What Is the Correct Partition Function for a Spin System?

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The discussion centers on the correct formulation of the partition function for a spin system, with participants analyzing different summation approaches. One participant presents a formula involving finite geometric series, while another points out an error in including the term for n=0 twice. There is confusion regarding the derivation of the hyperbolic sine function from the proposed equations. A helpful mathematical identity is shared to clarify the situation. The conversation highlights the complexities involved in accurately calculating the partition function in statistical mechanics.
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})##.
 
Thank you! I didn't see that I had included n=0 twice.
I spent hours trying to figure out what was wrong.
 
So is there some elegant way to do this or am I just supposed to follow my nose and sub the Taylor expansions for terms in the two boost matrices under the assumption ##v,w\ll 1##, then do three ugly matrix multiplications and get some horrifying kludge for ##R## and show that the product of ##R## and its transpose is the identity matrix with det(R)=1? Without loss of generality I made ##\mathbf{v}## point along the x-axis and since ##\mathbf{v}\cdot\mathbf{w} = 0## I set ##w_1 = 0## to...

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