I am trying to replicate a calculation from the linked paper but I am unsure if I understand their math. A locally invariant function defined as the characteristic polynomial is as follows: $$F_U(t) = det[\Re[M^{\dagger}UM] + t \cdot \Im[M^{\dagger}UM]]$$ Then it is said for $U$ with interaction coefficients $(x,y,z)$, we have $$F_U(t) = (t^2+1)(Ct^2 + Bt + A) -t^2$$ I'm having trouble understanding how $(x,y,z)$ are related to $t$, and because neither $M$ nor $U$ are defined as functions of $t$, then it seems like $F_U(t)$ is not a degree-4 polynomial.

In this paper, equation (9), they define $U$ with parameters $(\alpha, \beta, \gamma_1, \gamma_2, \delta_1, \delta_2)$ and find the corresponding coefficient $C=\frac{1}{16}(\cos{2\alpha}-\cos{2\beta})^2$. When I plug the same values in, since $U$ is not a function of $t$, then $F_U$ is not degree-4 and hence $C=0$. Any guidance would be appreciated, thank you.

Source: https://arxiv.org/abs/2105.06074

  • $\begingroup$ how are the coefficients $(x,y,z)$ related to $U$ exactly? $\endgroup$
    – glS
    May 17, 2022 at 20:06
  • 1
    $\begingroup$ I have no idea how all of these quantities are defined, but if $U$ is a $4\times 4$ matrix (two qubits as indicated by the title), then $F_U(t)$ is of course of degree 4. Recall $\mathrm{det}(t A) = t^d \mathrm{det}(A)$ if $A$ is $d \times d$. $\endgroup$ May 18, 2022 at 6:53
  • $\begingroup$ For $U \in SU(4)$, $(x,y,z)$ are defined as $U$'s coordinates in the Weyl chamber. I did not know about this rule $\text{det}(tA)=t^d \text{det}(A)$ which appears to be what I was looking for. Thanks! $\endgroup$ May 18, 2022 at 15:56

1 Answer 1


Resolved in comments, missing step was $\text{det}(tA) =t^d\text{det}(A)$


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