Timeline for Question about a finite time interval step in the derivation of the Feynman path integral in Sakurai
Current License: CC BY-SA 3.0
7 events
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| Apr 8, 2013 at 5:31 | comment | added | A friendly helper | @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 :) | |
| Apr 8, 2013 at 4:26 | comment | added | John M | 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? | |
| Apr 7, 2013 at 22:00 | comment | added | 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. | |
| Apr 7, 2013 at 21:58 | comment | added | A friendly helper | @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. | |
| Apr 7, 2013 at 21:50 | vote | accept | John M | ||
| Apr 7, 2013 at 21:54 | |||||
| Apr 7, 2013 at 21:50 | comment | added | John M | Actually, why would this not be a minus sign? | |
| Apr 7, 2013 at 21:18 | history | answered | A friendly helper | CC BY-SA 3.0 |