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I'm studying the proof for the decoherence of the off diagonal elements of a density matrix through scattering with the environment and I'm stuck at a certain point:

proof

My problem is A1.14 relation. (A1.13 as well to be honest, but I guess that the $(2\pi/L)^3$ is just sort of a normalization where $2\pi$ is, maybe $2\pi\hbar$ from the uncertainty relation and $L^3$ is the "volume of the scattering") Sadly I have no idea why $\delta^2(q-q')=\delta(q-q')L$. Oh, $q$ should be a generic module of a momentum (maybe before the scattering) and $q'$ should be the module associated to the momentum of the particle after the scattering.

Since both these relations are said to be "usual replacements" I hope somebody recognise them and can help me out.

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    $\begingroup$ Could you provide the source you're using? $\endgroup$ – Adomas Baliuka Feb 13 '17 at 14:43
  • $\begingroup$ That particular proof is from "Decoherence and the Appearance of a Classical World in Quantum Theory", appendix 1 by Joos. A very similar proof with the same replacements is also present in "The theory of open quantum systems" by Breuer and Petruccione $\endgroup$ – Mattz Feb 13 '17 at 21:37
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In one dimension we have

$$(2 \pi)\delta(p) = \int_{-\infty}^\infty \mathrm{d}x~e^{ipx}$$

And so, formally,

$$ (2\pi)\delta(0) = \int_{-\infty}^\infty \mathrm{d}x$$

If we imagine that we are actually working in a large but finite space with length $L$, then we can intepret $(2 \pi) \delta(0)$ as this length $L$. The relevance of this is that when we have the square of a delta function, the first will enforce the condition in the argument of the delta function, and this will leave the second delta function as $\delta(0)$. This is technically undefined, so we need to write it as something sensible – in this case, as the length of the space we're considering.

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The answer given by jg255 is rather heuristic, but it gets the correct answer. A better approach is to realize that in scattering theory, when we convert scattering amplitudes, which have an energy-momentum conservation delta function, into probablities by squaring them and interpreting the delta-squared as a delta times the volume of space time, we are secretly using Fermi's Golden Rule. In other words to do the calculation properly you need to work in a finite space-time box, square to convert amplitudes to probablities, and only then take the infinite volume and time limit. The answer comes out the same as the Heuristic approach.

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