# Does there exist a potential which realizes this strange quantum mechanical system?

I have done some courses on quantum mechanics and statistical mechanics in the past. Since I also do math, I wonder about converge issues which are usually not such a problem in physics. One of those questions is the following. I will describe the background, but in the end it boils down to a question about ordinary differential equations.

In quantum mechanics on the real line, we start with a potential $V: \mathbb{R} \to \mathbb{R}$ and try to solve the Schrödinger question $i\hbar \frac{\partial}{\partial t}\Psi(x,t) = - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t)+V(x)\Psi(x,t)$. In many cases this can be accomplished by seperating variables, in which case we obtain the equation $E\Psi(x,t) = - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t)+V(x)\Psi(x,t)$ which we try to solve for $E$ and $\Psi$ to obtain a basis for our space of states together with an associated energy spectrum. For example, if we have a harmonic oscillator, $V(x) = \frac{1}{2}m\omega^2x^2$ and we get $E_n = \hbar \omega (n+\frac{1}{2})$ and $\Psi_n$ a certain product of exponentials and Hermite polynomials. We assume that the energy in normalized such that the lowest energy state has energy $0$.

If the states of our system are non-degenerate, i.e. there is only one state for each energy level in the spectrum, then the partition function in statistical mechanics for this system is given by the sum $Z(\beta) = \sum_n \exp(-\beta E_n)$, where $\beta$ is the inverse temperature $\frac{1}{k_B T}$. It is clear that this sum can be divergent; in fact for a free particle ($V = 0$), it is not even well defined since spectrum is a continuum.

However, I was wondering about the following question: Is there a system such that $Z(\beta)$ diverges for $\beta < \alpha$ and converges for $\beta > \alpha$ for some $\alpha \in \mathbb{R}_{> 0}$? Am I correct in thinking that such a system is most likely an approximation of another system, which undergoes a phase transition at $\beta = \alpha$?

Anyway, an obvious candidate would be a potential $V$ such that the spectrum is given $E_n = C \log(n+1)$ for $n \geq 0$ and $C > 0$. This gets me to my main mathematical question: Does such a potential (or one with spectrum asymptotically similar) exist? If so, can you give it explicitly?

One the circle, the theory of Sturm-Liouville equations tells us that the eigenvalues must go asymptotically as $C n^2$, so in this case such problems can't occur. I don't know much about spectral theory for Sturm-Liouville equations on the real line though. The second question is therefore: What is known about the asymptotics of the spectrum of a Sturm-Liouville operator on the real line?

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Presumably you mean converges for $\beta>\alpha$. The closest thing I've seen to the spectrum you describe is this: en.wikipedia.org/wiki/Primon_gas –  j.c. Aug 12 '10 at 11:30
And as usual, John Baez has a good writeup with plenty of references math.ucr.edu/home/baez/week199.html –  j.c. Aug 12 '10 at 11:43
The divergence that skupers describes is known more generally as a Hagedorn temperature - more details here: en.wikipedia.org/wiki/Hagedorn_temperature –  j.c. Aug 12 '10 at 11:51
@jc Thank you, I fixed that mistake. The Primon gas indeed seems to behave in the way I want, but is there a quantum mechanical realization on the line of that system? –  skupers Aug 12 '10 at 12:13
@skupers I would check the reference by Bost and Connes cited on Baez's page - I don't have online access to it though. –  j.c. Aug 12 '10 at 12:25
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## 2 Answers

The answer is yes, such a system exists. Here is how to construct it:

Denote by $H = - \frac{d^2}{dx^2} + V$ the operator on $L^2(\mathbb{R})$ and by $H_{\pm}$ the operators on $L^2(R_{\pm})$ obtained by restricting $H$ to the corresponding half lines. It is then known that $$H = H_{-} \oplus H_{+} + \text{rank one}.$$ This implies that if $H_{+}$ and $H_{-}$ both have discrete spectrum, then also $H$ has discrete spectrum. Furthermore, one obtains that the sets $$\sigma(H),\quad and \quad \sigma(H_+) \cup \sigma(H_-)$$ interlace. This implies that if you describe the spectrum of $H_{+}$ and $H_{-}$ to satisfy an asymptotic formula like $$E_{\pm,n} = \alpha \log(n + 1)$$ Then also the one of $H$ will. (Prescribing the spectrum of $H_+$ and $H_-$ can be done by standard inverse spectral theory). Of course this does not give you an exact description of the spectrum of $H$ but it is good enough for your purposes.

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If I understand your first question correctly, then the answer is yes. In fact, all physical matter exhibits this behavior. Allow me to answer in the following mathematically nonrigorous way:

Consider that even in a lone hydrogen atom, the Hamiltonian operator for the nonrelativistic electron

$H = - \frac 1 2 \nabla^2 + \frac{1}{r}$

has a discrete spectrum of bound states corresponding to the 1s, 2s, 2p, 3s, ... atomic orbitals and a continuous spectrum of unbound states corresponding to an electron that is unbound for all practical purposes. Thus at sufficiently high temperature (probably at $\beta^{-1}$ = kT ~ 0.5) there will be significant population of the continuous spectrum and you would have to deal with counting the continuous spectrum in the partition function.

The same phenomenon exists for all atoms and collections of atoms, even when the nuclear and interactions terms are turned on.

I am not 100% confident that the same thing holds in the relativistic case too, but I would be surprised if it did not.

Regarding your discussion of the harmonic oscillator, and the comment that "such a system [exhibiting such divergence at a critical temperature] is most likely an approximation of another system", I would go so far as to say that it is the other way round, that almost all the time "nice" systems like the harmonic oscillator are in fact derived as asympotic approximations to messier Hamiltonians. For example, you could write down the molecular Hamiltonian

$H = \sum_i -\frac 1 2 \nabla_i^2 + \sum_{ij} \frac 1 {r_{ij}} - \sum_{Ki} \frac {Z_K} {r_{iK}} + \sum_K -\frac 1 2 \nabla_K^2$

which as mentioned above has both a discrete part and a continuous part to its spectrum, and assume that we are interested only in the regime where we care about slow atomic nuclear motions, and that they move very little, and from there derive an effective lattice Hamiltonian of coupled harmonic oscillators. While the phase transition can be observed in the original molecular Hamiltonian, it would not be possible to see this occur in the simplified Hamiltonian since the the discrete spectrum of the harmonic oscillators would go on forever without becoming continuous.

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