Probably not, but you could always define $a=\sum_n \sqrt{n}\left|n-1\right>\left<n\right|$ which acts on the eigenstates, and write $H=E(a^\dagger a)^2.$ This doesn't help you solve the problem to begin with, but it does let you write $H$ with $a,a^\dagger$ once you know the spectrum.
Assuming that you are quantizing a free massive complex scalar field, whose Hamiltonian is:
$H=\int d^{D-1} x (\partial_{\mu}\varphi^{\dagger} \partial^{\mu}\varphi+m^2\varphi^{\dagger}\varphi) $
you can see easily that if you plugin for the expression of the field in terms of the creation/annihilation operators stated above and do the integration over $x$, you will obtain the expression:
$H=\int d^{D-1} k \hspace{0.1cm}\omega_{k}(a^{\dagger}_{\mathbf{k}}a_{\mathbf{k}}+b^{\dagger}_{\mathbf{k}}b_{\mathbf{k}}) $
(the cross terms should vanish upon integration). Now the particles created by $a^{\dagger}$, $b^{\dagger}$ carry different charge under the U(1) symmetry and that's why we distinguish between the operators a and b unlike the case of the real scalar, where the particle is it's own antiparticle. Say that you wanted to quantize slightly differently with
$$\varphi(\vec{x},t) = \int\frac{d^Dk}{\sqrt{(2\pi)^D2\omega_k}}\left[a^{\dagger}(\vec{k})\mathrm{e}^{-i(\omega_kt-\vec{k}\cdot\vec{x})} + b(\vec{k})\mathrm{e}^{i(\omega_kt-\vec{k}\cdot\vec{x})}\right].$$
$a\rightarrow a^{\dagger}$, $b^{\dagger}\rightarrow b$,
you would get a Hamiltonian which is not normal-ordered, and thus the energy of the ground state would be infinite because the constant that you gain from ordering the Hamiltonian the right way is infinite, as you can see below.
$H=\int d^{D-1} k \hspace{0.1cm}\omega_{k}(a_{\mathbf{k}}a^{\dagger}_{\mathbf{k}}+b_{\mathbf{k}}b^{\dagger}_{\mathbf{k}})=\int d^{D-1} k \hspace{0.1cm}\omega_{k}(a^{\dagger}_{\mathbf{k}}a_{\mathbf{k}}+b^{\dagger}_{\mathbf{k}}b_{\mathbf{k}})+\int d^{D-1} k \hspace{0.1cm}\omega_{k}$
I guess your professor insisted on this interpretation of "positive" energy to avoid introducing normal ordering (the operation that puts all creation the left of annihilation operators).
Best Answer
In general the reason is that it was desired to find an operator solution to the equation, so that the fields could be operators in analogy to the observable operators that we have in nonrelativistic quantum mechanics.
I don't have an explicit proof of it, but a professor of mathematical physics who I spoke to told me that "putting operators in place of $a$ and $a^\dagger$ gives the most general operator solution to the equation of motion". At that point it can be shown that these operators must be annihilation and creation operators.