arivero
Mar1-04, 11:58 AM
... 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.
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.