Dear Juan D. Bulnes,

Yes, the space does not change

Dear I. Yu. Popov 

I want to venture a guess but I would suggest you checked this from other sources. If I understand you correctly you speak of perturbations to the state vectors in the Hilbert space of the system and not the perturbation to the Hamiltonian  which would give rise to such a situation.

In this context I would like to add to what Juan D Bulnes has stated. Basically if the perturbation to the Hamiltonian is itself Hermitian in nature as the potential given in Juan's example it should modify the eigenstates of the unperturbed Hamiltonian in such a way that the perturbed eigenstates form a complete basis set. This is a standard requirement for any Hermitian operator.

I would like whoever reads this to critically examine my argument and suggest any corrections if reqiured.

Dear I. Yu. Popov

There are old results that probably are helpful in this case. See, for  example, the book by F.V. Atkinson "Discrete and continuous boundary problems", Academic Press, 1964.

I send you some of them

Dear Pinchas Malits,

Thank you.

Igor

@I. Yu. Popov,

to make your question meaningful you have to define what a 'perturbation of a basis' is for you. It certainly is not such definite concept as 'perurbation of a Hamiltonian' and also in this case there are many technical aspects that need to be specified.

Consider the (uniquely defined) linear map such that

D : \psi_n --> \phi_n. 

What you are asking for is a condition on D to be invertible. Here are a few conditions:

A)  D is uniformly close to the identity, more precisely

|| 1 - D||_\infty^2  := sup_{ = 1}    <  1 

(proof:  D^{-1} = (1 - (1-D) )^{-1} = \sum_{n = 0}^\infty (1-D)^n

this is actually a quite effective way to compute the inverse if ||1-D|| is small.)

B) (1 - D)  is traceclass and the Fredholm determinant of D  is non zero. Here

det( D) = exp (log (det (1 - (1 - D))))

           =  exp (- \sum_{n = 1}^\infty ( 1 / n) tr (1-D)^n))

Dear Prof. Brussee,

Thank you.

As a continuation of my previous post: yet another condition:

C) 1- D is a  compact operator and D is injective. This is a less restrictive but maybe harder to check version of B. 

(proof: 1  -  (1-D)  is a Fredholm operator with index = index(1)  = 0, so just as in the finite dimensional case, injective implies surjective and therefore invertibility.)

Dear Yuri,

Please read the book by Gohberg & Krein, chapter about bases, theorem by  M.  Krein.

If I understand your question, you will find everything there.

Dear Boris Pavlovich,

Thank you. Of course, I know this book. I would like to find more fresh results.

Igor

Dear Igor,

With regard to your question, you might find the attached preprint by Boris Mityagin and Petr Siegl, as well as some references therein, of some interest.

Kind regards,

Silvestro

Silvestro,

Thank you!

Igor

Dear Popov

Please see the work of mine in 

JPhys.Soc.Japan in (1998     ).where I have discussed normalisation

condition .I hope this will give an idea to write completeness relation.

Further first write the orthogonality relation .

Hope you will able to find a clue.

Best wishes.

B.Rath