Understanding Isospin in Neutron/Proton System: T & T3

  • Thread starter Thread starter malawi_glenn
  • Start date Start date
  • Tags Tags
    Isospin
malawi_glenn
Science Advisor
Messages
6,735
Reaction score
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> &quot; T_{3}, <i>the total isospin</i> T <i>can take any value at least as great as </i>|T_{3} | [/tex].&amp;quot; The two proton system can therefore have T = 0 or 1. And the same thing regarding the 2N system.&lt;br /&gt; &lt;br /&gt; And also how can there be two combinations of P-N that gives T_{3} = 0? And why isn&amp;#039;t just T = 0 allowed?&lt;br /&gt; &lt;br /&gt; Should I try to think &amp;quot;backwards&amp;quot;: &lt;i&gt;Given a value on T, what values of T_{3} can I have, and what combinations of N and P do they represent?&lt;/i&gt;&lt;br /&gt; &lt;br /&gt; Cheers&lt;img src=&quot;https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f644.png&quot; class=&quot;smilie smilie--emoji&quot; loading=&quot;lazy&quot; width=&quot;64&quot; height=&quot;64&quot; alt=&quot;:rolleyes:&quot; title=&quot;Roll Eyes :rolleyes:&quot; data-smilie=&quot;11&quot;data-shortname=&quot;:rolleyes:&quot; /&gt;
 
Physics news on Phys.org
"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:
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.
 
okay, I think I got it now. Thanx a lot dude! =)
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top