1. Are these state eigenstates? How can i calculate their energy?
Act Hamiltonian on the states $\vert 00,00\rangle $ and $\vert 11,11\rangle $. The 4 number denotes occupation for site 1 spin-up, site 1 spin-down, site 2 spin-up, and site 2 spin-down:
$$
H=\sum_{\sigma=\uparrow,\downarrow}[\epsilon_1 c_{1\sigma}^\dagger\,c_{1\sigma}
+ \epsilon_2 c_{2\sigma}^\dagger\,c_{2\sigma}
+ w(c_{1\sigma}^\dagger\,c_{2\sigma}+c_{2\sigma}^\dagger\,c_{1\sigma})].
$$
$$
H \vert 00, 00\rangle = 0 \vert 00, 00\rangle.
$$
The state $\vert 00, 00\rangle$ is a eigen state of $H$ with eigen value $0$.
Next, examine $H \vert 11, 11\rangle$:
\begin{align}
\epsilon_1 c^\dagger_{1\uparrow}c_{1\uparrow} \vert 11, 11\rangle & = \epsilon_1 \vert 11, 11\rangle \\
\epsilon_2 c^\dagger_{2\uparrow}c_{2\uparrow} \vert 11, 11\rangle &= \epsilon_2 \vert 11, 11\rangle\\
w c^\dagger_{1\uparrow}c_{2\uparrow} \vert 11, 11\rangle &=w c^\dagger_{1\uparrow}
\vert 11, 01\rangle = 0 \vert 11, 01\rangle = 0\vert 11, 11\rangle;\\
w c^\dagger_{2\uparrow}c_{1\uparrow} \vert 11, 11\rangle &=w c^\dagger_{2\uparrow}
\vert 01, 11\rangle = 0 \vert 01, 11\rangle = 0\vert 11, 11\rangle;
\end{align}
The last two terms vanish, due to exclusion of fermion $c^\dagger\vert 1\rangle = 0 $.
Adding the cases for spin-down, we have
$$
H \vert 11, 11\rangle = (\epsilon_1+\epsilon_1+ \epsilon_2+\epsilon_2+0+0+0+0) \vert 11, 11\rangle =2 (\epsilon_1+ \epsilon_2) \vert 11, 11\rangle.
$$
The state $\vert 11, 11\rangle$ is eigen state of eigen value $2 (\epsilon_1+ \epsilon_2)$.
The basis changed
Writing the Hamitonian in matrix form:
$$
\sum_{\sigma=\uparrow,\downarrow}
\begin{bmatrix} c_{1\sigma}^\dagger & c_{2\sigma}^\dagger \\ \end{bmatrix}
\begin{bmatrix}
\epsilon_1 & w \\
w & \epsilon_2
\end{bmatrix}
\begin{bmatrix} c_{1\sigma} \\ c_{2\sigma} \end{bmatrix}
$$
To diagonlize the Hamitonian, by an unitary matrix
$$
U =\begin{bmatrix}
\cos \frac{\theta}{2} & \sin \frac{\theta}{2} \\
-\sin \frac{\theta}{2} & \cos \frac{\theta}{2}
\end{bmatrix};\,\,\,
U^T =\begin{bmatrix}
\cos \frac{\theta}{2} & -\sin \frac{\theta}{2} \\
\sin \frac{\theta}{2} & \cos \frac{\theta}{2}
\end{bmatrix}
$$
where $\tan\theta = \frac{2 \,w}{\epsilon_2 - \epsilon_1}$. The Hamiltonian will become diagonal by $U^T H U$. In the following process.
$$
\sum_{\sigma=\uparrow,\downarrow}
\begin{bmatrix} c_{1\sigma}^\dagger & c_{2\sigma}^\dagger \\ \end{bmatrix}
\mathbf{U} \mathbf{U^T}
\begin{bmatrix}
\epsilon_1 & w \\
w & \epsilon_2
\end{bmatrix}
\mathbf{U} \mathbf{U^T}
\begin{bmatrix} c_{1\sigma} \\ c_{2\sigma} \end{bmatrix}
=\sum_{\sigma=\uparrow,\downarrow}
\begin{bmatrix} a_{1\sigma}^\dagger & a_{2\sigma}^\dagger \\ \end{bmatrix}
\begin{bmatrix}
\lambda_1 & 0 \\
0 & \lambda_2
\end{bmatrix}
\begin{bmatrix} a_{1\sigma} \\ a_{2\sigma} \end{bmatrix}
$$
where $\lambda_{1, 2} = \frac{\epsilon_2+\epsilon_1}{2}\mp \sqrt{\left(\frac{\epsilon_2-\epsilon_1}{2}\right)^2 + w^2}$. The new bases:
$$ \tag{1}
\begin{bmatrix} a_{1\sigma} \\ a_{2\sigma} \end{bmatrix} =
\begin{bmatrix}
\cos \frac{\theta}{2} & -\sin \frac{\theta}{2} \\
\sin \frac{\theta}{2} & \cos \frac{\theta}{2}
\end{bmatrix}
\begin{bmatrix} c_{1\sigma} \\ c_{2\sigma} \end{bmatrix}
$$
The Hamiltonian with new bases is diagonal
$$
H=\sum_{\sigma=\uparrow,\downarrow}[\lambda_1 a_{1\sigma}^\dagger\,a_{1\sigma}
+ \lambda_2 a_{2\sigma}^\dagger\,a_{2\sigma}.
$$
Show in the equation (1), the new basis $a_1$ and $a_2$ mixed the site $1$ and $2$, but didn't mixed the spin states.
\begin{align}
a_{1\sigma} & = \cos\frac{\theta}{2}\, c_{1\sigma} - \sin\frac{\theta}{2} \, c_{2\sigma};\\
a_{2\sigma} & = \sin\frac{\theta}{2}\, c_{1\sigma} + \cos\frac{\theta}{2} \, c_{2\sigma};
\end{align}
Appendix
- Diagonal of 2x2 symmetry matrix, by finding its eigen vectors:
$$ \tag{A1}
\mathbf{A} =
\begin{bmatrix}
a & c \\
c & b
\end{bmatrix};
\,\,\,\text{ eigen vector }\,\,
\begin{bmatrix}
a & c \\
c & b
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
= \lambda
\begin{bmatrix}
x \\
y
\end{bmatrix};
$$
First find its eigen values, by solving determinant equation $\Vert \mathbf{A} - \lambda \mathbf{I} \Vert = 0$
The quadratic equation render two roots (eigenvalues) for $\lambda$
$$\tag{A2}
\lambda_{+, -} = \frac{a+b}{2}\pm \sqrt{\left(\frac{b-a}{2}\right)^2 + c^2}
$$
Substitue Eq. (A2) into Eq. (A1) to find the ratio between $x:y$
\begin{align}
& (a-\lambda) x + c y = 0;\\
\text{Let } & \,\,\,x = c; \,\,\, y = \lambda -a; \\
\text{Normalization } &\,\,\, N = \sqrt{x^2 + y^2} = 4 \cos^2\left(\frac{\theta}{2}\right)\\
\text{Define } & \,\,\,\tan\theta = \frac{c}{(b-a)/2}.
\end{align}
The eigen vector
$$
\begin{bmatrix}
x \\
y
\end{bmatrix};
= \begin{bmatrix}
\cos \frac{\theta}{2} \\
-\sin \frac{\theta}{2}
\end{bmatrix};
\,\,\,\text{ and }\,\,\,
\begin{bmatrix}
\sin \frac{\theta}{2} \\
\cos \frac{\theta}{2}
\end{bmatrix};
$$
The two eigen vectors corresponding to $\lambda_{\pm}$ are mutual orthogonal
These tqo eigen vectors form the column of the unitary matrix $\mathbf{U}$
$$
\mathbf{U} =
\begin{bmatrix}
\cos \frac{\theta}{2} & \sin \frac{\theta}{2} \\
-\sin \frac{\theta}{2} & \cos \frac{\theta}{2}
\end{bmatrix}
$$
- Non-eigen state
\begin{align}
H &\vert 11,10\rangle \\
= & \sum_{\sigma=\uparrow,\downarrow}[\epsilon_1 c_{1\sigma}^\dagger\,c_{1\sigma}
+ \epsilon_2 c_{2\sigma}^\dagger\,c_{2\sigma}
+ w(c_{1\sigma}^\dagger\,c_{2\sigma}+c_{2\sigma}^\dagger\,c_{1\sigma})]
\vert 11,10\rangle
\end{align}
The terms:
\begin{align}
c_{1\sigma}^\dagger\,c_{1\sigma} \,\vert 11,10\rangle &= 1\vert 11,10\rangle;\text{ for both spins;} \\
c_{2\uparrow}^\dagger\,c_{2\uparrow}\, \vert 11,10\rangle &= 1\vert11,10\rangle;\\
c_{2\downarrow}^\dagger\,c_{2\downarrow}\, \vert 11,10\rangle &= 0\vert11,10\rangle;\\
c_{1\sigma}^\dagger\,c_{2\sigma}\, \vert 11,10\rangle &= 0; \text{ due to the term } c_{1\sigma}^\dagger\\
c_{2\downarrow}^\dagger\,c_{1\downarrow})\, \vert 11,10\rangle &= 1 \vert 10,11\rangle;
\end{align}
Put the above results all together:
\begin{align}
H \vert 11,10\rangle =&
\,\left( 2\epsilon_1 + \epsilon_2 +0+0+0+0\right)\,\vert 11,10\rangle
+ \,w \, \vert 10,11\rangle\\
= & \,\left( 2\epsilon_1 + \epsilon_2 \right)\,\vert 11,10\rangle
+ \,w \, \vert 10,11\rangle
\end{align}
