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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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2 Answers 2

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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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  • $\begingroup$ Im sorry I don't follow from maths line 2 to 3 why has the denominator disappeared? $\endgroup$
    – user21119
    Apr 28, 2013 at 0:00
  • $\begingroup$ Sorry about that, does this change explain things better? $\endgroup$ Apr 28, 2013 at 1:08
  • $\begingroup$ 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 $dx$ $\endgroup$
    – user21119
    Apr 28, 2013 at 7:47
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    $\begingroup$ 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... $\endgroup$ Apr 28, 2013 at 17:44
  • $\begingroup$ Nice, clear explanation. $\endgroup$
    – gn0m0n
    Sep 8, 2015 at 17:27
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For \begin{align} x\rightarrow\frac{x_{k+1}+x_{k}}{2}, \end{align} it is often called mid-point prescription. It reveals itself in the operator ordering problem, and is important to lead to gauge invariant Schrodinger equation. You can find details around page 183 of the book: Stochastic Processes and Quantum Mechanics, Path Integrals in Physics (M. Chaichian and A. Demichev).

The ambiguity of choosing discretized method in quantum path integral also relates to that of path integral in stochastic process. The mid-point prescription is the Stratonovich's way, where the ordinary calculus can be applied (fails when other stochastic integral is used). You may find more discussions in this paper: http://arxiv.org/abs/1401.7682.

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