- 6,735
- 2,431
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 m_{t} = 1/2 and neutron has m_{t} = - 1/2.
The isospin obeys the same rules for angular momentum vecotrs.
The third component of a nucleus isospin is:
T_{3} = \frac{1}{2} (Z-N)
For any value on T_{3}, the total isospin T can take any value at least as great as |T_{3} | [/tex].<br /> <br /> 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: T_{3} = 1(two protons); T_{3} = - 1(two neutrons), and two combinations with T_{3} = 0(one neutron and one proton). The first two states must have T = 1, while the latter two can have T = 0 and T =1.<br /> <br /> - - - <br /> <br /> Now this is really confusing me. I am think that the according to the statement: <i>For any value on</i> " T_{3}, <i>the total isospin</i> T <i>can take any value at least as great as </i>|T_{3} | [/tex].&quot; The two proton system can therefore have T = 0 or 1. And the same thing regarding the 2N system.<br /> <br /> And also how can there be two combinations of P-N that gives T_{3} = 0? And why isn&#039;t just T = 0 allowed?<br /> <br /> Should I try to think &quot;backwards&quot;: <i>Given a value on T, what values of T_{3} can I have, and what combinations of N and P do they represent?</i><br /> <br /> Cheers<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f644.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":rolleyes:" title="Roll Eyes :rolleyes:" data-smilie="11"data-shortname=":rolleyes:" />
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 m_{t} = 1/2 and neutron has m_{t} = - 1/2.
The isospin obeys the same rules for angular momentum vecotrs.
The third component of a nucleus isospin is:
T_{3} = \frac{1}{2} (Z-N)
For any value on T_{3}, the total isospin T can take any value at least as great as |T_{3} | [/tex].<br /> <br /> 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: T_{3} = 1(two protons); T_{3} = - 1(two neutrons), and two combinations with T_{3} = 0(one neutron and one proton). The first two states must have T = 1, while the latter two can have T = 0 and T =1.<br /> <br /> - - - <br /> <br /> Now this is really confusing me. I am think that the according to the statement: <i>For any value on</i> " T_{3}, <i>the total isospin</i> T <i>can take any value at least as great as </i>|T_{3} | [/tex].&quot; The two proton system can therefore have T = 0 or 1. And the same thing regarding the 2N system.<br /> <br /> And also how can there be two combinations of P-N that gives T_{3} = 0? And why isn&#039;t just T = 0 allowed?<br /> <br /> Should I try to think &quot;backwards&quot;: <i>Given a value on T, what values of T_{3} can I have, and what combinations of N and P do they represent?</i><br /> <br /> Cheers<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f644.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":rolleyes:" title="Roll Eyes :rolleyes:" data-smilie="11"data-shortname=":rolleyes:" />