# A simple question on creation and annihilation operators

We know that the KG solution for a Spin-0 particle has the following Hamiltonian $$\hat{H}=∫ d^{3}p\frac{ω_{p}}{2}(\hat{a}_{p}\hat{a}^{\dagger}_{p}+\hat{a}^{\dagger}_{p}\hat{a}_{p})\hspace{2cm}[\hat{a}_{p},\hat{a}^{\dagger}_{p'}]=δ^{3}(p-p')\hspace{2cm}ω^{2}_{p}:=p^{2}+m^{2}$$ so we can rewrite ($$\hat{n}_{p}:=\hat{a}^{\dagger}_{p}\hat{a}_{p}$$) $$\hat{H}=∫ d^{3}p\frac{ω_{p}}{2}(2\hat{n}_{p}+[\hat{a}_{p},\hat{a}^{\dagger}_{p}])=∫ d^{3}pω_{p}(\hat{n}_{p}+\frac{1}{2})$$ My question is:-why we can use $$[\hat{a}_{p},\hat{a}^{\dagger}_{p}]=1$$ instead of the previous one $$[\hat{a}_{p},\hat{a}^{\dagger}_{p}]=δ^{3}(0)$$? Seems like we're using the commutation relations for a many particles system, so I assume that we're just omitting the discrete index because we have just one particle, right? We should have use explicitly used the notation $$[\hat{a}_{p,i},\hat{a}^{\dagger}_{p',j}]=δ^{3}(p-p')δ_{i,j}$$ but why do we omit the Dirac delta then? I think that the conclusion would be the same since the last term is still infinite, is it just a convention?

EDIT: I think the problem is that it's not the operator which is infinite, but only the expectation value right? So I should have written like $$∫ d^{3}pω_{p}(\hat{n}_{p}+\frac{δ^{3}(0)}{2})\Rightarrow <\tilde{p}|\hat{H}|\tilde{p}>=\infty$$ Please tell me if I'm wrong,

Where did you see this? You need to use: $$[a_p,a_p^\dagger]=\delta(0)$$ the other is wrong. Physically, your Hamiltonian is infinite because your volume is infinite. The relevant quantity is energy density that you obtain by “dividing” the infinite delta. This is to be anticipated by translation invariance.