Tewtor.ai

Thanks for sharing that—it’s tough when examples don’t match the exam. We’ll go step by step and keep it simple. First, can you confirm: are we talking about a single non-degenerate energy level (ground state) and possibly one or more excited states at much higher energy, with $T \to 0$ so $\beta = 1/(k_B T) \to \infty$?

Here’s the core idea at very low $T$:

• Canonical partition function: $Z = \sum_i g_i e^{-\beta E_i}$, where $g_i$ is degeneracy.
• Probability of state $i$: $p_i = \dfrac{g_i e^{-\beta E_i}}{Z}$.

Case A: Ground state energy $E_0 = 0$ (degeneracy $g_0$), next level $E_1 \gg k_B T$ (degeneracy $g_1$).

• $Z \approx g_0 e^{-\beta \cdot 0} + g_1 e^{-\beta E_1} + \cdots \approx g_0$ (excited terms vanish as $e^{-\beta E_1} \to 0$).
• $p_0 \approx \dfrac{g_0}{Z} \approx 1$, and for any excited level $p_{n>0} \approx 0$.

Case B: Ground state energy $E_0 = \varepsilon$ with $\varepsilon > 0$ but very small.

• $Z \approx g_0 e^{-\beta \varepsilon} + g_1 e^{-\beta E_1} + \cdots \approx g_0 e^{-\beta \varepsilon}$.
• $p_0 \approx \dfrac{g_0 e^{-\beta \varepsilon}}{Z} \approx 1$, and $p_{n>0} \approx 0$.

Key comparison:

• Shifting the zero of energy by a constant $C$ multiplies $Z$ by $e^{-\beta C}$, but probabilities $p_i$ are unchanged because the factor cancels in numerator and denominator.
• So $E_0 = 0$ vs $E_0 = \varepsilon$ gives different $Z$ (by $e^{-\beta \varepsilon}$) but the same $p_i$.
• At $T \to 0$, the system is in the ground state with probability $1$ (assuming non-degenerate or within degenerate manifold).

Quick check: Do you want to include ground-state degeneracy or specific excited energies, or keep it general like this?