@@ -139,7 +139,7 @@ Using the fact that $\alpha(t) = e^{-i \omega t} \alpha(0)$, we can see that the
In quantum optics, $\hat{x}$ and $\hat{p}$ are called quadratures.
The photon number operator $\hat{n}$ is $\hat{a}^\dagger \hat{a}$ (i.e. $\langle n | \hat{n} | n \rangle = n$), and the Hamiltonian then becomes $\hat{H} = (\frac{1}{2} + \hat{a}^\dagger \hat{a})$. Using this, we find that
The photon number operator $\hat{n}$ is $\hat{a}^\dagger \hat{a}$ (i.e. $\langle n | \hat{n} | n \rangle = n$), and the Hamiltonian then becomes $\hat{H} = \hbar \omega (\frac{1}{2} + \hat{a}^\dagger \hat{a})$. Using this, we find that