Why the ground-state energy of S-1/2 Anti-Ferromagnetic Heisenberg Chain is not$-\frac{N}{4}J$ The Hamiltonian of traditional Heisenberg model is
$$\hat H = J\sum_{<i,j>}\vec{S_i}\cdot\vec{S_j}=J\sum_{<i,j>}\left(S_i^zS_j^z+\frac{1}{2}\left(S_i^+S_j^-+S_i^-S_j^+\right)\right)$$
if J is positive, we can get a anti-ferromagnetic state. But why the ground-state energy is about -4.5154(I got it by using DMRG in ALPS, assuming J=1 and N=10) but not trivially $-\frac{1}{4}\times 1\times 10=-2.5$ for the state
$ |\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow> $or$|{\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow}>$.
Then what is the true ground state of it?
 A: Your candidate for the ground state, known as the Néel state, is not an eigenstate of the Hamiltonian. It is, however, a good starting point for finding the true (quantum mechanical) ground state. One systematic approach to this is ''linear spin-wave theory".
Here, we reformulate the Hamiltonian in terms of bosonic creation and annihilation operators ($a_n^\dagger$, $a_n$) via the Holstein–Primakoff transformation. The Néel state is then the vacuum of these operators. Linearizing the resulting Hamiltonian, so that it contains only terms proportional to $a_n^\dagger a_m$, $a_na_m$ or $a_n^\dagger a_m^\dagger$ then allows to diagonalize it using a Bogoliubov transformation ($a_n^\dagger\to b_n^\dagger$, $a_n\to b_n$). For details on the calculation see ref. [1].
The resulting Hamiltonian is then given by
\begin{equation}
H=-\frac12JNzS^2-\frac12JzS\sum_k\big(1-\sqrt{1-\gamma_k^2}\big)+JzS\sum_k\sqrt{1-\gamma_k^2}b_k^\dagger b_k\ ,
\end{equation}
where $z$ is the number of nearest neighbors and $\gamma_k$ depends on the precise form of the lattice (in the case of a one dimensional chain we have $\gamma_k=\cos(k)$). The ground state is then given by the vacuum of the Bogoliubov transformed operators $b_k|0\rangle=0$.
With $J=1$, $S=1/2$, $z=2$ and $N=10$, the first part of this Hamiltonian reproduces your result. There are, however, quantum corrections in the second part, further lowering the energy.

[1] A.J. Beekman, L. Rademaker, J. van Wezel, An Introduction to Spontaneous Symmetry Breaking, arXiv:1909.01820 [hep-th]
