What is the Limit of the Partition Function in the Low Temperature Regime?

AI Thread Summary
The discussion focuses on determining the limit of the partition function in the low temperature regime, specifically when kt is much less than ε. The partition function is initially defined, but when evaluated under low temperature conditions, it appears to diverge due to an infinite number of energy levels. A key insight is that by isolating the ground state (n=0), the partition function simplifies to Z=1, which aligns with expectations for low temperatures. Participants emphasize the necessity of considering the ground state to resolve the divergence issue. Ultimately, the consensus is that focusing on n=0 allows for the correct evaluation of the partition function in this context.
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Homework Statement


Ground state energy is set at 0.
E_n=\left(1-\frac{1}{n+1}\right)\in with no degeneracy (\Omega(n)=1); (n=0,1,2...)
Write down the partition function and look for its limit when kt \gg \in\\ kt \ll \in

Homework Equations

The Attempt at a Solution


Partition function for this is Z=\sum_{n=0}^\infty e^{-\beta\left(1-\frac{1}{n-1}\right)\in}
Consider Z when ##kt \ll \in## then ##\beta e \gg1## then ## e^{-\beta e} \rightarrow 0## This leads to the whole summation will go to 0. But we know that at low temperature, Z always goes to 1.
I have tried to calculate the summation but this series is divergent.
How can I change the calculation to reach Z =1?
 
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I agree - if you have an infinite number of energy levels, bounded above by \epsilon, then the partition function diverges.
 
What happens if you single out ##n=0##?
 
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DrClaude said:
What happens if you single out ##n=0##?
With n = 0 \left(1-\frac{1}{0+1}\right)=0 then Z=1 in two cases. But its supposed to be 1 just in the case that the temperature is low kt\ll\epsilon
I have tried some direct methods to find the limit of this function, but it turned out that the function is divergent. Then all of them became useless.
 
What I meant is take ##n=0## out of the sum in the low T limit, and you recover ##Z=1## as expected.
 
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DrClaude said:
What I meant is take ##n=0## out of the sum in the low T limit, and you recover ##Z=1## as expected.
Got it now. Thank you.
 
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