The Levinson's theorem states that the number of bound states $N_b$ is $$ N_b = \frac{1}{\pi} (\delta (0) - \delta (\infty)).$$

  1. My question is: can this number be negative? For example if the potential is positive within its range. If so, what is the meaning of negative numbers of bound states?

  2. Additional question: from what I have seen in graphs the phase $\delta(E)$ is always negative. Does it hold as a general statement that phase is always negative?

  • $\begingroup$ I have added an answer - does it answer your question? If not please point out any issues you may have with it, so that I can improve it. $\endgroup$
    – Martin C.
    Commented Jul 6, 2022 at 19:35

1 Answer 1


The short answer is no, the number of bound states in Levinson's theorem cannot be negative.

The validity of Levinson's theorem requires the potential $V$ to have certain properties.

Firstly, it must be radially symmetric, i.e. $V = V(r)$.

Secondly, the potential must satisfy certain asymptotic limits (see e.g. the wikipedia article) - these limits require the potential to be attractive at the origin, and at spatial infinity. Only attractive potentials can, potentially, have bound states.

I don't know which graphs you are referring to, but if the phase shift was negative, then the potential was presumably repulsive (and not attractive) in that regime, .e.g. the nuclear force at less than about 0.7fm. So Levinson's theorem does not hold.

Physics is full of counterintuitive results, but in this case the common sense notion of using positive integers to count a finite set of discrete things is sufficient!


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