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I'm reading through Griffiths Intro to QM 2nd Ed. and when it comes to bound/scattering states (2.5) they say:

$E<0 \implies$ bound state

$E>0 \implies$ scattering state

Why doesn't this change depending on whether you have a positive or negative delta-function potential?

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    $\begingroup$ This is determined by the behaviour at infinity. For a scattering state in 1d, the stationary state with energy $E$ goes as $\exp(\pm i kx)$ with $E=\hbar^2k^2/2m>0$ while for $E<0$, it goes as $\exp(- \kappa |x|)$ with $E=-\hbar^2\kappa^2/2m>0$. $\endgroup$
    – suresh
    Oct 13, 2014 at 3:09
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    $\begingroup$ To clarify, for bound state, it should be $E<V(\pm \infty)$, which is just happen to be zero in this case. $\endgroup$
    – unsym
    Oct 13, 2014 at 3:52

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This definition of bound and scattering states is not quite correct, although it holds for many potentials. There are counterexamples to this fact that have roots in a paper by von Neumann and Wigner. One is the spherical potential $$V(r)=32\sin r \frac{\sin r-(1+r)\cos r}{\left(2+2r-\sin 2r\right)^2}$$ It is not hard to check that $V(r)$ is a bounded continuous function that vanishes at infinity. Even though, the function $$\psi(x)=\frac{2\sin r}{r\left(2+2r-\sin 2r\right)}$$ is an eigenfunction of $H=-\Delta + V$, with eigenvalue $1>0$.

Mathematica graphics

So this is an example of a bound state for which these conditions do not hold. The precise definition of bound states is more subtle and is given by the elements of $$\mathcal{H}_{\text{bound}}(H)=\left\{\psi(x,0) \in L^2(\Bbb{R}^n):\lim_{r\to\infty}\sup_{t\in \Bbb{R}} \int_{\Bbb{R^n}\backslash B(0;r)}|\psi(x,t)|^2dx=0\right\},$$ where $\psi(x,t)=e^{-itH}\psi(x,0)$, that is, the states that are localized in space for any time $t$. It's always true that $\mathcal{H}_{\text{bound}}(H) \supset \mathcal{H}_{\text{p}}(H)$, the closure of the set of linear combinations of eigenvectors of $H$, and for some potentials the equality holds.

For the delta-function potential, the realization of a self-adjoint operator that has the right properties is not so simple, but can be done in some ways. But as Griffiths discuss, the change from a negative to positive delta potential kills the bound state, since its only eigenfunction is not normalizable anymore and all states are scattering states.

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  • $\begingroup$ Is the eigenstate you mention embedded in a continuum (and hence a resonance? or subject to a Fano coupling to the continuum?) Or does the continuum start above it? More generally, can you provide a reference to the original paper or a good review of that potential? $\endgroup$ Oct 13, 2014 at 14:32
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    $\begingroup$ Thanks for the image! It is embedded in the continuum spectrum that is in this case $\sigma_c(H)=[0,\infty)$. The original paper is von Neumann, John; Wigner, Eugene (1929). "Über merkwürdige diskrete Eigenwerte". Physikalische Zeitschrift 30: 465–467. But for good references about these states check: for a more physical look section 10.4 of Ballentine's Quantum Mechanics; for a mathematical-physical look section 11.4.2 of de Oliveira's Intermediate Spectral Theory and Quantum Dynamics. $\endgroup$ Oct 13, 2014 at 14:38
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If $E < V\left(-\infty\right)$, $E<V\left(+\infty\right)$, and $E > V_{min}$ (necessary for $\Psi$ to be normalizable), then it is a bound state, and the spectrum will be discrete: $$ \Psi\left(x,t\right) = \sum_n c_n \Psi_n\left(x,t\right). $$ Otherwise -- if $E > V\left(-\infty\right)$ or $E > V\left(+\infty\right)$ -- it is a scattering state, and the spectrum will be continuous: $$ \Psi\left(x,t\right) = \int dk \ c\left(k\right) \Psi_k\left(x,t\right). $$

$V\left(\pm \infty\right) = 0$ for both $V\left(x\right) = +\delta\left(x\right)$ and $V\left(x\right) = - \delta\left(x\right)$, so $E$ has to be negative to have a bound state.

$\min +\delta\left(x\right) = 0$, so there are no bound states for $V\left(x\right) = +\delta\left(x\right)$.

$\min -\delta\left(x\right) = -\infty$, so for $V\left(x\right) = -\delta\left(x\right)$, $E<0$ for a bound state and $E > 0$ for a scattering state, as you have.

EDIT: As @Mateus Sampaio pointed out though, apparently there are exceptions to the general rules above.

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