TITLE: The well-defined temperature and 0D Ising model (Ref. Shankar)
QUESTION [1 upvotes]: I’m reading Shankar’s book Quantum field theory and condensed matter. On page 17, these two bold sentences seem to contradict each other:

The system in contact with the heat bath and described by Z need not be large. It could even be a single atom in a gas. The fluctuations of the average energy relative to the average need not be small in general. However, for large systems a certain simplification arises.
Consider a very large system. Suppose we group all states of energy E (and hence the same Boltzmann weight) and write $Z$ as a sum over energies rather than states; we obtain
  $$Z=\sum_E e^{-\beta E}\Omega(E)$$
  where $\Omega$ is the number of states of energy $E$. For a system with $N$ degrees of freedom, $\Omega(E)$ grows very fast in $N$ (as $E^N$) while $e^{-\beta E}$ falls expotentially. The product then has a very sharp maximum at some $E^*$.The maximum of $$e^{−βE}\Omega(E) = e^{−(βE−ln\Omega(E))}$$
  is the minimum of $βE−ln\Omega(E)$, which occurs when its $E$-derivative vanishes:
  $$β = \frac{dln\Omega(E)}{dE}|_{E^∗}$$
  which is the familiar result that in the most probable state the temperatures of the two parts in thermal contact, the reservoir and the system, are equal. (This assumes the system is large enough to have a well-defined temperature. For example, it cannot contain just ten molecules. But then the preceding arguments would not apply anyway.)
We may write
  $$Z = Ae^{−β(E^∗−kT ln\Omega(E^∗))},$$
  where $A$ is some prefactor which measures the width of the peak and is typically some power of $N$. If we take the logarithm of both sides we obtain (upon dropping $lnA$ compared to the ln of the exponential),
  $$ln Z = −β F = β (E^∗ − kT ln\Omega(E^∗ )) ≡ β (E∗ − S(E^∗ )T ),$$
  where the entropy of the system is
  $$S=kln\Omega(E^∗)$$.
  In the limit $N → ∞$, the maximum at $E^∗$ is so sharp that the system has essentially that one
  energy which we may identify also with the mean $⟨E⟩ = U$


How should I understand those two bold sentences?

Chapter 2 of this book discusses Ising model in $d=0$, which has just two spin.
On page 20, It evaluates the partition function for this case.
$$E(s)=-Js_1s_2-B(s_1+s_2)$$
$$Z=\sum_{s_1,s_2}e^{\beta J s_1s_2+\beta B(s_1+s_2)}$$

How is the temperature $T$ well-defined in this case?

REPLY [1 votes]: It is indeed pretty poorly written. What he means seems to be the following:

only a large system has a well defined temperature;
since he wants to state the equality of the temperatures of the reservoir and of the system, he needs the system to be large;
in any case, his argument that the energy concentrates on its most probable value only works for large systems (he is letting $N$ become very large).

On the other hand, he also says that

the partition function can be used for small systems (even a single molecule);

This is fine, since the reservoir is macroscopic and therefore has a well-defined temperature. Note that, here, he is not saying that the small system has a well defined temperature, so there is no contradiction.
(That being said, in my opinion, one can argue that a small system at equilibrium with a macroscopic reservoir at temperature $T$ is also at temperature $T$. Indeed making many observations on the small system (e.g., determining the empirical distribution of its energy) will allow you to determine the temperature of the reservoir. However, a small system that is isolated does not have a well defined temperature.)
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