A generic $3\times 3$ Hamiltonian can be expressed in terms of eight Gell-Mann matrices ($\lambda$) as \begin{align} {\cal H} &= h_{0} I + H= h_{0} I + \sum_{\alpha=1}^{8} h_{\alpha} \lambda_{\alpha}, \\ {\cal H}|n \rangle &=( h_{0} + \epsilon_{n}) | n \rangle \qquad \text{For } n\in \{1,2,3 \}, \end{align} where I is the identity matrix, $|n \rangle $ are eigenvectors and $\epsilon_{n}$ denote eigenvalues of $H$. This decomposition enable one to determine different qunatities, e.g., Berry curvature, in terms of matrix elements of Gell-Mann matrices.

Using properties of $\lambda$ matrices, we can evaluate the non-diagonal matrix elements of $\langle m |[\lambda_{\alpha}, H]| n \rangle $, \begin{align} \lambda_{\alpha}^{mn} = 2 {\rm i} \sum_{\gamma, \beta} f_{\alpha \beta \gamma} h_{\beta} \frac{ \lambda_{\gamma}^{mn}}{\epsilon_{m} -\epsilon_{n}} = \sum_{\gamma} {\cal F}_{\alpha \gamma}^{mn} \lambda_{\gamma}^{mn}, \end{align} where $f_{\alpha \beta \gamma}$ are SU(3) structure constants and we have employed $[\lambda_{\alpha}, \lambda_{\beta}] = {\rm i} \sum_{\gamma} f_{\alpha \beta \gamma}\lambda_{\gamma} $. My question is whether it is possible to evaluate \begin{align} \frac{ \lambda_{\alpha}^{mn} }{\epsilon_{m} -\epsilon_{n}} = \sum_{\gamma} {\cal K}_{\alpha \gamma}^{mn} \lambda_{\gamma}^{mn}, \end{align} such that $\cal K$ can expressed merely in terms of $h$ and matrix elements of $\lambda$?

  • $\begingroup$ Do you fully understand how that denominator entered the matrix equation? The eigenvalues are a property of the h s. $\endgroup$ Commented Aug 10, 2020 at 13:19
  • $\begingroup$ Did you mean in \lambda_{\alpha}^{mn}/(\epsilon_{m}-\epsilon_{n})? These matrix elements appear in quantities constructed from position operators. $\endgroup$
    – Shasa
    Commented Aug 10, 2020 at 13:47
  • $\begingroup$ You evidently want $\epsilon_m-\epsilon_n$ in the numerator on the left of your last equation, not the denominator. You are asking if ${\cal F}^{mn}_{\alpha\gamma}(\epsilon_m-\epsilon_n)$ is independent of the parameter $\epsilon_m-\epsilon_n$? Of course not. That parameter is a property of h, as you should reassure yourself with simple examples. $\endgroup$ Commented Aug 10, 2020 at 16:35
  • $\begingroup$ I would like to have $\epsilon_{m}- \epsilon_{n}$ in the denominator, as I have presented in the question. The point, that you've also mentioned, is that I somehow should find a way to substitute $\epsilon$ with $h$ and matrix elements of $\lambda$ as diagonalizing my 3*3 matrix is not easy. $\endgroup$
    – Shasa
    Commented Aug 10, 2020 at 16:50
  • $\begingroup$ We indeed know that $\epsilon_{n} = \lambda_{\alpha}^{nn} h_{\alpha} $ but this keeps the matrix elements of $\lambda$ in the denominator which it might be problematic when they are zero. $\endgroup$
    – Shasa
    Commented Aug 10, 2020 at 16:52

1 Answer 1


You might be trapping yourself in your notation. First, appreciate that $[\lambda_\alpha,{\cal H}]= [\lambda_\alpha, H ]$, so w.l.o.g. set $h_0=0$. Moreover, since $H|n\rangle=\epsilon_n|n\rangle$, the $\epsilon_n$s are functionals of h.

You are considering $$ [\lambda_{\alpha}, H]= 2i \sum_{\beta, \gamma} f^{\alpha \beta\gamma}h_\beta \lambda_\gamma. $$

Defining $\Lambda ^\alpha (m,n)=\langle m |\lambda_{\alpha} | n \rangle $, the matrix elements of the above linear equation are $$ \lambda_{\alpha}(m,n) \equiv \langle m |[\lambda_{\alpha},H] | n \rangle =(\epsilon_n-\epsilon_m) \Lambda^\alpha (m,n)\\ = 2 {\rm i} \sum_{\gamma, \beta} f_{\alpha \beta \gamma} h_{\beta} \Lambda_{\gamma}(m,n) \equiv \sum_{\gamma} (\epsilon_n-\epsilon_m){\cal F}_{\alpha \gamma}(mn) \Lambda_{\gamma}(m,n). $$ (NB. You have interchanged n with m in your eigenvalues' expression, not clear why.)

For off diagonal elements, recalling (m,n) are just indices never summed over, you have the eigenvalue-1 equation $$ \Lambda_{\alpha} (m,n) = \sum_{\gamma} {\cal F}_{\alpha \gamma} (m,n) \Lambda_{\gamma} (m,n) . $$

Evidently ${\cal, F, K,...}$, etc, are all functionals of h, and hence the corresponding eigenvalues $\epsilon_m$.

It would be easiest for you to simply choose $h_1 =1$ and the rest vanishing, to check your expressions. E.g., $\vec \Lambda (1,-1)= (0,i,1,0,0,0,0,0)^T$, with $\epsilon_{-1}-\epsilon_{1}=-2$, hence ${\cal F}_{\alpha \gamma}(1,-1)=-i f_{\alpha 1\gamma}$, imaginary antisymmetric, etc... Recall $1=f_{ 3 1 2}=2f_{ 7 1 4}=2f_{ 5 1 6}$. So ${\cal F}_{ 23}(1,-1)=i$. Observe the evident eigenvector with eigenvalue 1. You might as well have found it this way, if you did not have it!

But I still don't fully fathom your objective. If you were interested in exponentiation of H, you might well consider this.

  • $\begingroup$ Thank you for your response and for recommending a reference. Well, as my matrix H is given, all $h_{\alpha}$ are also fixed. My original motivation for this question was finding a way to isolate one matrix element of $\lambda_{\alpha}$ without computing the whole matrix. This was because eigenvalues of $H$ are quite complicated and I hardly can find its eigenvectors; I was looking for an alternative such that I can compute the matrix elements of $\lambda_{\alpha}$ in the eigenbasis by using a sum $h$ and matrices $\lambda_{\alpha}$ in the original space. $\endgroup$
    – Shasa
    Commented Aug 12, 2020 at 9:39
  • $\begingroup$ The basis of eigenvectors itself you are using depends on the h s. Moreover, Tr$\lambda_\alpha H= 2 h_\alpha$. $\endgroup$ Commented Aug 12, 2020 at 12:10

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