I noticed that if a QM Hamiltonian is translationally&rotationally invariant (typically it is a single-particle version of some QFT Hamiltonian, like Klein-Gordon, Dirac or free Schrödinger in the non-relativistic case), then its solutions can be enumerated using ONLY quantum numbers, which are the eigenvalues of the symmetry operators from the canonical set of commuting observables: momentum, angular momentum (and spin in the non-relativistic setting) plus other symmetries (electric charge, color, ect.). In slightly other words, if we restrict ourselves to a block of the Hamiltonian matrix with fixed values of the symmetry operators' eigenvalues, each such block is one-dimensional.

If, however, the Hamiltonian does not have all of the spacetime symmetries (which typically happens if it contains non-local interactions), one may have quantum numbers, which are not associated with any symmetries [wrong, see the answer to this question]. For example, in the case of a 2D harmonic oscillator, the radial quantum number does not correspond to a conserved charge. In this case, the translational symmetry is broken. We can also say that the states in the blocks of the Hamiltonian matrix, corresponding to fixed values of charges, contain multiple states, which have to be enumerated with an additional label $n$.

What is the general rule we observe here? In which cases do we need such "additional" labels (and how many of those) and in which we don't? How is this related to Noether's theorem?


Quite interestingly, the two problems with broken spacetime symmetry, which immediately come to our minds — the 3D harmonic oscillator and the Coulomb potential — both have extra symmetry operators (the Runge-Lenz vector and the $SU(3)$ generators, correspondingly), which enables enumerating solutions using its.

Actually, the case of the 3D HO is just a particular case of the $N$-dimensional oscillator, which always has an additional $Sp(2N,\mathbb{R})$ symmetry, see this answer.


The motivation for this question comes from QFT. While in the second-quantized formulation we typically define Fock states using plane waves, this is not the only possible choice. The plane waves are the solutions of the wave equation for the free particle. In fact, we can define the modes in Fock states using solutions of any single-particle equation, as we often do in quantum chemistry.

The reason for doing so is straightforward — the usage of the plane wave basis may be far from being optimal for a given problem. For example, if we describe a confined system, instead of plane waves we can define multi-particle states based on the solutions of the harmonic oscillator. In particular, in the light-front formulation of QCD, the low-energy confinement of quarks is described by the transverse 2D HO confinement (this stems from AdS/QCD, and also from the fact that 2D HO transverse confinement on the light front corresponds to the linear confinement in equal time). However, we may also introduce the longitudinal confinement, which in the low-energy limit reduces the whole potential to the 3D HO.

Moreover, it may be beneficial to solve a problem with Hamiltonian $H$ in the basis of eigenfunctions of some $H^\prime$, where $H^\prime$ has seemingly nothing to do with $H$, as was done with the relativistic electron considered as a bound state of virtual electrons and photons.

Initially, I thought that in QFT the modes in the Fock state are necessarily labeled by the eigenvalues of charge operators. Then I was pointed out that this is not the case if one uses the solutions of such Hamiltonians as 2D HO or 3D HO. Well, it turns out that for these cases, my initial statement is correct due to some residual symmetries, yet it may not be true in the general case (apparently, if the single-particle equation is not integrable, as By-Symmetry mentioned in the comments) [wrong, see the answer to this question].

  • $\begingroup$ How much degeneracy is allowed sounds very model dependent. Kramers theorem is the only general rule I know. $\endgroup$ Feb 2, 2022 at 20:52
  • $\begingroup$ Intuitively this sounds like you are essentially asking if the system is integrable. Someone who knows more about integrable quantum systems than I do may be able to give you an answer $\endgroup$ Feb 2, 2022 at 21:35
  • $\begingroup$ Thank you, will take me some time to comprehend the notion of (quantum) integrability :) $\endgroup$
    – mavzolej
    Feb 2, 2022 at 22:59
  • $\begingroup$ Not sure what you mean by “both being cubic Casimir operators”. In the case of $SU(3)$, the Lie algebra is of rank 2 so there is one quadratic and one cubic Casimir. The generators are not invariant (they transform by the 8-dimensional irrep (1,1)) so are not Casimirs. $H$ in $\mathfrak{so}(4)$ is also not invariant. $\endgroup$ Feb 2, 2022 at 23:08
  • $\begingroup$ Thank you, that was misleading on my part. $\endgroup$
    – mavzolej
    Feb 2, 2022 at 23:44

1 Answer 1


One may have quantum numbers, which are not associated with any symmetries

This is not true: Given an assignment of real numbers $\lambda_i$ to eigenstates of the Hamiltonian, we can simply define the self-adjoint operator $\Lambda$ which multiplies every state by its corresponding $\lambda_i$. Since this operator is by construction diagonal in an eigenbasis of the Hamiltonian, it commutes with it and is hence the generator of a symmetry. Therefore, all assignments of numbers to eigenstates of the Hamiltonian are trivially "associated with a symmetry", unless you provide a definition of "symmetry" that distinguishes these "artificial" constructions from more "natural" operators.

  • $\begingroup$ This is a nice observation. I probably had in mind the distinction between the cases when such an operator can be constructed by some algebraic means vs the cases when this operator can only be found approximately by some brute-force study. Does this intuition match the definition of integrability? 🤔 $\endgroup$
    – mavzolej
    Feb 2, 2022 at 23:43
  • 1
    $\begingroup$ @mavzolej I don't know what "integrability" means in the quantum context. Classically, all systems are locally maximally superintegrable, but there are often global obstructions (see this and this answer by Qmechanic, in particular the first question is essentially the classical version of your question here). In the quantum case, there's no obvious such local/global distinction. $\endgroup$
    – ACuriousMind
    Feb 2, 2022 at 23:52

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