Given a symmetric (densely defined) operator in a Hilbert space, there might be quite a lot of selfadjoint extensions to it. This might be the case for a Schrödinger operator with a "bad" potential. There is a "smallest" one (Friedrichs) and a largest one (Krein), and all others are in some sense in between. Considering the corresponding Schrödinger equations, to each of these extensions there is a (completely different) unitary group solving it. My question is: what is the physical meaning of these extensions? How do you distinguish between the different unitary groups? Is there one which is physically "relevant"? Why is the Friedrichs extension chosen so often?
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1$\begingroup$ I am asking this question as a mathematician trying to understand the meaning and motivation of the objects I am working with. $\endgroup$– András BátkaiCommented Sep 15, 2011 at 19:49
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$\begingroup$ Related: What's the deal with momentum in the infinite square well? $\endgroup$– Emilio PisantyCommented Feb 17, 2019 at 19:44
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2 Answers
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The differential operator itself (defined on some domain) encodes local information about the dynamics of the quantum system . Its self-adjoint extensions depend precisely on choices of boundary conditions of the states that the operator acts on, hence on global information about the kinematics of the physical system.
This is even true fully abstractly, mathematically: in a precise sense the self-adjoint extensions of symmetric operators (under mild conditions) are classified by choices of boundary data.
More information on this is collected here
http://ncatlab.org/nlab/show/self-adjoint+extension
See the references on applications in physics there for examples of choices of boundary conditions in physics and how they lead to self-adjoint extensions of symmetric Hamiltonians. And see the article by Wei-Jiang there for the fully general notion of boundary conditions.
answered Sep 15, 2011 at 23:37
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A typical interpretation of the self-adjoint extensions for the free hamiltonian in a line segment is that you get a four parametric family of possible boundary conditions, to preserve unitarity. Some of them just "bounce" the wave, some others "teletransport" it from one wall to the other. So it is also traditional to imagine this segment as a circle where you have removed a point, and then you are in the mood of studying "point interactions" or generalisations of dirac-delta potentials. The topic resurfaces from time to time, but surely some old references can be digged starting from M. Carreau. Four-parameter point-interaction in 1d quantum systems. Journal of Physics A, 26:427, 1993. In some works, I quote also Seba and Polonyi.
Sometimes the extensions are linked to the question of the domain of definition for the operator and then to the existence of anomalies. Here Phys.Rev.D34: 674-677, 1986, "Anomalies in conservation laws in the Hamiltonian formalism", revisited by the same autor, J G Esteve, later in Phys.Rev.D66:125013,2002 ( http://arxiv.org/abs/hep-th/0207164 ). These topics have been live for years in the university of Zaragoza; some related material, perhaps more about boundary conditions than about extensions, is http://arxiv.org/abs/0704.1084, http://arxiv.org/abs/quant-ph/0609023, http://arxiv.org/abs/0712.4353
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$\begingroup$ I hadn't been aware of the reference by Esteve. I have added it to the references of the nLab entry ncatlab.org/nlab/show/quantum+anomaly (many more references are currently still missing there, of course). $\endgroup$ Commented Sep 16, 2011 at 11:14
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$\begingroup$ @Urs Schreiber Thanks for the add. The topic was common folklore in Zaragoza in the nineties and it was not infrequent in PhD theses, but I think that its main role was motivational, either aiming towards other topics, or used as a guide when exploring some other concept. For instance, it was very valuable to me in order to navigate Albeverio et al, who had got into a confusing notation/naming for some self adjoint extensions classifying these "1D point interactions". $\endgroup$– user135Commented Sep 16, 2011 at 11:35
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2$\begingroup$ Thank, I like both answers very much. The references are great. Unfortunately, I have to choose an answer to accept... $\endgroup$– András BátkaiCommented Sep 16, 2011 at 19:23