ajb Posted June 18, 2006 Share Posted June 18, 2006 In quantum mechanics it is assumed that any state can be expaned in terms of the eigenvectors of some operator. That is the eigenvectors form a basis for the Hilberts space. Does anyone know if this is automatic from the theory of Hilbert spaces and their operators or if it is a postulate of quantum mechanics? Link to comment Share on other sites More sharing options...

Perturbation Posted June 18, 2006 Share Posted June 18, 2006 It depends if the Hilbert space is separable, that is that any state-vector can be written as a sum of elements of a countable basis set. And whether you can expand it into a basis of an operator's eigenvectors depends whether the eigenvectors are linearally independant, i.e. orthogonal. For the eigenvectors of a Hermitian operator this is always the case, it's a simple property of Hermitian operators, at least if they're non-degenerate. It's usually assumed that the Hilbert space is separable, but I do believe the guys working on early quantum gravity came across problems with separability when they tried to build a Hilbert space, and was one reason why they moved to using spin networks. If you wanted to use a set of eigenvectors of some operator as an expansion consisting of purely diagonal terms for some other operator, the two operators would have to commute. Link to comment Share on other sites More sharing options...

timo Posted June 19, 2006 Share Posted June 19, 2006 I faintly remember that you need the "Auswahlaxiom" which I suppose is called "axiom of choice" in english. Other than that I would not assume that anything physics-specific (the axiom of choice of course is not really physics specific, either) comes in: Sketch of the reasoning I´d try to follow: In QM, operators are linear. => They can be written as a matrix. -> The eigenvectors of a matrix form a base for the space (here comes a small problem as the only people having told me so far are physicists so it might not be true in general - but I think it is). Link to comment Share on other sites More sharing options...

Perturbation Posted June 20, 2006 Share Posted June 20, 2006 Oh, yeah, axiom of choice. Ignore what I said about separability. http://en.wikipedia.org/wiki/Hilbert_space#Bases Link to comment Share on other sites More sharing options...

ajb Posted June 22, 2006 Author Share Posted June 22, 2006 Atheist, your last point is exactly my question. Even if we assume that the Hilbert space is seprable and therefore has at least one basis. Can this (or any other) basis be written in terms of eigenvectors and eigenvalues of "some" operator? It is true that in standard quantum mechanics we can do this. However, my question really is "does this come for free"? Things like Zorns lemma and the axiom of choice are needed to set up a basis on any vector space. My question was not really aimed at that. Link to comment Share on other sites More sharing options...

Severian Posted June 22, 2006 Share Posted June 22, 2006 Can this (or any other) basis be written in terms of eigenvectors and eigenvalues of "some" operator? Yes. This is a true statement since the operation can be simply projection onto the basis. So this does indeed 'come for free'. Link to comment Share on other sites More sharing options...

ajb Posted June 22, 2006 Author Share Posted June 22, 2006 I didn't think of that! Link to comment Share on other sites More sharing options...

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