Diagonalization of a Hamiltonian in Second Quantization I have the following Hamiltonian of a fermionic two particle system.
$H =2 \epsilon_m f^\dagger f + \epsilon_d d^\dagger d + t df +t f^\dagger d^\dagger + t f^\dagger d + td^\dagger f $
t $\in I R$
Now I want to diagonalize it.
$\rightarrow H = \epsilon_a a^\dagger a + \epsilon_b b^\dagger b$
What is the best way to do this?
My attempt was to rewrite the Hamiltonian with the first quantization:
$$
 \begin{pmatrix}
        -\epsilon_m - \epsilon_d/2 & 0 & 0 & t\\
        0 &  -\epsilon_m + \epsilon_d/2 & t & 0 \\
        0 & t &  \epsilon_m - \epsilon_d/2 &\\
        t & 0 & 0 &  \epsilon_m + \epsilon_d/2
        \end{pmatrix}
$$
How do the eigenvalues of this matrix correspond with $\epsilon_a$ and $\epsilon_b$? The eigenvectors should also somehow correspond to $a^\dagger$ and $b^\dagger$. But how can I figure out the coefficients of 
$a^\dagger = r d^\dagger + s f^\dagger$?
 A: The energies and states you get are correct. I will point out something else that may be useful. I will use the following basis: $$|00> = |0>;\quad|01> = d^\dagger|0>; |10> = f^\dagger|0>;\quad |11>=f^\dagger d^\dagger |0>.$$
The Hamiltonian at hand conserves parity, so you can write the Fock-space matrix in a more convenient way:
$$H=\begin{array}{c|cc|cc} \,&|00>&|11>& |01> & |10> \\ \hline <00| & -(\epsilon_m + \epsilon_d/2) & t & 0 & 0\\ <11| & t & (\epsilon_m + \epsilon_d/2) & 0 & 0 \\ \hline <01|&0 & 0& -(\epsilon_m-\epsilon_d/2) & t \\ <10| & 0 & 0 & t & \epsilon_m-\epsilon_d/2   \end{array},$$
which is block-diagonal, so you only have to deal with subspaces of size $2\times 2$. Let's label the parities of the subspaces as $P=0$ and $P=1$, respectively, as $P=(n_f+n_d)\text{mod}2$. The eigenvalues are
$$E_{0,\pm}=\pm\frac{\sqrt{(\epsilon_d+2\epsilon_m)^2+4t^2}}{2};\quad E_{1,\pm}=\pm\frac{\sqrt{(\epsilon_d-2\epsilon_m)^2+4t^2}}{2},$$ where the index $0$ or $1$ refers to the parity. You already got the normalized eigenstates, and all you need to do is reorganize them to put them in this particular ordering. To make things quicker, I'll simply write them as follows: $$|0\pm>=\alpha_0^{\pm}|00>+\beta_0^{\pm}|11>;\quad |1\pm>=\alpha_1^{\pm}|01>+\beta_1^{\pm}|10>,$$ with $\alpha_P^{\pm}$ and $\beta_{P}^{\pm}$ the coefficients you found. You can associate to these the operators $A_{0,\pm}^\dagger|0> = |0\pm>$ and $A_{1,\pm}^\dagger|0>=|1\pm>$.
Now, the operators $A_{1,\pm}$ have fermion commutation relations, because they are single-particle operators: $$\{A_{1,\sigma}^\dagger,A_{1,\sigma'}\}=\{\alpha_{1,\sigma}d^\dagger+\beta_{1,\sigma}f^\dagger,\alpha_{1,\sigma'}^*d+\beta_{1,\sigma'}^*f\}=\alpha_{1,\sigma}\alpha_{1,\sigma'}^*+\beta_{1,\sigma}\beta_{1,\sigma'}^*=\delta_{\sigma,\sigma'},$$
by definition (the coefficients meet this condition, because they are the coefficients of mutually orthonormal vectors). For the $P=0$ subspace, what you have are actually bosons, but not normal bosons, because they mix different parts of Fock space ($0$ particle and $2$ particle). As a consequence, you cannot get normal boson commutation relations:
$$\begin{split}[A_{0,\sigma}^\dagger,A_{0,\sigma'}]=&[\alpha_{0,\sigma}+\beta_{0,\sigma}f^\dagger d^\dagger,\alpha_{0,\sigma'}^*+\beta_{0,\sigma'}^*d f]\\ =& \beta_{0\sigma}\beta_{0\sigma'}^*\left(d^\dagger d + f^\dagger f -1 \right).\end{split}$$  What you can do, on the other hand, is define operators of the form $B=\gamma d + \delta f^\dagger$ and $C = \lambda d^\dagger + \mu f$, such that you can make products of the form $$B A_{1,\sigma}^\dagger = \gamma \alpha_{1,\sigma} (1-n_d) + \gamma \beta_{1,\sigma} d f^\dagger + \delta \alpha_{1,\sigma} f^\dagger d^\dagger + \delta \beta_{1,\sigma} (1-n_f).$$ Applied to the vacuum, the terms $d f^\dagger$, $n_f$ and $n_d$ vanish, and you get $$B A^\dagger_{1,\sigma}|0> = (\gamma \alpha_{1\sigma} + \delta \beta_{1\sigma})|00> + \delta \alpha_{1\sigma}|11>.$$ The coefficients $\gamma$ and $\delta$ can be fixed such that $$(\gamma \alpha_{1\sigma} + \delta \beta_{1\sigma})=\alpha_{0,\sigma'}\quad;\quad \delta \alpha_{1\sigma} = \beta_{0,\sigma'}.$$ The weird fermion $B$ satisfies $\{B,B^\dagger\}=|\gamma|^2+|\delta|^2$.
I'll leave the rest for you to work out. BTW, the Hamiltonian you are solving represents the coupling between a Majorana fermion $\gamma_1$, and an ordinary fermion $(d^\dagger, d)$, in the form $H=\epsilon_d d^\dagger d + \gamma_1 (d^\dagger - d)$. The Majorana fermion can be written in terms of the auxiliary fermion $(f^\dagger,f)$ as $\gamma_1 = f^\dagger + f$. You can define an independent Majorana $i\gamma_2 = f^\dagger - f$, and together they satisfy the Clifford algebra $\{\gamma_i, \gamma_j \}=2\delta_{ij}$. If you have further questions, or want to learn more, you can look up papers about Majoranas coupled to normal fermions. I hope this was of some use to you.
