In perturbation theory perturbed eigenstates expanded by unperturbed eigenstates, but we know when the system perturbed its Hilbert space altered and hence its basis changed, then we can't state this basis spans the new Hilbert space and we can't expand perturbed eigenstates by them. How this problem resolved? Please don't answer this question to say that it is approximately true since in textbooks this is done in exact manner.
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1$\begingroup$ For a similar question and answers, see physics.stackexchange.com/q/105164 $\endgroup$– UrgjeCommented Mar 31, 2014 at 10:01
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then we can't state this basis spans the new Hilbert space
There is no "new" Hilbert space. If the original basis of space $L_2(x\in \mathbb{R})$ consisting of eigenfunctions of $\hat{H}$ is complete, any admissible function $\psi$ can be expanded, eigenfunctions of $\hat{H}+\hat{H}{}'$ including.
answered Mar 30, 2014 at 11:01
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In perturbation theory perturbed eigenstates expanded by unperturbed eigenstates,
Yes, they are!
but we know when the system perturbed its Hilbert space altered and hence its basis changed,
Having a basis means that any vector is expanded by those. Thus, in general, the new eigenvectors are a linear combination of the former.
then we can't state this basis spans the new Hilbert space and we can't expand perturbed eigenstates by them.
The system is the same, your Hilbert space is unchanged. However, the eigenvectors have changed. What you must do is find the new set of eigenvectors, which can be expressed as combination of the former eigenvectors.
Please don't answer this question to say that it is approximately true since in textbooks this is done in exact manner.
You're wrong. It IS an approximate solution, it needs to be computed order by order. However, there are a few examples were the perturbative solution (up to certain order) is the exact solution, but it is not the rule! [a typical example is a charges harmonic oscillator in the presence of a constant electric field].
