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I am reading Peskin and Schroeder (path integrals) and it states that discretising the classical action gives:

$$S~=~\int \left(\frac{m}{2}\dot{x}^{2}-V(x)\right) dt ~\rightarrow~ \sum \left[\frac{m}{2}\frac{(x_{k+1}-x_{k})^{2}}{\epsilon}-\epsilon V(\frac{x_{k+1}+x_{k}}{2})\right] $$

Is it convention that $$\dot{x} ~\rightarrow~ (x_{k+1}-x_{k})/2$$

and

$$x~\rightarrow~ (x_{k+1}+x_{k})/2~?$$

I understand that $\epsilon$ is effectively $dt$, but I am not sure why $x$ behaves like this.

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Possible duplicate: physics.stackexchange.com/q/60372/2451 . Related question by OP: physics.stackexchange.com/q/61545/2451 . For a related discussion of operator ordering problems, see e.g. this Phys.SE post. –  Qmechanic Apr 27 '13 at 23:23
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1 Answer 1

Recall the definition for a derivative is $$\frac{\mathrm{d}x(t)}{\mathrm{d}t}=\lim_{\Delta t\to 0}\frac{x(t+\Delta t)-x(t)}{\Delta t}$$ So if we insist on having discrete values of time, we get $$\frac{\mathrm{d}x}{\mathrm{d}t}\approx\frac{x(t+\Delta t)-x(t)}{\Delta t}$$ Or if everything is in integer multiples of some interval, we have $$\dot{x}\approx \frac{x(k\Delta t+\Delta t) - x(k\Delta t)}{\Delta t} = \frac{x_{k+1}-x_{k}}{\epsilon}$$ where we define $x_{k}=x(k\Delta t)$ and introduce $\epsilon=\Delta t$.

Now, the potential is a numerical approximation to an integral: $$\int V(x)\,\mathrm{d}t \approx \sum V\left(\frac{x_{k+1}+x_{k}}{2}\right)\,\epsilon$$ where the argument is just the midpoint between $x_{k+1}$ and $x_{k}$. This is the rectangle method...

We can combine these terms together to get the desired result $$\int\left(\frac{m\dot{x}^{2}}{2}-V(x)\right)\mathrm{d}t\approx\sum_{k}\left(\frac{m}{2}\left[\frac{x_{k+1}-x_{k}}{\epsilon}\right]^{2}-V\left[\frac{x_{k+1}+x_{k}}{2}\right]\right)\epsilon$$ And by basic arithmetic, we see the right hand side simplifies (using distributivity) to $$\int\left(\frac{\dot{x}^{2}}{2m}-V(x)\right)\mathrm{d}t\approx\sum_{k}\left(\frac{m}{2}\frac{(x_{k+1}-x_{k})^{2}}{\epsilon}-\epsilon\cdot V\left[\frac{x_{k+1}+x_{k}}{2}\right]\right)$$ That concludes the derivation, I think...

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Im sorry I don't follow from maths line 2 to 3 why has the denominator disappeared? –  user21119 Apr 28 '13 at 0:00
    
Sorry about that, does this change explain things better? –  Alex Nelson Apr 28 '13 at 1:08
    
I thinks so, does this mean that $k\Delta t=t$?, Sorry i didnt put this in but it is actually $S=T-V dt$ not $dx$ –  user21119 Apr 28 '13 at 7:47
    
Well, we discretize $t$, so $t_{k}=t_{0}+k\Delta t$ where $t_{0}$ is "some initial starting time" (usually set to 0 for simplicity). NB: I am considering the action term-by-term, so you can recombine it to get the desired sum... –  Alex Nelson Apr 28 '13 at 17:44
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