Two state Hamiltonian in Feynman path integral

Question

In the typical constructions of the path integral I have seen, they always start from the assumption that the Hamiltonian one starts with is a function of $x$ and $p$, aka for $n$ degrees of freedom:

$$H=\sum_k^n \frac{p_k^2}{2m_k} + V(x_1,x_2,\cdots,x_n) \, ,$$

which allows one to utilize Trotter splitting to simplify the propagator. In the derivation of the path integral, this step is at

$$ \langle x_f |e^{-iHt/\hbar}|x_0\rangle = \int dx_1\cdots \int dx_{N-1}\prod_{k=1}^{N}\langle x_{k}|e^{-iH\Delta t_k/\hbar}|x_{k-1}\rangle$$

where you can write in the limit that $N \rightarrow \infty$ $$e^{-iH\Delta t_k/\hbar} \approx e^{-iV\Delta t_k/2\hbar}e^{-iT\Delta t_k/\hbar}e^{-iV\Delta t_k/2\hbar}$$

where the potential terms are now directly attached to position states, leading to nice things down the derivation.

How does the derivation change, and thus the path integral formulation change once you no longer assume you start with a Hamiltonian that depends on position and momentum?

Let's say there is a two state Hamiltonian $H = A\sigma_x + B\sigma_z$ (where $\sigma$ are the Pauli spin matrices), and now the Trotter splitting no longer makes sense. How does one proceed with calculating the propagator from here?

Answer 1

Shankar's review on renormalization group uses two-level system as an example for introducing Grassman numbers formalism, fermionic path integral, etc. (also available on arXive) Two-level system, of course, can be viewed as a single Fermionic state. However, as @Adam noted in the comments, you might need a spin-path integral, in which case good references are the books by Nagaosa or Auerbach.

Remark
The sources cited above all treat many-particle systems, i.e., mainly working in second quantization formalism. The OP seems to suggest a signle-particle path integral, along the lines of Feynmann-Hibbs book - these are rarely used, and the sources are indeed scarse.

Here, instead of momentum states, one could use spin eigenstates $|\sigma\rangle$ and calculate $$ \langle\sigma_f|e^{-iHt/\hbar}|\sigma_i\rangle = \sum_{\sigma_1}\sum_{\sigma_2}...\sum_{\sigma_{N-1}}\prod_{k=1}^N \langle\sigma_k|e^{-iH\Delta t_k/\hbar}|\sigma_{k-1}\rangle $$ and so on. This will however miss some important phase terms, which is why I recommend consulting the books cited above (or simply googling the full derivation).