Ground state degeneracy occurs whenever there exists a unitary operator which acts non-trivially on a ground state and commutes with the Hamiltonian of the system.

I just want to find a potential $V(\mathbf{r})$, not necessary the central potential, such that Schrodinger equation in d-dimensional (no internal degrees of freedom like spin) $$E\Psi (\mathbf {r})=\left[{\frac {-\hbar ^{2}}{2\mu }}\nabla ^{2}+V(\mathbf {r} )\right]\Psi (\mathbf {r} )$$ has degenerate ground state.

I've tried many ways, but failed.

1 For example, given a potential $V(\mathbf{r})$, and solve eigenenergy $E_0, E_1\cdots$. I want to construct $H'=H-(E_1-E_0)|1\rangle\langle1|$, but this part $(E_1-E_0)|1\rangle\langle1|$ in position representation is not a local potential.

2 Certainly, it's easy to construct the finite dimensional quantum system have ground state degeneracy, that is we can write Hamiltonian matrix as a diagonal matrix with multiple lowest eigenvalues $H=diag(E_0,E_0,E_1 )$. But I don't want this trivial way.

3 It's also easy to construct quantum mechanical system with internal degree of freedom like spin. And internal degrees of freedom don't have dynamics. For example, hydrogen model with spin degree of freedom. For lowest energy $n=1,l=0$, we can have $s=\pm1/2$ with same energy. This way is also trivial.

4 And we know the scattering state in 1-dim has continuous spectra and every state is double degenerate. I want to construct $V(\mathbf{r})\ge0$ such that $$\lim_{|\mathbf{r}|\rightarrow\infty}V(\mathbf{r})\rightarrow 0$$ Even though all $E>0$ have degeneracy, $E=0$ is still unique.

5 Certainly, in 1-dim if $V(x)$ is a double infinite deep potential, we can have degenerate ground state. But this example is also trivial.

6 The potential with spontaneous broken symmetries, e.g. $V(x)= -x^2 +x^4$, is also impossible. There is a enegry gap between even parity and odd parity.

So my question is, apart from above trivial examples whether we can construct a example, that is in d-dim a particle without internal d.o.f, like spin, can have degenerate ground state in some potential.

This question may be a question in partial differential function. If such $V(\mathbf{r})$ does not exist, how to prove.

  • $\begingroup$ I'm not sure I understand the question : would a symmetric double well potential do the job ? In any case you can describe a degenerate state as an internal degree of freedom if you like. I'm not sure nomenclature is completely under control in that case. What would be an internal degree of freedom ? What would be a degenerate state if not a state which transforms non-trivially under the action of an operator commuting with the Hamiltonian ? For the symmetric double well potential the operator is space-inversion: both wave-functions $\Psi(x)$ and $\Psi(-x)$ give the same energy. $\endgroup$
    – FraSchelle
    Dec 22, 2016 at 14:04
  • $\begingroup$ To construct a degenerate state as an internal degree of freedom, suppose you have a double well potential. It has a degenerate state in case of a symmetric potential, and the two associated states are non-degenerate when the potential is tilted (i.e. it becomes asymmetric). These states are separated from the rest of the eigenstates by a gap (since we are discussing quantum mechanics, we discuss only energy levels (not bands), and so gaps always exist ...). Now, project your model to the lowest eigenstates, you end up with a Bloch sphere picture that you can represent using Pauli matrices. $\endgroup$
    – FraSchelle
    Dec 22, 2016 at 14:09
  • $\begingroup$ Said differently, the space of twice degenerate states is isomorphic to the two-level system. I should think a bit more to write down an explicit isomorphism (and perhaps it's worth a question), but clearly this is a basic thing in quantum mechanics. Open any book on the two-level system you should see this statement. $\endgroup$
    – FraSchelle
    Dec 22, 2016 at 14:12
  • $\begingroup$ @FraSchelle I just want to construct a potential $V(\mathbf{r})$. Not necessary the central potential, and the problem are not necessary in 1-dim. In this potential, schrodinger equation can have a degenerate ground state. $\endgroup$
    – 346699
    Dec 22, 2016 at 16:34
  • 1
    $\begingroup$ @FraSchelle You can check this potential, the ground state is still unique. The ground state is even parity, and the odd parity's eigenenergy is larger than the even one's. $\endgroup$
    – 346699
    Dec 23, 2016 at 16:55

1 Answer 1


For Hamitonian operator like this form $-\Delta +V(x)$, the ground state is always non-degeneracy in $n$-dim if the potential is continuous and bounded from below and let $-\Delta +V(x)$ be essentially self-adjoint. You can see the proof in Page 51 James Glimm and Arthur Jaffe's Quantum Physics. Or see the proof.

If you don't limit the Hamitonian to this form( $-\Delta +V(x)$), then if you put magnetic field then it's easy to construct the degeneracy ground state. see Landau level.


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