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(From The Oxford Solid State Basics by Steven H. Simon [Chapter 2.1])

Reading through a condensed matter textbook and struggling to see how the author progresses with the calculation in the chapter where he discusses Einstein's calculation for a single harmonic oscillator in one dimension: $$ Z_{1D} = \sum_{n\geqslant0}e^{-\beta \hbar\omega(n+1/2)}$$ $$=\frac{e^{-\beta\hbar\omega/2}}{{1-e^{-\beta\hbar\omega}}}$$ $$=\frac{1}{2\sinh(\beta\hbar\omega/2)}$$

Struggling with seeing the logic behind line 1 to line 2 and from line 2 to line 3. I thought Taylor or Maclaurin may be used but I wasn't sure for how to go about this. Also for the last line I have no idea where the hyperbolic sine comes from.

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  • $\begingroup$ These are two good problems to struggle with. Instructive math manipulations. Keep at it, substituting, factoring out common terms (hints),... (I hope no one posts a solution too soon.) $\endgroup$
    – garyp
    Jun 25, 2019 at 17:19
  • $\begingroup$ At the first line, factor $e^{-\beta \hbar \omega /2}$ out, then see your summation carefully, n goes from 0 to infinity, so it's a geometric sum, like $\sum1/n^2$, how to you solve it? in second line, divide numerator and denominator by $e^{-\beta \hbar \omega /2}$, and check hyperbolic function. $\endgroup$
    – Paradoxy
    Jun 25, 2019 at 17:33

2 Answers 2

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For the first: Look at the maximal $n$, and apply the partial sum of the geometric sum.
For the second: Look at the definition of $\sinh(x)$.

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  • $\begingroup$ Isn't the maximal "n" equal to positive infinity as it is upwards from 0 to positive infinity? I am not sure how this helps to solve this. Also I am not sure how knowing the definition of sinh(x) would help to make the jump from line 2 to line 3. $\endgroup$ Jun 25, 2019 at 17:11
  • $\begingroup$ The sum does go to infinity. The hint here might be misleading. $\endgroup$
    – garyp
    Jun 25, 2019 at 17:21
  • $\begingroup$ My formulation (and thoughts) were not very helpful. Use the formula for geometric series en.wikipedia.org/wiki/…, after you have multiplied out the term independent from n. For the second, you really only need to use the connection between $sinh$ and $exp$. I hope this is better understandable. $\endgroup$ Jun 25, 2019 at 17:42
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Rewriting $\sum_{n=0}^{\infty} e^{-\beta\hbar\omega(n+1/2)}$ as $\sum_{n=0}^{\infty} (e^{-\beta\hbar\omega(1/2)} * r)$ [where r = $e^{-\beta\hbar\omega n}$] may expose the logic of the first step for you: As $0 \leq e^{-\beta\hbar\omega n} \leq 1$, this sum converges to the value in step 2.

For the next step, you're just multiplying both the numerator and the denominator by $e^{\frac{\beta\hbar\omega}{2}}$, to get $({e^{\frac{\beta\hbar\omega}{2}} - e^{\frac{-\beta\hbar\omega}{2}}})^{-1}$ and from there substituting the definition of sinh.

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