Hydrogen atom: Energies and eigenstates

• I
• LagrangeEuler
In summary, the energy levels of the hydrogen atom refer to the energies of the electron in the atom, and they are negative because the zero point of energy is chosen to be the energy of a free electron at rest. This convention is useful because it indicates the amount of energy required to ionize the atom. The energies of the proton are not assigned as the atom is studied as a whole. The negative energies are an artifact of the choice to take the energy at infinity to be zero and can be interpreted as the amount of energy needed to move from the free electron state to the bound state. The Schrodinger equation used to calculate the energy levels for the relative motion treats the electron and proton asymmetrically as the potential only comes from the interaction between
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?
$$E_n \propto \frac{-1}{n^2}$$

LagrangeEuler said:
Are that energies of the atom or of an electron in the atom?

They are the energies of the electron.

LagrangeEuler said:
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?

LagrangeEuler said:
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".

LagrangeEuler said:
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.

PeroK and LagrangeEuler
PeterDonis said:
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.

mattt, PeroK, dextercioby and 2 others
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?

LagrangeEuler said:
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 teh zero energy state to the -13.6 eV bound state.

Last edited:
dextercioby
DrClaude said:
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.

DrClaude said:
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 Schrodinger 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.

PeterDonis said:
Yes, I don't disagree with this view. But I think it's important to note that the Schrodinger 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##.

DrClaude said:
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.

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).

hilbert2 said:
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.

dextercioby
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.

LagrangeEuler said:
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.

mattt
LagrangeEuler said:
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.,
$$\mu=\frac{m_{\text{e}}m_{\text{p}}}{m_{\text{e}}+m_{\text{p}}}.$$

mattt and hutchphd
LagrangeEuler said:
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.

LagrangeEuler, mattt and vanhees71

1. What is a hydrogen atom?

A hydrogen atom is the simplest atom and consists of one proton and one electron. It is the most abundant element in the universe and plays a crucial role in the formation of stars and galaxies.

2. What are the energies of a hydrogen atom?

The energies of a hydrogen atom are quantized, meaning they can only take on certain discrete values. These energies are determined by the electron's distance from the nucleus, with higher energy levels corresponding to larger distances.

3. What are eigenstates in a hydrogen atom?

Eigenstates in a hydrogen atom refer to the specific energy states that the electron can occupy. These states are described by quantum numbers, which determine the energy, orbital shape, and orientation of the electron in relation to the nucleus.

4. How are the energies and eigenstates of a hydrogen atom calculated?

The energies and eigenstates of a hydrogen atom can be calculated using the Schrödinger equation, a fundamental equation in quantum mechanics. This equation takes into account the potential energy of the electron and the kinetic energy of the electron's motion.

5. What is the significance of the hydrogen atom's energies and eigenstates?

The energies and eigenstates of a hydrogen atom have significant implications in understanding the behavior of atoms, molecules, and matter in general. They also play a crucial role in the development of technologies such as lasers and nuclear energy.

• Quantum Physics
Replies
2
Views
1K
• Quantum Physics
Replies
9
Views
2K
• Quantum Physics
Replies
6
Views
1K
• Quantum Physics
Replies
6
Views
201
• Quantum Physics
Replies
11
Views
1K
• Quantum Physics
Replies
12
Views
1K
• Quantum Physics
Replies
38
Views
3K
• Quantum Physics
Replies
2
Views
912
• Quantum Physics
Replies
5
Views
730
• Quantum Physics
Replies
3
Views
1K