# Non-integrable Wavefunctions

Suppose first-order perturbation yields a credible correction to the energy, but a correction to the wave function that's not square-integrable. That can happen, I see no reason why it couldn't. Unless there is some proof that it can't happen (I haven't found any). And if it does happen, then would you believe the energy shift you calculated?

• Off the top of my head, that should be forbidden by the Kato-Rellich theorem. But I will check just to be sure. – DanielC Oct 11 '17 at 6:15
• Yes, thanks.... The point is that, in perturbation theory, the 1st-order correction to the eigenfunction is expressed as a linear combination of all the unperturbed eigenfunctions. Each of the unperturbed eigenfunctions is square-integrable, granted, but the series may yield a non-integrable function. – Jessica Oct 11 '17 at 6:47
• The corrections themselves depend on your potential. The question then is what potential do you allow to be added to your problem. Taking as an example the hydrogen atom, we get a correction that behaves like $\Sigma \frac{m^2n^2}{m^2-n^2} <n^{(0)}|V|m^{(0)}>$. If we can introduce a potential which has its matrix elements decay slower than $n^2 m^2$, we can get a divergent series. – Omry Oct 11 '17 at 7:13
• That is a delicate problem, it should be discussed in Chap VI of Kato, Perturbation Theory for Linear Operators, Springer 1966 – Valter Moretti Oct 11 '17 at 8:55
• Not exactly what you're asking, but consider the harmonic oscillator Hamiltonian $H$ with a $\lambda x^4$ perturbation. This has "credible" corrections for the eigenvalues and eigenvectors, but for $\lambda < 0$, $H+\lambda x^4$ does not have discrete eigenvalues, so you can't believe formal perturbation calculations. – Keith McClary Oct 12 '17 at 1:36

I think the wiki's proof for first order correction term is rather clear. https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)#First_order_corrections

The important sentence connect to your quesiton is "Let $$V$$ be a Hamiltonian representing a weak physical disturbance". The core of pertubation is the linearlization under small changes, so don't think $$V$$ as some usual potentional that contain singularities, if it does, it probabily meant your system is confined in a much smaller region by infinite wall.

Then, before you obtain the correction term, there is expression $$(E_n^{(0)}-E_k^{(0)} )\langle k^{(0)}|n^{(1)} \rangle=\langle k^{(0)}|V|n^{(1)} \rangle$$ which guarranteed the RHS is a small finite number, and each peatubated basis vector(at least the first one) is square integrable.(square integrable meant you can get a finite number, not Riemann integability.)

If you constructed a $$V$$ such that it's not square integrable, then you should not be using pertubation theory, because that's a "kick" not a pertubation.

From the basics of QM, the only wave functions which are square integrable are physical. Therefore, the wavefunctions which are not square integrable can only appeare on math but no physical system can have such a wavefunction.

Besides the original Hamiltonian, the corrections to energies and wavefunctions only depends on the perturbations(interaction Hamiltonian). By restricting on physically acceptable wavefunctions, we can essentially restrict the allowed perturbations on the given original system.

The great thing is that this type of restrictions on allowed interactions does naturally ocurr in the context of Quantum Field Theory (QFT). By invoking Gauge invariance on the original lagrangian.

For example, by invoking Gauge invariance on the free Dirac lagrangian we get the quantum electrodynamics, QED, Lagrangian which only have one interaction term corresponding to fermion-photon interaction.

Also, in QFT non-physical interactions can be ruled out by dimensional analysis and renormalizability of the theory.

• So no free particles for you? – lalala Sep 24 '18 at 2:38
• @lalala what do you mean? – Aman pawar Sep 24 '18 at 12:30
• Eigenfunctions of free particle Hamiltonian are not square integrable. – lalala Sep 24 '18 at 16:33
• @lalala No. It is square integrable. For a free particle: \psi_k(r) = [e^{ik.r}]/\sqrt{V} and \integral dr |\psi_k(r)|^2 = 1. – Aman pawar Sep 24 '18 at 20:37
• It is not integrable since $\psi^*\psi$ is a contant, so integrating over the whole space will give infinities... unless of course $V$ is also infinite, but then your $\psi_k(r)$ is zero everywhere... – ZeroTheHero Nov 15 '20 at 17:39