# What's the mass of $D_1(2430)^{\pm}$?

What's the mass of $$D_1(2430)^{\pm}$$? The standard reference http://pdglive.lbl.gov/ParticleGroup.action?init=0&node=MXXX035 only shows the data for $$D_1(2430)^0$$ but not for $$D_1(2430)^{\pm}$$. For some other particles that similarly have "cousins" with different electric charges, the listing will

1. either have a separate page for the charged ones and another for the neutral ones (e.g. a page labeled $$D_0^*(2400)^{\pm}$$ and another page labeled $$D_0^*(2400)^{0}$$), or
2. one page for all the differently charged "cousins" (e.g. one page labeled $$b_1(1235)$$ that supposedly should work for $$b_1(1235)^0$$, $$\overline{b}_1(1235)^0$$, $$b_1(1235)^+$$, $$b_1(1235)^-$$.)
• What makes you think it isn't 2430 MeV? That is how I understand the nomenclature. – rob Dec 29 '18 at 3:38
• The fact many particles whose name have such a parenthetical integer in Review of Particle Physics actually have a different and/or a more accurate mass value. – qazwsx Dec 29 '18 at 3:40
• For example, pdglive.lbl.gov/DataBlock.action?node=M185M – qazwsx Dec 29 '18 at 3:52

You have also asked a similar question for the b1 meson and were told in the comments that the accuracy of measuring the masses of these resonances, and possibly the data available , is less than the minimum sigmas (of PDG fits to values of different experiments) away from each other. In this case the mass in the parenthesis is the mass of the resonance, and neutral to charged cannot be separated with any accuracy.

To get a feeling for magnitudes in MeV look at the $$D^0$$ mass and errors versus $$D^+$$ ,page 7 here for acceptable errors .

The mass difference is 4.822+/-0.015 MeV . It means that there are many experiments measuring masses with accuracy.

For the ones where there is no separation of the masses ,charged from neutral, in general it means that there is no statistical separation in the measured values, also possibly not enough experiments. The resonance signal is there, but the calculated errors are too large to separate the masses of charged to neutral with any accuracy.

The particular D you are asking about seems to have only a zero charged mode anyway, there is one experiment that has seen it and its decay mode is to a $$D^*(2010)$$ $$π^-$$. The width is large and with large errors, so it needs more experimental evidence, though it has been published. You have to go to the original paper to see whether it is an only neutral resonance or more measurements are needed to clear up its structure.

• So the fact $D_1(2430)^0$ has its own page and explicitly labels the charge with a superscript 0 is not significant? The data on that page also applies to $D_1(2430)^{\pm}$? If that's true, I'd expect the particle's name in that page is simply $D_1(2430)$ as many other similar cases, e.g. pdglive.lbl.gov/DataBlock.action?node=M099M labeled with $K_1(1650)$ but applies for all the different charge states $K_1(1650)^0$, $\bar{K}_1(1650)^0$, $K_1(1650)^+$, and $K_1(1650)^-$ due to the experimental data's limited accuracy. – qazwsx Dec 29 '18 at 4:24
• No , it could also be a neutral only resonance as the ω or η. It means no experiment has seen a charged mode, so it is an open question for the next experiements – anna v Dec 29 '18 at 4:28
• So, basically $D_1(2430)^{\pm}$ have not been observed, and that's why PDG lists the mass explicitly for the neutral $D_1(2430)^{0}$ only. But for the example of $K_1(1650)$, all those four "cousins" have been observed, but the accuracy is not good enough to have their different masses, so there is only one page labeled $K_1(1650)$. Right? – qazwsx Dec 29 '18 at 4:44
• yes, thats it in a nutshell – anna v Dec 29 '18 at 4:45
• I added the two examples, that should clarify why it was confusing to me. – qazwsx Dec 29 '18 at 4:53