# Question about a finite time interval step in the derivation of the Feynman path integral in Sakurai

This may be a possible errata but Sakurai (pp 126 in the 2nd Edition) states that starting with $$S = \int dt \,\,\scr{L_{\mathrm{classical}}}$$ Looking at a finite-time-interval of the action:

$$S(n,n-1) = \int\limits_{t_{n-1}}^{t_n} dt \left( \frac{m \dot{x}^2}{2} - V(x)\right)\,\,\,\,\,\,(1)$$

Now using a straight-line approximation for points $(x_{n-1},t_{n-1})$ and $(x_n,t_n)$, which implies

$$\dot{x} = \frac{x_n-x_{n-1}}{t_n-t_{n-1}} = \frac{x_n-x_{n-1}}{\Delta t}$$ Sakurai then has Eq (1) become: $$S(n,n-1) = \Delta t\left( \left(\frac{m}{2}\right) \left(\frac{x_n-x_{n-1}}{\Delta t}\right)^2 - V\left(\frac{x_n+x_{n-1}}{2}\right)\right)$$

My question is why is the potential now dependent on $V\left(\frac{x_n + x_{n-1}}{2}\right) \mathrm{\,\,instead \,\,\,of\,\,\,} V\left(x_n-x_{n-1}\right)$?

I checked the most recent errata (4/5/13) posted by J. Napolitano ( http://homepages.rpi.edu/~napolj/ErrataMQM.pdf ) and there is none for this page. Could anyone clarify this step for me? Thanks.

• For a related discussion of operator ordering problems, see e.g. this Phys.SE post. – Qmechanic Apr 7 '13 at 21:29

You said it yourself. It's a straightline approximations. Hence the potential takes the value at the averaged (intermediate) point between $x_{n-1}$ and $x_{n}$, i.e $$\frac{x_n+x_{n-1}}{2}$$
• @JohnM: because -- as you wrote yourself -- it's the STRAIGHT LINE approximation :). Everything is discretized. So you need to think: "What's the value of the potential if I consider the interval $x_{n}$ $x_{n-1}$"? Now, because you assume a the interval to be a straight line, you just approximate the potential to take the value at the middle point of your interval. This becomes exact if $x_{n}$ tends to $x_{n-1}$. I'm sorry. I don't know how to make it clearer. It's really no rocket science :) – A friendly helper Apr 8 '13 at 5:31