Scale factor in conjugate scalar field inside conformally flat spacetime Consider a lagrangian density of a scalar field
$$ \mathscr{L} = \frac{1}{2} \partial_\alpha \phi \partial^\alpha \phi - \frac{1}{2} m^2 \phi^2 $$
inside a conformally flat spacetime with scale factor $a(t)$.
Expliciting $\mathscr{L}$ trough the auxiliary field $\chi\doteq a\phi$, and considering that $(g^{\alpha\beta})=a^{-2}\eta$ I obtain (full derivation at the bottom of the page)
$$ 2 a^4 \mathscr{L} = \partial_0 \left( \frac{\dot{a}}{a} \chi^2 \right) + \dot{\chi}^2 - \delta_{ij} \partial_i \chi \partial^j \chi - \left( a^2 m^2 - \frac{\ddot{a}}{a} \right) \chi^2 $$
Now I should discard the time derivative term so the conjugate field is simply
$$ \pi \doteq \frac{\partial\mathscr{L}}{\partial\dot{\chi}} = a^{-4} \dot{\chi} $$
but in the book I'm reading "Introduction to Quantum Effects in Gravity - Mukhanov, Winitzki" the term $a^{-4}$ doesn't appear. This has an important consequence: imposing the canonical equal time commutation relation between the auxiliary field and its conjugate expanded in their mode functions, I find
$$ \dot{v}_{\mathrm{k}} v_{\mathrm{k}}^\ast - v_{\mathrm{k}} \dot{v}_{\mathrm{k}}^\ast = 2 \mathrm{i} a^4$$
instead of
$$ \dot{v}_{\mathrm{k}} v_{\mathrm{k}}^\ast - v_{\mathrm{k}} \dot{v}_{\mathrm{k}}^\ast = 2 \mathrm{i}$$
For example, in De Sitter spacetime where $a=a_0 e^{Ht}$ my result causes mode functions to disappear when $t\to -\infty$.
I don't understand if my result is wrong, I repeated the calculation several times in these months and still obtain the same result. Where is the issue?
Here the full derivation:
$$
\mathscr{L} = \frac{1}{2} \partial_\alpha \left( a^{-1} \chi \right) \partial^\alpha \left( a^{-1} \chi \right) - \frac{1}{2} m^2 a^{-2} \chi^2
\\
\mathscr{L} = \frac{1}{2} \partial_0 \left( a^{-1} \chi \right) g^{00} \partial_0 \left( a^{-1} \chi \right) + \frac{1}{2} a^{-2} g^{ij} \partial_i \chi \partial_j \chi - \frac{1}{2} m^2 a^{-2} \chi^2
\\
\mathscr{L} = \frac{1}{2} a^{-2} \left( - \frac{\dot{a}}{a^2} \chi + a^{-1} \dot{\chi} \right)^2 - \frac{1}{2} a^{-4} \delta_{ij} \partial_i \chi \partial^j \chi - \frac{1}{2} m^2 a^{-2} \chi^2
$$
Multiplying by $2a^4$ and summing and subtracting an equal term
$$
2 a^4 \mathscr{L} = \frac{\dot{a}^2}{a^2} \chi^2 - 2 \frac{\dot{a}}{a} \chi \dot{\chi} + \dot{\chi}^2 - \delta_{ij} \partial_i \chi \partial^j \chi - m^2 a^2 \chi^2 + \left( - \frac{\ddot{a}}{a} \chi^2 + \frac{\ddot{a}}{a} \chi^2 \right)
\\
2 a^4 \mathscr{L} = \partial_0 \left( \frac{\dot{a}}{a} \chi^2 \right) + \dot{\chi}^2 - \delta_{ij} \partial_i \chi \partial^j \chi - \left( a^2 m^2 - \frac{\ddot{a}}{a} \right) \chi^2
$$
 A: This is why I prefer writing down the action of your field theory, which reads
$$S = \int d^4x \sqrt{-g}\mathscr{L}$$
One should better first simplify everything taking into account $\sqrt{-g}$ and then varying the action with respect to the fields. (In the derivation of the Euler-Lagrange equations using the above action, you need to keep $\sqrt{-g}$ when doing the integration by parts. From the derivation you will see that your canonical momentum should rather be $\pi = \sqrt{-g}\frac{\partial \mathscr{L}}{\partial \dot{\chi}}$.)
Since $\sqrt{-g}=\sqrt{-a^2(-a^2)^3}=a^4$, your issue is resolved.
More details added
Let us vary the above action with Lagrangian $\mathscr{L}(\phi, \partial \phi)$ with respect to $\phi$.
$$\delta S = \int d^4x \left[\sqrt{-g} \frac{\partial \mathscr{L}}{\partial \phi} \delta \phi + \sqrt{-g} \frac{\partial \mathscr{L}}{\partial (\partial_{\mu}\phi)} (\partial_{\mu} \delta\phi) \right]$$
Now we integrate the second term by parts:
$$\delta S = \int d^4x \left[\sqrt{-g} \frac{\partial \mathscr{L}}{\partial \phi} \delta \phi + \nabla_{\mu} \left(\sqrt{-g} \frac{\partial \mathscr{L}}{\partial (\partial_{\mu}\phi)} \delta\phi \right) -  \nabla_{\mu} \left(\sqrt{-g} \frac{\partial \mathscr{L}}{\partial (\partial_{\mu}\phi)}\right) \delta \phi \right]$$
Here we used $\nabla_{\mu}\delta \phi = \partial_{\mu} \delta \phi$ for scalar fields. Now we drop the boundary term (=second term). Demanding $\delta S=0$ yields
$$ \nabla_{\mu} \left(\sqrt{-g} \frac{\partial \mathscr{L}}{\partial (\partial_{\mu}\phi)}\right) - \sqrt{-g} \frac{\partial \mathscr{L}}{\partial \phi}=0$$
The term in the brackets is then canonical conjugate:
$$\Pi^{\mu}:=\sqrt{-g} \frac{\partial \mathscr{L}}{\partial (\partial_{\mu}\phi)}$$
In particular:
$$\Pi^{0}=\sqrt{-g} \frac{\partial \mathscr{L}}{\partial \dot{\phi}}$$
Alternatively, you can start computing the action $S$ more explicitly and you will get
$$S = \int d^4x a^4(t) \mathscr{L}_{\text{yourLagrangian}}$$
and you can work effectively with $\mathscr{L}^{\prime} :=a^4 \mathscr{L}$ using your "flat space Euler-Lagrange equation". This is why in some textbooks on QFT in curved spacetime the Lagrangian is defined by $\mathscr{L}_{\text{curved}} := \sqrt{-g}\mathscr{L}_{\text{flat}}$. I personally prefer to work with the action right away and then vary the action.
