When we calculate the total charge and energy of a quantum field by using Noether's theorem, we find that they are infinitely large, even if we consider a finite spacetime volume_ $$H = \int_V \mathrm{d }k^3 \frac{\omega_k}{(2\pi)^3 } a^\dagger(\vec k) a(\vec k)+ \frac{1}{2} \int_V \mathrm{d }k^3 \omega_k \delta(\vec 0) \, . $$ It it sometimes argued that these infinite contribution can be removed by renormalizing the ground state charge and energy.
Often this problem is dealt with by normal ordering the Hamiltonian. But I was wondering if it is also possible to introduce a bare ground state charge $Q_0$ and energy $E_0$ that allows us to absorb the infinite contributions by setting the value of the measurable ground state charge and energy to the observed value (zero) $$ Q \equiv Q_0 + C \int_V \delta(0) =0 \, . $$ $$ E \equiv E_0 + \tilde{C} \int_V \delta(0) =0 \, . $$ This would be a lot more similar to how we usually renormalize but I'm struggling to understand where exactly $Q_0$, $E_0$ appear in the theory.
Usually, the bare parameters are equal to parameters in the Lagrangian. This suggests that there areadditional term in the Lagrangian $$ \mathcal{L} = \ldots + Q_0 \times ? + E_0 \times ?$$ that yields constant contribution to the Noether charges. For example, instead of the formula given above for the Hamiltonian we find $$H = \int_V \mathrm{d }k^3 \frac{\omega_k}{(2\pi)^3 } a^\dagger(\vec k) a(\vec k)+ \frac{1}{2} \int_V \mathrm{d }k^3 \omega_k \delta(\vec 0) + E_0 \, . $$ This could be used to renormalize the ground state energy and charge as described above.
Does this kind of approach make sense? And if yes, is it discussed in more explicit terms somewhere?
