What happens to the partition functions in the limit $T\to 0$ or $\beta\to\infty$?

Question

Consider the canonical and grand canonical partition functions given by $$Z_C=\sum\limits_{i}g(E_i)e^{-\beta E_i}$$ and $$Z_G=\sum\limits_{i}g(E_i)e^{-\beta (E_i-\mu)}$$ respectively with $\beta=\frac{1}{k_BT}$.

Questions

$\bullet$ What happens to these partition functions in the limit $\beta\to\infty$? Does it become a constant (in the sense that independent of $E_i$)?

$\bullet$ What is the physical significance of the limiting result (whatever it turns out)?

Update: The existing answer doesn't include the role of $g(E)$ i.e., the degeneracy of the energy level $E$ which is crucial for taking the limit. It also doesn't mention what happens to the grand partition function in the same limit. It is trickier because $\mu$ itself changes with temperature $T$.

Answer 1

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.

Answer 2

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.