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This is from Krane, p 389:
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, called isospin.
Nucleon has isospin number t = 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, which 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![Roll Eyes :rolleyes: :rolleyes:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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, called isospin.
Nucleon has isospin number t = 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, which 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