I Four neutrons form a transient isolated entity - a tetraneutron?

Astronuc
Staff Emeritus
Science Advisor
Gold Member
Messages
22,340
Reaction score
7,138
I was reading another article when this headline from June 22 caught my attention.

Collisions hint that four neutrons form a transient isolated entity​

https://www.nature.com/articles/d41586-022-01634-x

An experiment firing helium-8 nuclei at a proton target has generated evidence that four neutrons can exist transiently without any other matter. But doubts remain, because the existence of such systems is at odds with theory.

Unfortunately, one has to subscribe or otherwise have access through one's institution. I haven't read the paper yet. Helium-8 is already pretty exotic.

Meanwhile, there is an apparently related paper with open access.

Observation of a correlated free four-neutron system​

https://www.nature.com/articles/s41586-022-04827-6

I've noticed relate threads on PF. In 2010, there was apparently no evidence of a tetraneutron system, but by 2016, there was some possible indications. Now apparently, there are claims of such as system.

Edit/update: I was reading a post in Nuclear Engineering which included a Nature article.
https://www.physicsforums.com/threads/what-is-the-ignition-cliff.1017073/
 
Last edited:
  • Like
Likes ohwilleke and vanhees71
Physics news on Phys.org
I'm not sure what - if anything - this means.

The tetraneutron is unbound. They don't seem to dispute that. Is there some unbound state where the 4 neutrons are briefly close together? If you roll a ball onto the top of the hill, so it has just barely enough energy to go over the top and down the other side, it will move slowly at the top and spend a lot of time there. I don't see anything beyond the analogy here.

That doesn't mean there's no interest in it. Just that I don't know what the point is to say "unbound but..."
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top