A typical statistical mechanics explanation for the probability ($p_i$) that a system (S) in contact with a reservoir (R) will take on a particular energy ($E_i$) is as follows (quoting from Greiner's textbook on the subject):

We ask for the probability $p_i$ of finding the system S in a certain microstate $i$ with the energy $E_i$. If S is a closed system, $p_i$ will be proportional to the number of microstates $\Omega_S(E_i)$. Analogously, $p_i$ is proportional to the number of microstates in the total closed system for which S lies in the microstate i with the energy $E_i$. Obviously this is just equal to the number of microstates of the heat bath for the energy $E - E_i$ , since S only assumes one microstate i:

$$p_i \propto \Omega_R(E_R) = \Omega_R(E_{total} - E_R)$$

But this is not "Obvious" to me at all since it seems to me that the total phase space (limited to a particular system energy $E_i$) will have additional degeneracy compared to the reservoir's phase space (limited to a particular reservoir energy $E_{total} - E_i$).

Consider the following minimal working example: Imagine a system comprising a single particle. The particle can only hold energy as x-direction momentum, so the system Hamiltonian is $p_{S,x}^2/2m$. Imagine the system is in thermal contact with a reservoir that has only two particles, each of which can only hold energy as x-direction momentum, so the reservoir Hamiltonian is $p_{R1,x}^2/2m + p_{R2x}^2/2m$ and the total system Hamiltonian is $p_{S,x}^2/2m+p_{R1,x}^2/2m + p_{R2x}^2/2m$. Energy of the total system is constant, of course. Then the phase space of the TOTAL system is visible as a 3D hollow spherical surface centered on the origin of the phase space axes $p_{S,x}$, $p_{R1,x}$, $p_{R2,x}$. Probability of the system being at any one location on this sphere is uniform. Thus surface area is proportional to probability. I'm trying to validate the truth of Greiner's textbook quote above by comparing the "surface area" on this phase-space sphere corresponding to a particular system energy $E_i$ (i.e., intersection of a plane at a particular value of $p_{S,x}$ with the sphere) to the "surface area" of the reservoir's phase-space (when reservoir has energy $E_{total} -E_i$). The curvature of the sphere creates a difference between the total system phase space and the reservoir's phase space, causing confusion to me, but I do see that "surface area" of phase space for single $E_i$ in this case has units of length rather than traditional length$^2$ units of area.

Is this a good example to consider? Does anyone know of a better way of understanding it more clearly / intuitively?


1 Answer 1


The wording is critical, and the source of endless confusion among students. In the canonical ensemble, the probability that system $S$ is in a specific microstate $\mu$ is given by $$\mathrm{Prob}(\mu) = e^{-\mathcal E(\mu)/kT} / Z(T) \qquad Z(T)= \sum_\mu e^{-\mathcal E(\mu)/kT}$$ where $\mathcal E(\mu)$ is energy of $S$ when it occupies the microstate $\mu$.

This can be derived by applying the microcanonical ensemble to the total system $T=S+R$. Let the total energy of $T$ be $E$. How many microstates of $T$ correspond to the subsystem $S$ being in microstate $\mu$? Answer: since we know $S$ is in microstate $\mu$, the reservoir $R$ can occupy any of its own microstates with energy $E-\mathcal E(\mu)$. Therefore, the total number of microstates of $T$ in which subsystem $S$ is in microstate $\mu$ is given by $$\Omega_R\big(E-\mathcal E(t)\big) =e^{S_R\big(E-\mathcal E(\mu)\big)/k} \approx e^{S_R(E)/k}e^{-S'_R(E) \mathcal E(\mu)} = Ce^{-\mathcal E(\mu)/kT}$$

where we have used that $S'_R(E) \equiv 1/T$ is the inverse temperature of the reservoir and $C$ is a constant which does not depend on $\mu$. Therefore, the probability that $S$ occupies the microstate $\mu$ is proportional to the Boltzmann factor, and the unknown constant can be obtained by normalization:

$$1 = \sum_{\mu} C e^{-\mathcal E(\mu)/kT} = C \sum_\mu e^{-\mathcal E(\mu)/kT} \implies C = \frac{1}{\sum_\mu e^{-\mathcal E(\mu)/kT}} \equiv 1/Z(T)$$

But this is not "Obvious" to me at all since it seems to me that the total phase space (limited to a particular system energy $E_i$) will have additional degeneracy compared to the reservoir's phase space (limited to a particular reservoir energy $E_{total}-E_i$).

This is your mistake. We are not restricting system $S$ merely to have energy $E_i$ - were are restricting system $S$ to be in a specific microstate, which incidentally happens to have energy $E_i$. There are generically many microstates of $S$ which have that energy, which is what I assume you mean when you say there will be additional degeneracy, but we are precisely specifying the microstate of $S$ from the outset.

  • $\begingroup$ A much cheaper way to say it is that, as long as microstate multiplicity $g(E)$ is neither zero nor infinity, then $\propto g(E)e^{-E/k_BT}$ is the same as $\propto e^{-E/k_BT}$. The behaviour that the text is intending to extract, is the boltzmann factor, and so any sysetm that gets it is ok. Later, when we actually care about the system itself, it is easy to multiply the microstate multiplicity manually. Or we can consider, temporarily, $g(E)=1$, say, the case of one single QHO. $\endgroup$ May 13 at 3:45
  • $\begingroup$ @naturallyInconsistent With respect, I think that kind of reasoning is precisely why OP is confused. We are not asking for the probability that $S$ has energy $E$ - we are asking for the probability that $S$ is in a specific microstate. Since that probability depends only on the energy of the microstate, we can obtain the probability that the system has energy $E$ by simply multiplying by the degeneracy factor $g(E)$, but this fact is derived from the statement we were trying to make in the first place. $\endgroup$
    – J. Murray
    May 13 at 4:01
  • $\begingroup$ Which is why I gave you the upvote the moment I saw that you covered this confusion $\endgroup$ May 13 at 4:05
  • $\begingroup$ @naturallyInconsistent Ah, my mistake - I thought you were proposing that I modify my answer to adopt that line of reasoning. $\endgroup$
    – J. Murray
    May 13 at 6:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.