Diagonalizing a Hamiltonian I am currently self studying many-body physics from Introduction to Many Body Physics and am trying to derive (3.153) from (3.152). The Bogoliubov transformation for fermions $a_1$ and $a_2$ are given by the equations:
$c_1 = ua_1 + va_2^{+}$ and $c_2^+ = -va_1 + ua_2^+$, where $u,v \in \mathbb{R}$ and $u^2 + v^2 = 1$. From this one can see that that
$a_1 = uc_1 - vc_2^{+}$ and $a_2^+ = vc_1 + uc_2^+$. From this we see that $a_1^+ = uc_1^+ - vc_2$ and $a_2 = vc_1^+ + uc_2$.
Now my goal is to show that the Hamiltonian (3.152) $H = \epsilon(a_1^+a_1 - a_2a_2^+) + \Delta(a_1^+a_2^+ + h.c)$ diagonalizes to (3.153)$H = \sqrt{\epsilon^2 + \Delta^2}(c_1^+c_1 + c_2^+c_2 - 1)$.
where h.c. means Hermitian complex
Using the formulae from the previous system of linear equations I found that $a_1^+ a_1 - a_2a_2^+ = (u^2 - v^2)(c_1^+ c_1 - c_2c_2^+) - 2uv(c_1^+ c_2^+ + c_2c_1)$
$a_1^+a_2^+ = uvc_1^+c_1 + u^2 c_1^+c_2^+ - v^2c_2c_1 - uvc_2c_2^+$
However, I cannot see how to go from one Hamiltonian to another.
My Question: Am I doing something wrong here? Can anyone show me how to diagonalize this Hamiltonian?
 A: We can write the initial Hamiltonian in the Fock basis of the $a_i$ operators. Realizing that $a_i=\vphantom{a}_i|0\rangle\langle 1|_i$ lets us write $a_1^+ a_1=|1\rangle\langle1|\otimes\mathbb{I}$, $a_2 a_2^+=\mathbb{I}\otimes|0\rangle\langle0|$,  $a_1^+ a_2^+=|1\rangle\langle0|\otimes|1\rangle\langle0|$, etc., we can write the Hamiltonian in the $|i\rangle_1\otimes|j\rangle_2$ basis as
$$H=\begin{pmatrix}-\epsilon&0&0&\Delta\\0&0&0&0\\0&0&0&0\\\Delta&0&0&\epsilon\end{pmatrix}.$$ This matrix is easy to diagonalize: there are two eigenvectors $|1\rangle\otimes|0\rangle$ and $|0\rangle\otimes|1\rangle$ with eigenvalue $0$, then there are two eigenvectors proportional to $$\left(-\epsilon\pm\sqrt{\epsilon^2+\Delta^2}\right)|0\rangle\otimes|0\rangle+ \Delta|1\rangle\otimes|1\rangle$$ with eigenvalues $\pm\sqrt{\epsilon^2+\Delta^2}$.
As for what the textbook does, I prefer to do it the other way around. We can express the second Hamiltonian as
\begin{align}
c_1^+c_1+c_2^+c_2-1=&(ua_1^++va_2)(ua_1+va_2^+)+(-va_1+ua_2^+)(-va_1^++ua_2)-1\\
=&u^2 a_1^+a_1 + v^2 a_2 a_2^++uv(a_1^+a_2^++a_2a_1)+v^2 a_1a_1^++u^2a_2^+a_2-uv(a_1a_2+a_2^+a_1^+)-1\\
=&u^2 a_1^+a_1 + v^2 a_2 a_2^++uv(a_1^+a_2^++a_2a_1)+v^2 (1-a_1^+a_1)+u^2(1-a_2a_2^+)+uv(a_2a_1+a_1^+a_2^+)-1\\
=&a_1^+a_1(u^2-v^2)+a_2a_2^+(v^2-u^2)+v^2+u^2+2uv(a_1^+a_2^++a_2a_1)-1\\
=&(u^2-v^2)(a_1^+a_1-a_2a_2^+)+2uv(a_1^+a_2^++h.c.).
\end{align} The crucial step is the third equality, where I used the ferminonic properties $a_i a_i^+=1-a_i^+a_i$ and $a_i a_j=-a_ja_i$ for $i\neq j$. We can immediately read off the required relationships $u^2-v^2=\epsilon$ and $2uv=\Delta$.
