This is a more conceptual question.

When we take the isotropic harmonic oscillator:

$$V = \frac{1}{2}m\omega^2(x^2+y^2+z^2)$$

the eigenvalue equation solves to:


where $\phi_l$, $\phi_m$ and $\phi_n$ are the solutions to the 1D QHO (Quantum Harmonic Oscillator) in $x$, $y$ and $z$ respectively and the quantized numbers $l$, $m$ and $n$ are the excitations, respectively.

The total energy is:

$$E = \hbar\omega\left(l+m+n+\frac{3}{2}\right)$$

It follows, that we can get to the "first excited state" by having $\{l,m,n\}$ equal $\{1,0,0\}$ or $\{0,1,0\}$ or $\{0,0,1\}$, corresponding to the excitation being in the $x$-oriented oscillator, $y$-oriented oscillator or $z$-oriented oscillator, respectively.

Yielding a degeneracy of "3" for the first excited state of the 3D QHO.

My question is: I don't feel like this makes any sense conceptually. The potential has a rotational symmetry of order infinite since it is isotropic in $x$, $y$ and $z$. That is to say: why can't the eigenstate have its "excitation" not along the $x$, $y$ or $z$, but rather along another "direction", for instance the diagonal line between $x$ and $y$?

I would expect the infinite rotational symmetry to yield an infinite number of degenerate states that are rotations of each other, since the potential itself is rotational symmetric, so why stop at 3?

It feels to me that the number 3 drops out as an artifact, from the way we solved the problem in 3D Cartesian coordinates. I know that the math works. But I need conceptual understanding and intuition.

  • 5
    $\begingroup$ In a sense you are right: the 3 states span a 3-dimensional Hilbert space, in which rotation acts basically as you described.. So there are infinitely number of states in this space, it's just that they are all superpositions of the three basis states. $\endgroup$
    – Meng Cheng
    Aug 5, 2023 at 14:54
  • 1
    $\begingroup$ So the way to view this is that "degeneracy" refers to the size of the set of linearly independent states that share the same energy, thus forming a basis for states that possess that energy? $\endgroup$
    – The Feadow
    Aug 5, 2023 at 14:57
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    $\begingroup$ Yes, it is the dimension of the eigenspace... $\endgroup$ Aug 5, 2023 at 14:59
  • 2
    $\begingroup$ yes, degeneracy is the dimension of the Hilbert space of states with the same energy. If you count the number of states, it's always uncountable infinity unless there is no degeneracy. $\endgroup$
    – Meng Cheng
    Aug 5, 2023 at 14:59
  • 2
    $\begingroup$ An aside, arguably. The full symmetry of you system is not SO(3), but SU(3)... $\endgroup$ Aug 5, 2023 at 15:04

1 Answer 1


There is an infinite number of states with energy - say - $\frac52 \hbar \omega$: there is an infinite number of possible normalized linear combination of the $3$ basis states $\vert 1,0,0\rangle, \vert 0,1,0\rangle,\vert 0,0,1\rangle$.

There’s a distinction between the number of basis states in a space and the number of states in that space. There’s an infinite number of vectors in the 2d plane, but still only two basis vectors (the choice of which is largely arbitrary).

Now what determines the number of independent basis states is actually tied to the symmetry of the system. For the $N$-dimensional harmonic oscillator, the symmetry group is $U(N)$ (not $SO(N)$ or $SO(2N)$; see this question about the $N=3$ case).

The number of basis states is then given by the dimensionality of some representations of the group $U(N)$. For $N=3$, this is $\frac12 (p+1)(p+2)$ where $p=l+m+n$. Thus, for $p=0$ (the ground state), there is only one state, for $p=1$ (first excited state), there are $3$ states and so forth.

For $N=4$, the dimensionality is $\frac16 (p+1)(p+2)(p+3)$ etc.


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