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Which is the "delta prime potential" nowadays?

+ 2 like - 0 dislike
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I am running across a series of questions in SE about the potentials with support in a single point and all my folklore on the topic is from work I did 20 years ago. At that time there was two set of boundary conditions that were called $\delta'(x)$ potential

  1. The one of Holden, that actually when into the book of Albeverio et al, and that is the mirror of the conditions of the delta: it asks the derivative of the wavefunction to be continuous, and the waveunction itself to have a jump proportional to its derivative.
  2. The one of Kurasov, asking for a proportionality factor between the sides of the derivative, and other proporionality between the sides of the wavefunction, with both factors being one the inverse of the other.

Which is the current status of the nomenclature? I am under the impression that some other denomination has been used for the case 1, something related to operators. Also, which should be the analytical notation to write case 1 in a Hamiltonian (assuming case 2 is the $\delta '(x)$; it it is not, also same question for case 2 :-)

Bonus question: an argument to favour case 2 was the scale-invariance, namely that the S-matrix does not depend of the momenta, and that thas was linked to the "dimensions" of the delta prime potencial being $L^{-2}$. Why do we claim that the dimensions are these?

asked Aug 24, 2015 in General Physics by al.rivero (15 points) [ revision history ]
edited Aug 24, 2015 by al.rivero

The lecture notes on operator theory by Jan Derezinski contain a detailed discussion of delta interactions and their renormalization.

@ArnodlNeumaier I see but for delta prime it only discusses, without detail, the operator $|\delta'><\delta'|$, and does it in the half line, so no questions about continuity. At least, it focus the question on the connection with self-adjoint extensions

I don't know more. The dimension statement comes from looking at what happens when you substitute in the problem $x\to sx$ and $p\to s^{-1}p$.

@ArnoldNeumeir this is a great hint. I had forgotten how to argue for the dimension.

You should use the automatic ping completion facility (directly below the edit box) to reach a user - in both cases you tried you didn't reach me because of a typo! 

@ArnoldNeumaier Great! Thanks, and sorry by the typo

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