My question is about the two versions of the path integral, Hamiltonian and Lagrangian, that show up in most derivation of path integral quantum mechanics, but specifically in this case the derivation presented in Altland and Simons pg. 98-101.
They consider the propagator $\langle x_f, t_f |\ U\left(t_f; t_i\right) | x_i, t_i \rangle$, discretise the path and the time evolution operator $U$ into $N$ steps and insert a lot of identities of the form $\int dx |x \rangle\langle x|$ and $\int dp |p \rangle\langle p|$ to arrive at an expression for the propagator given by:
$$\langle x_f, t_f |\ U\left(t_f; t_i\right) | x_i, t_i \rangle = \int \mathcal{Dx} \exp \left[ \frac{i}{\hbar} \int_0^t dt' ( p \dot q - H(p,q) ) \right] \ \ \text{with} H = T+V \tag{1}$$
where the $\mathcal{Dx}$ stands for a path integral over BOTH $p$ and $q$, i.e. $\mathcal{Dx} = \lim_{N \to \infty} dq_1...dq_N dp_1...dp_N$. They call this the Hamiltonian/Phase Space Path Integral and this seems to be the standard way of deriving it looking at different literature.
They then go on to draw a connection to the classical Hamiltonian and Lagrangian where $L = p\dot{q} - H(p,q)$ and inspired by this they set out to find a Lagrangian path integral. Restricting themselves to Hamiltonians quadratic in $p$ they perform the $p$ integrals of $\mathcal{Dx}$ and find a path integral with an expression in the exponent that can be identified as the classical Lagrangian:
$$\langle x_f, t_f |\ U\left(t_f; t_i\right) | x_i, t_i \rangle = \int \mathcal{Dq} \exp \left[ \frac{i}{\hbar} \int_0^t dt' L(q,\dot{q}) \right] \ \ \text{with} \ \ L = T - V \tag{2}$$
which they appropriately call the Lagrangian/Configuration Space Path Integral, which also seems to be the standard way of deriving it.
My question is:
Given that during the derivation we replaced all the operators $\hat p$ and $\hat q$ with "actual values" along a path in phase space, why can we not just directly identify $L = p\dot{q} - H(p,q)$ and apply what we have learned from Lagrangian mechanics to immediately write $L = T-V$ in equ. (1) without having to do the cumbersome $p$ integrals? Obviously this would give us something different from equ. (2):
$$\langle x_f, t_f |\ U\left(t_f; t_i\right) | x_i, t_i \rangle = \int \mathcal{Dx} \exp \left[ \frac{i}{\hbar} \int_0^t dt' ( L(q,\dot q ) \right] \tag{3}$$