# Deriving the path integral from the time-slice approach for a general hamiltonian

I am reading lecture notes Deriving the Path Integral by A. Wipf (pdf) trying to better understand how to derive the path integral. Specifically, equation (2.27) is:

$$K(t,q',q)=\int d\omega_1... d\omega_2\prod_{j=0}^{j=n-1}\langle \omega_{j+1} | e^{-itH/\hbar n} | \omega_j \rangle.$$

What I do not like about the author's proof is that midway in the proof he converts $$H$$ to a non-relativistic particle in a potential, then show that the prod part is equivalent to an integral over $$t$$ (equation 2.29) for this specific Hamiltonian. Then, simply 'suggests' that it works for a general action (equation 2.30).

How can I show that

$$\lim_{n\to \infty} \prod_{j=0}^{j=n-1}\langle \omega_{j+1} | e^{-itH/\hbar n} | \omega_j \rangle \to e^{\int L(t,q',q)dt}$$

Without having to use the special case of a non-relativistic particle in a potential. Can it be done in the general case for any Hamiltonian? One would have to consider $$H$$ as a matrix, and work out the correspondance leaving $$H$$ as a matrix.

• As a side node, I am not sure why the author changes the notation from $$q$$ to $$\omega$$ at 2.26 to 2.27. Could we had simply kept using $$q$$?

\begin{align} &\text{Operator formalism} \cr & \qquad\qquad \updownarrow \cr &\text{Path integral formalism} \end{align} \tag{1}
\begin{align} &\text{Operator formalism} \cr & \qquad\qquad \updownarrow \cr &\text{Hamiltonian phase space path integral formalism}\cr & \qquad\qquad \updownarrow \cr &\text{Lagrangian path integral formalism} \end{align} \tag{2}