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I'd like to know if there is a general numerical method of diagonalizing the bosonic quadratic Hamiltonian below

$$H=\sum_{i,j=1}^NT_{ij}b_i^\dagger b_j+\frac{1}{2}\sum_{i,j=1}^N\left(U_{ij}b_i^\dagger b_j^\dagger+U_{ij}^*b_ib_j\right)\!,$$

using the Bogoliubov transformation, where $T$ is an $N\times N$ Hermitian matrix and $U$ is $N\times N$ symmetric but can in general be complex. One may rewrite $H$ in $2N\times 2N$ matrix form

$$H=\frac{1}{2}\begin{pmatrix} b^\dagger & b \end{pmatrix}\begin{bmatrix} T & U\\ U^* & T^* \end{bmatrix}\begin{pmatrix} b\\ b^\dagger \end{pmatrix}+\mathrm{const},$$

where $b=(b_1,b_2,\ldots,b_N)$ is the vector of bosonic annihilation operators. The main challenge is that the Bogoliubov transformation for bosons

$$\begin{pmatrix} b\\b^\dagger \end{pmatrix}=\begin{bmatrix} M & N\\ N^* & M^* \end{bmatrix}\begin{pmatrix} \xi \\ \xi^\dagger \end{pmatrix}$$

is not unitary, but simplectic, as has been discussed in many similar questions. To preserve the bosonic commutation relations of $[b,b^\dagger]$ and $[\xi,\xi^\dagger]$, the transformation matrix satisfies

$$\begin{pmatrix} M & N\\ N^* & M^* \end{pmatrix}\begin{bmatrix} I & 0\\ 0 & -I \end{bmatrix}\begin{pmatrix} M^\dagger & N^T\\ N^\dagger & M^T \end{pmatrix}=\begin{bmatrix} I & 0\\ 0 & -I \end{bmatrix}\!.$$

For fermions, $-I\,$ becomes $I$ and the transformation matrix is unitary. So is there a general numerical procedure to find the bosonic Bogoliubov transformation that diagonalizes $H$ into

$$H=\frac{1}{2}\begin{pmatrix} \xi^\dagger & \xi \end{pmatrix}\begin{bmatrix} \Gamma & 0\\ 0 & \Gamma \end{bmatrix}\begin{pmatrix} \xi \\ \xi^\dagger \end{pmatrix}+\mathrm{const},$$

where $\Gamma$ is real-diagonal?

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  • $\begingroup$ Sure. Take your matrix with T's and U's in it, multiply it (on either side) by your I,-I matrix, and diagonalize that. $\endgroup$ – march Jun 24 '19 at 4:29
  • $\begingroup$ @march, I have obtained $$\begin{bmatrix} T & -U\\ -U^* & T^*\end{bmatrix}=\begin{pmatrix} M & N\\ N^* & M^*\end{pmatrix}\begin{bmatrix}\Gamma & 0\\0 & \Gamma\end{bmatrix}\begin{pmatrix} M^\dagger & N^T\\N^\dagger & M^T\end{pmatrix}\!,$$ but still the transformation matrices are not unitary. If we first diagonalize the Hamiltonian matrix using unitary matrices, will that help us find the simplectic transformation matrix that preserves the I,-I metric? $\endgroup$ – George Jun 24 '19 at 4:45
  • $\begingroup$ I’ll write something up tomorrow about this. Doing what I said will not yield a unitary matrix, but it will be diagonalizable with real eigenvalues, and the resulting eigenvectors, properly used, will yield the correct canonical transformation. $\endgroup$ – march Jun 24 '19 at 4:47
  • $\begingroup$ @march, Looks like I'm close but still cannot figure out. Since the Hamiltonian is Hermitian, eigenvectors with different eigenvalues must be orthogonal. So we have to make linear combinations of eigenvectors with the same eigenvalue to satisfy the I,-I metric? $\endgroup$ – George Jun 26 '19 at 1:48
  • $\begingroup$ Sorry I haven't posted something yet! I promise I'll do it tomorrow. But just real quick: the transformation matrices will not be unitary. The eigenvectors will not be orthogonal. That's okay, because that's how it works. $\endgroup$ – march Jun 26 '19 at 2:52
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Self answer:

We start with a simple identity $$\begin{bmatrix} I & 0\\ 0 & -I \end{bmatrix}\begin{pmatrix} M^\dagger & N^T\\ N^\dagger & M^T \end{pmatrix}\begin{bmatrix} I & 0\\ 0 & -I \end{bmatrix}=\begin{pmatrix} M^\dagger & -N^T\\ -N^\dagger & M^T \end{pmatrix}\!.\quad(*)$$ Then the symplectic condition that the canonical transformation matrix satisfies can be rewritten as $$\begin{pmatrix} M & N\\ N^* & M^* \end{pmatrix}\begin{pmatrix} M^\dagger & -N^T\\ -N^\dagger & M^T \end{pmatrix}=\begin{bmatrix} I & 0\\ 0 & I \end{bmatrix}\!.$$ Therefore, we see the transformation matrix and its inverse multiplied together to get the identity matrix. Plugging the transformation into the Hamiltonian, we obtain $$\begin{pmatrix} M^\dagger & N^T\\ N^\dagger & M^T \end{pmatrix}\begin{bmatrix} T & U\\ U^* & T^* \end{bmatrix}\begin{pmatrix} M & N\\ N^* & M^* \end{pmatrix}=\begin{bmatrix} \Gamma & 0\\ 0 & \Gamma \end{bmatrix}\!.$$ We now use again the equation $(*)$ to obtain $$\begin{bmatrix} I & 0\\ 0 & -I \end{bmatrix}\begin{pmatrix} M^\dagger & -N^T\\ -N^\dagger & M^T \end{pmatrix}\begin{bmatrix} I & 0\\ 0 & -I \end{bmatrix}\begin{bmatrix} T & U\\ U^* & T^* \end{bmatrix}\begin{pmatrix} M & N\\ N^* & M^* \end{pmatrix}=\begin{bmatrix} \Gamma & 0\\ 0 & \Gamma \end{bmatrix}\!,$$ which then simplifies to $$\begin{pmatrix} M^\dagger & -N^T\\ -N^\dagger & M^T \end{pmatrix}\begin{bmatrix} T & U\\ -U^* & -T^* \end{bmatrix}\begin{pmatrix} M & N\\ N^* & M^* \end{pmatrix}=\begin{bmatrix} \Gamma & 0\\ 0 & -\Gamma \end{bmatrix}\!,$$ and hence we have $$\begin{bmatrix} T & U\\ -U^* & -T^* \end{bmatrix}=\begin{pmatrix} M & N\\ N^* & M^* \end{pmatrix}\begin{bmatrix} \Gamma & 0\\ 0 & -\Gamma \end{bmatrix}\begin{pmatrix} M^\dagger & -N^T\\ -N^\dagger & M^T \end{pmatrix}\!.$$ By diagonalizing the non-Hermitian matrix on the left-hand side (which equals $\mathrm{diag}(I,-I)$ times the Hamiltonian), we obtain the transformation matrix whose columns are the eigenvectors of the non-Hermitian matrix.

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I would suggest the following reference: A. G. D. Maestro and M. J. Gingras, J. Phys. Condens. Matter 16, 3339 (2004). It has an appendix that fully describes how to do those steps (in k-space).

Please note that it is not enough to diagonalize the matrix [T, U;-U*,-T*], the eigenvectors of this matrix are not necessarily eigenvectors of [T, U;U*,T*] as explained in the reference as you the Bogoliubov matrix must satisfy three conditions at the same time. Letting H be the square matrix that you want to diagonalize, Z is the Bogoliubov transformation, and G=diag[I,-I], the conditions that Z needs to satisfy are

(1) $Z^{\dagger}HZ=\rm diag[e1,..,en,e1,..en]$

where ek is the energy dispersion depending on the system in hands. For example, in rare earth pyrochlores you have 4 bands and thus n=4.

(2) $ZGZ^{\dagger}=G$; this is basically the commutator relations in compact form [i.e. you want to impose that your new operators are bosonic as well].

(3) $Z^{-1} GM Z=-\rm diag[e1,...,en,-e1,..,-en]$

The requirement in (3) is actually comes from combining (1,2). Surprisingly, solving (3) is not always enough! Considering the energy dispersions for bosons you can get them from (3). For the Bogoliubov matrix you need first to construct a matrix $\tilde{Z}$ from the eigenvectors of GM, then impose that

\begin{equation} Z=\tilde{Z}P \end{equation}

where P is block diagonal matrix. Using this expression in (2), we find that $PGP^{\dagger}=W$, where $W=(\tilde{Z}^{\dagger}G\tilde{Z})^{-1}$ is block diagonal. The procedure is to evaluate W using the result obtained earlier for $\tilde{Z}$, then using the formula $PGP^{\dagger}=W$, we conclude that the ith block in P is related to the ith block in W as $\pm P_{i}P_{i}^{\dagger}=W_{i}$. Now, since W is Hermitian by construction (check the definition of W above), then we can use linear algebra to say that the block $W_{i}$ is Hermitian and diagonalizable, i.e. there exist an invertible matrix $X_{i}$ such that $W_{i}=X_{i}D_{i}X_{i}^{-1}$, where $D_{i}$ is a diagonal matrix with eigenvalues of $W_{i}$ on the diagonal. Using this expression of $W_{i}$ together with $\pm P_{i}P_{i}^{\dagger}=W_{i}$, one can easily verify that

\begin{equation} P_{i}=X_{i}\sqrt{\pm D_{i}}X_{i}^{-1} \end{equation}

This way you construct the matrix P and thus the Bogoliubov matrix becomes

\begin{equation} Z=\tilde{Z}P \end{equation}

This is basically the summary of the numerical (could be sometimes exact depending on the problem in hand) method for diagonalizing quadratic bosonic Hamiltonian in $\vec{k}$-space.

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    $\begingroup$ I guess the reason for P here is in case there is degeneracy? If there is no degeneracy, Z suffices to be the solution right? $\endgroup$ – Histoscienology Jun 2 '20 at 21:11

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