I am struggeling with how to tackle a specific Hamiltonian. I am working with a mean-field Hubbard model and after the introduction of a specific order parameter and transform to momentum space, it is on the form
\begin{equation}
\mathcal{H}=\sum_{k}
\begin{bmatrix}
c_{k,\uparrow}^\dagger \\ c_{k,\downarrow}^\dagger\\ c_{k+Q_1,\downarrow}^\dagger\\ c_{k+Q_2,\downarrow}^\dagger \\ c_{k-Q_3,\uparrow}^\dagger \\ c_{k-Q_3,\downarrow}^\dagger
\end{bmatrix}^{\boldsymbol\top}
\begin{bmatrix}
\epsilon_k & \alpha & ia & -ia & -a & 0 \vphantom{c_{k-Q_3,\downarrow}^\dagger}\\
\alpha^* & \epsilon_k & 0 & 0 & 0 & a \vphantom{c_{k-Q_3,\downarrow}^\dagger}\\
-ia & 0 & \epsilon_{k+Q_1} & 0 & 0 & 0 \vphantom{c_{k-Q_3,\downarrow}^\dagger}\\
ia & 0 & 0 & \epsilon_{k+Q_2} & 0 &0 \vphantom{c_{k-Q_3,\downarrow}^\dagger}\\
-a & 0 & 0 & 0 & \epsilon_{k-Q_3} & \alpha \vphantom{c_{k-Q_3,\downarrow}^\dagger}\\
0 & a & 0 & 0 & \alpha^* & \epsilon_{k-Q_3}\vphantom{c_{k-Q_3,\downarrow}^\dagger}
\end{bmatrix}
\begin{bmatrix}
c_{k,\uparrow} \vphantom{c_{k-Q_3,\downarrow}^\dagger}\\
c_{k,\downarrow}\vphantom{c_{k-Q_3,\downarrow}^\dagger}\\
c_{k+Q_1,\downarrow}\vphantom{c_{k-Q_3,\downarrow}^\dagger}\\
c_{k+Q_2,\downarrow} \vphantom{c_{k-Q_3,\downarrow}^\dagger}\\
c_{k-Q_3,\uparrow} \vphantom{c_{k-Q_3,\downarrow}^\dagger}\\
c_{k-Q_3,\downarrow}\vphantom{c_{k-Q_3,\downarrow}^\dagger}
\end{bmatrix}
\tag{A01}\label{A01}
\end{equation}
where $a$ is real. My problem is that I cannot seem to find a way to diagonalize this matrix analytically. I am aware of how to perform this numerically, so my question has to do with methods for obtaining analytical expressions for the eigenvalues.
Have any of you approached a similar problem and may point me in the direction of an appropriate technique? I have looked into using some variant of a Bogoliubov transformation, but the number of operators makes me think that it is not feasible, but I may be wrong. Is it possible to apply a Bogoliubov transformation to this problem?
Best Answer
$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\mb}[1]{\mathbf {#1}} \newcommand{\mc}[1]{\mathcal {#1}} \newcommand{\mr}[1]{\mathrm {#1}} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\greq}{\bl\ge} \newcommand{\leseq}{\bl\le} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\blr}[1]{\left[#1\right]} \newcommand{\vlr}[1]{\vert#1\vert} \newcommand{\Vlr}[1]{\Vert#1\Vert} \newcommand{\lara}[1]{\langle#1\rangle} \newcommand{\lav}[1]{\langle#1|} \newcommand{\vra}[1]{|#1\rangle} \newcommand{\lavra}[2]{\langle#1|#2\rangle} \newcommand{\lavvra}[3]{\langle#1|\,#2\,|#3\rangle} \newcommand{\vp}{\vphantom{\dfrac{a}{b}}} \newcommand{\hp}[1]{\hphantom{#1}} \newcommand{\x}{\bl\times} \newcommand{\qqlraqq}{\qquad\bl{-\!\!\!-\!\!\!-\!\!\!\longrightarrow}\qquad} \newcommand{\qqLraqq}{\qquad\boldsymbol{\e\!\e\!\e\!\e\!\Longrightarrow}\qquad} \newcommand{\tl}[1]{\tag{#1}\label{#1}} \newcommand{\hebl}{\bl{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}}$
At first consider the Hamiltonian of the $k$th$\m$series term
\begin{equation} \mc H_k\e \begin{bmatrix} \hp{\m i}\epsilon_k & \alpha & ia & \!\!\!\!\m ia & \!\!\!\!\m a & 0 \vp\\ \hp{\m i}\alpha^* & \epsilon_k & 0 & 0 & 0 & a \vp\\ \m ia & 0 & \epsilon_{k+Q_1} & 0 & 0 & 0 \vp\\ \hp\m ia & 0 & 0 & \epsilon_{k+Q_2} & 0 &0 \vp\\ \m a & 0 & 0 & 0 & \epsilon_{k-Q_3} & \alpha \vp\\ \hp{\m i}0 & a & 0 & 0 & \alpha^* & \epsilon_{k-Q_3}\vp \end{bmatrix} \tl{01} \end{equation} To simplify the notation we change to the expression \begin{equation} \mc K\e \begin{bmatrix} \begin{array}{c|c|c|c|c|c} r & g\p ih & ib & \m ib & \m b & 0 \vp\\ \hline g\m ih & r & \hp{gg} 0 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} b \hp{hh} \vp\\ \hline \m ib & 0 & s & 0 & 0 & 0 \vp\\ \hline ib & 0 & 0 & t & 0 & 0 \vp\\ \hline \m b & 0 & 0 & 0 & u & g\p ih \vp\\ \hline 0 & b & 0 & 0 & g\m ih & u\vp \end{array} \end{bmatrix} \tl{02} \end{equation} where $\;b,g,h,r,s,t,u\;$ are real numbers.
The matrix $\;\mc K\;$ is a $\;6\times 6\;$ hermitian matrix so it has 6 real eigenvalues $\;\lambda_\rho\:(\rho\e 1,\cdots, 6)$, roots of its characteristic polynomial \begin{equation} \mr P\plr{\lambda}\e a_6\lambda^6 \p a_5\lambda^5 \p a_4\lambda^4 \p a_3\lambda^3 \p a_2\lambda^2 \p a_1\lambda\p a_0 \tl{03} \end{equation} where $\;a_\sigma\:(\sigma\e 0,\cdots, 6)\;$ real coefficients.
The problem to obtain analytical expressions for the eigenvalues is equivalent to the problem of finding analytical solutions of a 6th degree polynomial equation. We know that the latter is impossible in general, except of special cases.
The least we could help here is to give below the expressions of the coefficients $\;a_\sigma\;$ in terms of the variables $\;b,g,h,r,s,t,u\;$ so that to obtain numerically the eigenvalues for given values of these variables. \begin{equation} \begin{split} a_6 & \e 1\\ a_5 & \e -2r-s-t-2u\\ a_4 & \e -4b^2-2g^2-2h^2+r^2+2rs+2rt+st+4ru+2su+2tu+u^2\\ a_3 & \e 4b^2r+2g^2r+2h^2r+3b^2s+2g^2s+2h^2s-r^2s+3b^2t+2g^2t+2h^2t-r^2t-2rst\\ & \hp\e +6b^2u+2g^2u+2h^2u-2r^2u-4rsu-4rtu-2stu-2ru^2-su^2-tu^2\\ a_2 & \e 3b^4+4b^2g^2+g^4+4b^2h^2+2g^2h^2+h^4-g^2r^2-h^2r^2-3b^2rs-2g^2rs-2h^2rs-3b^2rt\\ & \hp\e -2g^2rt-2h^2rt-2b^2st-2g^2st-2h^2st+r^2st-6b^2ru-4b^2su-2g^2su-2h^2su+2r^2su\\ & \hp\e -4b^2tu-2g^2tu-2h^2tu+2r^2tu+4rstu-2b^2u^2-g^2u^2-h^2u^2+r^2 u^2+2rsu^2+2rtu^2+stu^2\\ a_1 & \e -2b^2g^2r-2b^2h^2r-2b^4s-3b^2g^2s-g^4s-3b^2h^2s-2g^2h^2s-h^4s+g^2r^2s+h^2r^2s-2b^4t\\ & \hp\e -3b^2g^2t-g^4t-3b^2h^2t-2g^2h^2t-h^4t+g^2r^2t+h^2r^2t+2b^2rst+2g^2rst+2h^2rst-2b^4u\\ & \hp\e +4b^2rsu+4b^2rtu+2b^2stu+2g^2stu+2h^2stu-2r^2stu+2b^2ru^2+b^2su^2+g^2su^2+h^2su^2\\ & \hp\e -r^2su^2+b^2tu^2+g^2tu^2+h^2tu^2-r^2tu^2-2rstu^2\\ a_0 & \e b^2g^2rs+b^2h^2rs+b^2g^2rt+b^2h^2rt+b^4st+2b^2g^2st+g^4st+2b^2h^2st+2g^2h^2st+h^4st-g^2r^2st\\ & \hp\e -h^2r^2st+b^4su+b^4tu-2b^2rstu-b^2rsu^2-b^2rtu^2-g^2stu^2-h^2stu^2+r^2stu^2\\ \end{split} \tl{04} \end{equation}
Note that \begin{align} a_5 & \e -2r-s-t-2u \e \m \texttt{trace}\plr{\mc K}\e \m \sum\limits_{\rho\e 1}^{\rho\e 6}\lambda_\rho \tl{05.1}\\ a_0 & \e b^2g^2rs\p\cdots\cdots\p r^2stu^2 \e \texttt{det}\plr{\mc K}\e \prod\limits_{\rho\e 1}^{\rho\e 6}\lambda_\rho \tl{05.2} \end{align}
$\hebl$
EXAMPLE A
Consider that \begin{equation} h\e 0\,, \qquad b\e g\e r\e s\e t\e u \e 1 \tl{A-01} \end{equation} that is \begin{equation} \mc K\e \begin{bmatrix} \begin{array}{c|c|c|c|c|c} 1 & 1 & i & \m i & \m 1 & 0 \vp\\ \hline \hp{gg} 1 \hp{hh} & \hp{gg} 1 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} b \hp{hh} \vp\\ \hline \m i & 0 & 1 & 0 & 0 & 0 \vp\\ \hline i & 0 & 0 & 1 & 0 & 0 \vp\\ \hline \m 1 & 0 & 0 & 0 & 1 & 1\vp\\ \hline 0 & 1 & 0 & 0 & 1 & 1\vp \end{array} \end{bmatrix} \tl{A-02} \end{equation} The characteristic polynomial is \begin{equation} \mr P\plr{\lambda}\e \lambda^6 \m 6\lambda^5 \p 9\lambda^4 \p 4\lambda^3 \m 13\lambda^2 \p 2\lambda\p 3 \tl{A-03} \end{equation} As shown in Figure-01 a graphical solution yields the following eigenvalues \begin{equation} \begin{bmatrix} \hp g \lambda_1 \hp h \vp\\ \hp g \lambda_2 \hp h \vp\\ \hp g \lambda_3 \hp h \vp\\ \hp g \lambda_4 \hp h \vp\\ \hp g \lambda_5 \hp h \vp\\ \hp g \lambda_6 \hp h \vp \end{bmatrix}\e \begin{bmatrix} \hp g \m 1.00000000 \hp h \vp\\ \hp g \m 0.41421356 \hp h \vp\\ \hp g \hp\m\: 1.00000000 \hp h \vp\\ \hp g \hp\m\: 1.00000000 \hp h \vp\\ \hp g \hp\m\: 2.41421356 \hp h \vp\\ \hp g \hp\m\: 3.00000000 \hp h \vp \end{bmatrix} \tl{A-04} \end{equation} while the result with MATHEMATICA is \begin{equation} \begin{bmatrix} \hp g \lambda_1 \hp h \vp\\ \hp g \lambda_2 \hp h \vp\\ \hp g \lambda_3 \hp h \vp\\ \hp g \lambda_4 \hp h \vp\\ \hp g \lambda_5 \hp h \vp\\ \hp g \lambda_6 \hp h \vp \end{bmatrix}\e \begin{bmatrix} \hp g \m 1 \hp h \vp\\ \hp g \hp\m 1\m\sqrt{2} \hp h \vp\\ \hp g \hp\m\: 1 \hp h \vp\\ \hp g \hp\m\: 1 \hp h \vp\\ \hp g \hp\m\: 1\p\sqrt{2} \hp h \vp\\ \hp g \hp\m\: 3 \hp h \vp \end{bmatrix} \tl{A-05} \end{equation} in full agreement with \eqref{A-04}.
$\hebl$
EXAMPLE B
Consider that \begin{equation} b\e g\e s\e u\e 1\,,\: h\e -1\,,\: r\e 2\,,\: t\e 0.5 \tl{B-01} \end{equation} that is \begin{equation} \mc K\e \begin{bmatrix} \begin{array}{c|c|c|c|c|c} 2 & 1-i & i & \m i & \m 1 & 0 \vp\\ \hline 1+i & \hp{gg} 2 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} 0 \hp{hh} & \hp{gg} b \hp{hh} \vp\\ \hline \m i & 0 & 1 & 0 & 0 & 0 \vp\\ \hline i & 0 & 0 & 0.5 & 0 & 0 \vp\\ \hline \m 1 & 0 & 0 & 0 & 1 & 1-i \vp\\ \hline \hp{gg} 0 \hp{hh} & 1 & 0 & 0 & 1+i & 1\vp \end{array} \end{bmatrix} \tl{B-02} \end{equation} The characteristic polynomial is \begin{equation} \mr P\plr{\lambda}\e \lambda^6 \m 7.5\lambda^5 \p 14.5\lambda^4 \p 2\lambda^3 \m 16.5\lambda^2 \p 1.5\lambda\p 4 \tl{B-03} \end{equation} As shown in Figure-02 a graphical solution yields the following eigenvalues \begin{equation} \begin{bmatrix} \hp g \lambda_1 \hp h \vp\\ \hp g \lambda_2 \hp h \vp\\ \hp g \lambda_3 \hp h \vp\\ \hp g \lambda_4 \hp h \vp\\ \hp g \lambda_5 \hp h \vp\\ \hp g \lambda_6 \hp h \vp \end{bmatrix}\e \begin{bmatrix} \hp g \m 0.807861\hp h \vp\\ \hp g \m 0.503229 \hp h \vp\\ \hp g \hp\m\:\, 0.698461 \hp h \vp\\ \hp g \hp\m\: 1.213848 \hp h \vp\\ \hp g \hp\m\: 2.907977 \hp h \vp\\ \hp g \hp\m\: 3.990804 \hp h \vp \end{bmatrix} \tl{B-04} \end{equation} while the result with MATHEMATICA is \begin{equation} \begin{bmatrix} \hp g \lambda_1 \hp h \vp\\ \hp g \lambda_2 \hp h \vp\\ \hp g \lambda_3 \hp h \vp\\ \hp g \lambda_4 \hp h \vp\\ \hp g \lambda_5 \hp h \vp\\ \hp g \lambda_6 \hp h \vp \end{bmatrix}\e \begin{bmatrix} \hp g \m 0.807861 \hp h \vp\\ \hp g \m 0.503229 \hp h \vp\\ \hp g \hp\m\:\, 0.698461 \hp h \vp\\ \hp g \hp\m\:\, 1.21385 \hp{hh} \vp\\ \hp g \hp\m\:\, 2.90798 \hp{hh} \vp\\ \hp g \hp\m\:\, 3.9908 \hp{hhh} \end{bmatrix} \tl{B-05} \end{equation} in full agreement with \eqref{B-04}.