Exact diagonalization by Bogoliubov transformationby arojo Tags: bogoliubov, diagonalization, exact, transformation 

#1
Dec613, 04:31 AM

P: 16

Hello all,
I am developing a model of multiple gaps in a square lattice. I simplified the associated Hamiltonian to make it quadratic. In this approximation it is given by, [tex] H = \begin{pmatrix} \xi_\mathbf{k} & \sigma U_1 & U_2 & U_2\\ \sigma U_1 & \xi_{\mathbf{k}+(\pi,\pi)} & 0 & 0\\  U_2 & 0 & \xi_{\mathbf{k}+(\pi/2,0)} & 0\\  U_2 & 0 & 0 & \xi_{\mathbf{k}+(0,\pi/2)} \end{pmatrix} [/tex] And my Nambu operator is given by, [tex] ψ_\mathbf{k} = \begin{pmatrix} c_{\mathbf{k},\sigma} \\ c_{\mathbf{k}+(\pi,\pi),\sigma} \\ c_{\mathbf{k}+(\pi/2,0),\sigma} \\ c_{\mathbf{k}+(0,\pi/2),\sigma} \end{pmatrix} [/tex] I tried to diagonalized by making three Bogoliubov transformations, the first to diagonalize the upper right submatrix of H, and then the two others (a sort of nested transformations). But I get a lengthy result, what I would like to know if there is a smart transformation which allows me to write [tex] H = A_1^\dagger A_2^\dagger A_3^\dagger D A_3 A_2 A_1 [/tex] or simply [tex] H = U^\dagger D U [/tex] Or the only way is to use just brute force? Thanks 



#2
Dec613, 07:57 AM

Sci Advisor
P: 3,377

Diagonalizing a 4x4 matrix can be done analytically, as the eigenvalues result as solutions of a fourth order polynomial which may be of special form. Have you tried?




#3
Dec613, 09:11 AM

P: 16

Hi DrDu,
Actually I started by doing precisely that, but I got a messy result. Which certainly is analytical but hard "to read", at least from the point of view of getting an idea of what is going on without doing the numerics. Actually I should rephrase my question as is there any elegant representation or expression of the diagonalization as for example the one obtained in BCS Superconductivity? Thanks 


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