• Quadrature state
• Fock state - Number state
• Thermal state
• Coherent state
  • LASER(Light Amplification by Stimulated Emission of Radiation)：history
• squeeze state
• Light Modes

The wave-like, classcial properties of a light mode are in the mode function; all we need to know for deducing the quantum physics of the mode is the Bose commutation relation for the mode operators

Bose commutation relations $\begin{bmatrix}\hat{a}k,\hat{a}{k'}^\dagger\end{bmatrix}=\delta_{kk'},\quad\begin{bmatrix}\hat{a}k,\hat{a}{k'}\end{bmatrix}=0.$

For single mode, it is just $\begin{bmatrix}\hat{a},\hat{a}^\dagger\end{bmatrix}=1$

? single mode 是否假设单色波模？单色模有哈密顿量正比光子数，其他波模的哈密顿量是什么形式【未计算】 single mode 根据分解方法确定模式，傅里叶变化算符对应频率（单色）；

The key operators of the single mode:

• photon number operator $\hat n=\hat a ^\dagger\hat a.$

• phase shifting operator $\hat U(\theta)\equiv\exp(-i\theta \hat n)$,$\hat{U}^\dagger(\theta)\hat{a}\hat{U}(\theta)=\hat{a}\exp(-\mathrm{i}\theta).$.

  when the observer wish to change the phase of the mode, then (in Schrodinger picture)$\hat{\mathbf{A}}_k(\mathbf{r},t)=\mathbf{A}_k(\mathbf{r},t)\exp(-\mathrm{i}\theta)\hat{a}$ ↔$\hat{\rho}(\theta)=\hat{U}\hat{\rho}\hat{U}^\dagger.$

• quadratures $\hat{q}=\frac{1}{\sqrt{2}}\left(\hat{a}^\dagger+\hat{a}\right)$,$\hat{p}=\frac{\mathrm{i}}{\sqrt{2}}\left(\hat{a}^\dagger-\hat{a}\right)$; $\hat{x}_{\lambda}=\frac{1}{\sqrt{2}}[\hat{a}\exp(-\mathrm{i}\lambda)+\hat{a}^{\dagger}\exp(\mathrm{i}\lambda)]$.

  For monochromatic modes $\hat q,\hat p$ correspond to the in-phase & out-of-phase component of the field amplitude[$\exp(-i\omega_k t)$and $\pi/2$ phase later][$\hat x_\lambda,\hat x_{\lambda+\pi/2}$]

  $\hat a=\frac 1 {\sqrt 2}(\hat q + i\hat p)$