简单解释做了什么：The principle of interferometry is to recombine coherently the beams from two or more independent telescopes and measure the so-called complex visibilities of the fringe patterns produced by the interferences.
【两个望远镜收集的光进行recombine，对这两束光的干涉图案进行采集】
物理学基础：According to the van Cittert-Zernicke theorem, for an ideal interferometer, the complex visibility \(V_{j_{1}, j_{2}}(t)\) of the fringes produced by the interferences of the telescopes \(j_{1}\) and \(j_{2}\) at time \(t\) is proportional to the Fourier transform of the object brightness distribution \(\hat{I}\left(\boldsymbol{v}_{j_{1}, j_{2}}(t)\right)\) at spatial frequency \(\boldsymbol{v}_{j_{1}, j_{2}}(t)=\boldsymbol{B}_{j_{1}, j_{2}}^{\perp}(t) / \lambda\), where \(\lambda\) is the wavelength, and the socalled baseline \(\boldsymbol{B}_{j_{1}, j_{2}}^{\perp}(t)\) represents the separation between the two telescopes projected on a plane perpendicular to the line of sight .
【1.干涉图案 \(V_{j_{1}, j_{2}}(t)\) 和 \(\hat{I}\left(\boldsymbol{v}_{j_{1}, j_{2}}(t)\right)\)成正比；2. \(\hat{I}\left(\boldsymbol{v}_{j_{1}, j_{2}}(t)\right)\)是物体亮度信息的傅里叶变换在某一空间频率的分量（存疑）；3.这一空间频率是\(\boldsymbol{v}_{j_{1}, j_{2}}(t)=\boldsymbol{B}_{j_{1}, j_{2}}^{\perp}(t) / \lambda\)，其中\(\lambda\)是波长，baseline \(\boldsymbol{B}_{j_{1}, j_{2}}^{\perp}(t)\)是两个望远镜的距离在垂直于入射光的平面的投影。】
简写：
\[\begin{aligned}&V_{m} \stackrel{\text { def }}{=} V_{j_{1, m}, j_{2, m}}\left(t_{m}\right) \\ &\boldsymbol{v}_{m} \stackrel{\text { def }}{=} \boldsymbol{B}_{j_{1, m}, j_{2, m}}^{\perp}\left(t_{m}\right) / \lambda \end{aligned} \]
image model
\[I(\boldsymbol{\theta})=\sum_{n=1}^{N} x_{n} b_{n}(\boldsymbol{\theta})\]
\(\boldsymbol{x}=\left\{x_{n}\right\}_{n=1}^{N}\) 是图像系数，比如pixel values，\(\left\{b_{n}(\boldsymbol{\theta})\right\}_{n=1}^{N}\)是选择的basis of functions比如每个pixel的response function。要重建的信息就是x。这样的话可以得到傅里叶变换的表达式：
\[\hat{I}(v)=\sum_{n} x_{n} \hat{b}_{n}(v)=\hat{b}(v) \sum_{n} x_{n} \mathrm{e}^{-\mathrm{i} 2 \pi \theta_{n} \cdot v}\]
把b也带入有：
\[\begin{aligned}\hat{I}_{m}=\hat{I}\left(\boldsymbol{v}_{m}\right)=\sum_{n} A_{m, n} x_{n}=(\mathbf{A} \cdot \boldsymbol{x})_{m} \\ A_{m, n}=\hat{b}_{n}\left(v_{m}\right)=\hat{b}\left(v_{m}\right) \mathrm{e}^{-\mathrm{i} 2 \pi \theta_{n} \cdot v_{m}} \end{aligned} \]
由于望远镜数量少+欠缺相位信息，这必然是一个病态的问题，因此需要先验的限制（正则化项）。
几种正则化方式：
1.Quadratic smoothness: \(f_{\text {prior }}(\boldsymbol{x})=\|\boldsymbol{x}-\mathbf{S} \cdot \boldsymbol{x}\|^{2}\)，where S is a smoothing operator implemented via finite differences。
2.Compactness：\(f_{\text {prior }}(x)=\sum_{n} w_{n}^{\text {prior }} x_{n}^{2}\)，to enforce compactness, the weights \(w_{n}^{\text {prior }}>0\) have to increase with the distance from the center of the image.（比如\(w_{n}^{\text {prior }}=\left\|\boldsymbol{\theta}_{n} / \Delta \theta\right\|^{2}\)）
3.Total variation：\(f_{\text {prior }}(\boldsymbol{x})=\sum_{n_{1}, n_{2}} \sqrt{\left\|\nabla x_{n_{1}, n_{2}}\right\|^{2}+\epsilon^{2}}\) where \(\left\|\nabla x_{n_{1}, n_{2}}\right\|^{2}=\left(x_{n_{1}+1, n 2}-x_{n_{1}, n_{2}}\right)^{2}+\left(x_{n_{1}, n_{2}+1}-x_{n_{1}, n_{2}}\right)^{2}\)
4.Maximum entropy methods：\(f_{\text {prior }}(\boldsymbol{x})=-\sum_{n} h\left(x_{n} ; \bar{x}_{n}\right)\)，文中写到了三种熵函数：\(\begin{array}{ll}\text { MEM-sqrt: } & h(x ; \bar{x})=\sqrt{x} \\ \text { MEM-log: } & h(x ; \bar{x})=\log (x) \\ \text { MEM-prior: } & h(x ; \bar{x})=x-\bar{x}-x \log (x / \bar{x})\end{array}\)
在有什么收获之前就已经看累了...最近找能看懂的论文的时间远大于看论文的时间了，感觉效率很低。收获应该算是，发现这里对病态问题的处理方法是变成一个最小化损失的问题，损失函数由原函数损失和正则函数一起组成，然后这篇论文略过了如何去最小化损失，专心讲regularization。
