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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?

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    $\begingroup$ 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. $\endgroup$
    – UVphoton
    Jun 23 at 18:17
  • $\begingroup$ @UVphoton So the total energy in this case is relative? $\endgroup$
    – CBBAM
    Jun 23 at 18:25
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    $\begingroup$ 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. $\endgroup$
    – UVphoton
    Jun 23 at 18:34
  • $\begingroup$ See Potential energy curve for intermolecular distance $\endgroup$
    – mmesser314
    Jun 23 at 23:01

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The total energy can be negative because, away from infinity, the potential can be negative. If the total energy is negative, then regions with zero potential are unaccessible for the particle, since it can't have negative kinetic energy. That's why Griffiths calls that situation "bounded": the cases with negative energy are constrained to the region of space in which the potential is negative enough for the total energy to be whatever value it is while the kinetic energy is still non-negative.

Particles with positive total energy, on the other hand, won't have those restrictions at infinity (for potentials vanishing at infinity). Hence, they can keep going on forever, without being restricted to a finite region. That's why Griffiths calls this case scattering.

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  • $\begingroup$ Thank you for the reply. What would be an example of a negative potential and what is its interpretation? At the moment I think of a negative potential as a sort of "attractive force", for lack of a better term. Also, from what I can gather the general rule is that you cannot access regions where the potential energy is greater than your total energy. So in this case it makes sense as to why a particle with negative energy cannot access a region of 0 potential. $\endgroup$
    – CBBAM
    Jun 23 at 18:21
  • $\begingroup$ Expanding on my last comment, the intuition I am building is that having a positive total energy means you have the ability to move freely (assuming the potential vanishes at infinity), where as a negative total energy means you are bounded to a certain region (in my analogy, you are subject to some kind of attractive force and hence negative). Is this the correct physical picture to have? $\endgroup$
    – CBBAM
    Jun 23 at 18:25
  • $\begingroup$ @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. $\endgroup$ Jun 23 at 18:26
  • $\begingroup$ @CCBAM Even with positive energy, there could be "hills" in the potential which the particle can't access (apart from tunnelling), so there are also some more details. For example, there could be a region where the interaction is repulsive. Nevertheless, that is indeed the general idea. $\endgroup$ Jun 23 at 18:28

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