I'm attempting to understand how to motivate quantum mechanics through the use of path integrals. Because the path integral approach provides such a direct connection to classical mechanics (via the principle of least action), and because it provides such a direct physical understanding of the difference between classical and quantum mechanics, I think it's a really valuable lens through which to understand quantum theory. However, I'm having some conceptual difficulties regarding the introduction of spin.
The problem is as follows: suppose we know nothing about quantum mechanics except that the space of states is a complex Hilbert space whose inner product maps quantum states to probability amplitudes. Given that a system is in some quantum state $|\psi,t_0\rangle$, we can write the state of the system at some future time $t$ as $$|\psi,t\rangle = \hat{U}(t,t_0)|\psi,t_0\rangle.$$ We know that $\hat{U}$ must be a unitary operator due to the probabilistic interpretation of the inner product $\langle\psi|\psi\rangle$, and therefore that $$U(t_0,t) = U^\dagger(t,t_0).$$ Taking a time derivative, we have that $$\frac{\partial}{\partial t}|\psi,t\rangle = \frac{\partial\hat{U}(t,t_0)}{\partial t}|\psi,t_0\rangle = \frac{\partial\hat{U}(t,t_0)}{\partial t}\hat{U}^\dagger(t,t_0)|\psi,t\rangle.$$ If we define the operator $$\Lambda(t,t_0) = \frac{\partial\hat{U}(t,t_0)}{\partial t}\hat{U}^\dagger(t,t_0),$$ we can show (using only the unitarity of $\hat{U}$ that $\Lambda(t,t_0)$ is a.) independent of $t_0$ and b.) anti-Hermitian. We then define the operator $$H(t) = i \hbar \Lambda(t),$$ such that the time evolution of a quantum state is given by the Schrodinger equation as $$i\hbar\frac{\partial}{\partial t}|\psi,t\rangle = H(t)|\psi,t\rangle.$$ Now all that remains is to determine the operator $H(t)$. We could simply postulate the form of this operator, but instead we will use the path integral approach as a postulate and derive the form of $H(t)$ from there. We assume that the matrix elements of the unitary time evolution operator $\hat{U}(t,t_0$ are given by $$\langle\vec{r}|\hat{U}(t,t_0)|\vec{r}_0\rangle = \int\mathcal{D}\vec{r}\,e^{\frac{i}{\hbar}S[\vec{r}(t)]},$$ where $S$ is the classical action and $\int\mathcal{D}\vec{r}$ denotes integration (however it may be defined) over the space of paths connecting the spacetime points $(t_0,\vec{r}_0)$ and $(t,\vec{r})$. Using this postulate, we can write the matrix elements of the operator $H(t)$ as \begin{align}\langle\vec{r}|H(t)|\vec{r}_0\rangle &= i\hbar\int \mathrm{d}\vec{r}'\frac{\partial}{\partial t}\left[\langle\vec{r}|\hat{U}(t,t_0)|\vec{r}'\rangle\right]\langle\vec{r}'|\hat{U}^\dagger(t,t_0)|\vec{r}_0\rangle,\\ &= i\hbar\int \mathrm{d}\vec{r}'\frac{\partial}{\partial t}\left[\int\mathcal{D}\vec{r}\,e^{\frac{i}{\hbar}S[\vec{r}(t)]}\right]\langle\vec{r}'|\hat{U}^\dagger(t,t_0)|\vec{r}_0\rangle,\\ &= i\hbar\int \mathrm{d}\vec{r}'\left[\int\mathcal{D}\vec{r}\frac{\partial}{\partial t}\,e^{\frac{i}{\hbar}S[\vec{r}(t)]}\right]\langle\vec{r}'|\hat{U}^\dagger(t,t_0)|\vec{r}_0\rangle,\\ &= i\hbar\int \mathrm{d}\vec{r}'\left[\int\mathcal{D}\vec{r}\frac{i}{\hbar}\frac{\partial S}{\partial t}\,e^{\frac{i}{\hbar}S[\vec{r}(t)]}\right]\langle\vec{r}'|\hat{U}^\dagger(t,t_0)|\vec{r}_0\rangle,\\ &= \int \mathrm{d}\vec{r}'\left[\int\mathcal{D}\vec{r}\left(-\frac{\partial S}{\partial t}\right)\,e^{\frac{i}{\hbar}S[\vec{r}(t)]}\right]\langle\vec{r}'|\hat{U}^\dagger(t,t_0)|\vec{r}_0\rangle.\end{align}
The Hamilton-Jacobi equations tells us that $-\frac{\partial S}{\partial t}$ is simply equal to the Hamiltonian. Now I have two related questions:
1.) Generally speaking, functions on phase space which appear inside a path integral correspond to operators which appear outside the path integral. Can the same be done for the quantity $-\frac{\partial S}{\partial t}$ via the appropriate use of the resolution of the identity?
2.) If so, then this equation would seem to derive the functional form of the operator $H(t)$ as the classical Hamiltonian with the usual substitution $(\vec{r},\vec{p})\rightarrow(\hat{\vec{r}},-i\hbar\vec{\nabla})$. However, the quantum Hamiltonian often contains spin operators, which (to my knowledge) do not have classical analogues. How do spin operators enter into $H(t)$ in this formalism?
