# Step potential bound states not bound

According to Griffiths, if the energy is less than the potential at −∞ and +∞ then the state is bound. For the step potential this would be if the energy is less than the step height. But there are no bound solutions Is the definition given by Griffiths then formulated incomplete? Here this was discussed, but my question remains.

Edit If $$V(-\infty)< E < V(\infty)$$ is neither bound or scatter, what is it?

• Why is it a problem if there are no bound states? Aug 4, 2022 at 21:13

For a step potential, the energy at $$-\infty$$ is $$V(-\infty) = 0$$ and the energy at $$+\infty$$ is $$V(+\infty) = V_0 > 0$$. A particle with $$0 < E < V_0$$ has $$E > V(-\infty)$$ by definition. Therefore, it does not satisfy the requirement for a bound state, which is that $$E < V(-\infty)$$ and $$E < V(+\infty)$$.
• Ok so you would say the definiton is $E < \min{\left(\lim_{x\to\infty}{V(x)}, \lim_{x\to-\infty}{V(x)}\right)}$. But on Wikipedia it states $E < \max{\left(\lim_{x\to\infty}{V(x)}, \lim_{x\to-\infty}{V(x)}\right)}$ to be positon-bound. Than this Wikipedia articel ist wrong? Aug 5, 2022 at 7:58
• @user18722294: Yeah, I'm pretty sure that's wrong. In the equation they write down, they need $\psi''/\psi$ to be exponentially suppressed for both $x \to \infty$ and $x \to - \infty$, which implies that we must have $\lim_{x \to \pm \infty} (V(x) - E) > 0$. Aug 5, 2022 at 11:43