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

Question

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 $E<V_0$ 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?

Answer 1

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 $E< E_{\text{dissociation}}$.

Note that (as alluded to in a comment), the bound state energies are always greater than the minimum energy of the potential.