- #1
Ravi Mohan
- 196
- 21
Hi,
I am studying path integral formulation from Ballentine. Till equation 4.50, I follow quiet well.
[tex]
G(x,t;x_0,t_0) = \lim_{N \to \infty}\int\ldots\int\left(\frac{m}{2\pi i\hbar\Delta t}\right)^{\frac{N+1}{2}}\exp{\sum_{j=0}^{N}\left(\frac{im(x_{j+1}-x_j)^2}{2\hbar\Delta t}-V(x_j)\right)}dx_1\ldots dx_N
[/tex]
I also follow that in continuum limit, summation converts to integral (argument of exponent). I am wondering what happens to the [itex]\Delta t[/itex] in the expression [itex]\left(\frac{m}{2\pi i\hbar\Delta t}\right)^{\frac{N+1}{2}}[/itex].
I am studying path integral formulation from Ballentine. Till equation 4.50, I follow quiet well.
[tex]
G(x,t;x_0,t_0) = \lim_{N \to \infty}\int\ldots\int\left(\frac{m}{2\pi i\hbar\Delta t}\right)^{\frac{N+1}{2}}\exp{\sum_{j=0}^{N}\left(\frac{im(x_{j+1}-x_j)^2}{2\hbar\Delta t}-V(x_j)\right)}dx_1\ldots dx_N
[/tex]
I also follow that in continuum limit, summation converts to integral (argument of exponent). I am wondering what happens to the [itex]\Delta t[/itex] in the expression [itex]\left(\frac{m}{2\pi i\hbar\Delta t}\right)^{\frac{N+1}{2}}[/itex].