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  What is the relation between renormalization and self-adjoint extension?

+ 1 like - 0 dislike

What is the relation between renormalization and self-adjoint extension? It seems that a renormalization scheme can be rigorously treated mathematically using the self-adjoint extension theory.

Is there any good reference on it?

This post imported from StackExchange Physics at 2015-05-13 19:01 (UTC), posted by SE-user Jiang-min Zhang
asked Nov 30, 2014 in Theoretical Physics by Jiang-min Zhang (40 points) [ no revision ]
retagged May 13, 2015


There are relations between the renormalization of singular potentials in quantum mechanics and self-adjoint extensions of associated Hamiltonians. See, e.g., http://arxiv.org/abs/hep-th/9305052 . But these relations are not simply to state precisely, as there are also exceptions. See, e.g., http://arxiv.org/abs/hep-th/0604018 and references there. Perhaps I'll write a more thorough answer, but not now. 

Renormalization in quantum field theory doesn't seem to follow the same pattern, though.

1 Answer

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Quantum dynamics is commonly known to be generated by self-adjoint operators. Therefore in order to properly define the dynamics of a system it is necessary to introduce a suitable self-adjoint Hamiltonian operator.

In quantum field theories, this task is extremely difficult, because the formal operators that emerge quantizing a classical field theory are---borrowing a term used by Glimm---plagued with divergences. In practice, apart from exceptionally easy situations, interacting Hamiltonians of QFT make sense only as quadratic forms, and their domain of definition as operators is only the vector zero.

The first task is then to obtain at least a densely defined symmetric operator that describes the system. This is already extremely difficult, and can be achieved only in few situations. This operation requires the subtraction from the operator of infinite quantities, but often also a renormalization of the vectors of the Hilbert space: the final space will be, in general, different from the original one.

After all that (most of the times it has been done independently, and with different tools such as functional integration), it is also necessary to prove that such densely defined operator is self-adjoint, or it has self-adjoint extensions. The "easy way" is to prove that it is bounded from below, so it has at least one self-adjoint extension.

Suppose you have boundedness from below, then you have a well-defined renormalized dynamics for the system. Now it is suitable that the aforementioned dynamics is also unique, because in principle different extensions of the Hamiltonian generate different dynamics with different properties. In order to do that, it is necessary either to prove essential self-adjointness of the operator, or that it is bounded from below and the corresponding form is closed.

This is the ambitious program of the Constructive QFT started in the sixties---some of the most influential contributors were Wightman, Glimm, Jaffe, Nelson---and it has achieved some definitive result only in few dimensions (at most 2+1), and in simple situations. Most of the time it is already a great achievement to obtain a renormalized densely defined symmetric operator!

In this brief description, I have only outlined the non-perturbative approach to renormalization that is related to your question on self-adjointness. To sum it up: the problems with renormalization of the Hamiltonians of QFT are very difficult already at the fundamental level of "definition of a symmetric operator". Existence of self-adjoint extensions is a subsequent question, in some sense unrelated (because boundedness from below can be proved "a priori" giving cut off independent informations about the regularized operators). Good references on constructive quantum field theory are the books by Glimm and Jaffe, e.g. their collected papers.

This post imported from StackExchange Physics at 2015-05-13 19:01 (UTC), posted by SE-user yuggib
answered Dec 1, 2014 by yuggib (360 points) [ no revision ]

The question was about the relation to self-adjoint extensions! This is completely missing from your answer.

@ArnoldNeumaier Well, my point was that first of all it is necessary to have a densely defined symmetric operator as a candidate to be the Hamiltonian, then you may start thinking about the eventual self-adjoint extensions, and which is the best suited to describe the dynamics (if there are more than one). And also, I was really thinking about QFT where there are problems even in defining the operator in the first place. Now, looking at the linked google search, I see there are some interpretations that link self-adjoint extensions of "QM" Hamiltonians with singular potentials, and a renormalization procedure to define them. I was not thinking about that particular subject (and actually do not know it very well); so in this context I agree that my answer is not so useful...

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