Consider RG transformation in the vicinity of a fixed point: $$ K_n^\prime=K_n^* + \delta K_{n}^\prime = K_n^* + \sum_m \frac{\partial K^\prime_n}{\partial K_m} \delta K_m + O(\delta K^2) $$ where $M_{nm} = \frac{\partial K^\prime_n}{\partial K_m}$ is the matrix of linearized RG.

Using semigroup property of RG: $$ R_{l_1} R_{l_2} = R_{l_2} R_{l_1} = R_{l_1 l_2} $$ where $l$ is the scale we choose to integrate out.

we have $$ M^{l_1} M^{l_2} = M^{l_2} M^{l_1} = M^{l_1 l_2} \\ \Lambda^{(\sigma)}_{l_1} \Lambda^{(\sigma)}_{l_2} = \Lambda^{(\sigma)}_{l_1 l_2} $$ where $\Lambda_l^{(\sigma)}$ is the eigenvalues of $M^l$, $\sigma$ used to label different eigenvalues.

Using above equation of eigenvalues, it can be shown that $$ \Lambda_l^{(\sigma)} = l^{y_{\sigma}} $$

Here is my question, from above form it seems that $\Lambda_l$ cannot be negative, but in some problem, the eigenvalues are negative through concrete calculation. How can it be possible?


1 Answer 1


The eigenvalues are non-negative, but the y's can be negative.


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