lot of problems, although have people discovered various ways to ignore this infinity and still obtain interesting results.
A: Because the vacuum-energy (lowest energy that's being neglected) doesn't affect the anaylsis that you want to do if you're using the Rabi Hamiltonian.
A: In general the term cannot be neglected as it causes the Casimir-Polder effect. In most cases it is just a constant and then it can be ignored, except for the fact that its present theoretical  value is something like $10^{123}$. This value is ludicrously large. ZPE is an open problem of theoretical physics.    
A: The "quantization" procedure is ambiguous: it may contain terms proportional to $\hbar$ disappearing in the classical limit. So, we can consider Rabi Hamiltonian as a result of quantization too. The rest has been already explained in the previous posts.
