LayMuon said:
Thanks for reply.
Why I3 is zero? What type of transitions should we understand? Why is there Delta?
I don’t understand what you mean by “what type of transitions”. I also don’t know how much you know about the subject. However, I can tell you what you need to know.
The Lagrangian of the quarks iso-doublet has three symmetries:
i) The invariance under a simultaneous phase change of the up and down quark fields, U_{ B } (1), given by
\delta \Psi = - i \alpha \Psi ,
leads to the conserved current
S^{ \mu } ( x ) = \bar{ \Psi } ( x ) \gamma^{ \mu } \Psi ( x ) ,
and to the constant charge
<br />
B = \int d^{ 3 } x \ S^{ 0 } ( x ) = \int d^{ 3 } x \ \left( u^{ \dagger } (x) u ( x ) + d^{ \dagger } ( x ) d ( x ) \right) ,<br />
which we call the Baryon Number.
ii) The invariance of the theory under iso-spin transformations, SU( 2 ),
\delta \Psi = - i \vec{ \beta } \cdot \frac{ \vec{ \tau } }{ 2 } \Psi ,
leads to the conserved isotopic spin current
\vec{ J }_{ \mu } ( x ) = \bar{ \Psi } ( x ) \gamma_{ \mu } \frac{ \vec{ \tau } }{ 2 } \Psi ( x ) .
Its associated constant charges (iso-vector) are given by
<br />
T^{ a } = \frac{ 1 }{ 2 } \int d^{ 3 } x \Psi^{ \dagger } ( x ) \tau^{ a } \Psi ( x ) .<br />
iii) The invariance under U_{ em } (1) transformation, given by,
<br />
\delta \Psi = - i \epsilon ( \frac{ I }{ 6 } + \frac{ \tau^{ 3 } }{ 2 } ) \Psi ,<br />
gives us the conserved electromagnetic current
J_{ em }^{ \mu } = \bar{ \Psi } \gamma^{ \mu } \left( \frac{ I }{ 6 } + \frac{ \tau^{ 3 } }{ 2 } \right) \Psi .
This can be written as
J_{ em }^{ \mu } ( x ) = \frac{ 1 }{ 6 } S^{ \mu } ( x ) + J^{ \mu }_{ 3 } ( x )
Integrating the time component of this current leads to the well-known relation
Q_{ em } = \frac{ B }{ 6 } + T_{ 3 } .
In order to study the collisions of hadrons, we need to know the properties of all possible intermediate states. Therefore, we are led to consider the matrix elements \langle 0 | J_{ em }^{ \mu } ( 0 ) | n \rangle, where | n \rangle is a set of ( small mass) intermediate states. From the properties of the vacuum, | 0 \rangle, and J_{em}^{ \mu } (under the above symmetries), we can prove the following properties of | n \rangle:
1) | n \rangle has zero electric charge and zero Baryon number.
2) | n \rangle is an eigenstate of the charge conjugation operator C, with eigenvalue C = - 1.
3) The total angular momentum and parity (J^{ P }) of | n \rangle is 1^{ - 1 }.
4) The total isotopic spin of | n \rangle, T, is either 0 or 1.
Try to prove these properties yourself. If you got stuck on any one, you can ask me for help.
Regards
Sam
