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In a course that I follow, we use the perturbation method to find the eigenvectors and energies to an Hamiltonian written $ H_0 + V $ where $V$ is a weak perturbation.

It is written that as $V$ is a weak perturbation, we can write the perturbed eigenvectors as a combination of the eigenvectors of the unperturbed hamiltonian.

But I thought that as we know the eigenvectors of the unperturbed Hamiltonian, then we have a basis of the Hilbert space so there is no need of this assumption (we always can write any vector as a linear combination of eigenvectors of the unperturbed hamiltonian).

Can you help me to understand this?

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    $\begingroup$ The original Hilbert space is defined as the vector space spanned by the eigenvectors of $H_0$. The new Hilbert space is defined as the vector space spanned by the eigenvectors of $H$. In general, these two Hilbert spaces need not be the same. By making the assumption that $V$ is a weak perturbation, we are assuming that the old and new Hilbert spaces are identical. If this is true, then we can choose as a basis, either set of eigenvectors. This is an important assumption in the validity of perturbation theory. $\endgroup$ – Prahar Feb 1 '17 at 17:37
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    $\begingroup$ For instance, an importance case where this breaks down is QCD at low energies, where the eigenvectors of the free Hamiltonian $H_0$ are quarks and gluons, whereas the eigenvectors of the interacting Hamiltonian $H$ are mesons, baryons and glueballs. The latter spans a completely different Hilbert space (for instance, quarks and gluons are not states in the interacting Hilbert space). $\endgroup$ – Prahar Feb 1 '17 at 17:38
  • $\begingroup$ The assumption involves a particular form of linear combination, is that right? Also, @Prahar, OP is almost surely taking a course in non-relativistic quantum mechanics (NRQM), where one can safely assume that $H$ and $H_0$ act in the same Hilbert space. $\endgroup$ – pppqqq Feb 1 '17 at 17:43
  • $\begingroup$ @pppqqq - I don't see what this assumption has to do with NRQM. For instance, suppose $H_0 = \frac{p^2}{2m_1}$. This is the free Hamiltonian corresponding to a free particle of mass $m_1$ and this has a corresponding Hilbert space ${\cal H}_0$. Now, suppose I choose $V = \frac{p^2}{2m_2}$. The new Hilbert space defined w.r.t. $H$ now contains new states corresponding to free particle $m_2$ which were not present in ${\cal H}_0$. Therefore, in this example, $V$ cannot be assumed to be a weak perturbation on $H_0$......... $\endgroup$ – Prahar Feb 1 '17 at 17:46
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    $\begingroup$ @Prahar sorry but I don't understand what you're talking about. Take the last example: both $H_0$ and $H$ are self-adjoint operators on, say, $\mathcal H =L^2(\text d ^3\mathbf x _1 \text d ^3 \mathbf x _2)$. For sure, $H$ can have a point spectrum, but that doesn't mean that it acts in another Hilbert space. I think that you might have in mind some QFT-related issues, but this would be an unnecessary distraction for the OP, in my opinion. $\endgroup$ – pppqqq Feb 1 '17 at 18:04