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I've seen a few references that say that in quantum mechanics of finite degrees of freedom, there is always a unique (i.e. nondegenerate) ground state, or in other words, that there is only one state (up to phase) of the Hamiltonian with the minimum eigenvalue.

My questions:

  1. Is it true?

  2. Under what condition is it true?

  3. I can easily construct a Hermitian operator, in a finite dimensional space, which has two lowest eigenvectors. For example, if $ \left\{ {\left| a \right\rangle ,\left| b \right\rangle ,\left| c \right\rangle } \right\} $ is an orthonormal basis of a 3-dimensional Hilbert space, define a Hamiltonian $$H = 1 \cdot \left| a \right\rangle \left\langle a \right| + 1 \cdot \left| b \right\rangle \left\langle b \right| + 2 \cdot \left| c \right\rangle \left\langle c \right|.$$ Then $\left| a \right\rangle $ and $\left| b \right\rangle $ are two ground states. If the answer to Q1 was 'yes', how is that consistent with this hamiltonian?

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  • $\begingroup$ Comments to the question formulation (v2): It seems that only Q2 is not already answered by yourself. Perhaps you should include your references so we may check independently what precisely is being said there. $\endgroup$ – Qmechanic Jun 29 '13 at 18:49
  • $\begingroup$ Well, this probably will be of little help, but my references are my QFT lecturer and a hard-copy of an unpublished draft of QFT notes that I stumbled upon, and it was told to me that it's by a very prominent physicist (I rather not mention the name since the draft is unpublished...) If nobody is familiar with this issue of uniqueness of the bound state, then forget my question... In any case, I don't think I answered myself, and if so, I would like you to restate my answer so I understand it. $\endgroup$ – Lior Jun 29 '13 at 19:02
  • $\begingroup$ If you like this question, you may also enjoy reading this Phys.SE post. $\endgroup$ – Qmechanic Jun 29 '13 at 19:44
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    $\begingroup$ @Lior: The stated claim in your question doesn't really make any sense, as you realize. It is possible instead that the claim was (i) every potential in finite dimensional ordinary QM has a unique ground state? (ii) the general sense that there are no "accidental" degeneracies in QM (because they require measure-zero fine tuning)? (iii) Something in relationship to a more specific system. $\endgroup$ – BebopButUnsteady Jun 30 '13 at 2:40
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    $\begingroup$ This is certainly not true as stated. For example, a nucleus has finitely many degrees of freedom, but nuclear ground states often have nonzero spin, so they're degenerate. $\endgroup$ – Ben Crowell Aug 15 '14 at 13:57
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To be clear:

Is the ground state of a quantum system always non-degenerate?

the answer is an unequivocal no. Real quantum systems can and do have degenerate ground states.

Some examples:

  • For a three-level system with hamiltonian $$H=\begin{pmatrix}1& 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{pmatrix},$$ as given by the OP, the ground state is degenerate. This should be all that's necessary to show that the claim is false in general.

  • Pretty much all atoms in a field-free vacuum have degenerate ground states, with the simplest examples being boron and carbon, which have $p$-shell electrons that fit multiple orthogonal magnetic-quantum-number states at exactly the same energy. The same is true for pretty much all of the periodic table, with the exception of atoms with full subshells. Thus, the alkaline earth metals, the noble gases, and the rightmost columns of the transition metals and the rare earths, have nondegenerate ground states, and everything else is degenerate.

    (On the other hand, it is important to note that these kinds of degenerate ground states can be relatively fragile, so e.g. if the atom wanders into a stray bit of magnetic field, that will lift the degeneracy, often by a nontrivial amount. However, that doesn't mean that the free-atom ground state isn't degenerate.)

  • This is exactly the same situation as that pointed out in a comment, regarding atomic nuclei, whose ground state will generically have nonzero angular momentum and will therefore be spatially degenerate.

  • A slew of ferromagnetic and anti-ferromagnetic materials in lattices that exhibit geometrical frustration, which is best exhibited graphically:

    That is, if three spins are linked with pairwise anti-ferromagnetic couplings, they try to point in opposite directions to each other, but there's no global solution that will avoid high-energy parallel alignments. This then leads naturally to a degenerate ground-state manifold.

Now, there is a large class of hamiltonians for which the ground state can be shown to be nondegenerate ─ they're explored in some depth, and with good references, in this MathOverflow thread ─ which includes many hamiltonians of the form $-\nabla^2 +V$, regardless of the dimension, for distinguishable quantum particles. However, but this class does not include all possible systems, particularly once you include fermionic particle statistics with strict antisymmetry requirements on the wavefunction.

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  • $\begingroup$ Here is a source which says that the ground state of $C$ is $^3P_0$. Unique? quantummechanics.ucsd.edu/ph130a/130_notes/node392.html $\endgroup$ – mithusengupta123 Nov 24 '18 at 10:30
  • $\begingroup$ @mithusengupta No. That's a degenerate state. $\endgroup$ – Emilio Pisanty Nov 24 '18 at 10:53
  • $\begingroup$ Since $L=1$, it is (2S+1)(2L+1)=3x3=9-fold degenerate? What about $J$ value? It says that that the degeneracy is (2J+1)=1. After SO interaction taken into account 2J+1 gives the degeneracy? $\endgroup$ – mithusengupta123 Nov 27 '18 at 2:49
  • $\begingroup$ @mithusengupta Ask separately. $\endgroup$ – Emilio Pisanty Nov 27 '18 at 7:54
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I believe that it is true as long as there does not exist a non-trivial unitary operator $U$ that commutes with the Hamiltonian ($[H,U] = 0$) in the subspace of ground states. If such an operator exists then for a ground state $|\phi_0\rangle$ with energy $E_0$ we have $$HU|\phi_0\rangle = UH|\phi_0\rangle = E_0\left(U|\phi_0\rangle\right)$$ and so $U|\phi_0\rangle$ also has the lowest possible energy $E_0$ and it thus also a ground state. Note that the statement of non-triviality of $U$ is important. It needs to be non-trivial in the subspace of ground states, that is $U|\phi_0\rangle \neq e^{i\theta}|\phi_0\rangle$ for any phase $\theta$, otherwise there is no degeneracy. (Unitarity is needed so that $U|\phi_0\rangle$ is a state with norm 1)


More succinctly, if there exists a unitary operator $U$ such that $[H,U]=0$ and $U|\phi_0\rangle\neq e^{i\theta}|\phi_0\rangle$ for any phase $\theta$ then we have ground state degeneracy.


In the example you have given we see that the matrix elements in the basis given $\{|a\rangle,|b\rangle,|c\rangle\}$ is $$H = \begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix}$$ from which we see there exists a unitary operator, with matrix elements $$U = \begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}$$ which commutes with $H$ and is non-trivial in the ground-state space.


Proof that non-existence of $U$ implies non-degenerate ground state:

Assume $\nexists U$ s.t. $\{[H,U]=0 ~~\mbox{and}~~ U|\phi_0\rangle \neq e^{i\theta}|\phi_0\rangle\}$

Now, for every state $|a\rangle$ and $|b\rangle$, $\exists U_{ab}$ which is unitary that takes us from $|a\rangle\rightarrow|b\rangle$. We are interested in the operator that take us from $|\phi_0\rangle$ to any $|a\rangle$ in our Hilbert space (which obviously includes all possible ground states), which we denote by $U_{a0}$. This means that any state $|a\rangle$ can be written as $|a\rangle = U_{a0}|\phi_0\rangle$. By our starting assumption $U_{a0}$ either satisfies $$(1)~~~~~~ [H,U_{a0}]\neq 0,~~~~~~~~\mbox{or}~~~~~~~~(2)~~~~~U_{a0}|\phi_0\rangle = e^{i\theta}|\phi_0\rangle$$ If (1), then we have $$H|a\rangle = H U_{a0}|\phi_0\rangle \neq U_{a0}H|\phi_0\rangle = E_0|a\rangle~~~\implies~~~H|a\rangle \neq E_0|a\rangle$$ and so $|a\rangle \neq |\phi_0\rangle$ is not a ground state.

If (2), then $|a\rangle = e^{i\theta}|\phi_0\rangle$ and so $|a\rangle$ and $|\phi_0\rangle$ represent the same state.

Thus the non-existence of $U$ implies the non-existence of a second ground state and thus non-degeneracy.

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    $\begingroup$ Thus, the non-existence of U implies the non-existence of a second state. $\endgroup$ – Jerry Schirmer Jun 29 '13 at 20:34
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    $\begingroup$ Thank you @JerrySchirmer, I should have added that one line - now I have :) $\endgroup$ – Will Jun 29 '13 at 20:37
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    $\begingroup$ Ah, it's not incorrect, just lacking a bit of detail (sorry about that, I could have been more clear). First of all, I was intending $U_{ab}$ to swap states $a$ and $b$, not just one way. So, $U_{a0}$ swaps $|a\rangle$ and $|\phi_0\rangle$. Thus when we say $[H,U_{a0}]\neq 0$ it is zero everywhere except on the subspace of eigenstates $|a\rangle$ and $|\phi_0\rangle$. And so $[H,U_{a0}]\neq 0$ does in fact imply $[H,U_{a0}]|\phi_0\rangle\neq 0$. I can make changes to my solution to make it more coherent if you like, but I think this should be done tomorrow once I have had some sleep :) $\endgroup$ – Will Jun 30 '13 at 5:52
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    $\begingroup$ Don't we have to be careful here, as there can exist nontrivial unitary operators $U$ such that $U|\phi_0\rangle = |\phi_0\rangle$, and hence is not a physically different state so there is no degeneracy. For example, represent the ground state as $(1,0)^T$ and consider $any$ unitary diagonal matrix. The result of this action is still $(1,0)^T$. I feel that there needs to be another condition on $U$, since just because it does not change a $specific$ state doesn't make it a trivial operator. $\endgroup$ – InertialObserver Jul 3 '17 at 19:34
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    $\begingroup$ This answer is little more than a tautology. The $U$ you have constructed is pretty much synonymous with the statement "the ground state is degenerate", and this answer provides no meaningful arguments either for or against the existence of nontrivial $U$s, to the actual content in this post is pretty much vacuous. $\endgroup$ – Emilio Pisanty Sep 18 '17 at 10:40
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Not sure about this but if you use that unitary operator to generate another ground state, then (if theres no infinite potential between the ground states) you can find some amplitude for tunneling between the two states and hence the ground state is some linear combination of the two with one of the combinations being lower than the original (and the other higher than the original). For people who want a real explanation see chp21 of Shankar Principles of QM, the section on imaginary time formalism.

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protected by Qmechanic Aug 15 '14 at 10:56

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