In non-relativistic quantum mechanics, what are the necessary conditions on the potential (or on the hamiltonian in general) for the ground state to be non-degenrate?
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4 Answers
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If a finite number of non-relativistic particles are moving in an infinite potential well, then the combined system has a nondegenerate ground state, regardless of the symmetry of the hamiltonian. I remember this from a long time ago, and I always thought it was impressive. I also remember I was always annoyed that I didn't know how to prove it, or know a reference where I can look it up. If you find one, let me know!
There's probably some sort of fancy entropic argument that you could use to get this result, if that's your thing.
If the potential was bounded above, I can't see immediately why this should create degeneracy on the ground state --- so it's plausible that the theorem holds in this case as well.
Systems containing infinite systems of particles can, and often do, exhibit degeneracy in their ground state.
answered Jun 4, 2010 at 9:01
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I think you can find an answer to your question in the book of Simon/Reed, "Methods of Mathematical Physics", vol.4 "Analysis of Operators". They have a chapter devoted to the question of the existence of nondegenerate ground states, chapter XIII.12.
One relevant theorem would be XIII.47, which says that the Schrödinger operator has a nondegenerate strictly positive ground state if the potential V is in $L^2_{loc}(\mathbb{R}^n)$ and $lim_{|x| \to \infty} V(x) = \infty$.
I don't think that there is a simple necessary condition on the potential, but only several sets of sufficient conditions, but could be wrong about that.
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$\begingroup$ Do you have any insight into what role unboundedness of the potential plays? I mean physically, rather than mathematically? $\endgroup$ Commented Jun 4, 2010 at 17:02
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1$\begingroup$ That's tricky! Allow some handwaving: If the potential goes to $\infty$ that means particles are confined to a bounded region (the probability to find them outside is very low/ can be made arbitrarily low). In a bounded region in classical physics there can be different locations which minimize the potential energy, but in quantum mechanics the ground state describes the probability to find it in any of those -> therefore it is unique in QM. But note that you can get degenerate ground states by dropping the other assumption on V ($V \in L^2_{loc}$). $\endgroup$ Commented Jun 4, 2010 at 18:03
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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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1$\begingroup$ @lcv Even the $V(x)$ is constant, the ground state is still unique. Because the eigenstate of $-\Delta^2$ is $e^{i k x}$ for $k\neq 0$ there is double degenerate. For $k=0$, it's unique up to a phase. $\endgroup$ Commented Jun 1, 2017 at 20:02
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$\begingroup$ @lcv Certainly, we always assumed the potential bounded from below. For detailed requirement and proof, you can consult the book I cite. $\endgroup$ Commented Jun 1, 2017 at 20:11
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$\begingroup$ @ChristianRemling Sorry, a typo. Thanks $\endgroup$ Commented Jun 1, 2017 at 20:57
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1$\begingroup$ Also, by a "ground state" one usually means an eigenfunction with eigenvalue equal to the minimum of the spectrum. In general, there is of course no reason why the bottom of the spectrum should be an eigenvalue, and $V=0$ is a simple example where it isn't. The natural way to understand the question seems to be to assume that $\min\sigma$ indeed is an eigenvalue (which will for example follow from the assumption that $V\to\infty$ from one of the other answers). $\endgroup$ Commented Jun 1, 2017 at 20:59
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A stronger result is true: the ground state of $H=-\Delta+V(x)$ (if one exists, that is, if the spectrum is bounded below and its minimum is an eigenvalue) is positive pointwise.
For a sketch of a proof, recall that an eigenfunction $y(x)$ with eigenvalue $\min\sigma$ minimizes the quadratic form $Q(y)=\int (|\nabla y|^2+V|y|^2)$. We can in fact assume that $y$ is real valued (take the real or imaginary part otherwise).
Notice that if $y\in H^1$ minimizes $Q$, then so do $|y|$ and $y\pm |y|$. If we now repeat Riemann's error from his proof of the Riemann mapping theorem and just assume that a minimizer has enough smoothness so that it will be a classical solution of the Euler-Lagrange equation, which here is just the original eigenvalue equation $-\Delta y+Vy=E y$, then we're already done (or close at least), because $y\pm |y|$ will not have second derivatives if the original function took values of both signs. One can make a rigorous proof out of this, based on the maximum principle for second order elliptic equations; see Theorem 6.5.2 of Evans's PDE book.
Finally, to make the connection to the original question more explicit, let me state more clearly what a rigorous version of this argument proves: any minimizer $y$ of $Q$ satisfies $y(x)>0$ after multiplication by a suitable constant. Since minimum energy eigenfuctions are minimizers, we can't have more than one, or we could take a linear combination that violates this condition.
answered Jun 2, 2017 at 0:11