Recall the general Hubbard Hamiltonian \[H=\sum_{xy\sigma}t_{xy}\hat{c}_{x\sigma}^\dagger
\hat{c}_{y\sigma}^{\phantom{\dagger}}+\sum_xU_x\hat{n}_{x\uparrow}\hat{n}_{x\downarrow}\]
where \(x\in\Lambda\) where \(\Lambda\) is a collection of points, like
lattice sites. Then \(|\Lambda|\) is
the number of points in \(\Lambda\)
(what we used to call \(N\)). Note that
\(U_x\) can now depend on the lattice
site.
Further note that \(t_{xy}=t_{yx}\) are
elements of a real hopping matrix which is also connected. This
means that there is a path of bonds \(t_{xy}\ne 0\) between any two points in
\(\Lambda\).
Theorem 1: (Attractive Case) Assume \(U_x\le 0\) for all \(x\in\Lambda\) and \(M\) is an even number of electrons. Then,
the ground state includes one with \(s=0\). If \(U_x<0\) for all \(x\in\Lambda\), then the ground state is
unique.
Some comments: Note, as \(U\rightarrow 0\) this is true, since we
have discrete levels from the band structure \(\epsilon(k)\) and we fill with \(\uparrow\downarrow\) in each level. We only
have degeneracies with higher spins states if the band structure has
degeneracies at the Fermi level.
As \(U\rightarrow-\infty\) it is true
since the ground state is constructed out of the bound \(\uparrow\downarrow\) states on each site
which are lowest in energy.
Proof: Since \(S^2\)
and \(S^z\) are conserved, we can work
in the \(S^z=0\) subspace with \(N_\uparrow=N_\downarrow=\frac{M}{2}\).
Then, let \(\{\psi_\alpha\}\) be basis
functions for \(n\) spinless electrons.
There are \(\begin{pmatrix}|\Lambda| \\
n\end{pmatrix}=m\) such basis functions. We choose these basis
functions to all be real.
The ground state \(\left|\psi\right\rangle\) can be written as
\[\left|\psi\right\rangle=\sum_{\alpha\beta}W_{\alpha\beta}
\ \psi_\uparrow^\alpha\otimes\psi_\downarrow^\beta\] where \(W\) is an \(m\times m\) matrix. Without loss of
generality, we can say \(W\) is
Hermitian so that \(W_{\alpha\beta}=W_{\beta\alpha}^*\). \[\begin{aligned}
\left\langle\psi\middle|\psi\right\rangle=& \
\sum_{\alpha\beta\gamma\delta}(\psi_\uparrow^\alpha\otimes\psi_\downarrow^\beta)^\dagger
\ W_{\alpha\beta}^*W_{\gamma\delta} \
(\psi_\uparrow^\gamma\otimes\psi_\downarrow^\delta) \\
=& \
\sum_{\alpha\beta}|W_{\alpha\beta}|^2=\sum_{\alpha\beta}W_{\alpha\beta}W_{\beta\alpha}=\text{Tr}(W^2)
\end{aligned}\] where we made use of orthonormality of the basis
functions and the Hermiticity of \(W\).
Then, \[\begin{aligned}
\left\langle\psi\middle|\hat{T}\middle|\psi\right\rangle=& \
\sum_{\alpha\beta\gamma\delta}W_{\alpha\beta}^*(\psi_\uparrow^\alpha\otimes\psi_\downarrow^\beta)^\dagger\sum_{xy}t_{xy}(\hat{c}_{x\uparrow}^\dagger
\hat{c}_{y\uparrow}^{\phantom{\dagger}}+\hat{c}_{x\downarrow}^\dagger
\hat{c}_{y\downarrow}^{\phantom{\dagger}})W_{\gamma\delta}(\psi_\uparrow^\alpha\otimes\psi_\downarrow^\beta)
\\
=& \
\sum_{\alpha\beta\gamma\delta}\left(W_{\alpha\beta}^*\left\langle\psi_\uparrow^\alpha\middle|\sum_{xy}t_{xy}\hat{c}_{x\uparrow}^\dagger
\hat{c}_{y\uparrow}^{\phantom{\dagger}}\middle|\psi_\uparrow^\gamma\right\rangle
W_{\gamma\delta}\delta_{\beta\delta}+W_{\alpha\beta}^*\left\langle\psi_\downarrow^\beta\middle|\sum_{xy}t_{xy}\hat{c}_{x\downarrow}^\dagger
\hat{c}_{y\downarrow}^{\phantom{\dagger}}\middle|\psi_\downarrow^\delta\right\rangle
W_{\gamma\delta}\delta_{\alpha\gamma}\right)
\end{aligned}\] Define: \[K_{\alpha\beta}=\left\langle\psi^\alpha\middle|\sum_{xy}t_{xy}\hat{c}_x^\dagger
\hat{c}_y^{\phantom{\dagger}}\middle|\psi^\beta\right\rangle\]
Then, \[\begin{aligned}
\left\langle\psi\middle|\hat{T}\middle|\psi\right\rangle=& \
\sum_{\alpha\beta\gamma}\left(W_{\alpha\beta}^*K_{\alpha\gamma}W_{\gamma\beta}+W_{\alpha\beta}^*K_{\beta\gamma}W_{\alpha\gamma}\right)
\\
=& \ \text{Tr}(KW^2)+\text{Tr}(W^2K^T)=2\text{Tr}(KW^2)
\end{aligned}\] since \(K^T=K^\dagger\) and \(W^2K^\dagger=(KW^{\dagger2})=(KW^2)^\dagger\).
We can also calculate \[\begin{aligned}
\left\langle\psi\middle|\hat{{U}}\middle|\psi\right\rangle=& \
\sum_xU_x\sum_{\alpha\beta\gamma\delta}(\psi_\uparrow^\alpha\otimes\psi_\downarrow^\beta)^\dagger
W_{\alpha\beta}^*\hat{n}_{x\uparrow}\hat{n}_{x\downarrow}(\psi_\uparrow^\gamma\otimes\psi_\downarrow^\delta)W_{\gamma\delta}
\\
=& \
\sum_xU_x\sum_{\alpha\beta\gamma\delta}W_{\alpha\beta}^*\left\langle\psi_\uparrow^\alpha\middle|\hat{n}_{x\uparrow}\middle|\psi_\uparrow^\gamma\right\rangle\left\langle\psi_\downarrow^\beta\middle|\hat{n}_{x\downarrow}\middle|\psi_\downarrow^\delta\right\rangle
W_{\gamma\delta}.
\end{aligned}\] Define \((L_x)_{\alpha\beta}=\left\langle\psi^\alpha\middle|\hat{n}_x\middle|\psi^\beta\right\rangle\).
Note \((L_x)_{\alpha\beta}=(L_x)_{\beta\alpha}\)
since all \(\psi\)’s are real. \[\begin{aligned}
\left\langle\psi\middle|\hat{{U}}\middle|\psi\right\rangle=& \
\sum_xU_x\sum_{\alpha\beta\gamma\delta}W_{\alpha\beta}^*(L_x)_{\alpha\gamma}W_{\gamma\delta}(L_x)_{\beta\delta}
\\
=& \
\sum_xU_x\sum_{\alpha\beta\gamma\delta}W_{\beta\alpha}(L_x)_{\alpha\gamma}W_{\gamma\delta}(L_x)_{\delta\beta}
\\
=& \ \sum_xU_x \ \text{Tr}(WL_xWL_x).
\end{aligned}\] So \(E(W)=\left\langle\psi\middle|H\middle|\psi\right\rangle=2\text{Tr}(KW^2)+\sum_xU_x
\ \text{Tr}(WL_xWL_x)\) when \(\text{Tr}(W^2)=1\). Now consider a positive
semidefinite matrix \(|W|\) where \(|W|=\sqrt{W^2}\), which is determined by
diagonalizing \(W\) and forming \[|W|=\begin{pmatrix}|w_1| & 0 & 0 & \\
0 & |w_2| & 0 & \\ 0 & 0 & |w_3| & \\ &
& & \ \ddots \end{pmatrix}.\] In general, \(|W|\ne|W_{\alpha\beta}|\). That holds only
in the basis where \(|W|\) is diagonal.
Let’s examine \(E(W)\) in the diagonal
basis. Obviously \(\text{Tr}(KW^2)=\text{Tr}(K|W|^2)\) and
\[\begin{aligned}
\text{Tr}(WL_xWL_x)=& \ \sum_{ij}W_iW_j(L_x)_{ij}(L_x)_{ji} \\
=& \ \sum_{ij}W_iW_j|(L_x)_{ij}|^2\le
\sum_{ij}|W_i||W_j||(L_x)_{ij}|^2 \\
\le& \ \text{Tr}(|W|L_x|W|L_x).
\end{aligned}\] Since \(U_x\le
0\), we have \(E(|W|)\le E(W)\).
Among all ground states, there is one with \(W=|W|\). Note that normalization says \[\text{Tr}(W^2)=\sum_{i=1}^\infty W_i^2=1\implies
\text{Tr}|W|=\sum_i|w|_{i}\ne 0,\] so we work in coordinate
representation for \(\psi_\alpha\)
\[\psi_\alpha=\hat{c}_{x1}^\dagger
\hat{c}_{x2}^\dagger\cdots\left|0\right\rangle.\] Then \(\text{Tr}(W)=\sum_\alpha W_{\alpha\alpha}\ne
0\). Therefore, the vector \(\psi_\uparrow^\alpha\otimes\psi_\downarrow^\alpha\)
is in the ground state expansion. But \(S=0\) for this state so the ground state
has a spin singlet state.
The proof of uniqueness is straightforward, but we don’t have enough
time to do so here.
Theorem 2: Assume \(U_x=U>0\) is independent of \(x\). Assume \(|\Lambda|\) is even \(\Lambda\) is bipartite. Let \(M=|\Lambda|=\) half-filled band. Then \(S=\frac{1}{2}(|B|-|A|)=0\) (for most
bipartite lattices)
Proof: Need to do the partial particle-hole
transformation which changes \(U\mapsto
-U\), \(N_\uparrow\mapsto|\Lambda|-N_\uparrow\),
\(N_\downarrow\mapsto N_\downarrow\).
This gives us \(N_\uparrow+N_\downarrow=|\Lambda|\implies|\Lambda|-N_\uparrow+N_\downarrow=|\Lambda|\implies
N_\uparrow=N_\downarrow\) but the ground state for the attractive
case with \(N_\uparrow=N_\downarrow\)
has \(S=0\) and is unique by theorem 1.
Hence \(J=0\) is the unique ground
state for the repulsive case.
Consider the case of a very large \(U\). No double occupancy is allowed. Then,
at \(U=\infty\) all spin configurations
are degenerate. But what about finite, but large \(U\)?
The singlet state is shifted down in energy proportional to \(t^2/U\). In general, we find that the
Hamiltonian for large \(U\) maps onto
\[\hat{H}(U\to\infty)\to\frac{2}{U}\sum_{xy}t_{xy}^2\left(\mathbf{S}_x\cdot\mathbf{S}_y-\frac{1}{4}\right)\]
for large \(U\) at half filling. The
ground state of this Hamiltonian is known to have \(S=\frac{1}{2}(|B|-|A|)\) and since the
ground state is nondegenerate, \(S\)
cannot change.
\[\implies S=\frac{1}{2}(|B|-|A|)\] at
half-filling for the Hubbard model!