of double bosons term
------------------------------------

The last term in the Hamiltonian is the term $$H_{2}=(1-b^{2})\sum_{n}n\tilde{a}_{-n}\tilde{a}_{n}\,,$$ fermionization leads to an expression $$\begin{aligned}
H_{2} & = & (1-b^{2})\left(\sum_{r,k\in\mathbb{Z}+\frac{1}{2}}n:\psi_{k}\psi_{-n-k}^{*}\psi_{r}\psi_{n-r}^{*}:+\text{Contractions}\right)\label{H2}\\
 & = & (1-b^{2})\left(\sum_{\begin{subarray}{c}
r+s<0,k+l>0\\
r+s+k+l=0\\
r,s,k,l\in\mathbb{Z}+\frac{1}{2}
\end{subarray}}(k+l):\psi_{r}\psi_{s}^{*}\psi_{k}\psi_{l}^{*}:+\text{Contractions}\right)\,.\nonumber\end{aligned}$$ The contractions in (\[H2\]) now can be calculated as $$\begin{aligned}
\text{Contractions} & = & \sum_{\begin{subarray}{c}
n>0\\
k,l>0
\end{subarray}}n\left((\psi_{-k}\psi_{k-n}^{*}-\psi_{-n-k}^{*}\psi_{k})(\psi_{-l}\psi_{n+l}^{*}-\psi_{n-l}^{*}\psi_{l})\right)\nonumber \\
 &  & -\sum_{r,k\in\mathbb{Z}+\frac{1}{2}}n:\psi_{k}\psi_{-n-k}^{*}\psi_{r}\psi_{n-r}^{*}:\nonumber \\
 & = & \sum_{k>l>0}(k-l)\psi_{-k}\psi_{k}^{*}+\sum_{0<k<l}(l-k)\psi_{-l}^{*}\psi_{l}\nonumber \\
 & = & \sum_{k>0}(\frac{(k-\frac{1}{2})(k+\frac{1}{2})}{2})(\psi_{-k}\psi_{k}^{*}+\psi_{-k}^{*}\psi_{k})\nonumber \\
 & = & \sum_{k>0}\frac{1}{2}(k^{2}-\frac{1}{4})(\psi_{-k}\psi_{k}^{*}+\psi_{-k}^{*}\psi_{k})\,.\end{aligned}$$

Full expression
---------------

Combining all the expressions, we obtain the total $H$ $$\begin{aligned}
\label{eq:full_Hamiltonian}
H & = & H_{0}+H_{1}+H_{2}=\sum_{k>0,k\in\mathbb{Z}+\frac{1}{2}}\left(k^{2}+\frac{3}{4}\right)(\psi_{-k}\psi_{k}^{*}-\psi_{-k}^{*}\psi_{k})\\
 &  & +\frac{2}{3}(1-b^{2})\left(\sum_{\begin{subarray}{c}
r,s,k,l\in\mathbb{Z}+\frac{1}{2}\\
r+s<0,l+k>0\\
r+s+k+l=0
\end{subarray}}:\psi_{r}\psi_{s}^{*}\psi_{k}\psi_{l}^{*}:\left(\frac{1}{2}(s-r)+k-l\right)\right.\nonumber \\
 &  & -\left.\frac{1}{16}\sum_{k>0,k\in\mathbb{Z}+\frac{1}{2}}(6k-1)(2k-1)(\psi_{-k}\psi_{k}^{*}-\psi_{-k}^{*}\psi_{k})\right)\nonumber \\
 &  & +(1-b^{2})\left(\sum_{\begin{subarray}{c}
r+s<0,k+l>0\\
r+s+k+l=0\\
r,s,k,l\in\mathbb{Z}+\frac{1}{2}
\end{subarray}}(k+l):\psi_{r}\psi_{s}^{*}\psi_{k}\psi_{l}^{*}:\right.\nonumber \\
 &  & +\left.\sum_{k>0}\frac{1}{2}(k^{2}-\frac{1}{4})(\psi_{-k}\psi_{k}^{*}+\psi_{-k}^{*}\psi_{k})\right)\nonumber \\
 & = & H_{0}+(1-b^{2})H_{d}+(1-b^{2})H_{t}\,.\nonumber\end{aligned}$$ Here we introduce $$\begin{aligned}
H_{d} & = & \sum_{k>0,k\in\mathbb{Z}+\frac{1}{2}}\frac{1}{3}(k-\frac{1}{2})\psi_{-k}\psi_{k}^{*}+(k-\frac{1}{2})(k+\frac{1}{6})\psi_{-k}^{*}\psi_{k}\\
 &  & +\sum_{\begin{subarray}{c}
k+l>0\\
k,l\in\mathbb{Z}+\frac{1}{2}
\end{subarray}}\frac{2}{3}(2k+l):\psi_{-l}\psi_{-k}^{*}\psi_{k}\psi_{l}^{*}:\,,\nonumber \\
H_{t} & = & \sum_{\begin{subarray}{c}
k+l>0\\
s,k,l\in\mathbb{Z}+\frac{1}{2}
\end{subarray}}(2k+\frac{2}{3}(s+l)):\psi_{-s-k-l}\psi_{s}^{*}\psi_{k}\psi_{l}^{*}:\end{aligned}$$

Fermionic Triangularization
---------------------------

### $H_{d}$ shifts eigen-energy

Now it is clear that $$H_0^\beta\equiv H_{0}+(1-b^{2})H_{d}$$ preserves the Schur state, but $H_{d}$ changes the eigen-energy. The aim here is to prove the eigen-energy $$E_{\lambda}^{\beta}=E_{\lambda}^{1}+(1-b^{2})\sum_{i}^{\ell(\lambda^{t})}(\lambda_{i}^{t})^{2}\, ,$$ where $$E_{\lambda}^{1}=\sum_{i}^{d(\lambda)}(\lambda_{i}-i+\frac{1}{2})^{2}-(\lambda_{i}^{t}-i+\frac{1}{2})^{2}$$ is the eigen-energy of free fermions excitation labeled by Young diagram $\lambda$. For simplicity, we define $$\begin{aligned}
n_{i}(\lambda)=\lambda_{i}-i+\frac{1}{2},\,\, m_{i}(\lambda)=\lambda_{i}^{t}-i+\frac{1}{2}\,.\end{aligned}$$

The eigen-energy of $H_{d}$ 