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  What's the definition of a short range potential and why is it defined this way?

+ 2 like - 0 dislike
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This question is related to this one on SE. However, I'm still not satisfied with the answers. My question can be divided as these 2:

  1. Is this just a convention (as Claudius said) or is there any profound reason? 
  2. Why is $r^{-2}$ also long range? (As an example, on the P.5 of Quantum Mechanics: Selected Topics, "The centrifugal potential cannot be regarded as a short-range potential." and $V(r)=\frac{l(l+1)}{r^2}$)
asked May 18, 2016 in Theoretical Physics by forthebest (25 points) [ revision history ]
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@Arnold Neumaier Sorry for my late response. You've said that depend on the specific cases we study, we may have different definitions of short or long range potential. Could you give a concrete example? Or for the second part of my OP, why centrifugal potential is treated as long range?

@forthebest : The centrifugal potential can well be "short-range" if $r$ is within $(a,b)$ with finite $a$ and $b$. In order to tell what potential is short- or long-range, one has to compare the potential term with the others at $r\to\infty$. With kinetic term, for example.

@Vladimir Kalitvianski. I think your criterion about the centrifugal potential is wrong. Since for any non-singular potential it will definetely finite, even for exponentially growing potential. This is absurd. As to compare with kinetic term, in Schordinger equation, there are no r dependent kinematic term at all, then how to compare them?

@forthebest : I gave a reason why tan exponentially falling potential is considered short range. jjscale gave on SE a reason why a potential $O(r^{-2}$ is considered short-range in the context of a scattering calculation. Your reference on the centrifugal potential probably gives a reason why in its context it cannot be considered as short range. I can't check since you didn't give acomplete reference.

@forthebest : You have to compare the term with a given solution $\psi$, not just $d^2/dr^2$. And a given $\psi$ is what Arnold Neumaier means when speaking of particular problem. If the potential term prevails the kinetic one at finite $r$ and when $r\to\infty$, then the potential in this problem is short-range. If it starts gradually to prevail only when $r\to\infty$, then it is a long-range potential at the given circumstances.

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As jjcale mentioned in the answer on SE, what counts as short-scale depends on the application. In relativistic quantum field theory, massive particle generate short-range forces with an exponentially decaying potential, while massless particles generate (in the absense of confinement) long range forces with a modified Coulomb potential, essntially $1/r$ for large $r$. The two cases behave very differently in many respects.
 

@ArnoldNeumaier I think what @Forthebest is asking for is why we call it short- or long-range potentials. So why is it that a exponentially dying potential results in "short-range" and 1/r potential results in "long-range." I would have to think more about it, but my first guess would be to look at the power spectrum. In my mind I think of a long-range force as one who can emit radiation as you go towards infinity (dipole, quadrupole, etc.), while short-range forces are always screened due to the exponentially decaying potential.

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