Erratum to: Spin-1 diquark contributing to the formation of tetraquarks in light mesons

1 Erratum to: Eur. Phys. J. C (2017) 77:173 DOI 10.1140/epjc/s10052-017-4736-6

In the paper above, we have proposed a tetraquark picture with the mixing scheme for the $I_z=1$ members of the isovector ($I=1$) resonances, $a^+_0 (980)$, $a^+_0 (1450)$. In particular, their mass splittings fit relatively well with the hyperfine mass splittings if they are viewed as mixtures of two spin-configurations of diquark–antidiquark constituents, $|J,J_{12},J_{34}\rangle = |000\rangle ,|011\rangle$, where J is the tetraquark spin, $J_{12}$ the diquark spin, $J_{34}$ the antidiquark spin. The second configuration involving the spin-1 diquark, $|011\rangle$, is found to be an important ingredient in explaining the resonances of our concern in this tetraquark picture. However, the existence of the $|011\rangle$ component requires additional tetraquarks to be found in $J=1$ and $J=2$ resonances with the spin configurations, $|J,J_{12},J_{34}\rangle =|111\rangle$ and $|211\rangle$, respectively.

In this erratum, we point out that our assignment of $a^+_1(1260)$ as a candidate for the $J=1$ tetraquark with the $|111\rangle$ configuration is incorrect because of the C -parity for its corresponding member in $I_z=0$. Specifically, we would like to demonstrate that the $|111\rangle$ state with $I=1, I_z=0$ must have the C -parity odd and, in this regard, a relevant candidate for the $|111\rangle$ state should be $b^0_1(1235)$ ($J^{PC}=1^{+-}$) instead of $a^0_1(1260)$ ($J^{PC}=1^{++}$). So its charged member ($I=1,I_z=1$), which in fact was considered in our paper, must be $b^+_1(1235)$ instead of $a^+_1(1260)$. Nevertheless, since their experimental masses are almost the same, $M[b_1(1235)]=1229.5$ MeV, $M[a_1(1260)]=1230$ MeV, our discussion in the paper, which is mostly based on the mass splittings, is unaltered except that $a_1(1260)$ is replaced with $b_1(1235)$. The other tetraquarks with $J=0,J=2$, with the spin configurations $|000\rangle$, $|011\rangle$, $|211\rangle$, are found to have $C=+$ so their assignments to the physical resonances are not contradictory with their C -parity.

To demonstrate that $C|111\rangle =-|111\rangle$ for the isospin member of $I=1,I_z=0$, we take the state with $J=1$ and the spin projection $M=1$ among three spin states in $|111\rangle$, and we denote this state as $|JM\rangle =|11\rangle$. The same proof can be done for the other spin states, $|JM\rangle =|10\rangle , |1-1\rangle$. The flavor structure of the member $I=1,I_z=0$ is $\frac{1}{\sqrt{2}} \left( [su][\bar{s}\bar{u}]-[ds][\bar{d}\bar{s}] \right)$. For our purpose, it would be enough to consider one specific combination of the flavor, $[su][\bar{s}\bar{u}]$. If we rewrite the state $|JM\rangle =|11\rangle$ with respect to the spins and their projections of diquark and antidiquark, $|J_{12}M_{12}\rangle _{[su]} |J_{34}M_{34}\rangle _{[\bar{s}\bar{u}]}$, we find

$$\begin{aligned} |11\rangle= & {} \frac{1}{\sqrt{2}}\left\{ |1_{12}1_{12}\rangle _{[su]}|1_{34}0_{34}\rangle _{[\bar{s}\bar{u}]}\right. \nonumber \\&\left. -\, |1_{12}0_{12}\rangle _{[su]}|1_{34}1_{34}\rangle _{[\bar{s}\bar{u}]}\right\} . \end{aligned}$$

(1)

Now it is straightforward to prove that the state above has $C=-$ by applying the charge conjugation [Eq. (2)], exchanging the diquark and antidiquark parts [Eq. (3)], and renaming the dummy indices $12 \leftrightarrow 34$ [Eq. (4)], i.e.,

$$\begin{aligned} C|11\rangle= & {} \frac{1}{\sqrt{2}}\left\{ |1_{12}1_{12}\rangle _{[\bar{s}\bar{u}]}|1_{34}0_{34}\rangle _{[su]}\right. \nonumber \\&\left. -\, |1_{12}0_{12}\rangle _{[\bar{s}\bar{u}]}|1_{34}1_{34}\rangle _{[su]}\right\} \ \end{aligned}$$

(2)

$$\begin{aligned}= & {} \frac{1}{\sqrt{2}}\left\{ |1_{34}0_{34}\rangle _{[su]} |1_{12}1_{12}\rangle _{[\bar{s}\bar{u}]}\right. \nonumber \\&\left. -\, |1_{34}1_{34}\rangle _{[su]}|1_{12}0_{12}\rangle _{[\bar{s}\bar{u}]}\right\} \ \end{aligned}$$

(3)

$$\begin{aligned}= & {} \frac{1}{\sqrt{2}}\left\{ |1_{12}0_{12}\rangle _{[su]} |1_{34}1_{34}\rangle _{[\bar{s}\bar{u}]}\right. \nonumber \\&\left. -\, |1_{12}1_{12}\rangle _{[su]}|1_{34}0_{34}\rangle _{[\bar{s}\bar{u}]}\right\} \ \end{aligned}$$

(4)

$$\begin{aligned}= & {} -|11\rangle . \end{aligned}$$

(5)