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

+ 0 like - 0 dislike
130 views

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:

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

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.

This post imported from StackExchange Physics at 2014-04-15 16:40 (UCT), posted by SE-user John M
asked Apr 7, 2013 in Theoretical Physics by John M (0 points) [ no revision ]
For a related discussion of operator ordering problems, see e.g. this Phys.SE post.

This post imported from StackExchange Physics at 2014-04-15 16:41 (UCT), posted by SE-user Qmechanic

1 Answer

+ 2 like - 0 dislike

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 $$\frac{x_n+x_{n-1}}{2}$$

This post imported from StackExchange Physics at 2014-04-15 16:41 (UCT), posted by SE-user A friendly helper
answered Apr 7, 2013 by A friendly helper (320 points) [ no revision ]
Actually, why would this not be a minus sign?

This post imported from StackExchange Physics at 2014-04-15 16:41 (UCT), posted by SE-user John M
@JohnM: As you said, and as i explained: You take the average of two points. Taking averages of a quantity is done by adding all quantities and dividing by the number of quantities. Eg: You have a cake and six friends. Each thus gets 1/6 of that cake. Same is done here.

This post imported from StackExchange Physics at 2014-04-15 16:41 (UCT), posted by SE-user A friendly helper
@JohnM: The - sign for the derivative comes in because it's about the CHANGE of that value between two points. THat's something different altogether.

This post imported from StackExchange Physics at 2014-04-15 16:41 (UCT), posted by SE-user A friendly helper
Additional: Why should the input variable of the potential be the average distance? Why would we not care about the average value of potential between the two points?

This post imported from StackExchange Physics at 2014-04-15 16:41 (UCT), posted by SE-user John M
@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 :)

This post imported from StackExchange Physics at 2014-04-15 16:41 (UCT), posted by SE-user A friendly helper

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