What Is the Correct Partition Function for a Spin System?

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SUMMARY

The correct partition function for a spin system is derived from the sums of two finite geometric series, represented as Z = (1 - e^(hβ(i+1))) / (1 - e^(hβ)) + (1 - e^(-hβ(i+1))) / (1 - e^(-hβ)). The discussion highlights the importance of avoiding double counting, specifically the inclusion of n=0, which can lead to errors in calculations. The relationship 1 - e^x = e^(x/2)(e^(-x/2) - e^(x/2)) is also noted as a helpful identity in resolving issues related to the partition function.

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  • Understanding of statistical mechanics and partition functions
  • Familiarity with geometric series and their summation
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happyparticle
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Homework Statement
Consider a spin i, integer or half integer with the states n = -i, -(i-1),..(i-1), i
Z component of the spin is ##S_z = n\hbar## and the energy eigenvalues of this system in a magnetic field are given by: ##E_n = nh##
Find the partition function in term of 2 sinh ratio
Relevant Equations
##Z = \sum_{-i}^{i} = e^{-E_n \beta}##
##Z = \sum_{-i}^{i} = e^{-E_n \beta}##

##Z = \sum_{0}^j e^{nh\beta} + \sum_{0}^j e^{-nh\beta}##
Those sums are 2 finites geometric series
##Z = \frac{1- e^{h\beta(i+1)}}{1-e^{h\beta}} + \frac{1-e^{-h\beta(i+1)}}{1-e^{-h\beta}}##
I don't think this is ring since from that I can't get 2 sinh. However, I'm not sure where is my error.
 
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Note that you've included ##n=0## twice!
 
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happyparticle said:
I don't think this is ring since from that I can't get 2 sinh. However, I'm not sure where is my error.
This might help: ##1-e^x = e^{x/2}(e^{-x/2}-e^{x/2})##.
 
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Thank you! I didn't see that I had included n=0 twice.
I spent hours trying to figure out what was wrong.
 

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