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 $q$?

1 Answer 1


Any textbook/lecture notes derivation of the correspondence between

$$\begin{align} &\text{Operator formalism} \cr & \qquad\qquad \updownarrow \cr &\text{Path integral formalism} \end{align} \tag{1}$$

is a formal derivation, which discards contributions in the process. It is naturally to split the derivation into 2 parts:

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

Very briefly, in the first leg one discretizes phase space paths by alternate between introducing complete positions and momenta bases. In the second leg one integrates out the momenta. For more details, see e.g. this, this, this related Phys.SE posts and links therein.


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