Recently, I have been learning about path integrals and I was wondering, can the probability of a certain path be weighted more in a path integral? Said in another way, can certain paths have more probability in a path integral?
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1$\begingroup$ That's the whole point of the path integral. $\endgroup$– DanuCommented Dec 15, 2014 at 0:31
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$\begingroup$ What do you mean by "probability of a path"? The $\mathrm{e}^{\mathrm{i}S[\gamma]}$, for $\gamma$ a path? And what do you mean by them having "more probability"? $\endgroup$– ACuriousMind ♦Commented Dec 15, 2014 at 0:33
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$\begingroup$ @Danu well, in one of my other questions, it was mentioned that the probabiliy is always the same for each path... $\endgroup$– TanMathCommented Dec 15, 2014 at 0:34
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1$\begingroup$ @TAbraham the whole point of the path integral is that the particle takes (in Feynman's words "sniffs out") all paths between its start and end point. Not that it takes only one and chooses with various probabilities. You absolutely shouldn't think of it taking any one path (because this is wrong). $\endgroup$– or1426Commented Dec 15, 2014 at 0:43
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1$\begingroup$ @TAbraham Please try to at least reach a minimum level of understanding on what it is you're talking about, before asking your questions. Read about path integrals, here e.g., or grab one among the so-many nicely written basic QM textbooks and get started! You cannot learn everything just by asking questions one by one in a forum... $\endgroup$– EllieCommented Dec 15, 2014 at 0:47
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1 Answer
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In general the "weighting" of each path $q$ in a path integral is given by $e^{\frac{i}{\hbar}S[q]}$. Then paths for which the action $S$ is stationary with respect to small deviations from the path are the only ones which really contribute because the contributions from those with non stationary $S$ get averaged out as the phase changes very rapidly (because $\hbar$ is very small).
The number $S[q]$ is defined as:
$$S[q] = \int_{t_0}^{t_1}L(q(t),\dot{q}(t), t)dt$$
Where $t$ is some parameter that varies along the path and $L$ is the lagrangian. The Langrangian will depend on the details of your system but for a free particle it looks like the classical kinetic energy $L = \frac{1}{2}\dot{q}^2$.
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$\begingroup$ @TAbraham it is the action along the path q. $\endgroup$– or1426Commented Dec 15, 2014 at 0:44
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$\begingroup$ u mean what's calculated by the Euler-Lagrange equations? but isn't that for classical mechanics? $\endgroup$– TanMathCommented Dec 15, 2014 at 0:45
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$\begingroup$ @TAbraham, you use the same action $S[q]$ in classical mechanics and in the path integral formulation of quantum mechanics. The difference is that in classical mechanics, the particle takes the path of stationary action; while in quantum mechanics, the particle takes all of the paths. But it can be showed that the path of stationary action is usually the dominant part of the path integral due to positive interference, which is why the principle of stationary action works so well in classical mechanics. $\endgroup$– jabiraliCommented Dec 16, 2014 at 1:20