What happens to the partition functions in the limit $Tto 0$ or $betatoinfty$?

In the limit that $beta to infty $, all the $e^{-beta E_i}$'s go to zero FAST. But the slowest one to go to zero is the lowest $E_i$. This is the ground state, $E_0$.

For large $beta$, $Z$ is very small:

$Z = e^{-beta E_0} + e^{-beta E_1} + e^{-beta E_2} + ... approx e^{-beta E_0} $

So as the temperature goes to absolute zero, the probability that the system will go into its ground state approaches 1.

$rm Prob(E_0) approx frac{e^{-beta E_0}}{e^{-beta E_0}} = 1$

$rm Prob(E_1) approx frac{e^{-beta E_1}}{e^{-beta E_0}} = 0$

Edit

The degeneracy plays a subtle role but does not change the physical interpretation much. $g(epsilon_i)$ is simply going to be an integer that multiplies each $e^{-beta E_i}$. For example, let's consider a two state system with $E_0; g(epsilon_0) = 2 $ and $E_1; g(epsilon_1) = 4$. Our partition function is

$Z = 2e^{-beta E_0} + 4e^{-beta E_1}$.

In the $beta to 0$ ($T to infty$) limit, the degeneracy plays a huge role! The probability of being in state with $E_0$ is $frac{2}{2 + 4} = 33%$ and $P(E_1) = 67%$. (Since $e^{0} to 1)$

But in the $beta to infty$ low temp limit, the degeneracy (of the system - see fermi gas for why that distinction is important) has basically no effect, since muliplying $E_1$ by a constant will not saving it from going to 0 quickly due to the $e^{-beta E_1}$ term.

I will want to think a bit more about the $Z_G$ before giving an answer - if anyone wants to chime in feel free.

I resolved the issue myself.

Indeed as $beta to infty$, $e^{-beta E_i} to infty$ as well since E_i is negative. So the partition function will tend to $infty$.

However, check this out.

Let $E^m$ be defined as min({ $E_i$ }). Then we can write the partition function as begin{align} Z = e^{-beta E^m} sum_i e^{-beta underbrace{(E_i - E^m)}_{Delta E_i}} end{align} Notice that $Delta E_i$ is positive, so when $beta to infty$, $Z = e^{-beta E^m} to infty$. However, the probability of the configuration with energy $E^m$ will be 1 and the rest of configurations will have probability zero, which is an expected fact that at zero temperature, the distribution is a delta function about the minimum energy configuration.

Best,

Shankha.