# Isospin question

1. Jun 23, 2007

### malawi_glenn

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 $|T_{3} | [/tex]. We consider as an example the two-nucleon system, wich can have T of 0 or 1. There are thus four possible 3-axis components: [itex] T_{3} = 1$(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.

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Now this is really confusing me. I am think that the according to the statement: For any value on " $T_{3}$, the total isospin $T$ can take any value at least as great as $|T_{3} | [/tex]." The two proton system can therefore have T = 0 or 1. And the same thing regarding the 2N system. And also how can there be two combinations of P-N that gives [itex] T_{3} = 0$? And why isn't just T = 0 allowed?

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

Cheers

2. Jun 23, 2007

### Meir Achuz

"The two proton system can therefore have T = 0 or 1. And the same thing regarding the 2N system."

The two p system has T_3=+1, so T cannot equal zero.
The two n system has T_3=-1, so T cannot equal zero.

Last edited: Jun 23, 2007
3. Jun 23, 2007

### Meir Achuz

T_3=0 can come from the two different combinations
(pn+np)/sqrt{2} for T=1, and
(pn-np)/sqrt{2} for T=0.

4. Jun 24, 2007

### malawi_glenn

okay, I think I got it now. Thanx a lot dude! =)