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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.

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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

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# Isospin question

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