Hydrogen atom: Energies and eigenstates

[LagrangeEuler]

When we say energy levels of the hydrogen atom. Are that energies of the atom or of an electron in the atom? Also corresponding states?
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html
Why energies are negative?
E_n \propto \frac{-1}{n^2}

 

[PeterDonis]

Are that energies of the atom or of an electron in the atom?

They are the energies of the electron.

Why energies are negative?

Because the "zero" point of energy is chosen to be the energy of an electron that is free (not bound in a hydrogen atom) and at rest. The energy of any bound electron will be less than that, so it appears as negative. This convention is useful because it tells you directly how much energy would have to be added to the electron to remove it from the atom (i.e., to ionize the atom). For example, since the energy of the electron in the ground state is - 13.6 eV, that tells you that it takes 13.6 eV of energy to ionize the hydrogen atom from its ground state.

 

[LagrangeEuler]

Thanks. But why they called it the wave function of the hydrogen atom? In the hydrogen atom, we also have a proton. I am confused about that. And what are the energies of a proton?

 

[PeterDonis]

why they called it the wave function of the hydrogen atom?

For historical reasons, and because it's easier to say or write than "the wave function of the electron in the hydrogen atom".

what are the energies of a proton?

When analyzing the energy levels of an electron in a hydrogen atom, in simpler treatments the proton is assumed to be at rest at the origin. Strictly speaking, this is not correct, but the proton is so much more massive than the electron that it is a reasonably good approximation to ignore its motion.

A more rigorous treatment reformulates the problem to separate out the center of mass motion of the combined proton-electron system (the atom) from the internal relative motion of the proton and electron; the energy levels are then the allowed energies of the internal relative motion. This gives a more accurate prediction of the actual energy levels because the effective mass of the electron in this formulation is now the "reduced mass" $M m / \left( M + m \right)$, where $M$ is the mass of the proton and $m$ is the mass of the electron.

Either way, no energy levels are assigned to the proton itself. In nuclear physics, nuclei of isotopes other than hydrogen-1, which have multiple nucleons, can be analyzed in terms of energy levels, but that is a separate question and should be discussed in a separate thread.

 

[DrClaude]

They are the energies of the electron.

I disagree. It is the internal energy of the hydrogen atom, of the electron and the proton together.

While the derivation is often done with a fixed nucleus, as you point out the correct derivation is based on the separation of the centre-of-mass motion from the relative motion. If you compare the two, the numerical approximation due to the fixed nucleus is very small.

Conceptually, I think it is wrong to discuss the energy eigenstates in terms of electrons only, even though I admit that I do it often (like "putting electrons in atomic orbitals"). I try to drive the point to my students that the atom must be taken as a whole. This is particularly important to properly understand the coupling between the atom and the electromagnetic field. It is the atom as a whole that absorbs/emits photons, not just the electron.

 

[LagrangeEuler]

That is actually my problem. You solved it in the system of the center of mass. And you get those levels and states. However how then you get -13,6eV. Why is negative then? End why that ground state case actually defines ionization energy of electron?

 

[Nugatory]

However how then you get -13,6eV. Why is negative then? End why that ground state case actually defines ionization energy of electron?

The negative energies are just an artifact of the choice (which has no physical significance) to take the energy at infinity (that's classical wording - the equivalent quantum mechanical formulation is harder to write in English but comes down to the same thing) to be zero. Interpret -13.6 eV as saying that we would have to add 13.6 eV to the system to get it into the zero-energy state where the electron is free; or equivalently that if a hydrogen nucleus captures a free electron 13.6 eV will be radiated away to get from the zero energy state to the -13.6 eV bound state.

 

[Nugatory]

I disagree. It is the internal energy of the hydrogen atom, of the electron and the proton together.

I must confess that I agree with DrClaude here - it's best to attribute the energy to the entire system rather than trying to attach it to any single component of the system. This doesn't matter if we take the position of the nucleus to be fixed, but still a good habit when working with any system that includes a potential.

 

[PeterDonis]

It is the internal energy of the hydrogen atom, of the electron and the proton together.

Yes, I don't disagree with this view. But I think it's important to note that the Schrödinger Equation we solve to get the energy levels for the relative motion has a Coulomb potential in it due to the nucleus, but no potential due to the electron, even though the electron also has a charge. So we are not treating the proton and the electron symmetrically.

 

[DrClaude]

Yes, I don't disagree with this view. But I think it's important to note that the Schrödinger Equation we solve to get the energy levels for the relative motion has a Coulomb potential in it due to the nucleus, but no potential due to the electron, even though the electron also has a charge. So we are not treating the proton and the electron symmetrically.

The potential comes from the interaction between the electron and the nucleus. For a single-electron atom is literally
$$
V(r) = \frac{q_1 q_2}{4 \pi \epsilon_0 r}
$$
with $q_1 = Z e$ ($Z$ the atomic number) and $q_2 = -e$.

 

[PeterDonis]

The potential comes from the interaction between the electron and the nucleus.

Yes, you're right, both the electron and the nuclear charge appear in the potential.

 

[hilbert2]

I would call it "energy of the electron assuming the nucleus is a point charge of much larger mass". The basic-level theory of hydrogenic atoms ignores the finite size of the nucleus, and in addition to that the nuclei have internal degrees of freedom that aren't even understood completely (but don't have effect on the chemical properties of an atom).

 

[hutchphd]

I would call it "energy of the electron assuming the nucleus is a point charge of much larger mass". The basic-level theory of hydrogenic atoms ignores the finite size of the nucleus, and in addition to that the nuclei have internal degrees of freedom that aren't even understood completely (but don't have effect on the chemical properties of an atom).

Use of the Center of Mass frame provides an exact solution regardless. It demands the reduced mass be used. You can make positronium (a positron and electron) and solve it exactly by using the reduced mass. Its not a big deal.

 

[LagrangeEuler]

Yes but in energy expression only mass of electron appears
E_n=-\frac{m_ee^4}{2 (4\pi\epsilon_0)^2\hbar^2}\frac{1}{n^2}
Also if you write down
$\psi(r)=\frac{1}{\sqrt{\pi}a_0^{\frac{3}{2}}}e^{-\frac{r}{a_0}}$
from this you get probability of finding the electron on some distance from the nucleus.

 

[vanhees71]

When we say energy levels of the hydrogen atom. Are that energies of the atom or of an electron in the atom? Also corresponding states?
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html
Why energies are negative?
E_n \propto \frac{-1}{n^2}

They are the energy levels of the atom as a hole in the center-momentum frame (i.e., in the reference frame where the atom is at rest).

Since, however, $m_{\text{p}} \simeq 1830 m_{\text{e}}$, you can use the approximation that the proton is fixed, and in this approximation it's the energy of the electrons in the rest frame of the proton.

 

[vanhees71]

Yes but in energy expression only mass of electron appears
E_n=-\frac{m_ee^4}{2 (4\pi\epsilon_0)^2\hbar^2}\frac{1}{n^2}
Also if you write down
$\psi(r)=\frac{1}{\sqrt{\pi}a_0^{\frac{3}{2}}}e^{-\frac{r}{a_0}}$
from this you get probability of finding the electron on some distance from the nucleus.

Strictly speaking what should be in this formula is the reduced mass of the proton-electron system, i.e.,
$$\mu=\frac{m_{\text{e}}m_{\text{p}}}{m_{\text{e}}+m_{\text{p}}}.$$

 

[DrClaude]

Also if you write down
$\psi(r)=\frac{1}{\sqrt{\pi}a_0^{\frac{3}{2}}}e^{-\frac{r}{a_0}}$
from this you get probability of finding the electron on some distance from the nucleus.

Which is also the probability of finding the proton and the electron at a certain distance from each other.

 

[vanhees71]

Another interesting aspect of the treatment of the (non-relativistic) hydrogen problem as a two-body system is entanglement:

https://arxiv.org/abs/quant-ph/9709052v2