Determining the Sign of $E$ When Solving the Time-Independent Schrödinger Equation

I am having trouble understanding how to choose the sign of $$E$$ when solving the time-independent Schrödinger equation. I understand that for potentials where $$V(\pm\infty) < E$$, we want scattering solutions, and for $$V(\pm\infty) > E$$, we want bound state solutions. However, there are cases where this criterion does not apply

A good example for my question is the step potential question from Griffiths (Problem 2.34). Consider a step potential: $$V(x) = \begin{cases} V_0 & \text{if } x > 0, \\ 0 & \text{if } x \leq 0. \end{cases}$$ Given that $$E I want to find the solutions to the Schrödinger equation for each region and then apply the boundary conditions.

I am having trouble deciding how to solve the Schrödinger equation for the left region where $$V=0$$

The time-independent Schrödinger equation for this region is: $$\frac{d^2 \Psi}{dx^2} = -\frac{2mE}{\hbar^2} \Psi$$

My confusion about the choosing the sing of E:

If I assume E > 0, I find scattering state solutions. $$k^2=\frac{2mE}{\hbar^2}$$ then the solution is $$\Psi=Ae^{ikx}+Be^{-ikx}$$

If I assume E < 0, I find bound state solutions. $$k^2=\frac{-2mE}{\hbar^2}$$ then the solution is $$\Psi=Ae^{kx}$$

These two lead to different types of wavefunctions. How do I make the correct choice for E in this situation or in general?

• The energy can never be less than the minimum of the potential energy. So, in this case, it has to be $E>0$. I remember that this is on Griffiths book, in chapter 2 i suppose. Commented Jun 16 at 23:26
• You are correct that to have normalizable solutions, you need $E > V_{min}$ . In this case, $V_{min} = 0$ . However, for a Dirac delta potential, we can look for both scattering and bound-state solutions. For that, we assume E > 0 for scattering solutions and E < 0 for bound-state solutions, depending on what we want as a solution. Commented Jun 16 at 23:34

Normally, if the potential is finite at infinity, one often set this potential to be $$0$$: $$V(\infty)=0$$. In this case, bound states have $$E<0$$. Thus, for hydrogen, all $$E<0$$ for bound states. This is the same choice as one would do in classical mechanics. Likewise in a finite well (finite depth), one chooses the potential to be $$0$$ outside the well and all bound states have $$E<0$$.
On the other hand, for the harmonic oscillator and the infinite well, the potential is unbounded so the bound state have $$E>0$$.
In principle you can always "shift up" or "shift down" the reference energy by some finite amount. For instance, in the Morse potential, one may choose the dissociation energy to be $$0$$ and then all bound states have $$E<0$$. On the other hand, one can also choose the bottom of the potential to be $$0$$, in which case all $$E>0$$ but the bound states appear when $$E< E_{\text{dissociation}}$$.