I have a lagrangian
$$ L(x^{a}, \dot{x}^{a}, t), $$
which is non-degenerate, quadratic in the fields, and contains an explicit dependence on the evolution parameter $t$.
If $L$ was time-independent, I would follow the following algorithm to deduce the quantum theory:
- Define the canonical momenta as $p_a = \partial L / \partial \dot{x}^{a}$ and the phase space as a decompactified limit of the symplectic manifolds with the canonical symplectic form given by $\left\{ x^{a}, p_{b} \right\} = \delta^{a}_{b}$.
- Find a linear combination $a^{a} = \alpha x^{a} + \beta p^{a}$ such that $\left\{a^{a}, a^{*\,b}\right\} = (i \hbar)^{-1} \delta^a_b.$
- Promote $a$ and $a^{*}$ to operators $\hat{a}$ and $\hat{a}^{\dagger}$ and define $\left|0 \right>$ as an element of the Hilbert space annihilated by all $\hat{a}$.
- Build the direct representation of $\hat{a}$, $\hat{a}^{\dagger}$ on $\mathcal{H}$ using the commutation relation $[\hat{a}, \hat{a}^{\dagger}] = 1$ and re-interpret $\hat{a}$, $\hat{a}^{\dagger}$ as annihilation and creation operators.
- For any classical observable use the Weyl quantization map to assign an operator to it. Because $a$ and $a^{*}$ are linear in $x$ and $p$, the Weyl quantization map ensures that the same relation holds for operators, thus proving consistency of the quantization scheme. Re-express all resulting operators in terms of $\hat{a}$ and $\hat{a}^{\dagger}$ to get their explicit representation on $\mathcal{H}$.
That works well for non-degenerate quadratic in $x$, $\dot{x}$ lagrangians with no explicit $t$ dependence.
I want to know how the recipe above changes (or if it is possible to use it at all) when $L$ contains an explicit time dependence. For simplicity, let's assume that it is still non-degenerate and quadratic in $x$, $\dot{x}$.