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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].<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: [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.<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> " [itex]T_{3}[/itex], <i>the total isospin</i> [itex]T[/itex] <i>can take any value at least as great as </i>[itex]|T_{3} | [/tex]." 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 [itex]T_{3} = 0[/itex]? And why isn't just T = 0 allowed?<br /> <br /> Should I try to think "backwards": <i>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?</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:" />[/itex][/itex]
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].<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: [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.<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> " [itex]T_{3}[/itex], <i>the total isospin</i> [itex]T[/itex] <i>can take any value at least as great as </i>[itex]|T_{3} | [/tex]." 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 [itex]T_{3} = 0[/itex]? And why isn't just T = 0 allowed?<br /> <br /> Should I try to think "backwards": <i>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?</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:" />[/itex][/itex]