A free quantum field is usually defined as a distribution, a linear map from a space of test functions $\mathcal{F}$ into an algebra of operators $\mathcal{A}$. We can write $\hat\phi:\mathcal{F}\rightarrow\mathcal{A}; f\mapsto\hat\phi_f=a_{f^*}+a^\dagger_f$. We take the vacuum vector to be annihilated by $a_f$, $a_f|0\rangle=0$. Using $\hat\phi_f$ we can write down, for example, a one--particle vector state $\hat\phi_f|0\rangle=|f\rangle$. We could construct a free quantum field at a point as $\hat\phi(x)=\hat\phi_{\delta_x}$, except that this is *improper *insofar as the space of test functions is usually taken not to include distributions such as the Dirac delta function $\delta_x(z)$, or we can work with $\hat\phi(x)$ provided we keep clear in our minds that it is a distribution, not a function, with correspondingly different rules.

For a mass $m$ real Klein-Gordon quantum field operator $\hat\phi_f$, we can write the commutator as $[\hat\phi_f,\hat\phi_g]=(f^*,g)-(g^*,f)$, where the two--particle vacuum expectation value (VEV) $\langle 0|\hat\phi^\dagger_f\hat\phi_g|0\rangle=\langle 0|a_fa^\dagger_g|0\rangle=(f,g)$ is a positive semi--definite inner product $$(f,g)=\int \tilde f^*(k)2\pi\delta(k\cdot k-m^2)\theta(k_0)\tilde g(k)\frac{\mathrm{d}^4k}{(2\pi)^4}.$$

This is a manifestly Lorentz and translation invariant construction, which, loosely, projects the test functions $f$ and $g$ to specific, positive-frequency (because of the ${\small \theta(k_0)}$) solutions of the Klein--Gordon equation (because of the ${\small 2\pi\delta(k\cdot k-m^2)}$). Without the introduction of components (as for the electromagnetic field, say) $(f,g)$ is the only positive semi--definite inner product with the required properties that we can write down. The properties of the Klein--Gordon quantum field are fixed by this construction (together with common ideas about algebras and Hilbert spaces).

Whether you will like this construction better than what you will find in most textbooks, and whether you will find it helpful, will depend on how you like your mathematics. The problem with this construction, however, is that interacting quantum fields are constructed heuristically using $\hat\phi(x)$, they cannot be constructed using $\hat\phi_f$.