# Why does this condition guarantees there exists only a finite number of discrete energy levels?

I'm reading section 2.2.1 of the book Solitons, Instantons and Twistors by Maciej Dunajski. The section is on the subject of direct scattering.

It is claimed that, considering Schrodinger's equation with the class of potentials $u(x)$ such that $|u(x)|\rightarrow 0$ as $x\rightarrow \pm\infty$, the integral condition

$$\int_{\mathbb{R}} \left(1+|x|\right) \left|u(x)\right| dx<\infty$$

guarantees that there exists only a finite number of discrete energy levels.

I'm failing to understand in what way this condition guarantees this. Help is really appreciated.

It follows from Bargmann's limit. The Bargmann bound is well known for a three-dimensional central potential [1, Thm. XIII.9] [2,3]: $$N(\ell) < \frac{1}{2\ell + 1} \int_0^\infty r V^-(r) dr$$ Here, $\ell$ is the azimuthal quantum number, $N(\ell)$ the number of bound eigenstates with this quantum number and $V^-(r) = \max\{0,-V(r)\}$.

With a trick, we can apply this to a one-dimensional potential [2, Section III] [1, Problem XIII.22]. The trick is that the 1D Schrödinger equation $$(-\partial_x^2 + u) \psi(x) = E\psi(x)$$ is equal to the $\ell=0$ radial Schrödinger equation $$(-\partial_r^2 + V(r)) \psi_{rad}(r) = E\psi_{rad}(r)$$ except for the $\psi_{rad}(0) = 0$ boundary condition.

Also, we need to know that the number of negative energy bound states equals the number of nodes of the zero-energy wave function, and the same holds in the 3D case for the number of nodes of the zero-energy radial wave function. Long story short, we need to count the nodes of the solution to $$(-\partial_x^2 + u) \psi(x) = 0 .$$ And we can do so by splitting the problem into two parts that look like counting the negative energy bound states of 3D radial problems. This is explained in detail in [2, Section III], the result is that $$\boxed{N < 1 + \int_{-\infty}^\infty |x| u^-(x) dx }$$ in the 1D-case.

Finally, apply this to your case. Note that $0 \leq |x| \leq |x|+1$ and $0 \leq u^-(x) \leq |u(x)|$. Hence, $$N < 1 + \int_{-\infty}^\infty (1 + |x|)\, |u(x)|\; dx < \infty ,$$ there is a finite number of bound eigenstates.

[1] M. Reed, B. Simon: Methods of Modern Mathematical Physics 4: Analysis of Operators.
[2] K. Chadan, N.N. Khuri, A. Martin, T.T. Wu: Bound States in one and two Spatial Dimensions (arXiv:math-ph/0208011)
[3] https://en.wikipedia.org/wiki/Bargmann%27s_limit

• Let me comment on your additional conditions that I didn't use: $u \in L^1$ and $u \to 0$ for $x \to \pm\infty$. I assume that those guarantee applicability of Bargmann's limit. But to be honest, I'm not sure about the precise conditions, and neither [1] nor [2] state them for the 1D case. I think, $u \in L^1$ could imply $V(\vec x)$ is in $R + L^\infty$ as required for the 3D theorem. $u \to 0$ is probably needed to have negative discrete and positive continuous spectrum in the first place. Maybe someone else knows more about that... – Noiralef May 9 '17 at 18:27