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

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 $T$.

• What problems did you encounter in trying to take the limit yourself? $\mathrm{e}^{-x},x\to \infty$ is not a hard limit to take. – ACuriousMind Sep 18 '17 at 9:47
• @ACuriousMind Of course, it's not hard mathematically. But there is another factor sitting in front of it $g(E)$ which is the degeneracy factor, and it requires to know how that behaves. I have also expressed my doubts in a comment to Senor O's answer. If it were really a mathematical problem of taking limits I wouldn't have asked. :-) There is also a question about grand partition function, and I think it's tricky because $\mu$ also varies with $T$. – SRS Sep 18 '17 at 10:06

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.

• 1. You meant $\rm Prob(E_1)$ in the last line of your answer. Right? 2. Could you include in your answer the role of $g(E)$ i.e., the degeneracy of the energy level $E$? You didn't take them into account in your answer. 3. Also, can you comment on the grand partition function as well? The non-trivial fact is that $\mu$ itself changes with $T$ or $\beta$. @Senor O – SRS Sep 18 '17 at 4:56