Suppose a system is under the influence of a potential which vanishes at $\pm \infty$. Now we know that if the energy of the system is negative ($E<0$) then the system is in a bound state and the bound state energy spectrum is discrete. For $E>0$, the system is in scattering states (with continuous energy spectrum). My question is:

What happens at $E=0$? Can we consider it a discrete bound state or should we consider it as a scattering state and why?

  • 1
    $\begingroup$ Well, have you tried to see what happens when you put in $E=0$? $\endgroup$ – Aaron Stevens Aug 20 at 20:59
  • $\begingroup$ You mean if we put $E=0$ in the Schrodinger equation? I have a question about that, if I get a non-normalizable state then does it mean that the state is a scattering state and if it is normalizable then it is a bound state? $\endgroup$ – abhijit975 Aug 21 at 14:21
  • $\begingroup$ I have a $\delta$-function potential. So at $E=0$, the wave function is just a constant meaning it is non-normalizable. $\endgroup$ – abhijit975 Aug 21 at 15:42
  • $\begingroup$ Doesn't the delta function potential only have one bound state anyway? $\endgroup$ – Aaron Stevens Aug 22 at 1:42
  • $\begingroup$ Yes it has only one bound state but I was trying to see what will happen to the bound state in the limit the coupling constant (i.e. the constant that multiplies $\delta (x)$) goes to zero. $\endgroup$ – abhijit975 Aug 22 at 15:47

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