Consider a collection of H atoms. As we bring the atoms together they
will solidify into a solid lattice (ignoring any molecular bonding
effects for the moment).
When in a solid form, the 1s electrons spread into an energy band and
there is one electron per lattice site.
This can be described by a tight-binding model for the 1s electrons,
which allows electrons to “hop” between nearest neighbors. \[\hat{T}=-t\sum_{\langle
ij\rangle\sigma}(c_{i\sigma}^\dagger
c_{j\sigma}^{\phantom{\dagger}}+c_{j\sigma}^\dagger
c_{i\sigma}^{\phantom{\dagger}})\] where \(\langle ij\rangle\) counts nearest
neighbors once. Because the electrons are negatively charged, we also
expect Coulomb repulsion. But because the electrons are mobile, and can
screen each other out, we approximate the repulsion by an on-site \(U\) only: \[\hat{U}=U\sum_in_{i\uparrow}n_{i\downarrow}.\]
The Hubbard model is the sum of these two terms \[\boxed{\hat{H}_{\text{Hubbard}}=-t\sum_{\langle
ij\rangle\sigma}(c_{i\sigma}^\dagger
c_{j\sigma}^{\phantom{\dagger}}+c_{j\sigma}^\dagger
c_{i\sigma}^{\phantom{\dagger}})+U\sum_in_{i\uparrow}n_{i\downarrow}.}\]
Let’s examine \(\hat{H}\) in momentum space (Bloch basis). Define: \[\begin{aligned} a_{k\sigma}^\dagger=& \ \frac{1}{\sqrt{V}}\sum_je^{ik\cdot r_j}c_{j\sigma}^\dagger \\ a_{k\sigma}^{\phantom{\dagger}}=& \ \frac{1}{\sqrt{V}}\sum_je^{-ik\cdot r_j}c_{j\sigma}^{\phantom{\dagger}} \end{aligned}\] where \(V\) is the number of lattice sites. Then, \[\begin{aligned} c_{j\sigma}^\dagger=& \ \frac{1}{\sqrt{V}}\sum_{k}e^{-ik\cdot r_j}a_{k\sigma}^\dagger \\ c_{j\sigma}^{\phantom{\dagger}}=& \ \frac{1}{\sqrt{V}}\sum_{k}e^{ik\cdot r_j}a_{k\sigma}^{\phantom{\dagger}}, \end{aligned}\] which we can plug into the Hubbard Hamiltonian. \[\begin{aligned} \hat{H}&=-t\sum_{\langle ij\rangle\sigma}\frac{1}{V}\sum_{kk'}\left(e^{-ik\cdot r_i+ik'\cdot r_j}a_{k\sigma}^\dagger a_{k'\sigma}^{\phantom{\dagger}}+e^{ik\cdot r_i-ik'\cdot r_j}a_{k'\sigma}^\dagger a_{k\sigma}^{\phantom{\dagger}}\right)\\&+U\sum_i\frac{1}{V^2}\sum_{k_1,k_2,k_3,k_4}e^{-i(k_1-k_2)\cdot r_i-i(k_3-k_4)\cdot r_i}a_{k_1\uparrow}^\dagger a_{k_2\uparrow}^{\phantom{\dagger}}a_{k_3\downarrow}^\dagger a_{k_4\downarrow}^{\phantom{\dagger}}. \end{aligned}\] Note that for nearest neighbor pairs \(r_j=r_i+\delta\) with \(\delta\) being the nearest neighbor translation vector. So, \[\begin{aligned} \frac{1}{V}\sum_{\langle ij\rangle}e^{-ik\cdot r_i+ik'\cdot r_j}=& \ \frac{1}{V}\frac{1}{2}\sum_i\sum_\delta e^{-i(k-k')\cdot r_i+ik'\cdot\delta} \\ =& \ \frac{1}{2}\sum_\delta e^{ik\cdot\delta}\delta_{kk'}. \end{aligned}\] So the first term (kinetic energy) becomes \(\sum_{k\sigma}=\varepsilon_ka_{k\sigma}^\dagger a_{k\sigma}^{\phantom{\dagger}}\) with \[\varepsilon_k=-t\frac{1}{2}\sum_\delta(e^{ik\cdot\delta}+e^{-ik\cdot\delta})=-t\sum_\delta\cos(k\cdot\delta)\] and the second term becomes \[\frac{U}{V}\sum_{kk'q}a_{k+q\uparrow}^\dagger a_{k\uparrow}^{\phantom{\dagger}}a_{k'-q\downarrow}^\dagger a_{k'\downarrow}^{\phantom{\dagger}}.\] So we find that the Hubbard Hamiltonian takes the following form in momentum space: \[\hat{H}=\sum_{k\sigma}=\varepsilon_ka_{k\sigma}^\dagger a_{k\sigma}^{\phantom{\dagger}}+\frac{U}{V}\sum_{kk'q}a_{k+q\uparrow}^\dagger a_{k\uparrow}^{\phantom{\dagger}}a_{k'-q\downarrow}^\dagger a_{k'\downarrow}^{\phantom{\dagger}}.\] Note that in real space the first term is complicated but the second term is diagonal, while the opposite occurs in momentum space. The key problem is to find the eigenvalues and properties of the ground state for arbitrary \(U\) values.
Suppose the lattice is bipartite \(\implies\) \(t_{ij}\ne 0\) only if \(i\in A\) and \(j\in B\) or \(i\in B\) and \(j\in A\) (never \(AA\) or \(BB\)). Some examples with nearest neighbor hopping:
Simple cubic lattice
Square lattice
Body-centered cubic lattice
NOT face-centered cubic lattice
NOT triangular lattice
Then, \(t\rightarrow -t\) is a
symmetry. Proof: Define \((c_{i\sigma}')^\dagger=(-1)^{\varepsilon(i)}c_{i\sigma}^\dagger\)
and \(c_{i\sigma}'=(-1)^{\varepsilon(i)}c_{i\sigma}^{\phantom{\dagger}}\)
with \(\varepsilon(i)=1\) when \(i\in A\) and \(0\) when \(i\in
B\). Then, \[\hat{H}=t\sum_{\langle
ij\rangle \sigma}((c_{i\sigma}')^\dagger
c_{j\sigma}'(c_{j\sigma}')^\dagger
c_{i\sigma}')+U\sum_in_{i\uparrow}'n_{i\downarrow}'\]
since \(c_{i\sigma}^\dagger
c_{j\sigma}=-(c_{i\sigma}')^\dagger c_{j\sigma}'\) when
\(i\) and \(j\) are on different sublattices.
Therefore, the eigenvalues are symmetric with respect to \(t\rightarrow -t\).
The Hubbard model also has partial particle-hole symmetry. Let \[d_{i\uparrow}^\dagger=c_{i\uparrow}(-1)^{\varepsilon(i)},
\quad d_{i\uparrow}=c_{i\uparrow}^\dagger(-1)^\varepsilon(i)\]
\[d_{i\downarrow}^\dagger=c_{i\downarrow}^\dagger,
\quad d_{i\downarrow}=c_{i\downarrow}\] Then, \[c_{i\uparrow}^\dagger
c_{j\uparrow}=d_{i\uparrow}d_{j\uparrow}^\dagger(-1)^{\varepsilon(i)+\varepsilon(j)}=-d_{i\uparrow}d_{j\uparrow}^\dagger=d_{j\uparrow}^\dagger
d_{i\uparrow}\] So, \[c_{i\uparrow}^\dagger
c_{j\uparrow}+c_{j\uparrow}^\dagger c_{i\uparrow}=d_{i\uparrow}^\dagger
d_{j\uparrow}+d_{j\uparrow}^\dagger d_{i\uparrow}\] and \[n_{i\uparrow}\rightarrow
d_{i\uparrow}d_{i\uparrow}^\dagger=-d_{i\uparrow}^\dagger
d_{i\uparrow}+1\implies N_\uparrow=V-N_\uparrow\] So, \[\hat{H}\rightarrow-t\sum_{\langle
ij\rangle\sigma}(d_{i\sigma}^\dagger d_{j\sigma}+d_{j\sigma}^\dagger
d_{i\sigma})-U\sum_{i}d_{i\uparrow}^\dagger
d_{i\uparrow}d_{i\downarrow}^\dagger
d_{i\downarrow}+U\sum_id_{i\downarrow}^\dagger d_{i\downarrow}\]
\[\implies
E(U,N_\uparrow,\downarrow)=E(-U,V-N_\uparrow,N_\downarrow)+UN_\downarrow\]
If \(N_\uparrow=N_\downarrow=\frac{V}{2}\)
(half-filling case), then \[\boxed{E(U,V/2,V/2)=E(-U,V/2,V/2)+\frac{UV}{2}}\]
Therefore, up to a constant, energies are symmetric for \(U\rightarrow-U\) at half-filling. Let’s
look at spin next. \[\left[\hat{H},\sum_kn_{k\sigma'}\right]=-t\sum_{\langle
ij\rangle\sigma}[c_{i\sigma}^\dagger
c_{j\sigma}^{\phantom{\dagger}}+c_{j\sigma}^\dagger
c_{i\sigma}^{\phantom{\dagger}},n_{k\sigma'}]\] and since
\([n_{k\sigma},n_{k'\sigma'}]=0\)
\[\begin{aligned}
=& \ -t\sum_{\langle
ij\rangle}\sum_k\left(\delta_{ik}(-c_{i\sigma'}^\dagger
c_{j\sigma'}^{\phantom{\dagger}}+c_{j\sigma'}^\dagger
c_{i\sigma'}^{\phantom{\dagger}})+\delta_{jk}(c_{i\sigma'}^\dagger
c_{j\sigma'}^{\phantom{\dagger}}-c_{j\sigma'}^\dagger
c_{i\sigma}^{\phantom{\dagger}})\right) \\
=& \ -t\sum_{\langle ij\rangle}(-c_{i\sigma'}^\dagger
c_{j\sigma'}^{\phantom{\dagger}}+c_{j\sigma'}^\dagger
c_{i\sigma'}^{\phantom{\dagger}}+c_{i\sigma'}^\dagger
c_{j\sigma'}^{\phantom{\dagger}}-c_{j\sigma'}^\dagger
c_{i\sigma'}^{\phantom{\dagger}}) \\
=& \ 0.
\end{aligned}\] So \(\hat{S}_z=\frac{1}{2}\sum_i(n_{i\uparrow}-n_{i\downarrow})\)
and \(\hat{N}=\sum_i(n_{i\uparrow}+n_{i\downarrow})\)
both commute with \(\hat{H}\). Hence,
we can have simultaneous eigenstates with definite \(S_z\) and \(N\). \[\left[H,\sum_kc_{k\uparrow}^\dagger
c_{k\downarrow}^{\phantom{\dagger}}\right]=-t\sum_{<ij>\sigma}\sum_k[c_{i\sigma}^\dagger
c_{j\sigma}^{\phantom{\dagger}}+c_{j\sigma}^\dagger
c_{i\sigma}^{\phantom{\dagger}},c_{k\uparrow}^\dagger
c_{k\downarrow}^{\phantom{\dagger}}]+U\sum_i\sum_k[n_{i\uparrow}n_{i\downarrow},c_{k\uparrow}^\dagger
c_{k\downarrow}]\] but \([c_{i\sigma}^\dagger
c_{j\sigma}^{\phantom{\dagger}}+c_{j\sigma}^\dagger
c_{i\sigma}^{\phantom{\dagger}}, c_{k\uparrow}^\dagger
c_{k\downarrow}^{\phantom{\dagger}}]=\delta_{ik}(c_{j\uparrow}^\dagger
c_{i\downarrow}^{\phantom{\dagger}}-c_{i\uparrow}^\dagger
c_{j\downarrow})^{\phantom{\dagger}}+\delta_{jk}(c_{i\uparrow}^\dagger
c_{j\downarrow}^{\phantom{\dagger}}-c_{j\uparrow}^\dagger
c_{i\downarrow}^{\phantom{\dagger}}).\) When summed over \(\langle ij\rangle\), this vanishes. Then,
\[\begin{aligned}
[n_{i\uparrow}n_{i\downarrow},c_{k\uparrow}^\dagger
c_{k\downarrow}^{\phantom{\dagger}}]=& \
n_{i\uparrow}[n_{i\downarrow}, c_{k\uparrow}^\dagger
c_{k\downarrow}^{\phantom{\dagger}}]+[n_{i\uparrow},c_{k\uparrow}^\dagger
c_{k\downarrow}^{\phantom{\dagger}}]n_{i\downarrow} \\
=& \ -n_{i\uparrow}c_{i\uparrow}^\dagger
c_{i\downarrow}^{\phantom{\dagger}}\delta_{ik}+c_{i\uparrow}^\dagger
c_{i\downarrow}^{\phantom{\dagger}}\delta_{ik} \\
=& \ 0.
\end{aligned}\] This means that \([\hat{H},\hat{S}^+]=[\hat{H},\hat{S}^-]=0\)
so \(S^2\) and \(S_z\) are good quantum numbers.
Now, look at pseudospin. We already showed that \([\hat{H},\hat{J}_z]=0\) where \(\hat{J}_z=\frac{1}{2}(\hat{N}-V)\). Define:
\[\begin{aligned}
\hat{J}^+=& \ \sum_ic_{i\uparrow}^\dagger
c_{i\downarrow}^\dagger(-1)^{\varepsilon(i)} \ \text{(pair creation
operator)} \\
\hat{J}^-=& \
\sum_ic_{i\downarrow}^{\phantom{\dagger}}c_{i\uparrow}^{\phantom{\dagger}}(-1)^{\varepsilon(i)}
\ \text{(pair destruction operator)}
\end{aligned}\] \[\begin{aligned}
[\hat{T},\hat{J}^+]=& \ -t\sum_{\langle ij\rangle
\sigma}\sum_k[c_{i\sigma}^\dagger
c_{j\sigma}^{\phantom{\dagger}}+c_{j\sigma}^\dagger
c_{i\sigma}^{\phantom{\dagger}},c_{k\sigma}^\dagger
c_{k\downarrow}^\dagger (-1)^{\varepsilon(k)}] \\
=& \
-t\sum_{<ij>}\sum_k\left(\delta_{ik}(c_{j\uparrow}^\dagger
c_{k\downarrow}^\dagger(-1)^{\varepsilon(k)}-c_{j\downarrow}^\dagger
c_{k\uparrow}^\dagger(-1)^{\varepsilon(k)})+\delta_{jk}(c_{i\uparrow}^\dagger
c_{k\downarrow}^{\phantom{\dagger}}(-1)^{\varepsilon(k)}-c_{i\downarrow}^\dagger
c_{k\uparrow}^\dagger(-1)^{\varepsilon(k)})\right) \\
=& \ -t\sum_{\langle ij\rangle}(c_{j\uparrow}^\dagger
c_{i\downarrow}^\dagger-c_{j\downarrow}^\dagger
c_{i\uparrow}^\dagger)(-1)^{\varepsilon(i)}+(c_{i\uparrow}^\dagger
c_{j\downarrow}^\dagger-c_{i\downarrow}^\dagger
c_{j\uparrow}^\dagger)(-1)^{\varepsilon(j)} \\
=& \ 0
\end{aligned}\] and \[\begin{aligned}
[\hat{U},\hat{J}^+]=& \
U\sum_{ij}[n_{i\uparrow}n_{i\downarrow},c_{j\uparrow}^\dagger
c_{j\downarrow}^\dagger(-1)^{\varepsilon(j)}] \\
=& \ U\sum_{ij}\delta_{ij}\left(c_{i\uparrow}^\dagger
c_{i\downarrow}^\dagger
n_{i\downarrow}(-1)^{\varepsilon(i)}+n_{i\uparrow}c_{i\uparrow}^\dagger
c_{i\downarrow}^\dagger(-1)^{\varepsilon(i)}\right) \\
=& \ U\sum_i(c_{i\uparrow}^\dagger c_{i\downarrow}^\dagger
c_{i\downarrow}^\dagger
c_{i\downarrow}^{\phantom{\dagger}}(-1)^{\varepsilon(i)}+c_{i\uparrow}^\dagger
c_{i\uparrow}^{\phantom{\dagger}}c_{i\uparrow}^\dagger
c_{i\downarrow}^\dagger(-1)^{\varepsilon(i)}) \\
=& \ UJ^+.
\end{aligned}\] So \([\hat{H},\hat{J}^+]=U\hat{J}^+\).
Therefore, \(\hat{J}^+\) is a raising
operator for \(\hat{H}\) and \(j,m_j\) are good quantum numbers. One can
also show that \[\begin{aligned}
&[J^z,J^\pm]=\pm J^\pm \\
&[J^+,J^-]=2J^z,
\end{aligned}\] which means that \(\hat{J}\) acts like an angular momentum
operator just like \(\hat{S}\) does.
Also, \[E(m_j)=E(-j)+(m_j+J)U\] You
will examine this on the homework and also see in the next lecture.
\(m_j\) governs the number of particles
as \(m_j\) increases by 1, the number
of particles increases by 2 as we have added a pair with \(J^+\).
When \(U=0\), use the momentum space
representation and bath tub principle to fill in the noninteracting
levels (always a metal).
When \(U\rightarrow\infty\), use the
real space representation. There is no double occupancy and at
half-filling there is one electron per site which are frozen and cannot
move (insulator). So we will have a metal-insulator phase transition as
a function of \(U\).
For \(d=1\), \(U_{mit}=0^+\) and for \(d\rightarrow\infty\), \(U_{mit}\approx\) bandwidth.