Consider the Hubbard model on a 4-site ring. \[\hat{H}=\sum_{ij\sigma}t_{ij}\hat{c}_{i\sigma}^\dagger
\hat{c}_{j\sigma}^{\phantom{\dagger}}+U\sum_{i=1}^4\hat{c}_{i\uparrow}^\dagger
\hat{c}_{i\uparrow}^{\phantom{\dagger}}\hat{c}_{i\downarrow}^\dagger
\hat{c}_{i\downarrow}^{\phantom{\dagger}}\] with hopping matrix
\[t_{ij}=\begin{pmatrix}\phantom{-}0 & -1
& \phantom{-}0 & -1 \\ -1 & \phantom{-}0 & -1 &
\phantom{-}0 \\ \phantom{-}0 & -1 & \phantom{-}0 & -1 \\ -1
& \phantom{-}0 & -1 & \phantom{-}0\end{pmatrix}.\]
Find the eigenvalues of \(\hat{H}\) and
plot them as a function of \(U/(4+U)\)
for positive and negative \(U\) for
\(N=0,1,2,3,4\).
\(\underline{N=0}\): There is only
one state here. Determine the energy and \(J,S\).
\(\underline{N=1}\): There are 8
states. Use the basis \[\begin{aligned}
&\left|1\right\rangle=1\uparrow+2\uparrow+3\uparrow+4\uparrow \\
&\left|2\right\rangle=1\uparrow-2\uparrow+3\uparrow-4\uparrow \\
&\left|3\right\rangle=1\uparrow+2\uparrow-3\uparrow-4\uparrow \\
&\left|4\right\rangle=1\uparrow-2\uparrow-3\uparrow+4\uparrow
\end{aligned}\] to calculate the energy (Note: \(\uparrow\) is degenerate with \(\downarrow\)). Determine \(S\) and \(J\) for each state as well.
\(\underline{N=2}\): There are 28
total states. 18 are \(S=1\) and 10 are
\(S=0\).
Calculate the energies for each of these \(S=1\) states \[\begin{aligned}
&\left|1\right\rangle=1\uparrow2\uparrow-2\uparrow3\uparrow+3\uparrow4\uparrow-4\uparrow1\uparrow
\\
&\left|2\right\rangle=1\uparrow2\uparrow+2\uparrow3\uparrow+3\uparrow4\uparrow+4\uparrow1\uparrow
\\
&\left|3\right\rangle=1\uparrow4\uparrow+2\uparrow3\uparrow \\
&\left|4\right\rangle=1\uparrow3\uparrow+2\uparrow4\uparrow \\
&\left|5\right\rangle=1\uparrow2\uparrow-3\uparrow4\uparrow \\
&\left|6\right\rangle=1\uparrow3\uparrow-2\uparrow4\uparrow
\end{aligned}\] which are each three-fold degenerate. Determine
\(J\) for each state as well.
Then, calculate the energy for each of these \(S=0\) states. \[\begin{aligned}
&\left|1\right\rangle=1\uparrow1\downarrow+2\uparrow2\downarrow+3\uparrow3\downarrow+4\uparrow4\downarrow
\\
&\left|2\right\rangle=1\uparrow2\downarrow+2\uparrow1\downarrow+1\uparrow4\downarrow+4\uparrow1\downarrow+2\uparrow3\downarrow+3\uparrow2\downarrow+3\uparrow4\downarrow+4\uparrow3\downarrow
\\
&\left|3\right\rangle=1\uparrow3\downarrow+2\uparrow4\downarrow+3\uparrow1\downarrow+4\uparrow2\downarrow
\\
&\left|4\right\rangle=1\uparrow2\downarrow-1\uparrow4\downarrow+2\uparrow1\downarrow-2\uparrow3\downarrow-3\uparrow2\downarrow+3\uparrow4\downarrow-4\uparrow1\downarrow+4\uparrow3\downarrow
\\
&\left|5\right\rangle=1\uparrow1\downarrow-2\uparrow2\downarrow+3\uparrow3\downarrow-4\uparrow4\downarrow
\\
&\left|6\right\rangle=1\uparrow3\downarrow-2\uparrow4\downarrow+3\uparrow1\downarrow-4\uparrow2\downarrow
\\
&\left|7\right\rangle=1\uparrow1\downarrow+2\uparrow2\downarrow-3\uparrow3\downarrow-4\uparrow4\downarrow
\\
&\left|8\right\rangle=1\uparrow1\downarrow-2\uparrow2\downarrow-3\uparrow3\downarrow+4\uparrow4\downarrow
\\
&\left|9\right\rangle=1\uparrow2\downarrow+2\uparrow1\downarrow-3\uparrow4\downarrow-4\uparrow3\downarrow
\\
&\left|10\right\rangle=1\uparrow4\downarrow+4\uparrow1\downarrow-2\uparrow3\downarrow-3\uparrow2\downarrow
\end{aligned}\] Be sure to determine \(J\) of these states as well.
\(\underline{N=3}\): There are 56
states, 4 are \(S=\frac{3}{2}\) and 20
are \(S=\frac{1}{2}\). The \(S=\frac{3}{2}\) states are: \[\begin{aligned}
&\left|1\right\rangle=-1\uparrow2\uparrow3\uparrow+1\uparrow2\uparrow4\uparrow-1\uparrow3\uparrow4\uparrow+2\uparrow3\uparrow4\uparrow
\\
&\left|2\right\rangle=1\uparrow2\uparrow3\uparrow+1\uparrow2\uparrow4\uparrow+1\uparrow3\uparrow4\uparrow+2\uparrow3\uparrow4\uparrow
\\
&\left|3\right\rangle=1\uparrow2\uparrow3\uparrow-1\uparrow2\uparrow4\uparrow-1\uparrow3\uparrow4\uparrow+2\uparrow3\uparrow4\uparrow
\\
&\left|4\right\rangle=1\uparrow2\uparrow3\uparrow+1\uparrow2\uparrow4\uparrow-1\uparrow3\uparrow4\uparrow-2\uparrow3\uparrow4\uparrow
\end{aligned}\] where each of these states is four-fold
degenerate. The \(S=\frac{1}{2}\)
states are: \[\begin{aligned}
&\left|1\right\rangle=-1\uparrow2\uparrow1\downarrow+1\uparrow2\uparrow2\downarrow-1\uparrow4\uparrow1\downarrow+1\uparrow4\uparrow4\downarrow-2\uparrow3\uparrow2\downarrow+2\uparrow3\uparrow3\downarrow-3\uparrow4\uparrow3\downarrow+3\uparrow4\uparrow4\downarrow
\\
&\left|2\right\rangle=-1\uparrow2\uparrow3\downarrow+1\uparrow2\uparrow4\downarrow1\uparrow4\uparrow2\downarrow-1\uparrow4\uparrow3\downarrow+2\uparrow3\uparrow1\downarrow-2\uparrow3\uparrow4\downarrow-3\uparrow4\uparrow1\downarrow+3\uparrow4\uparrow2\downarrow
\\
&\left|3\right\rangle=-1\uparrow3\uparrow1\downarrow+1\uparrow3\uparrow3\downarrow-2\uparrow4\uparrow2\downarrow+2\uparrow4\uparrow4\downarrow
\\
&\left|4\right\rangle=-1\uparrow2\uparrow1\downarrow+1\uparrow2\uparrow2\downarrow+1\uparrow4\uparrow1\downarrow-1\uparrow4\uparrow4\downarrow+2\uparrow3\uparrow2\downarrow2\uparrow3\uparrow3\downarrow-3\uparrow4\uparrow3\downarrow+3\uparrow4\uparrow4\downarrow
\\
&\left|5\right\rangle=-1\uparrow2\uparrow3\downarrow+1\uparrow2\uparrow4\downarrow-2(1\uparrow3\uparrow2\downarrow)+2(1\uparrow3\uparrow4\downarrow)-1\uparrow4\uparrow2\downarrow+1\uparrow4\uparrow3\downarrow-2\uparrow3\uparrow1\downarrow+2\uparrow3\uparrow4\downarrow
\\&-2(2\uparrow4\uparrow1\downarrow)+2(2\uparrow4\uparrow3\downarrow)-3\uparrow4\uparrow1\downarrow+3\uparrow4\uparrow2\downarrow
\\
&\left|6\right\rangle=1\uparrow2\uparrow1\downarrow+1\uparrow2\uparrow2\downarrow+1\uparrow4\uparrow1\downarrow+1\uparrow4\uparrow4\downarrow-2\uparrow3\uparrow2\downarrow-2\uparrow3\uparrow3\downarrow+3\uparrow4\uparrow3\downarrow+3\uparrow4\uparrow4\downarrow
\\
&\left|7\right\rangle=1\uparrow2\downarrow3\downarrow+1\uparrow2\uparrow4\downarrow+1\uparrow4\uparrow2\downarrow+1\uparrow4\uparrow3\downarrow-2\uparrow3\uparrow1\downarrow-2\uparrow3\uparrow4\downarrow+3\uparrow4\uparrow1\downarrow+3\uparrow4\uparrow2\downarrow
\\
&\left|8\right\rangle=1\uparrow3\uparrow1\downarrow-1\uparrow3\uparrow3\downarrow-2\uparrow4\uparrow2\downarrow+2\uparrow4\uparrow4\downarrow
\\
&\left|9\right\rangle=1\uparrow2\uparrow1\downarrow+1\uparrow2\uparrow2\downarrow-1\uparrow4\uparrow1\downarrow-1\uparrow4\uparrow4\downarrow+2\uparrow3\uparrow2\downarrow+2\uparrow3\uparrow3\downarrow+3\uparrow4\uparrow3\downarrow+3\uparrow4\uparrow4\downarrow
\\
&\left|10\right\rangle=1\uparrow2\uparrow3\uparrow+1\uparrow2\uparrow4\uparrow+2(1\uparrow3\uparrow2\downarrow)-2(1\uparrow3\uparrow4\downarrow)-1\uparrow4\uparrow2\downarrow-1\uparrow4\uparrow3\downarrow+2\uparrow3\uparrow1\downarrow+2\uparrow3\uparrow4\downarrow\\&-2(2\uparrow4\uparrow1\downarrow)+2(2\uparrow4\uparrow3\downarrow)+3\uparrow4\uparrow1\downarrow+3\uparrow4\uparrow2\downarrow
\\
&\left|11\right\rangle=1\uparrow2\uparrow1\downarrow-1\uparrow2\uparrow2\downarrow-3\uparrow4\uparrow3\downarrow+3\uparrow4\uparrow4\downarrow
\\
&\left|12\right\rangle=1\uparrow4\uparrow1\downarrow+1\uparrow4\uparrow4\downarrow+2\uparrow3\uparrow2\downarrow+2\uparrow3\uparrow3\downarrow
\\
&\left|13\right\rangle=1\uparrow2\uparrow3\downarrow-1\uparrow2\uparrow4\downarrow+2(1\uparrow3\uparrow2\downarrow)+2(1\uparrow3\uparrow4\downarrow)+1\uparrow4\uparrow2\downarrow+1\uparrow4\uparrow3\downarrow+2\uparrow3\uparrow1\downarrow
+2\uparrow3\uparrow4\downarrow\\&+2(2\uparrow4\uparrow1\downarrow)+2(2\uparrow4\uparrow3\downarrow)-3\uparrow4\uparrow1\downarrow+3\uparrow4\uparrow2\downarrow
\\
&\left|14\right\rangle=1\uparrow2\uparrow3\downarrow-1\uparrow2\uparrow4\downarrow-1\uparrow4\uparrow2\downarrow-1\uparrow4\uparrow3\downarrow-2\uparrow3\uparrow1\downarrow-2\uparrow3\uparrow4\downarrow-3\uparrow4\uparrow1\downarrow+3\uparrow4\uparrow2\downarrow
\\
&\left|15\right\rangle=1\uparrow3\uparrow1\downarrow+1\uparrow3\uparrow3\downarrow+2\uparrow4\uparrow2\downarrow+2\uparrow4\uparrow4\downarrow
\end{aligned}\] where states \(\left|11\right\rangle\) through \(\left|15\right\rangle\) are four-fold
degenerate. Do NOT compute \(J\)
for these states. (you can infer the \(J\) of the eigenvalues by comparing to the
\(S=\frac{1}{2}\),\(N=1\) states and using your knowledge of
the \(\hat{J}_+\) operator.)
\(\underline{N=4}\): (EXTRA CREDIT -
it does not need to be completed)
\(N=4\) has 70 states: 1 is \(S=2\) (5-fold), 15 are \(S=1\) (3-fold) and \(20\) are \(S=0\).
\(S=2:\) \[\left|1\right\rangle=1\uparrow2\uparrow3\uparrow4\uparrow\]
\(S=1:\) \[\begin{aligned}
&\left|1\right\rangle=-1\uparrow 2\uparrow 3\uparrow
1\downarrow+1\uparrow 2\uparrow 3\uparrow 3\downarrow+1\uparrow
2\uparrow 4\uparrow 2\downarrow-1\uparrow 2\uparrow 4\uparrow
4\downarrow+1\uparrow 3\uparrow 4\uparrow 1\downarrow-1\uparrow
3\uparrow 4\uparrow 3\downarrow\\&-2\uparrow 3\uparrow 4\uparrow
2\downarrow+2\uparrow 3\uparrow 4\uparrow 4\downarrow \\
&\left|2\right\rangle=-1\uparrow 2\uparrow 3\uparrow
1\downarrow-1\uparrow 2\uparrow 3\uparrow 3\downarrow+1\uparrow
2\uparrow 4\uparrow 2\downarrow+1\uparrow 2\uparrow 4\uparrow
4\downarrow-1\uparrow 3\uparrow 4\uparrow 1\downarrow-1\uparrow
3\uparrow 4\uparrow 3\downarrow\\&+2\uparrow 3\uparrow 4\uparrow
2\downarrow+2\uparrow 3\uparrow 4\uparrow 4\downarrow \\
&\left|3\right\rangle=-1\uparrow 2\uparrow 3\uparrow
2\downarrow+1\uparrow 2\uparrow 4\uparrow 1\downarrow-1\uparrow
3\uparrow 4\uparrow 4\downarrow+2\uparrow 3\uparrow 4\uparrow
3\downarrow \\
&\left|4\right\rangle=1\uparrow 2\uparrow 3\uparrow
1\downarrow-1\uparrow 2\uparrow 3\uparrow 3\downarrow+1\uparrow
2\uparrow 4\uparrow 2\downarrow+1\uparrow 2\uparrow 4\uparrow
4\downarrow+1\uparrow 3\uparrow 4\uparrow 1\downarrow+1\uparrow
3\uparrow 4\uparrow 3\downarrow\\&+2\uparrow 3\uparrow 4\uparrow
2\downarrow+2\uparrow 3\uparrow 4\uparrow 4\downarrow \\
&\left|5\right\rangle=1\uparrow 2\uparrow 3\uparrow
1\downarrow+1\uparrow 2\uparrow 3\uparrow 3\downarrow+1\uparrow
2\uparrow 4\uparrow 2\downarrow+1\uparrow 2\uparrow 4\uparrow
4\downarrow+1\uparrow 3\uparrow 4\uparrow 1\downarrow+1\uparrow
3\uparrow 4\uparrow 3\downarrow\\&+2\uparrow 3\uparrow 4\uparrow
2\downarrow+2\uparrow 3\uparrow 4\uparrow 4\downarrow \\
&\left|6\right\rangle=1\uparrow 2\uparrow 3\uparrow
2\downarrow+1\uparrow 2\uparrow 4\uparrow 1\downarrow+1\uparrow
3\uparrow 4\uparrow 4\downarrow+2\uparrow 3\uparrow 4\uparrow
3\downarrow \\
&\left|7\right\rangle=1\uparrow 2\uparrow 3\uparrow
4\downarrow+1\uparrow 2\uparrow 4\uparrow 3\downarrow+1\uparrow
3\uparrow 4\uparrow 2\downarrow+2\uparrow 3\uparrow 4\uparrow
1\downarrow \\
&\left|8\right\rangle=1\uparrow 2\uparrow 3\uparrow
1\downarrow-1\uparrow 2\uparrow 4\uparrow 2\downarrow-1\uparrow
3\uparrow 4\uparrow 3\downarrow+2\uparrow 3\uparrow 4\uparrow
4\downarrow \\
&\left|9\right\rangle=1\uparrow 2\uparrow 3\uparrow
3\downarrow-1\uparrow 2\uparrow 4\uparrow 4\downarrow-1\uparrow
3\uparrow 4\uparrow 1\downarrow+2\uparrow 3\uparrow 4\uparrow
2\downarrow \\
&\left|10\right\rangle=1\uparrow 2\uparrow 3\uparrow
2\downarrow-1\uparrow 2\uparrow 4\uparrow 1\downarrow-1\uparrow
3\uparrow 4\uparrow 4\downarrow+2\uparrow 3\uparrow 4\uparrow
3\downarrow \\
&\left|11\right\rangle=-1\uparrow 2\uparrow 3\uparrow
4\downarrow+1\uparrow 2\uparrow 4\uparrow 3\downarrow+1\uparrow
3\uparrow 4\uparrow 2\downarrow-2\uparrow 3\uparrow 4\uparrow
1\downarrow
\end{aligned}\] States \(\left|8\right\rangle-\left|11\right\rangle\)
are six-fold degenerate, \(\left|1\right\rangle-\left|7\right\rangle\)
are three-fold degenerate.
\(S=0:\) \[\begin{aligned}
&\left|1\right\rangle=1\uparrow 2\uparrow 1\downarrow
2\downarrow+1\uparrow 4\uparrow 1\downarrow 4\downarrow+2\uparrow
3\uparrow 2\downarrow 3\downarrow+3\uparrow 4\uparrow 3\downarrow
4\downarrow \\
&\left|2\right\rangle=1\uparrow 2\uparrow 2\downarrow
3\downarrow-1\uparrow 2\uparrow 2\downarrow 4\downarrow+1\uparrow
3\uparrow 1\downarrow 2\downarrow+1\uparrow 3\uparrow 1\downarrow
4\downarrow+1\uparrow 3\uparrow 2\downarrow 3\downarrow-1\uparrow
3\uparrow 3\downarrow 4\downarrow\\&+1\uparrow 4\uparrow 1\downarrow
3\downarrow+1\uparrow 4\uparrow 2\downarrow 4\downarrow+2\uparrow
3\uparrow 1\downarrow 3\downarrow+2\uparrow 3\uparrow 2\downarrow
4\downarrow-2\uparrow 4\uparrow 1\downarrow 2\downarrow+2\uparrow
4\uparrow 1\downarrow 4\downarrow+2\uparrow 4\uparrow 2\downarrow
3\downarrow\\&+2\uparrow 4\uparrow 3\downarrow 4\downarrow-3\uparrow
4\uparrow 1\downarrow 3\downarrow+3\uparrow 4\uparrow 2\downarrow
4\downarrow \\
&\left|3\right\rangle=-1\uparrow 2\uparrow 1\downarrow
4\downarrow+1\uparrow 2\uparrow 2\downarrow 3\downarrow-1\uparrow
4\uparrow 1\downarrow 2\downarrow-1\uparrow 4\uparrow 3\downarrow
4\downarrow+2\uparrow 3\uparrow 1\downarrow 2\downarrow+2\uparrow
3\uparrow 3\downarrow 4\downarrow\\&-3\uparrow 4\uparrow 1\downarrow
4\downarrow+3\uparrow 4\uparrow 2\downarrow 3\downarrow \\
&\left|4\right\rangle=1\uparrow 2\uparrow 3\downarrow
4\downarrow-1\uparrow 4\uparrow 2\downarrow 3\downarrow-2\uparrow
3\uparrow 1\downarrow 4\downarrow+3\uparrow 4\uparrow 1\downarrow
2\downarrow \\
&\left|5\right\rangle=1\uparrow 3\uparrow 1\downarrow
3\downarrow+2\uparrow 4\uparrow 2\downarrow 4\downarrow \\
&\left|6\right\rangle=1\uparrow 2\uparrow 1\downarrow
2\downarrow-1\uparrow 4\uparrow 1\downarrow 4\downarrow-2\uparrow
3\uparrow 2\downarrow 3\downarrow+3\uparrow 4\uparrow 3\downarrow
4\downarrow \\
&\left|7\right\rangle=1\uparrow 2\uparrow 1\downarrow
3\downarrow-1\uparrow 2\uparrow 2\downarrow 4\downarrow+1\uparrow
3\uparrow 1\downarrow 2\downarrow-1\uparrow 3\uparrow 1\downarrow
4\downarrow-1\uparrow 3\uparrow 2\downarrow 3\downarrow-1\uparrow
3\uparrow 3\downarrow 4\downarrow\\&-1\uparrow 4\uparrow 1\downarrow
3\downarrow-1\uparrow 4\uparrow 2\downarrow 4\downarrow-2\uparrow
3\uparrow 1\downarrow 3\downarrow-2\uparrow 3\uparrow 2\downarrow
4\downarrow-2\uparrow 4\uparrow 1\downarrow 2\downarrow-2\uparrow
4\uparrow 1\downarrow 4\downarrow-2\uparrow 4\uparrow 2\downarrow
3\downarrow\\&+2\uparrow 4\uparrow 3\downarrow 4\downarrow-3\uparrow
4\uparrow 1\downarrow 3\downarrow+3\uparrow 4\uparrow 2\downarrow
4\downarrow \\
&\left|8\right\rangle=1\uparrow 2\uparrow 3\downarrow
4\downarrow+2(1\uparrow 3\uparrow 2\downarrow 4\downarrow)+1\uparrow
4\uparrow 2\downarrow 3\downarrow+2\uparrow 3\uparrow 1\downarrow
4\downarrow+2(2\uparrow 4\uparrow 1\downarrow 3\downarrow)+3\uparrow
4\uparrow 1\downarrow 2\downarrow \\
&\left|9\right\rangle=-1\uparrow 2\uparrow 1\downarrow
3\downarrow-1\uparrow 2\uparrow 2\downarrow 4\downarrow-1\uparrow
3\uparrow 1\downarrow 2\downarrow-1\uparrow 3\uparrow 1\downarrow
4\downarrow-1\uparrow 3\uparrow 2\downarrow 3\downarrow+1\uparrow
3\uparrow 3\downarrow 4\downarrow\\&-1\uparrow 4\uparrow 1\downarrow
3\downarrow+1\uparrow 4\uparrow 2\downarrow 4\downarrow-2\uparrow
3\uparrow 1\downarrow 3\downarrow+2\uparrow 3\uparrow 2\downarrow
4\downarrow-2\uparrow 4\uparrow 1\downarrow 2\downarrow+2\uparrow
4\uparrow 1\downarrow 4\downarrow+2\uparrow 4\uparrow 2\downarrow
3\downarrow\\&+2\uparrow 4\uparrow 3\downarrow 4\downarrow+3\uparrow
4\uparrow 1\downarrow 3\downarrow+3\uparrow 4\uparrow 2\downarrow
4\downarrow \\
&\left|10\right\rangle=-1\uparrow 2\uparrow 1\downarrow
4\downarrow-1\uparrow 2\uparrow 2\downarrow 3\downarrow-1\uparrow
4\uparrow 1\downarrow 2\downarrow+1\uparrow 4\uparrow 3\downarrow
4\downarrow-2\uparrow 3\uparrow 1\downarrow 2\downarrow+2\uparrow
3\uparrow 3\downarrow 4\downarrow\\&+3\uparrow 4\uparrow 1\downarrow
4\downarrow+3\uparrow 4\uparrow 2\downarrow 3\downarrow \\
&\left|11\right\rangle=-1\uparrow 3\uparrow 1\downarrow
3\downarrow+2\uparrow 4\uparrow 2\downarrow 4\downarrow \\
&\left|12\right\rangle=-1\uparrow 2\uparrow 1\downarrow
3\downarrow-1\uparrow 2\uparrow 2\downarrow 4\downarrow-1\uparrow
3\uparrow 1\downarrow 2\downarrow+1\uparrow 3\uparrow 1\downarrow
4\downarrow+1\uparrow 3\uparrow 2\downarrow 3\downarrow+1\uparrow
3\uparrow 3\downarrow 4\downarrow\\&+1\uparrow 4\uparrow 1\downarrow
3\downarrow-1\uparrow 4\uparrow 2\downarrow 4\downarrow+2\uparrow
3\uparrow 1\downarrow 3\downarrow-2\uparrow 3\uparrow 2\downarrow
4\downarrow-2\uparrow 4\uparrow 1\downarrow 2\downarrow-2\uparrow
4\uparrow 1\downarrow 4\downarrow-2\uparrow 4\uparrow 2\downarrow
3\downarrow\\&+2\uparrow 4\uparrow 3\downarrow 4\downarrow+3\uparrow
4\uparrow 1\downarrow 3\downarrow+3\uparrow 4\uparrow 2\downarrow
4\downarrow\\
&\left|13\right\rangle=-1\uparrow 2\uparrow 1\downarrow
2\downarrow+3\uparrow 4\uparrow 3\downarrow 4\downarrow \\
&\left|14\right\rangle=-1\uparrow 2\uparrow 1\downarrow
3\downarrow+1\uparrow 2\uparrow 2\downarrow 4\downarrow-1\uparrow
3\uparrow 1\downarrow 2\downarrow-1\uparrow 3\uparrow 3\downarrow
4\downarrow+2\uparrow 4\uparrow 1\downarrow 2\downarrow+2\uparrow
4\uparrow 3\downarrow 4\downarrow\\&-3\uparrow 4\uparrow 1\downarrow
3\downarrow+3\uparrow 4\uparrow 2\downarrow 4\downarrow \\
&\left|15\right\rangle=1\uparrow 3\uparrow 1\downarrow
4\downarrow-1\uparrow 3\uparrow 2\downarrow 3\downarrow+1\uparrow
4\uparrow 1\downarrow 3\downarrow-1\uparrow 4\uparrow 2\downarrow
4\downarrow-2\uparrow 3\uparrow 1\downarrow 3\downarrow+2\uparrow
3\uparrow 2\downarrow 4\downarrow\\&-2\uparrow 4\uparrow 1\downarrow
4\downarrow+2\uparrow 4\uparrow 2\downarrow 3\downarrow \\
&\left|16\right\rangle=1\uparrow 2\uparrow 1\downarrow
4\downarrow-1\uparrow 2\uparrow 2\downarrow 3\downarrow+1\uparrow
4\uparrow 1\downarrow 2\downarrow-1\uparrow 4\uparrow 3\downarrow
4\downarrow-2\uparrow 3\uparrow 1\downarrow 2\downarrow+2\uparrow
3\uparrow 3\downarrow 4\downarrow\\&-3\uparrow 4\uparrow 1\downarrow
4\downarrow+3\uparrow 4\uparrow 2\downarrow 3\downarrow
\end{aligned}\] States \(\left|13\right\rangle-\left|16\right\rangle\)
are two-fold degenerate. Find all energy levels.
Once you have calculated and plotted all energy levels for each
filling, be sure to state the quantum numbers (\(S\) and \(J\)) of the ground state. Do the quantum
numbers ever change due to a level crossing?
You should normalize each state and calculate \(\hat{H}\).
You should find the biggest matrix you have to diagonalize is a \(5\times5\) matrix. Mathematica should make
this calculation simple to do.
Recall \(\left\langle
m\middle|\hat{H}\middle|n\right\rangle=\left\langle
n\middle|\hat{H}\middle|m\right\rangle\) as well to reduce
labor.
You will need to use an efficient scheme to calculate the nonzero matrix
elements.
I have ordered the states by spatial symmetry (generated from a computer
program) so you should find states of \(\hat{H}\) where \(H_{nm}\ne0\) lie close to each other in
this list.
Plot energies \(E/(1+|U|)\) versus
\(U/(4+|U|)\) so they can be included
in one plot each for \(N=2,3,4\).
Otherwise, the energies will get too large to appear on your plot.