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**We don't *need* to, it just makes life easier (at least some of the time)**

As $\mathbf{k}$ is a 3D vector and $x$ a 4D vector, the answer doesn't appear quite as simple as 'Fourier Transforming makes them equivalent ways of doing the same thing, just in different spaces', but that essentially is the idea nonetheless. At $t=0$, these are just Fourier Transforms of each other, so $\varphi$ is, at least in some sense, the time evolved Fourier Transform of $a$.

Let's stick with the initial examples in the textbook and question: the hermitian free field $\varphi\left(x\right) = \varphi^+\left(x\right) + \varphi^-\left(x\right)$ and your example $H_1 = a^\dagger\left(\mathbf{k}\right) + a\left(\mathbf{k}\right)$. We can write the hermitian free field as \begin{align*}
\varphi\left(x\right) = \int e^{ikx}a\left(\mathbf{k}\right) + e^{-ikx}a^\dagger\left(\mathbf{k}\right)\\,d\tilde{k}
\end{align*} and we can also write $H_1$ as \begin{align*}
H_1 &= i\int e^{-ikx}\overleftrightarrow{\partial_0}\varphi\left(x\right) - e^{ikx}\overleftrightarrow{\partial_0}\varphi^\dagger\left(x\right)\\, d^3x.
\end{align*}

Both are fine, both make sense, having jumped over the hurdle of 'Hermiticity' in quantum physics. However, We have yet to add the extra bit of Lorentz invariance that allows things to be considered as proper QFTs. Interestingly enough, this places limits on the *spatial* interaction - the Hamiltonian must commute at spacelike separated points.

Srednicki shows that this works when $\varphi$ is a real scalar field, in which case it is 'fundamental' as it is easier to describe valid models in this way. However, creation and annihilation operators as functions of frequency have no such restrictions by default and so, require *additional* restrictions in order to be valid. Interestingly enough, working through (the Fourier Transformed version of) $\left[H_1\left(k\right),H_1\left(k'\right)\right]_{\pm}$ at $t=0$ gives terms

$$\int e^{-i\left(kx-k'x'\right)}\left[\varphi\left(x\right),\varphi^\dagger\left(x'\right)\right]_{\pm}\\,d^3x\\,d^3x'$$

which again leads us back to the requirement that $\varphi$ must be a commuting field.

In essence, you can just write whatever Hamiltonian in terms of $a\left(\mathbf{k}\right)$ and $a^\dagger\left(\mathbf{k}\right)$ using time-evolved Fourier Transforms but Lorentz invariance gives restrictions on $\varphi$ (although you could argue that this is because we've already imposed similar conditions on $a$). This means that whatever we write with a valid $\varphi$ is valid by default, so is considered 'more fundamental' and is also generally easier to work with.