I was recently told in the physics forum (https://physics.stackexchange.com/questions/616186/deriving-the-path-integral-from-the-time-slice-approach-for-a-general-hamiltonia) that it is not possible to derive the path integral formulation for an arbitrary hamiltonian. Specifically, in all cases one has to eventually replace the Hamiltonian $H$ with a specific implementation, such as that of a non-relativistic particle in a potential. However, I can't help but feel like this should work. Can one show, in the general case that:
$$ K(t,q',q)=\langle q' | e^{-itA} | q \rangle \equiv\int D[q]e^{-iS} $$
where $S$ is generally the 'action' in physics, but at step I am not sufficiently advanced in my derivation to produce an expression of $S$ in terms of $tA$.
For instance, I see the general proof as starting as follows:
$$ \begin{align} \langle q' | e^{-itA} | q \rangle &= \lim_{n\to \infty} \langle q' | \left( e^{-itA/n} \right)^n| q \rangle \end{align} $$
Then I can inject $1=\int dq |q \rangle \langle q| $:
$$ \begin{align} \langle q' | e^{-itA} | q \rangle &= \lim_{n\to \infty} \prod_{j=0}^{j=n-1} \int dq_{j+1} \langle q_{j+1} | e^{-itA/n} |q_j \rangle \end{align} $$
The next step that I see, is to actually perform the distribution of the terms in the bracket. This is extremely verbose. So to make it easier I decided to try it with a simplified case just to at least confirm that it works. So I assume that $A$ is a scalar and that the bra and the kets have only 1 element. This reduces to a single equation:
$$ \begin{align} \langle q' | e^{-itA} | q \rangle &= \lim_{n\to \infty} \prod_{j=0}^{j=n-1} \int dq_{j+1} e^{-itA/n} q_{j+1}q_j \end{align} $$
Here, we have to assume that $t$ is infinitesimal, but if we do we recover a Riemann sum integral for $itA$. And we also recover a functional integral for the qs:
$$ K(t,q',q)=q_{n} \left( \int q_{1}^2 d q_1 q_2^2 d q_2 ... e^{-i\int_0^t A dt} \right) q_0 $$
- I do not understand the meaning of the squared $q$, such as $q_1^2$ and $q_2^2$, etc. As this is a functional integral, is $A$ not supposed to be depended on the square $q$. After all the standard definition of the functional integral is:
$$ \int G[f]D[f] = \int_{-\infty}^\infty ... \int_{-\infty}^\infty G[f] \prod_x df(x) $$
should I have assume that $A$ is dependant on $q$, and wrote $A(q)$?
- Furthermore I do not understand why all the $qs$ are squared... what is the function then. Do I have to rewrite $D[f]$ as $D[\frac{1}{3}q^3]$ so that I get $q^2dq$ in my product? But then it means that $A$ should be dependant on $1/3q^3$ instead of q, no? However, I have never seem such a requirement in physics that $A$ depends on $1/3q^3$.