# Can bound states have zero energy?

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?

• Well, have you tried to see what happens when you put in $E=0$? – Aaron Stevens Aug 20 at 20:59
• 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? – abhijit975 Aug 21 at 14:21
• I have a $\delta$-function potential. So at $E=0$, the wave function is just a constant meaning it is non-normalizable. – abhijit975 Aug 21 at 15:42
• Doesn't the delta function potential only have one bound state anyway? – Aaron Stevens Aug 22 at 1:42
• 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. – abhijit975 Aug 22 at 15:47