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In the part "Quantum Brownian motion" of the book, The theroy of open quantum systems written by Breuer, the author investigates on the Caldeira-Leggett model: The Hamiltonian of the particle is

$H_{S}=\frac{1}{2m}p^{2}+V(x)$

The particle is coupled to a bath consisting of a large number of harmonic oscillators with masses $m_{n}$ and frequencies $\omega_{n}$

$H_{B}=\sum_{n}\hbar\omega_{n}(b_{n}^{\dagger}b_{n}+\frac{1}{2})=\sum_{n}\frac{1}{2m_{n}}p_{n}^{2}+\frac{1}{2}m_{n}\omega_{n}^{2}x_{n}^{2}$

where $b_{n}$ and $b_{n}^{\dagger}$ are teh annihilation and creation operators of the bath modes, while $x_{n}$ and $p_{n}$ are corresponding coordinates and canonically momenta.

The interaction is assumed to be:

$H_{I}=-x\sum_{n}k_{n}x_{n}$

where $k_{n}$ are coupling constants.

The following is my question:

The author says this type of interaction will yield a renormalization of the potential $V(x)$ of the Brownian particle. Hence we introduce a counter-term

$H_{c}=x^{2}\sum_{n}\frac{k_{n}^{2}}{2m_{n}\omega_{n}^{2}}$

to compensate for the renormalization resulted from interaction.

So my question is: In this case, there seems to be no infinities to be absorbed by counter-terms like that in QFT. And the renormalization caused by the interaction seems to have real physical meanning(in most case in QFT, they have not), like Lamb-shift. So how could we introduce counter-terms to cancel the renormalization in this case?

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There is nothing quantum or explicitly divergent here, but even at the classical level the extra "push" from the interaction with the bath oscillators changes the potential seen by the particle so that it is no longer $V(x)$. The counterterm is designed to cancell this effect so that the parameters in $V(x)$ coincide what would be measured experimentally.

See the discussion on page 155 and the footnote on page 156 of the follwoing lecture notes: https://courses.physics.illinois.edu/phys508/fa2018/amaster.pdf

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  • $\begingroup$ I understand it is no longer $V(x)$. My point is the shift should has real physically meaning, thus it should not be cancelled. Your claim,"so that the parameters in V(x) coincide what would be measured experimentally" puzzles me. In my understanding, $V(x)$ should be known beforehand, and there is nothing to change as renormalization done in QFT $\endgroup$ – xiang sun Mar 18 '19 at 16:04

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