# Can't a hydrogen electron have net positive energy?

http://www.physicspages.com/2011/06/07/hydrogen-atom-series-solution/

is the 2nd half of a solution to the hydrogen atom Schrödinger equation.

They derive that $E = -1/n^2 *$ (bunch of positive constants) (equation 29).

So this seems to be saying that the energy of the electron, that is the net of potential energy plus kinetic energy, must be negative.

But couldn't a hydrogen electron be excited somehow so that its kinetic energy outweighs its potential energy (its potential energy is negative since infinite separation of electron from nucleus is defined to be 0 potential energy) and thus have a net positive energy?

• Related: physics.stackexchange.com/q/133121/50583. An electron that has positive energy is not bound to the atom anymore (by convention). Is that what you're asking? – ACuriousMind Aug 8 '15 at 13:12
• Whenever a system has negative energy, it is a bound system. When it has positive energy, it is a free system. If you give positive energy to an electron it will become free and escape. This is obvious because if the kinetic energy is greater than the potential, it can escape the potential. This can be compared with the classical system of a staellite orbiting a planet. – Kartik Aug 8 '15 at 13:20
• CuriousMind: Thank you for the link. I see that I should check the sidebar suggested previous questions more often when asking a question. – a00 Aug 8 '15 at 14:30
• Kartik: Indeed the fact that if you choose the potential at infinity to be 0, then the magnitude of the potential energy (for example, -13.7 eVolts) telling you exactly how much kinetic energy is needed to escape, is a great reason for choosing the potential at infinity to be 0, and yet this was somehow never sunk into my head from previous classes. Thank you for reiterating the point. – a00 Aug 8 '15 at 14:31

The energy is defined as $$E = \frac{p^2}{2m} + V(\vec r)$$ where the first term is the kinetic energy and the second term is the potential energy calibrated so that $V(\vec r)=0$ for $|\vec r|\to\infty$.

Consequently, you may say that the energy in a given state (an analogy of an orbit in classical physics) is equal to the kinetic energy $T_\infty$ that the electron has when it escapes to infinity, $|\vec r|\to\infty$.

When this $T_\infty$ is positive, it means that the electron may – and almost certainly will – escape to infinity and never return back, approaching some constant nonzero velocity $\vec v$ as it flies away from the nucleus. If the electron escapes in this way, we say that the atom has been ionized. It is no longer a bound state. It doesn't make sense to call it a hydrogen atom. The corresponding classical trajectories would be hyperbolae, not ellipses as they are for the bound states.

In other words, for $E=T_\infty \gt 0$, we have the system of the electron and the nucleus, but they are not bound and shouldn't be described as "one object". The spectrum of this system – a part of the spectrum that one gets from solving the equations for the "hydrogen atom" – is continuous. This is characteristic for a composite particle that has been separated to pieces.

We need $E=T_\infty \lt 0$ which guarantees that the particle doesn't have enough energy to escape to $|\vec r|\to \infty$ at all. Consequently, the probability is basically 100 percent that the electron (in this case) is closer than a millimeter from the nucleus. They're bound. They're a bound state. One may effectively say that the electron is confined (by the potential energy profile and by its limited extra kinetic energy) to a finite region around the hydrogen atom.

If that's so, if we have a bound state, the energy is negative but the spectrum is discrete – much like it is discrete for a particle in a potential well or any other "compact" space.

So the spectrum of the Hamiltonian for the hydrogen atom is "mixed". It has the discrete states with $E\lt 0$ that you mentioned and that described an actual atom, a bound state of the nucleus and the electron. And then the spectrum has a continuous part for $E\gt 0$. For all values $E\gt 0$, there exists an eigenstate and it's some deformed plane wave for the electron describing the electron moving from space mostly away from the nucleus.

There are also eigenstates for $E=0$ (the marginal case corresponding to the parabolic trajectories in classical physics) but they're a measure-zero "edge" of the continuous spectrum – or, equivalently, an $n\to \infty$ limit of the discrete bound states – and the subtleties with the exact value $E=0$ may be pretty much neglected.

• Your answer has connected what I know about potential wells: we can think of negative energy magnitude representing depth of well. My thought was also provoked by your statement that the bound part (energy < 0) is discrete and the escaped part (energy > 0) is continuous. Overall your answer was clear but also provoked new insights which is uncommon. Thank you. – a00 Aug 8 '15 at 14:25
• Luboš Motl: also, your answer inspired me to check out your blog. Reading your Aug.7th 2015 post about quantum mechanics and Christianity, I find you an interesting (topics) and entertaining (clarity, well written) expositor. – a00 Aug 8 '15 at 14:29