Diagonalising a fermionic Hamiltonian via Bogoliubov transformation

Question

I am trying to diagonalise the fermionic Hamiltonian

$$ H= \sum_k f(k) a_k^\dagger a_k + \frac{ig(k)}{2} \left(a^\dagger_k a^\dagger_{-k} + a_k a_{-k} \right)$$

where $f(-k) = f(k)$ and $g(-k) = -g(k)$. I am aware that the standard procedure to diagonalise this Hamiltonian is to use a Bogoliubov transformation. In order to do this, we define the spinor $\psi_k = (a_k,a_{-k},a_k^\dagger,a_{-k}^\dagger)$, in which case we can write the Hamiltonian as

$$ H = \frac{1}{2} \sum_{k > 0} \psi^\dagger_k h(k) \psi_k + E_0, \quad h(k) = \begin{pmatrix} f(k) & 0 & 0 & ig(k) \\ 0 & f(k) & -ig(k) & 0 \\ 0 & ig(k) & -f(k) & 0 \\ -ig(k) & 0 & 0 & -f(k) \end{pmatrix}$$

where $E_0$ is some constant.

Bogoliubov transformations

In order to diagonalise $H$, we apply a Bogoliubov transformation of the form

$$ b_i = \sum_k u_{ik}a_k + v_{ik}a_k^\dagger, \quad b_i^\dagger = \sum_k u^*_{ik}a_k^\dagger + v_{ik}^*a_k$$

which can be captured in matrix notation as $\phi = M \psi$, where $\phi = ( b_1,b_2,\ldots;b_1^\dagger,b_2^\dagger,\ldots)$, $\psi = ( a_1,a_2,\ldots;a_1^\dagger,a_2^\dagger,\ldots)$,

$$ M = \begin{pmatrix} U & V \\ V^* & U^* \end{pmatrix}, $$

and $U$ and $V$ are the matrices of coefficients $u_{ik}$ and $v_{ik}$. In order for this to be canonical, so that $\{ b_i, b_j^\dagger \} = \delta_{ij}$ and $ \{ b_i,b_j \} = \{b_i^\dagger,b_j^\dagger \} = 0$, these two conditions impose

$$ \sum_{k} u_{ik}u_{jk}^* + v_{ik}v_{jk}^* = \delta_{ij}, \quad \sum_k u_{ik}v_{jk} + v_{ik}u_{jk} = 0 \\ \Rightarrow \quad UU^\dagger + VV^\dagger = \mathbb{I}, \quad UV^T + VU^T = 0$$

respectively. I believe these two conditions mean that $M$ is unitary.

My attempt

In my problem, we are working with a four-dimensional case with $\psi_k = (a_k,a_{-k},a_k^\dagger,a_{-k}^\dagger)$. As the matrix $M$ which performs the Bogoliubov transformation in this four-dimensional space must be unitary, one would expect we could choose $M$ to be the matrix whose columns are the eigenvectors of $h(k)$, as this indeed with diagonalise $h(k)$, i.e.

$$ H = \sum_k \psi^\dagger_k h(k) \psi_k = \sum_k \psi^\dagger_k (\Gamma(k)D(k) \Gamma^\dagger(k)) \psi_k \equiv \sum_k \phi^\dagger_k D(k) \phi_k$$

so I expect $M = \Gamma^\dagger$, where $\Gamma$ is the matrix whose columns are the normalised eigenvectors of $h(k)$. According to Wolfram, the eigenvectors of $h(k)$ are given by $$ v_1 = N_1 \left( \frac{i(f-E)}{g} , 0 , 0 , 1 \right), \quad v_2 = N_2 \left(0, -\frac{i(f-E)}{g}, 1 ,0 \right),\\v_3 = N_3 \left( \frac{i(f+E)}{g},0,0,1\right), \quad v_4 = N_4 \left( 0, -\frac{i(f+E)}{h},1,0 \right) $$

where $N_i$ are just normalisation factors, $E = + \sqrt{f^2 + g^2}$ and the eigenvalues of $v_1$ and $v_2$ are $-E$, while for $v_3$ and $v_4$ the eigenvalues are $+E$. Constructing $\Gamma$ from this and then taking its Hermitian conjugate, we get

$$ \Gamma^\dagger = \begin{pmatrix} -iN_1 \frac{(f-E)}{g} & 0 & 0 & N_1 \\ 0 & i N_2 \frac{(f-E)}{g} & N_2 & 0 \\-iN_3 \frac{(f+E)}{g} & 0 & 0 & N_3 \\ 0 & iN_4 \frac{(f+E)}{h} & N_4 & 0\end{pmatrix} $$

This matrix is what I hoped $M$ would be equal to, however it does not obey the required conditions above in order to be canonical, so I cannot choose this. I thought maybe I could multiply the normalisation constants by different phases, but this still doesn't work. For example, the top left block and bottom right block are diagonal and off-diagonal, however the above condition says they should be complex conjugates.

My question

I am aware there are many questions similar to mine such as this, this and this, however I would like to know how one can do it via matrix diagoanlisation instead of brute force substituting it into the Hamiltonian and demanding the off diagonal elements are zero. In many textbooks and literature, explicit calculation of the Bogoliubov transformation is not provided, and the answer is simply stated, e.g. appendix A of this paper.

So my question is the following: can one perform a Bogoliubov transformation by simply solving for the eigenvectors of the kernel matrix $h(k)$ and diagonalising it? If so, why has it not worked here?

Answer 1

So I think I managed to complete this problem with the helpful comment of Vadim, so here is my solution if anyone is interested:

When constructing the matrix of eigenvectors $\Gamma$, there is not a unique order in which the eigenvectors $\{ v_i \}$ should be arranged, so we have the freedom to permute the columns to ensure that $\Gamma^\dagger$ takes the form of

$$ M = \begin{pmatrix} U & V \\ V^* & U^* \end{pmatrix} $$

As it stands above, this is not true for the $\Gamma^\dagger$ I took in my question, however if I swap $v_1$ and $v_2$, then this has the effect of swapping the first and second rows of $\Gamma^\dagger$, mapping it to

$$ \Gamma^\dagger = \begin{pmatrix} 0 & i N_2 (f-E) & N_2g & 0 \\ -iN_1 (f-E) & 0 & 0 & N_1 g\\ -iN_3 (f+E) & 0 & 0 & N_3 g \\ 0 & iN_4 (f+E) & N_4 g & 0\end{pmatrix} $$

where I have multiplied the whole matrix by $g$ to tidy it up a bit. This already looks more promising. Now, the magnitude of $\{ N_i \}$ are fixed by the constraint that the eigenvectors $\{v_i \}$, essentially the rows of the matrix, must be normalised to unity, so

$$ |N_1| = |N_2| = \frac{1}{\sqrt{2E(E-f)}}, \quad |N_3| = |N_4| = \frac{1}{\sqrt{2E(E+f)}}$$

However, we can still have the freedom to choose the phases of the normalisation factors $\{ N_i \}$. If I choose the phases

$$ N_1 = |N_1|, \quad N_2 = i|N_2|, \quad N_3 = |N_3|, \quad N_4 = -i|N_4| $$

then $\Gamma^\dagger$ takes the desired form, with

$$ U = \begin{pmatrix} 0 & \sqrt{\frac{E-f}{2E}} \\ i\sqrt{\frac{E-f}{2E}} & 0\end{pmatrix}, \quad V = \begin{pmatrix} \frac{ig}{\sqrt{2E(E-f)}} & 0 \\ 0 & \frac{g}{\sqrt{2E(E-f)}} \end{pmatrix}$$

where to show that this is in the form of $M$ I use the fact that $E = +\sqrt{f^2 + g^2}$. which allows me to see that the diagonal blocks and off-diagonal blocks have the correct relationship.