The $K$ quantum number in the Shell Model---$^{50}$Cr

It was suggested~[1,2] that the $10^+_1$ state in $^{50}$Cr at 6.340~MeV does not belong to the $K=0^+$ g.s. band. In~[1] it is noted that the static quadrupole moments of the $J=2^+_1$--$8^+_1$ states are all negative, but that of $10^+_1$ is positive. While Ref.~[1] suggested that the $10^+_1$ state belongs to a high $K$ prolate band, in Ref.~[2] they assign it as $K=10^+$. There is a nearby second $10^+$ state. However, the $B(E2)_{10^+_2 \rightarrow 10^+_1}$ was not quoted by either group. In this work, we performed full $fp$ shell model calculations using four different interactions: FPD6, KB3, GXPF1, and GXPF1A. The results for $B(E2)_{10^+_2 \rightarrow 10^+_1}$ are robust around 135~e$^2$fm$^4$ and suggest strong $K$ mixing. It is not clear what the $K$ value for the $10_2^+$ state is. With FPD6, $Q(10^+_2)$ is negative, suggesting it is a member of the $K=0^+$ band. But it is hard to understand how to get strong mixing of $K=0^+$ and $K=10^+$. With the other interactions, $Q(10^+_2)$ is positive and thus inconsistent with a $K=0^+$ (prolate) band. If we assume that the $10^+_1$ state has $K=10^+$ and the $8^+_1$ state has $K=0^+$, then the $B(E2)_{10^+_1 \rightarrow 8^+_1}$ would vanish. However, for the last three interactions, the corresponding $B(E2)$ is about 75~e$^2$fm$^4$, which implies substantial $K$ mixing. Thus, while a $K=10^+$ assignment for the $10^+_1$ states makes the most sense in terms of energy systematics, in detail the situation is more complicated. [1] L.~Zamick et al., Phys. Rev. C {\bf 53}, 188 (1996); Phys. Rev. C {\bf 54}, 956 (1996). [2] F.~Brandolini et al., Phys. Rev. C {\bf 66}, 021302(R) (2002).