# Two state Hamiltonian in Feynman path integral

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?

• You are looking for (spin) coherent state path integral.
Jun 22, 2021 at 6:43
• Note that in the derivation of path integration that you're used to, you're just inserting $N$ resolutions of identity and then taking $N \rightarrow \infty$. You can do exactly the same thing with spins... it's just a different resolution of identity. Jun 22, 2021 at 7:36
• @DanielSank So to be easily compatible with the Pauli spin matrices in the Hamiltonian, our strategy should be to insert complete sets of states which are eigenstates of the Hamiltonian? That way, when you insert the states and come across something like $\langle \Omega |e^{-iHt/\hbar}| \Omega' \rangle$, where $|\Omega \rangle$ is some eigenstate the time-evolution operator can easily be taken out of the bra and ket? Jun 22, 2021 at 8:04
• More or less, yes. See Adam's comment. Jun 22, 2021 at 15:15

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).
• In this reference (researchgate.net/profile/Mekki-Aouachria/publication/…), The authors seem to take your spin eigenstate strategy above, only with these coherent states, $|\Omega \rangle = |\theta, \psi \rangle = e^{-i\psi S_z}e^{-i\theta S_y} |\uparrow \rangle$ ... does this make sense? Jun 22, 2021 at 9:56
• I understand formally they are just rotating the spin up state, but I suppose I am confused about why they would do this? Can this generate a spin down state so that $\langle\Omega|e^{-iHt/\hbar}|\Omega '\rangle$ is the probability amplitude to find the state in either the spin up or down state after time t? Jun 22, 2021 at 9:58