Can anyone help me with understanding how the $\tilde{\beta}$ and ${\tilde\mu_t{(x_t, x_0)}}$ are derived?
It seems to me that exponential term is a 2nd order polynomial term and it doesn't really look like a gaussian in the form of $\frac{(x - \mu)^2}{2\sigma^2}$.
Source:https://lilianweng.github.io/posts/2021-07-11-diffusion-models/#reverse-diffusion-process
It seems you've already understood up to the step of the exact conditional probability of the reverse diffusion process $q(\mathbf{x}_{t-1}|\mathbf{x}_t, \mathbf{x}_0)$ which is proportional to the exponential of your mentioned 2nd order polynomial with 3 terms per Bayes theorem along with conditional probabilities defined in the forward diffusion process with some mentioned nice property.
Since the Gaussian distribution is $f(x)=\frac{1}{\sqrt{2\pi\tilde{\beta}}}\text{exp}(-\frac{(x-\tilde{\mu})^2}{2\tilde{\beta}})$ (consistent with your reference where we use $\tilde{\beta}$ as the variance parameter), apparently to match the above proportional functional form with the Gaussian, one has to match the corresponding coefficients of the 3 terms of the 2nd order polynomial inside $\text{exp}$. Thus you clearly have $-\frac{1}{2\tilde{\beta}}=-\frac{1}{2}(\frac{\alpha_t}{\beta_t}+\frac{1}{1-\bar{\alpha_{t-1}}})$ from matching the leading term. This is nothing but the next line in your reference where you see $\tilde{\beta}_t=1/(\frac{\alpha_t}{\beta_t}+\frac{1}{1-\bar{\alpha_{t-1}}})$.
Similarly from matching the second term you have the functional form of $\tilde{\mu}$ which can also be expressed as $\tilde{\mu_t}(\mathbf{x}_t, \mathbf{x}_0)$, which can be also expressed as $\tilde{\mu_t}(\mathbf{x}_t, \mathbf{\epsilon}_t)$ with the mentioned nice property in the forward diffusion process. Note there's no utility from matching the third constant term here.
