# Fun with the free propagator

1. Mar 1, 2004

### arivero

... and perturbative quantum mechanics a la Feynman.

The free propagator is
$$K(1,2,t_1,t_2)=e^{i{(x_2-x_1)^2\over{t_2-t_1}}}$$
but I find more insightful to write it as
$$K^{12}_t=e^{ip_{12} x_{12}}$$
so I can see that it is unambiguously the product of momentum and displacement.

But when using it in a perturbative term with an interacting potential V
$$\int .. \int K^{1,2}_{t_{12}} V(x_2) K^{2,3}_{t_{23}} V(x_3).. V(x_{n-1}) K^{n-1,n}_{t_{n-1\; n}}}$$
it could be better to separate $$x_{ij}=x_j-x_i$$ so that the interaction is composed of terms
$$V(x_7) e^{i {(p_{78}-p_{67}) x_7\over \hbar}}$$ or, generically, $$V(x_j) e^{i {(\Delta p_j}) x_j \over \hbar}}$$
In this way we are half way to recover Newton's second law $$F= dp/dt$$, ie
$$V'(x_j) = {\Delta p_j \over \Delta t_j}$$
At first glance it seems that there is some ambiguity to define the interval $$\Delta t_j$$; a wider or narrower interval should give a different coupling constant. Guess one must examine the whole limiting procedure of the path integral to get the right value.

Last edited: Mar 1, 2004