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Commit 7f4a0a17 authored by T. van der Sar's avatar T. van der Sar
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typo Debye

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......@@ -123,11 +123,11 @@ As discussed in the previous lecture, the quantum mechanical excitations of a ha
The expected value of the total energy stored in the oscillators (which, from now on, we will simply denote as the total energy $E$) is given by the sum of the energy stored in the individual oscillators. These oscillators are characterized by their wavevector $\mathbf{k}$:
\begin{align}
E &= 3 \sum_\mathbf{k} \left(\frac{1}{2}\hbar\omega(\mathbf{k})+\hbar \omega(\mathbf{k}) n_{B}(\hbar \omega(\mathbf{k}))\right)\\
E &= 3 \sum_\mathbf{k} \left(\frac{1}{2}\hbar\omega(\mathbf{k})+\hbar \omega(\mathbf{k}) n_{B}(\beta \hbar \omega(\mathbf{k}))\right)\\
&= 3 \sum_\mathbf{k} \left(\frac{1}{2}\hbar\omega(\mathbf{k})+\frac{\hbar\omega(\mathbf{k})}{ e^{\hbar\omega(\mathbf{k})/{k_BT}}-1}\right).
\end{align}
Here we used that the expected occupation number is $n_B(\hbar \omega(\mathbf{k}))$.
Here we used that the expected occupation number is $n_B(\beta \hbar \omega(\mathbf{k}))$.
??? question "Where does the factor 3 come from?"
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