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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?

Qmechanic
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StarBucK
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    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. – Prahar Feb 01 '17 at 17:37
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    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). – Prahar Feb 01 '17 at 17:38
  • 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. – pppqqq Feb 01 '17 at 17:43
  • @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 $H_0$......... – Prahar Feb 01 '17 at 17:46
  • The assumption doesn't assume any form on the vector, we wrtie $|\Psi \rangle = \sum_k c_k |\phi_k \rangle$ where $|\phi_k \rangle$ are basis of eigenvectors of the unperturbed Hamiltonian. The assumption is only here to say "we can write $|\Psi \rangle$ like this (so it implicitly means that the $|\phi_k \rangle$) "still" are a base of the Hilbert space. – StarBucK Feb 01 '17 at 17:46
  • @pppqqq - ......Of course, in this trivial example, the resolution is to simply choose $H_0 = \frac{p^2}{2m_1} + \frac{p^2}{2m_2}$, but in more complicated examples this is a subtle question that requires care to study. For another example, suppose $H_0 = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2}$ and then I add $V = \lambda |r_1-r_2|$ and make $\lambda$ extremely large. Then, in ${\cal H}_0$, there exists a state in which the two particles exist separately, whereas due to immensely large interaction potential, in the full theory (w.r.t. $H$), one can only have bound states of the two particles. – Prahar Feb 01 '17 at 17:50
  • I feel like your confusion is coming from the fact that you think that any basis is Ok, as long as it is orthonormal and complete. But this is not true. The observables that you want to get out of the theory have to "live" in the Hilbert space built by diagonalization of the Hamiltonian of the system. – MsTais Feb 01 '17 at 17:53
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    @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. – pppqqq Feb 01 '17 at 18:04

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