typo fix (closes #13)

docs/lecture_8.md

@@ -65,10 +65,12 @@ $$
 {\mathcal H}=g\mu_{\rm B}{\bf B}\cdot{\bf \sigma}
 $$
 
-where $\mu_{\rm B}=\frac{e\hbar}{2m_{\rm e}}=10^{-23}$ J/T is the _Bohr magneton_ and $g$ is the _g-factor_, which for an electron is almost exactly 2. Its two eigenstates are $+\mu_{\rm B}B$ and $-\mu_{\rm B}B$. The statistical magnetisation for $n$ spins can be found using the Boltzmann distribution:
+where $\mu_{\rm B}=\frac{e\hbar}{2m_{\rm e}}=10^{-23}$ J/T is the _Bohr magneton_ and $g$ is the _g-factor_, which for an electron is almost exactly 2. Its two eigenstates are $+\mu_{\rm B}B$ and $-\mu_{\rm B}B$.
+Now using the Bolzmann weights we compute the expectation value of the magnetisation of $n$ spins:
 
 $$
-M=n\frac{\mu_{\rm B}{\rm e}^{-\beta\mu_{\rm B}B}-\mu_{\rm B}{\rm e}^{+\beta\mu_{\rm B}B}}{{\rm e}^{-\beta\mu_{\rm B}B}+{\rm e}^{+\beta\mu_{\rm B}B}}=n\mu_{\rm B}{\rm tanh}(\beta\mu_{\rm B}B)
+M = -n \frac{ \mu_{\rm B}{\rm e}^{-\beta\mu_{\rm B}B}-\mu_{\rm B}{\rm e}^{+\beta\mu_{\rm B}B} }{ {\rm e}^{-\beta\mu_{\rm B}B}+{\rm e}^{+\beta\mu_{\rm B}B}}
+= n\mu_{\rm B}{\rm tanh}(\beta\mu_{\rm B}B)
 $$
 
 For small $B$ this function is linear, allowing us to extract the susceptibility