# Einstein's Calculation (Condensed Matter Physics) [closed]

(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.

• 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.) Jun 25, 2019 at 17:19
• 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. Jun 25, 2019 at 17:33

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 $$\sinh(x)$$.
• 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. Jun 25, 2019 at 17:42
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.