Currently self-studying QFT and The Standard Model by Schwartz and I'm stuck at equation 1.5 in Part 1 regarding black-body radiation

Question

So basically the equation is basically a derivation of Planck's radiation law and I can't somehow find any resources as to how he derived it by adding a derivative inside. Planck says that each mode of frequency $\omega_n$ can be excited by $j$ times which gives the energy of the form $jE_n = j(\hbar\omega_n)$

This is equation 1.5:

$$\langle E_n \rangle = \frac{\sum_{j=0}^\infty (jE_n)e^{-jE_n\beta}}{\sum_{j=0}^\infty e^{-jE_n\beta}} = \frac{-\frac{d}{d\beta}\frac{1}{1-e^{-\hbar\omega_n\beta}}}{\frac{1}{1-e^{-\hbar\omega_n\beta}}} = \frac{\hbar\omega_n}{e^{\hbar\omega_n\beta}-1}\tag{1.5}$$

Where $\beta = \frac{1}{k_BT}$.

In short, I just don't get where the $-\frac{d}{d\beta}$ comes from. I also don't get how the summations disappear. Hope to get answers before I continue this book.

Answer 1

It just a clever use of the geometrical series:

$$\frac{1}{1-q} = \sum_{j=0}^\infty q^j$$

which is valid for any real number $q<1$.

Here $q = e^{-E_n\beta} \equiv e^{-\hbar\omega_n\beta}$. As long as the energies $E_n >0$ which is certainly fulfilled, we can make use of the geometrical series. And only this. The parameter $q=q(\beta)$ is a function of $\beta$, so we can take the derivative of it:

$$-\frac{d}{d\beta} \frac{1}{1-q(\beta)} = -\frac{d}{d\beta} \sum_{j=0}^\infty q(\beta)^j=-\sum_{j=0}^\infty \frac{d}{d\beta}q(\beta)^j =-\sum_{j=0}^\infty q'(\beta) j q(\beta)^{j-1}$$

where in the last step the chain rule was used. We compute the derivative of $q(\beta)$: $q'(\beta) = -E_n e^{-E_n\beta} = -E_n q(\beta)$.

In the next step we plug this result into the sum:

$$-\sum_{j=0}^\infty q'(\beta) j q(\beta)^{j-1} = -\sum_{j=0}^\infty (-E_n) q(\beta)j q(\beta)^{j-1} = \sum_{j=0}^\infty jE_n q(\beta)^j = \sum_{j=0}^\infty jE_n e^{-jE_n\beta}$$

quod est demonstrandum. One only has to plug it now into the expression for $\langle E_n\rangle$.