Understanding Electromagnetic Transitions: |ΔI|=1, ΔIz=0

[LayMuon]

We know that electromagnetic current can be written as $$j^{\mu}_{em} = \frac{1}{6} \bar{Q} \gamma^\mu Q + \bar{Q} \gamma^\mu \frac{\tau^3}{2} Q$$ where $Q = \begin{pmatrix} u \\ d \end{pmatrix}$. We say this induces a transition with $|\Delta I|=1, \Delta I_z =0$. What should we understand under that last statement? How do we see that?

 

[samalkhaiat]

It means that the hadronic em-current (to lowest order in e) contains a part which transforms like a SCALAR ( iso-scalar, $I = 0$) under $SU(2)$ and a part which transforms like the 3rd component of an ISO-VECTOR ($I = 1 , I_{ 3 } = 0$):
$$J^{ \mu }_{ \mbox{ em } } ( x ) = J^{ \mu }_{ \mbox{ em } , I = 0 } ( x ) + J^{ \mu }_{ \mbox{ em } , I = 1 , I_{ 3 } = 0 } ( x )$$
or
$$J^{ \mu }_{ \mbox{ em } } ( x ) = S^{ \mu } ( x ) + V^{ \mu }_{ 3 } ( x )$$
The two pieces are separately conserved, reflecting conservation of Baryon number and 3rd component of iso-spin. Indeed, it is easy to see that
$$I^{ 2 } V^{ \mu }_{ 3 } ( 0 ) | 0 \rangle = I_{ i } [ I_{ i } , V^{ \mu } ( 0 ) ] | 0 \rangle = \epsilon_{ i k 3 } \epsilon_{ i k l } V^{ \mu }_{ l } ( 0 ) | 0 \rangle = 2 V^{ \mu }_{ 3 } ( 0 ) | 0 \rangle .$$
This means that $V^{ \mu }_{ 3 } ( 0 ) | 0 \rangle$ is an eigenstate of $I^{ 2 }$ with eigenvalue $1 ( 1 + 1 ) = 2$, i.e. the iso-vector part of the hadronic em-current connects the vacuum to states with $I = 1 , I_{ 3 } = 0$.

I leave you to conclude that
$$I ( I + 1 ) \langle 0 | S^{ \mu } ( 0 ) | I , I_{ 3 } = 0 \rangle = 0 ,$$
which means that the iso-scalar part of the em-current, only connects the vacuum to states of total iso-spin zero.
Good luck

Sam

 

[LayMuon]

Thanks for reply.

Why I3 is zero? What type of transitions should we understand? Why is there Delta?

 

[samalkhaiat]

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

$$B = \int d^{ 3 } x \ S^{ 0 } ( x ) = \int d^{ 3 } x \ \left( u^{ \dagger } (x) u ( x ) + d^{ \dagger } ( x ) d ( x ) \right) ,$$
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

$$T^{ a } = \frac{ 1 }{ 2 } \int d^{ 3 } x \Psi^{ \dagger } ( x ) \tau^{ a } \Psi ( x ) .$$

iii) The invariance under $U_{ em } (1)$ transformation, given by,

$$\delta \Psi = - i \epsilon ( \frac{ I }{ 6 } + \frac{ \tau^{ 3 } }{ 2 } ) \Psi ,$$

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

 

[LayMuon]

thanks.