I am currently dealing with the perturbation theory. For the moment in my lectures we are considering the time independent perturbation theory. Degeneracy might be relevant or not. I have a set of questions, which deal more with an intuitive understanding of this theory:

  • Why is the perturbation theory releated to the hamilton operator of a quantum mechanical system, and not another operator of it? Is it because every type of disturbance results in an energy change for the system, hence why we consider the Hamilton operator?

Let's consider an unperturbed quantum mechanical system and focus on its Hamiltonian. The set of its eigenvectors (unperturbed eigenvectors) form a basis: $\{|n\rangle_0\}$. Then we have $_0\langle n|m\rangle_0=\delta_{n,m}$. Now if a small external disturbance acts on the system, we will consider the perturbed Hamiltonian, eigenvalues and eigenstates. In the lecture the following was written: $$|n\rangle_1=\sum_{m\neq n}c_m|m\rangle_0$$

What the formula implies is that the correction term of the first order for an arbitrary eigenstate of the perturbated system can be expressed as a linear combination of all the unperturbed eigenvectors of the system excluding the unperturbed eigenvector, which is the $0^{th}$ term in the expression for the perturbed eigenstate, for which we are trying to find the first order correction term. And the reason why we exclude the above mentioned unperturbed eigenstate is because:

$c_m=_0\langle m|n\rangle_1$

And if $m=n$ then:

$c_{m=n}=_0\langle n|n\rangle_1=0$.

The last equation implies that the inner produce between the 0th order correction term and the first for the same eigenstate is zero.

I have the following questions about what's going on here:

  1. For this specific example which involves 0th and 1st order correction terms. Under what ground or rather why can we be confident to say that the first order correction term of an arbitrary eigenstate can or is expressed as a linear combination of the 0th order correction terms of all the other eigenstates? Where does this come from?

  2. Can the formula above be generalized? What I mean by that is: How would the above expression look like if we were dealing with the 2nd order correction term, would it be a linear combination of the 1st order correction terms? Or in more general terms if we consider the $k^{th}$ correction term of an arbitrary eigenstate $|n\rangle$ then is this correction term a linear combination of the one-order-lower correction terms, meaning $(k-1)^{th}$ of all the eigenvectors, excluding the one which belongs to the eigenstate for which we try to find the $k^{th}$ term? Or is a summation of all the correction terms from the 0th to the $(k-1)^{th}$ of every single eigenstate?

  • $\begingroup$ could you please produce a more specific title? $\endgroup$ Dec 4, 2022 at 14:46

1 Answer 1


These two questions are related. A fundamental "assumption" about perturbation theory is that it's always possible to write the $k^{th}$ order correction as a linear combination of the unperturbed states. \begin{equation} \left | n \right >_k = \sum_m c_m \left | m \right >_0 \end{equation} Note that $m = n$ can come back for orders $k > 1$. The standard results in perturbation theory use this to express everything in terms of $\left | m \right >_0$ and you can find them on Wikipedia.

Now why do the free eigenstates span the interacting Hilbert space as well? I won't claim to have a rigorous proof ready. And indeed there are some cases like quantum field theory (see Haag's theorem) where it's not true. But the intuition is that square integrable functions (which are the sensible wavefunctions) can always be decomposed using the Fourier transform. This is an expansion in plane waves which are energy eigenstates of the free particle Hamiltonian on $\mathbb{R}^d$. A simple way to extend this result is to consider the "particle in a box" where the box is an arbitrarily shaped bounded domain $\Omega$. This would replace "plane waves" above with eigenfunctions of the Laplacian on $\Omega$. And happily, these again form a basis for square integrable wavefunctions on $\Omega$.

  • $\begingroup$ are you saying that while $_0\langle m|n\rangle_1 = 0$ if $m=n$, for $_0\langle m|n\rangle_i \neq 0$ for $m=n$ AND $i>1$, i.e $i=2$ and $n=m$ then $_0\langle n|n\rangle_2 \neq 0$ ? $\endgroup$
    – imbAF
    Dec 4, 2022 at 13:41
  • $\begingroup$ Yes. The system would have to be special to give $0$ for all $i$. $\endgroup$ Dec 4, 2022 at 14:44

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