# Physical interpretation of negative total energy

I am going through Griffith's text on Quantum Mechanics, in which he states $$\begin{cases} E < V(-\infty) \text{ and }V(\infty) \implies \text{bounded state}\\ E > V(-\infty) \text{ and }V(\infty) \implies \text{scattering state} \end{cases}$$ where $$E$$ is the total energy of the particle. This makes sense. However, he notes that most potentials tend to 0 as you approach infinity, and so the above simplifies to $$\begin{cases} E < 0 \implies \text{bounded state}\\ E > 0 \implies \text{scattering state} \end{cases}$$ This is where I am having some trouble with the physical interpretation. How can one have negative total energy? Can someone provide some intuition and maybe an example?

• It’s a matter of where you assign the reference. Usually what you care about is the potential difference and the amount of work done as you move from one place to another. Examples are gravity and orbits, or electrostatics as you move a charge in a coulomb potential. So if you assigned the reference differently you could change how you label the energy axis, but the work done or difference between energy levels etc. would still be the same. Jun 23 at 18:17
• @UVphoton So the total energy in this case is relative? Jun 23 at 18:25
• They are defining the reference to be at infinity. So the analogy is for gravity if E >0 the particle would not be able to orbit. For a quantum well potential or an atom the electron would not fall into the potential well and be trapped at an energy well unless E<0. Note even with E>0 it would still be influence by the potential and for example change direction -so for E>0 scattering. E<0 could be trapped in a bound state or energy level or orbit depending on the nature of the potential. Jun 23 at 18:34
• Jun 23 at 23:01

• @CCBAM A negative potential does correspond in these cases to an attractive force (although there can be more details). An example of a negative potential is that of the Kepler problem, for example (which also corresponds to the potential for the hydrogen atom). And precisely, a particle can't access regions where the potential is larger than the total energy. That would require the kinetic energy $\frac{p^2}{2m}$ to become negative, which is not possible. Jun 23 at 18:26