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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?

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  • $\begingroup$ If this is just about the diagonalization of the matrix, have you tried the Laplace expansion en.wikipedia.org/wiki/Laplace_expansion? There's a lot of zeros in the matrix so I'd expect that only a few terms survive. $\endgroup$
    – Wihtedeka
    Mar 1, 2022 at 18:19
  • $\begingroup$ Thank you for answering. Would I not end up with a polynomial of degree 6 across the diagonal if I employ that technique? As far as I know, I cannot obtain the roots of that polynomial? $\endgroup$
    – Eriksen
    Mar 1, 2022 at 18:45
  • $\begingroup$ Yes I guess you are right, unless it happens to factor nicely you still wouldn't be able to find a solution. $\endgroup$
    – Wihtedeka
    Mar 1, 2022 at 19:02
  • $\begingroup$ Bogoliubov transformation in this context simply means diagonalizing the 6x6 matrix. I don't see any good reason to expect that there are nice analytical expressions for eigenvalues/eigenvectors. $\endgroup$
    – Meng Cheng
    Mar 2, 2022 at 2:15
  • $\begingroup$ Can you elaborate on the relative magnitude of the parameters? $\endgroup$
    – loewe
    Mar 2, 2022 at 10:22

1 Answer 1

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$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\mb}[1]{\mathbf {#1}} \newcommand{\mc}[1]{\mathcal {#1}} \newcommand{\mr}[1]{\mathrm {#1}} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\greq}{\bl\ge} \newcommand{\leseq}{\bl\le} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\blr}[1]{\left[#1\right]} \newcommand{\vlr}[1]{\vert#1\vert} \newcommand{\Vlr}[1]{\Vert#1\Vert} \newcommand{\lara}[1]{\langle#1\rangle} \newcommand{\lav}[1]{\langle#1|} \newcommand{\vra}[1]{|#1\rangle} \newcommand{\lavra}[2]{\langle#1|#2\rangle} \newcommand{\lavvra}[3]{\langle#1|\,#2\,|#3\rangle} \newcommand{\vp}{\vphantom{\dfrac{a}{b}}} \newcommand{\hp}[1]{\hphantom{#1}} \newcommand{\x}{\bl\times} \newcommand{\qqlraqq}{\qquad\bl{-\!\!\!-\!\!\!-\!\!\!\longrightarrow}\qquad} \newcommand{\qqLraqq}{\qquad\boldsymbol{\e\!\e\!\e\!\e\!\Longrightarrow}\qquad} \newcommand{\tl}[1]{\tag{#1}\label{#1}} \newcommand{\hebl}{\bl{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}}$

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}

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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}.

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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}.

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