This is from Krane, p 389:(adsbygoogle = window.adsbygoogle || []).push({});

The neutron and the proton are treated as two different states of a single particle, the nucleon. The nucleon is assigned with a fictious spin vector, calledisospin.

Nucleon has isospin numbert = ½, a proton has [itex] m_{t} = 1/2[/itex] and neutron has [itex] m_{t} = - 1/2 [/itex].

The isospin obeys the same rules for angular momentum vecotrs.

The third component of a nucleus isospin is:

[tex] T_{3} = \frac{1}{2} (Z-N) [/tex]

For any value on [itex] T_{3} [/itex], the total isospin [itex] T [/itex] can take any value at least as great as [itex] |T_{3} | [/tex].

We consider as an example the two-nucleon system, wich can have T of 0 or 1. There are thus four possible 3-axis components: [itex] T_{3} = 1[/itex](two protons); [itex] T_{3} = - 1[/itex](two neutrons), and two combinations with [itex] T_{3} = 0 [/itex](one neutron and one proton). The first two states must have T = 1, while the latter two can have T = 0 and T =1.

- - -

Now this is really confusing me. I am think that the according to the statement:For any value on" [itex] T_{3} [/itex],the total isospin[itex] T [/itex]can take any value at least as great as[itex] |T_{3} | [/tex]." The two proton system can therefore have T = 0 or 1. And the same thing regarding the 2N system.

And also how can there be two combinations of P-N that gives [itex] T_{3} = 0 [/itex]? And why isn't just T = 0 allowed?

Should I try to think "backwards":Given a value on T, what values of [itex] T_{3} [/itex] can I have, and what combinations of N and P do they represent?

Cheers

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Isospin question

**Physics Forums | Science Articles, Homework Help, Discussion**