This came up in the context of the inverse scattering transform for the KdV equation. My primary reference, a set of lecture notes on integrable systems by Maciej Dunajski, makes the claim that the one-dimensional Schrodinger equation has only finitely many bound states if the potential satisfies

$$ \int_{-\infty}^{\infty} (1 + \lvert x \rvert) \lvert V(x) \rvert dx < \infty. $$

I tried looking at different sources, but everything I managed to find states this without proof. Why does the above condition imply finitely many bound states?

These are the places I found this statement repeated:

S. Novikov, S.V. Manakov, L.P. Pitaevskii, & V.E. Zakharov (1984) Theory of Solitons: The Inverse Scattering Method, Consultants Bureau, New York.

T. Aktosun, M. Klaus & C. van der Mee (1998) On the number of bound states for the one-dimensional Schrodinger equation, J. Math. Phys., 39 (9) (1998) (implicitly assumed, I think)

  • $\begingroup$ What is the meaning of $1+|x|$ if $x$ has a dimension? In other terms how do you adimemsion it ? $\endgroup$
    – Jhor
    Jul 26, 2019 at 6:54
  • $\begingroup$ If $x$ is dimensionful, one can add a constant of the same dimension instead of adding $1$. $\endgroup$
    – Styg
    Jul 26, 2019 at 7:08

1 Answer 1


Here is a partial answer.

  1. Ref. 1 essentially writes:

    Consider the class of potentials $u(x)$ such that $$ \int_{\mathbb{R}} \! dx~(1 +|x|)|u(x)| ~<~\infty, \tag{2.5a}$$ which of course implies that $$|u(x)| ~\to ~0 \quad\text{as}\quad |x|~\to~\infty.\tag{2.5b}$$ This condition (2.5a) guaranties that $$\text{there exists only a finite number of discrete energy levels}\tag{2.5c} $$ (thus it rules out both the harmonic oscillator and the hydrogen atom).

  2. Condition (2.5a) is called the Faddeev condition, cf. Ref. 2. Eq. (2.5b) does not follow from eq. (2.5a), as counterexamples shows (e.g. in the spirit of this Phys.SE post). In this answer we will assume eq. (2.5b) independently.

  3. Let us split the potential $u=u^+-u^-$ in its positive & negative parts. We may assume w.l.o.g. that $u=-u^-\leq 0$ since the positive part only (weakly) decreases the number of bound states, cf. this Phys.SE post.

  4. Ref. 3 lists a sufficient condition for eq. (2.5c) as $$ \int_{\mathbb{R}} \! dx \sqrt{u^-(x)} ~<~\infty. \tag{A}$$ This can heuristically be understood from the WKB estimate for the number of bound states $$ N ~\approx~ \iint_{H(x,p)\leq 0} \frac{dx~dp}{h} ~=~\frac{2}{h}\int_{\mathbb{R}} \! dx \sqrt{2m u^-(x)}.\tag{B}$$ Eq. (B) can be trusted in the large-$N$ limit, cf. the correspondence principle.

  5. The Cauchy-Schwarz inequality $$ \left(\int_{\mathbb{R}} \! dx \sqrt{u^-(x)}\right)^2 ~\leq~ \underbrace{\int_{\mathbb{R}} \! \frac{dx}{(1 +|x|)^{1+\varepsilon}}}_{<\infty} \int_{\mathbb{R}} \! dx~(1 +|x|)^{1+\varepsilon}u^-(x) \tag{C}$$ yields the following sufficient condition $$ \exists \varepsilon>0:~~\int_{\mathbb{R}} \! dx~(1 +|x|)^{1+\varepsilon}u^-(x) ~<~\infty \tag{D}$$ for eq. (A). Speaking in heuristic terms, it is reassuring that both conditions (2.5a) & (A) indicate that $u^-(x)$ should morally fall off quicker than $|x|^{-2}$ as $|x|\to\infty$.


  1. M. Dunajski, Integrable systems, lecture notes, 2012, p. 28.

  2. P.G. Drazin & R.S. Johnson, Solitons: An Introduction, 2nd edition, 1989; eq. (3.2).

  3. A. Messiah, QM, Vol. 1, 1967; p. 106.

  4. S. Novikov, S.V. Manakov, L.P. Pitaevskii & V.E. Zakharov Theory of Solitons: The Inverse Scattering Method, 1984.


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