related to Schur state $\lambda$ is $$E_{\lambda}^{d}=\sum_{i}^{\lambda_{1}}(\lambda_{i}^{t})^{2}\,.$$ \[Theo:Hdenergy\]

![The (2,2,1,1) Young diagram[]{data-label="fig:Fig1"}](Fig1.eps "fig:"){width=".2\textwidth"}\

Before proving this theorem, we now consider an example $\lambda=\{2,2,1,1\}$ as shown in Fig. \[fig:Fig1\]. The corresponding Schur state is $$|\lambda\rangle=(-)\psi_{-3/2}\psi_{-7/2}^{*}\psi_{-1/2}\psi_{-1/2}^{*}|vac\rangle\,.$$ From the formalism of $H_{d}$, the energy eigenvalue is contributed by the following terms $$\begin{aligned}
 &  & \frac{2}{3}\left(\frac{3}{2}-1\right)\psi_{-3/2}\psi_{3/2}^{*}\,,\,\left((\frac{7}{2}-\frac{1}{2})(\frac{7}{2}+\frac{1}{6})\right)\psi_{-7/2}^{*}\psi_{7/2}\\
 &  & \frac{2}{3}\left\{ \left(7+\frac{3}{2}\right)\psi_{-3/2}\psi_{-7/2}^{*}\psi_{7/2}\psi_{3/2}^{*}+\left(1+\frac{1}{2}\right)\psi_{-1/2}\psi_{-1/2}^{*}\psi_{1/2}\psi_{-1/2}^{*}\right.\\
 &  & \hspace{0.5cm}+\left(7+\frac{1}{2}\right)\psi_{-1/2}\psi_{-7/2}^{*}\psi_{7/2}\psi_{1/2}^{*}+\left(1+\frac{3}{2}\right)\psi_{-3/2}\psi_{-1/2}^{*}\psi_{1/2}\psi_{3/2}^{*}\\
 &  & \left.\hspace{0.5cm}+\left(1-\frac{3}{2}\right)\psi_{-3/2}\psi_{-1/2}\psi_{-1/2}^{*}\psi_{3/2}^{*}-\left(7-\frac{1}{2}\right)\psi_{-1/2}^{*}\psi_{-7/2}^{*}\psi_{7/2}\psi_{1/2}\right\} \,.\end{aligned}$$ Combine all of them, we have the eigenvalue of $H_{d}$ $$E_{\lambda}^{d}=\frac{1}{3}+11+\frac{2}{3}(10+10-7)=20=16+4\,.$$ [ **Proof of Theorem \[Theo:Hdenergy\]**]{}

For a generic Schur state, the proof of Theorem \[Theo:Hdenergy\] is following. A generic Schur state is denoted by $$|\lambda\rangle=(-)^{\sum_{i}^{d(\lambda)}\left(m_{i}-\frac{1}{2}\right)}\prod_{i}^{d(\lambda)}\psi_{-n_{i}}\psi_{-m_{i}}^{*}|vac\rangle\,.$$ Acting on it, the first two terms of $H_{d}$ contribute $$\begin{aligned}
\sum_{i}^{d(\lambda)}\frac{1}{3}(n_{i}-\frac{1}{2})+\left(m_{i}-\frac{1}{2}\right)\left(m_{i}+\frac{1}{6}\right)\,.\end{aligned}$$ The third term contains $$\left(\begin{array}{c}
2d(\lambda)\\
2
\end{array}\right)=d(\lambda)(2d(\lambda)-1)$$ terms. We can have three independent picking strategies. Among them there are $d(\lambda)^{2}$ terms picked from a pair of $\psi$ and $\psi^{*}$ which we call $\psi\psi^{*}$-strategy. Other terms are from two fermionic modes picked from either the set of all $\psi$’s or $\psi^{*}$’s (the $\psi\psi$-strategy and $\psi^{*}\psi^{*}$-strategy).

We first consider the $\psi\psi^{*}$ -strategy. The contribution is $$E_{\psi\psi^{*}}=\frac{2}{3}\sum_{i,j=1}^{d(\lambda)}(2m_{i}+n_{j})\,.$$ The $\psi\psi$ strategy has a signature contribution (-1). Its contribution to the total energy is $$E_{\psi\psi}=\frac{2}{3}\sum_{i>j}^{d(\lambda)}(2n_{i}-n_{j})\,.$$ Similarly, the $\psi^{*}\psi^{*}$ -strategy gives rise to $$E_{\psi^{*}\psi^{*}}=\frac{2}{3}\sum_{i<j}^{d(\lambda)}(-2m_{i}+m_{j})\,.$$ 