# Degeneracy of Hydrogen atomic orbitals with different l-values but same n-value

• octopus
In summary, the conversation discusses the degeneracy of energy levels in the hydrogen atom, particularly the 2s and 2p orbitals and the 3s, 3p, and 3d orbitals. It is suggested that this degeneracy is due to a hidden symmetry of the 1/r potential, resulting in a conserved quantity known as the Laplace-Lenz-Runge vector. This allows for the energy eigenvalue problem to be solved algebraically without using the Schrödinger equation. However, the conversation also notes that this reasoning may not fully explain the degeneracy and further research is needed.
octopus
I am terribly confused. I have always been hearing that in the hydrogen atom, 2s and 2p orbitals have the same energy. Similarly, the 3s, 3p and 3d orbitals have the same energies. This is also suggested by the hydrogen spectrum, my professor also believes the same, and I am unable to find anything against this on the internet.

But what is the basis for this degeneracy?
Upon solving the radial equations for 2s and 2p orbitals, we get the same eigenvalue for Energy, that depends only on the principal quantum number n. However, the wave functions also have an angular part and upon solving the angular equations for 2s and 2p we get a zero value for the 2s (angular momentum=0) and a finite value for 2p (angular momentum=root(2)*hbar). This angular momentum will contribute an extra value of root(2)*hbar/(2*I) to the energy. This will immediately give 2s and 2p different energy values, so they cannot be degenerate.

Have I gone wrong somewhere?

The l-degeneracy, i.e. the fact that E(n,l) = E(n) is l-independent is due to a hidden dynamical symmetry of the 1/r potential which results in an additional conserved quantity, the so-called Laplace-Lenz-Runge vector. The 1/r potential has not only the obvious SO(3) symmetry for spatial rotations but a larger SO(4) symmetry. The existence of the Laplace-Lenz-Runge vector and the l-degeneracy allows one to solve the energy eigenvalue problem algebraically w/o solving the Schrödinger equation (W. Pauli)

Just google for hydrogen atom SO(4) and you will find numerous articles, scripts and presentations. I am pretty sure that we had this disucussion here a couple of times.

Your reasoning regarding the additional l-term in Veff(r) giving the Ylm(Ω) functions a different energy is not correct b/c different l-values also affect the Rnl(r) functions. I think you can't understand the l-degeneracy by just solving the Schrödinger equation (you can derive it, but you don't see the deeper reason)

Thanks! :)
I get it, it has got something to do with the symmetry, I'll go and look that up.

However, I still find my reasoning contradictory to this, and I am unable to see any flaw in it.

tom.stoer said:
Your reasoning regarding the additional l-term in Veff(r) giving the Ylm(Ω) functions a different energy is not correct b/c different l-values also affect the Rnl(r) functions. I think you can't understand the l-degeneracy by just solving the Schrödinger equation (you can derive it, but you don't see the deeper reason)

I agree that different l-values affect the R(r) functions, however they don't affect the eigenvalues obtained when the separable hamiltonian acts on R, which depend only on 'n'.

To be more specific about the difference of radial functions for orbitals with different l, the 2s orbital has 1 radial node, 2p zero nodes. In general ns has n-1 nodes, np n-2, nd n-3 etc. These additional nodes make up for the lower centrifugal potential of states with lower l as compared to states with higher l.

Oh...
thanx! :)
I suppose the hamiltonian need not be separable, after all

Tom is correct. The symmetry is more explicit in parabolic coordinates. However, working in parabolic coordinates is not simple.

## 1. What is degeneracy in the context of hydrogen atomic orbitals?

Degeneracy refers to the equal energy levels of different quantum states. In the case of hydrogen atomic orbitals with the same n-value but different l-values, they have the same energy level and are considered degenerate.

## 2. How does degeneracy occur in hydrogen atomic orbitals?

Degeneracy occurs in hydrogen atomic orbitals because they have similar energy levels due to their similar distances from the nucleus. The energy levels are determined by the principal quantum number n, and since it is the same for all orbitals in this case, they are degenerate.

## 3. What is the significance of degeneracy in hydrogen atomic orbitals?

The degeneracy of hydrogen atomic orbitals allows for multiple orbitals to have the same energy level, which can affect the electron configuration and the behavior of electrons in these orbitals. It also plays a crucial role in determining the spectral lines of hydrogen.

## 4. How does the degeneracy of hydrogen atomic orbitals with different l-values affect the energy levels?

The degeneracy of hydrogen atomic orbitals with different l-values does not affect the energy levels, as they still have the same n-value. However, it does affect the shapes of the orbitals and the number of nodes in the wave function, which can affect the energy of the electron in that orbital.

## 5. Can the degeneracy of hydrogen atomic orbitals with different l-values be broken?

Yes, the degeneracy of hydrogen atomic orbitals with different l-values can be broken by applying an external magnetic field. This changes the energy levels of the orbitals and causes them to split, known as the Zeeman effect. This effect is used in spectroscopy to study the fine structure of spectral lines.

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