The summation conditions $i>j$ or $i<j$ reflects the condition $k+l>0$. Thus we have translated the Theorem \[Theo:Hdenergy\] into the form that $$\begin{aligned}
 &  & \sum_{i}^{\ell(\lambda)}(\lambda_{i}^{t})^{2}=\sum_{i}^{d(\lambda)}\frac{1}{3}(n_{i}-\frac{1}{2})+\left(m_{i}-\frac{1}{2}\right)\left(m_{i}+\frac{1}{6}\right)\label{imptcombine}\\
 &  & +\frac{2}{3}\left(\sum_{i,j=1}^{d(\lambda)}(2m_{i}+n_{j})+\sum_{i>j}^{d(\lambda)}(2n_{i}-n_{j})+\sum_{i<j}^{d(\lambda)}(-2m_{i}+m_{j})\right)\,.\nonumber\end{aligned}$$ We use mathematical induction to prove it here. A direct proof can be found in Appendix.

Suppose the relation holds for a Young diagram $\lambda$, with $|\lambda|=n$. We will prove it holds for adding one more box to $\lambda$ under or right to the certain position $(i,j)$. These two different cases are following.

**Case I**: $\lambda_{j}^{t}\geq j$ $ \lambda^t_{j-1}> i$ and the box is attached just under the $(i,j)$ box, so that $$m_{j}\rightarrow m_{j}+1\,.$$

The change of R.H.S of (\[imptcombine\]) will be $$\begin{aligned}
\Delta_{RHS} & = & \left(2m_{j}+\frac{2}{3}\right)+\frac{2}{3}\left(2d(\lambda)+j-1-2(d(\lambda)-j)\right)\\
 & = & 2m_{j}+2j=2\lambda_{j}^{t}+1=\Delta_{LHS}\,.\nonumber\end{aligned}$$

**Case II**: $\lambda_{i}\geq i$ $\lambda_{i-1}>j$, the box is attached to the right of the box $(i,j)$, so that $$n_{i}\rightarrow n_{i}+1\,.$$

The change of R.H.S of (\[imptcombine\]) will be $$\begin{aligned}
\Delta_{RHS} & = & \frac{1}{3}+\frac{2}{3}(3i-2)=2i-1\,.\end{aligned}$$ Notice that the $j+1$-th column has $\lambda_{j+1}^{t}=i-1$, so the change of L.H.S of (\[imptcombine\]) is $$\Delta_{L.H.S}=i^{2}-(i-1)^{2}=2i-1=\Delta_{R.H.S}\,.$$ Now we conclude that the identity (\[imptcombine\]) holds for any Young diagram and the Theorem \[Theo:Hdenergy\] is proved. Q.E.D.

### $H_{t}$ squeezes the state

As argued previously, the $H_t$ always 