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The free propagator is

[tex]K(1,2,t_1,t_2)=e^{i{(x_2-x_1)^2\over{t_2-t_1}}}[/tex]

but I find more insightful to write it as

[tex]K^{12}_t=e^{ip_{12} x_{12}}[/tex]

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

[tex]\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}}}[/tex]

it could be better to separate [tex]x_{ij}=x_j-x_i[/tex] so that the interaction is composed of terms

[tex] V(x_7) e^{i {(p_{78}-p_{67}) x_7\over \hbar}}[/tex] or, generically, [tex] V(x_j) e^{i {(\Delta p_j}) x_j \over \hbar}}[/tex]

In this way we are half way to recover Newton's second law [tex]F= dp/dt[/tex], ie

[tex]V'(x_j) = {\Delta p_j \over \Delta t_j} [/tex]

At first glance it seems that there is some ambiguity to define the interval [tex]\Delta t_j[/tex]; 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.

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# Fun with the free propagator

Can you offer guidance or do you also need help?

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