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In canonical ensemble, the probability is defined as \begin{equation} P(E)=\frac{g(E)\exp(-E/T)}{Z}, \end{equation} and the partition function is defined as \begin{equation} Z(T)=\int_0^{\infty}dE\,g(E)\exp(-E/T), \end{equation} where $g(E)$ is the phase space density of the energy states. The usual argument that one uses is that the Boltzmann factor sharply decreases with increasing energy while $g(E)$ sharply increasing with increasing energy, such that the product is a Bell-shaped function. What if $g(E)$ is not known? I assume this corresponds to nonequilibrium statistical mechanics, because I don't necessary have a maximum in the probability that corresponds to a maximum of the entropy. Can anybody of you point me to a book or literature, where this problem is treated in a general way: canonical ensemble, where $g(E)$ is not known. I get lost in the literature, because it looks to me that in non equilbrium statistical mechanics there are several different approaches for different physical systems. Thanks a lot in advance!

Update:

Let's try with an example. We have a system in canonical distribution and in thermodynamical equilibrium. Everything is as usual. The partition function $Z(T)$ is just the Laplace transform of the density of states $g(E)$, so I can think of several kind of $g(E)$. If I take $g(E)=\theta(E−E_0)/(E−E_0)^4$, am I still describing a thermodynamical equilibrium? To my eyes $g(E)=\theta(E−E_0)/(E−E_0)^4$ is a possible, well defined, degeneracy of the energy states, $Z(T)$ guarantees the probability to be normalized (by definition). To my eyes everything seems to be fine: I have a normalized probability positive definite. I am thinking I could remove the divergency of $g(E)$ at $E_0$ by including an $\epsilon$, so this is not a problem for me. The only problem I have is: g(E) is not sharply increasing with increasing energy, so the probability is not a Bell-shaped function. Can I still interpret this as a thermodynamical equilibrium? And if so, how? I mean, is there anything similar in the literature with gas,... Or is there anything wrong in my reasoning? A reference or book to understand this would be also helpful.

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3 Answers 3

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The only assumption that you have to use when writting the canonical distribution in terms of $g(E)$ is that the spectrum of energies is continuous. So either the system is classical or the quantum discrete espectrum can be approximated for a continuum. In all the examples that I've seen, the density turns out to be of the form

$g(E)\propto E^\alpha$

Independently of $\alpha$, $e^{-\beta E}$ goes to zero much faster than $g(E)$ goes to infinity, so the canonical partition function is well-defined. But this is a result rather than a requirement on $g(E)$!

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    $\begingroup$ You don't even need something as strong as continuity. In a quantum well $g(E)$ is a series of step functions, whilst in a quantum wire it a series of "steps" with $E^{-1/2}$ van-Hove singularities at the edges. $\endgroup$ Jun 7, 2021 at 16:40
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Density of states is known, if the problem is defined - i.e., if we are given the Hamiltonian and the appropriate boundary conditions. This will remain unchange (or largely unchanged) in non-equilibrium - however, in non-equilibrium the equations that you use will not apply, since they were obtained for equilibrium system. In simpler words: in non-equilibrium system the probability of an energy state is not proportional to $e^{-\beta E}$.

Update
If we have a system described by a Hamiltonian $\mathcal{H}(p,q)$, then the partition function is given by $$ Z = \int dp\int dqe^{-\beta\mathcal{H}(p,q)}=\\ \int dp\int dq\int dE\delta\left( E-\mathcal{H}(p,q)\right)e^{-\beta\mathcal{H}(p,q)}=\\ \int dEg(E)e^{-\beta E}, $$ where $$ g(E)=\int dp\int dq\delta\left( E-\mathcal{H}(p,q)\right) $$ In other words: if Hamiltonian $\mathcal{H}(p,q)$ is defined, then the density-of-states $g(E)$ is defined as well.

This may require greater care when some variables are discrete. E.g., let us consider a quantum system with energies $E_{n,\nu}$ degenerate over the index $\nu$, i.e., $E_{n, \nu}=E_n$. Writing $$ g(E)=\sum_\nu \delta(E-E_{n,\nu}) $$ is not very useful. Indeed, here $g(E)$ is undefined between the energy levels and singular at the levels themselves. Most of the derivations done for continuous $g(E)$ are not practical here. A better way is to define numbers $g_n$ counting the degeneracy of the energy states and write the partition function as $$ Z=tr\left[e^{-\beta H}\right]=\sum_{n,\nu}e^{-E_{n,\nu}}=\sum_ng_ne^{-E_n} $$

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  • $\begingroup$ Hi, thanks for your contribution. I guess you mean that in non-equilibrium statistical mechanics a canonical distribution can only be used, say, to define the initial state of a system; i.e., non-equilibrium statistical mechanics could study how a system evolve from an equilibrium configuration (in canonical distribution) to a non equilibrium configuration (whatever distribution). What about if nothing depends on time (I guess it would still be equilibrium then), but $g(E)$ is not sharply increasing with increasing energy? Then, the probability is not peaked... $\endgroup$
    – Voltrack
    Jun 1, 2021 at 20:20
  • $\begingroup$ ... fluctuations are not well defined; i.e., the probability is not a Bell-shaped function! How do I interpret such an equilibrium? Is it really an equilibrium? $g(E)$ describes the degeneracy of the energy levels. Let's think about the following example: we start with equilibrium, a well defined canonical ensemble, with Bell-shaped probability, everything is as usual. What happens physically if I change $g(E)$ to be an arbitrary function, such that the probability is not a Bell-shaped function anymore? $\endgroup$
    – Voltrack
    Jun 1, 2021 at 20:39
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    $\begingroup$ If nothing depends on time, it is a stationary state, not necessarily an equilibrium one. E.g., a current flow is a stationary state. $\endgroup$ Jun 2, 2021 at 5:36
  • $\begingroup$ I added more material to teh question - perhaps, it will help to clarify some points. $\endgroup$ Jun 2, 2021 at 7:06
  • $\begingroup$ Thanks for all details. What I am thinking is the following. The partition function $Z(\beta)$ is just the Laplace transform of the density of states $g(E)$, so I can think of several kind of $g(E)$. For example, $g(E)=\theta(E-E_0)/(E-E_0)^4$ is a possible degeneracy of the energy states, $Z(\beta)$ guarantees the probability to be normalized (by definition). So, I have a probability positive definite, but $g(E)$ is not sharply increasing with increasing energy. I could remove the divergency of $g(E)$ at $E_0$ by including an $\epsilon$ ... $\endgroup$
    – Voltrack
    Jun 2, 2021 at 10:36
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I'm not able to follow your arguments and further, you have written the wrong distribution.

$$Z=\int dE\ g(E)e^{-\beta E}$$ We can consider the density of the state to increase as in the classical limit, everything will be smooth out.

Once you know $\mathcal{H}$, you can compute $g(E)$ using $$g(E)=\frac{d\Omega}{dE}$$ Not knowing the $g(E)$ corresponds to not knowing the $\mathcal{H}(q,p)$.

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  • $\begingroup$ Hi, thanks for your answer. Why is my distribution wrong? I am using $\beta\equiv 1/k T$ in the $k=1$ units. I guess it was just a misunderstanding here... And yes, I don't know the Hamiltonian. I would like to face the problem in a very general way, trying to understand how the system changes its equilibrium with the heath bath, by changing the density of the states. A Bell-shape probability distribution corresponds to equilibrium, but if I change the shape of the probability distribution the equilibrium will become a nonequilibrium, when the distribution is not symmetric anymore. $\endgroup$
    – Voltrack
    May 19, 2021 at 19:05

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