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This question is from the book "Introductory Statistical Mechanics" by Bowley and Sanchez. The question is as follows:

Calculate the free energy of a system with N particles, each with spin 3/2 with one particle per site, given that the levels associated with the four spin states have energies $\frac{3}{2}\epsilon$, $\frac{1}{2}\epsilon$, $\frac{-1}{2}\epsilon$, and $\frac{-3}{2}\epsilon$ and degeneracies 1, 3, 3, and 1 respectively.

The answer given at the end of the book is $-3Nk_BT\ln(e^{\epsilon/2k_BT}+e^{-\epsilon/2k_BT})$

The way I proceeded to write the partition function is as follows: The degeneracy due to spin $g_n^{spin}=2s+1=4$

Hence the partition function is: $$Z=4e^{-3\epsilon/2k_BT}+12e^{-\epsilon/2k_BT}+12e^{\epsilon/2k_BT}+4e^{3\epsilon/2k_BT}$$ Note that I have also multiplied by the degeneracy indicated in the body of the question. However, since $F=-k_BT\ln(Z)$ the answer is not the same as quoted in the appendix of the book. I would be very grateful if you could provide insight on this issue.

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    $\begingroup$ You have already the spin degeneracies $1,3,3,1$. There is no more additional spin degeneracy $*4$. So, divide your $Z$ by $4$, and, with some little algebra with $\cosh(3x)$ and $\cosh(x)$, you will get the correct result. $\endgroup$
    – Trimok
    Commented Jan 20, 2014 at 17:47
  • $\begingroup$ Notice that $(x+y)^3 = 1x^3 + 3x^2y + 3xy^2 +1y^3$. The degeneracies given are binomial coefficients, after you divide by $4$ as Trimok suggests, so the four terms in the single-site partition function can be simplified into the cube of a binomial. $\endgroup$
    – joshphysics
    Commented Jan 20, 2014 at 18:32
  • $\begingroup$ Well thanks for the answers provided. I did not recognize the binomial coefficients. $\endgroup$
    – Vesnog
    Commented Jan 20, 2014 at 22:42

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