# Quantum number and group theory

• I

## Summary:

Is there any link between principal quantum number and symmetry ?

## Main Question or Discussion Point

Hi all,
Group theory show us that irreducible representation of SO(3) have dimension 2j+1. So we expect to see state with 2j+1 degeneracy.
But does group theory help to understand the principle quantum number n ? And in the case of problems with SO(3) symmetry does it explain its strange link with j :

$$n>j$$
?

(I well aware that n and n>j come "easily" with classic solving of the Hamiltonian I just want to know the limit of the group theory in this case).

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Yes the degeneracy of energy eigenstates comes from the symmetries of the Hamiltonian. The hydrogen atom in particular has not only the usual SO(3) rotational symmetry but also an SO(4) symmetry which explains the degeneracy in $l$.

Yes the degeneracy of energy eigenstates comes from the symmetries of the Hamiltonian. The hydrogen atom in particular has not only the usual SO(3) rotational symmetry but also an SO(4) symmetry which explains the degeneracy in $l$.
Thank you ! I will look for this SO(4) symmetry of hydrogen atom (sadly I never heard about it !).

That mean in a only SO(3) problem we expect only 2j+1 state degeneracy without $n$ ?

Also does group theory give us the relative value of the energy of different representations ?
$$E_{n}<E_{n+1}$$

Still a bit strange that a 3D problem has SO(4) symmetry. Hard to not ask : but what is the fourth dimension ? !

king vitamin
Gold Member
Of the central force problems with $V(r) = r^{\alpha}$, there are two special cases. The first is the Kepler problem, $\alpha = -1$, and the second is the simple harmonic oscillator, $\alpha = 2$. In these two cases, the SO(3) symmetry of the central force problem enlarges to SO(4) and SU(3) respectively. As a result, the eigenspectrum of these two problems has a much larger degeneracy than the $2\ell+1$ degeneracy coming from SO(3). But yes, without these extra symmetries, you only expect the $2 \ell + 1$ degeneracies.

The SO(4) symmetry for the Kepler problem comes from the conservation of the Runge-Lenz vector. This three-component vector can be combined with the three components of angular momentum to form an antisymmetric 4x4 tensor which generates SO(4) rotations. But you need to study the Hamiltonian in detail to figure out which irreps show up in the spectrum, and what the spectrum is. For example, if you algebraically solve the Hydrogen atom using the conserved Runge-Lenz vector (as was first done by Pauli in 1926), you find that only the "symmetric traceless tensor" irreps of SO(4) show up in the spectrum, and it is these irreps which have dimension $n^2$ for integer $n$. There are other irreps of SO(4) which do not appear in this system.

Similarly, the irreps of SU(3) are conventionally labelled by two integers, $(p,q)$ , as reviewed here. But only the $(0,q)$ irreps appear in the spectrum of the 3d simple harmonic oscillator.

dextercioby
Homework Helper
Actually, SO(4) is only the symmetry group for the Hamiltonian of the H-atom, iff one ignores the continuous spectrum. The continuous spectrum itself shows an SO(1,3) symmetry and the smallest connected symmetry group which is valid for the Hamiltonian with its full spectrum is SO(1,4), also called the de Sitter group.

A. Neumaier
The full structure of the hydrogen spectrum is best described by the dynamical symmetry group SO(2,4). There is a whole book about the classical and quantum Kepler problem describing all this.

But you need to study the Hamiltonian in detail to figure out which irreps show up in the spectrum, and what the spectrum is.
Is there "a simple" argument pointing out that not all irreps will show up ? (in the vein of totalitarian principle)

There is a whole book about the classical and quantum Kepler problem describing all this.
Do you have the reference ?

And, nothing about the relation beetween energy and irreps ( $E_n< E_{n+1}$), why higher dimension irresps have higher energy (more states need more volume in the phase state so higher energy ? but this kind of reasoning does not come from group theory...) ? I don't need an extansive answer, maybe juste a yes or a no . Anycase I will look myself for a more exhaustive treatment of the problem.

A. Neumaier
Do you have the reference ?
• B. Cordani, The Kepler problem: group theoretical aspects, regularization and quantization, with application to the study of perturbations, Birkhäuser, Basel 2003.
• A.O. Barut and G.L. Bornzin, SO(4,2) Formulation of the Symmetry Breaking in Relativistic Kepler Problems, J. Math. Phys., 12 (1971), 841.

king vitamin
Gold Member
First of all, thanks to the others for mentioning the symmetry group of the full Hydrogen spectrum including the continuous part. I'll add that the fact that the spectrum for $E>0$ is continuous can be related to the non-compactness of the Lie group describing it: the states connected by the generators of a non-compact algebra generally form a continuum.

Is there "a simple" argument pointing out that not all irreps will show up ? (in the vein of totalitarian principle)
I don't think there is one, but I'm happy to be corrected there.

And, nothing about the relation beetween energy and irreps ( $E_n< E_{n+1}$), why higher dimension irresps have higher energy (more states need more volume in the phase state so higher energy ? but this kind of reasoning does not come from group theory...) ? I don't need an extansive answer, maybe juste a yes or a no . Anycase I will look myself for a more exhaustive treatment of the problem.
I think it makes sense somewhat heuristically that higher irreps will turn out to have higher energy, but I don't know of any general principle saying that it has to be so. For example, usually in rotationally-symmetric problems your Hamiltonian takes the form $H \propto \mathbf{L}^2$, so the spectrum naturally scales as $\ell(\ell+1)$ with higher SO(3) irreps $\ell$. But I don't see anything wrong with formally defining a Hamiltonian like
$$H = \mathbf{L}^2 + \frac{a}{1 + \mathbf{L}^2},$$
where $a$ is some constant, and now depending on its value, the ground state could be some nontrivial irrep $\ell>0$.

vanhees71
Gold Member
Actually, SO(4) is only the symmetry group for the Hamiltonian of the H-atom, iff one ignores the continuous spectrum. The continuous spectrum itself shows an SO(1,3) symmetry and the smallest connected symmetry group which is valid for the Hamiltonian with its full spectrum is SO(1,4), also called the de Sitter group.
For the bound states, i.e., the energy eigenstates $E<0$, it's SO(4), for $E=0$ it's a group like the Galilei group, and for $E>0$ the Lorentz group SO(1,3).

You can get it from considering the Runge-Lenz vector from the classical Kepler problem and quantizing it. That's for the non-relativistic approximation only of course.

A. Neumaier
for $E=0$ it's a group like the Galilei group