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

In summary, the hadronic em-current contains a part that transforms like a SCALAR (iso-scalar, I = 0) and a part that transforms like the 3rd component of an ISO-VECTOR (I = 1, I_{ 3 } = 0). The two pieces are separately conserved, reflecting conservation of Baryon number and 3rd component of iso-spin.
  • #1
LayMuon
149
1
We know that electromagnetic current can be written as [tex] j^{\mu}_{em} = \frac{1}{6} \bar{Q} \gamma^\mu Q + \bar{Q} \gamma^\mu \frac{\tau^3}{2} Q [/tex] where [itex] Q = \begin{pmatrix} u \\ d \end{pmatrix} [/itex]. We say this induces a transition with [itex] |\Delta I|=1, \Delta I_z =0 [/itex]. What should we understand under that last statement? How do we see that?
 
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  • #2
It means that the hadronic em-current (to lowest order in e) contains a part which transforms like a SCALAR ( iso-scalar, [itex]I = 0[/itex]) under [itex]SU(2)[/itex] and a part which transforms like the 3rd component of an ISO-VECTOR ([itex]I = 1 , I_{ 3 } = 0[/itex]):
[tex]
J^{ \mu }_{ \mbox{ em } } ( x ) = J^{ \mu }_{ \mbox{ em } , I = 0 } ( x ) + J^{ \mu }_{ \mbox{ em } , I = 1 , I_{ 3 } = 0 } ( x )
[/tex]
or
[tex]
J^{ \mu }_{ \mbox{ em } } ( x ) = S^{ \mu } ( x ) + V^{ \mu }_{ 3 } ( x )
[/tex]
The two pieces are separately conserved, reflecting conservation of Baryon number and 3rd component of iso-spin. Indeed, it is easy to see that
[tex]
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 .
[/tex]
This means that [itex]V^{ \mu }_{ 3 } ( 0 ) | 0 \rangle[/itex] is an eigenstate of [itex]I^{ 2 }[/itex] with eigenvalue [itex]1 ( 1 + 1 ) = 2[/itex], i.e. the iso-vector part of the hadronic em-current connects the vacuum to states with [itex]I = 1 , I_{ 3 } = 0[/itex].

I leave you to conclude that
[tex]
I ( I + 1 ) \langle 0 | S^{ \mu } ( 0 ) | I , I_{ 3 } = 0 \rangle = 0 ,
[/tex]
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
 
  • #3
Thanks for reply.

Why I3 is zero? What type of transitions should we understand? Why is there Delta?
 
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  • #4
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, [itex]U_{ B } (1)[/itex], given by

[tex]\delta \Psi = - i \alpha \Psi ,[/tex]

leads to the conserved current

[tex]S^{ \mu } ( x ) = \bar{ \Psi } ( x ) \gamma^{ \mu } \Psi ( x ) ,[/tex]

and to the constant charge

[tex]
B = \int d^{ 3 } x \ S^{ 0 } ( x ) = \int d^{ 3 } x \ \left( u^{ \dagger } (x) u ( x ) + d^{ \dagger } ( x ) d ( x ) \right) ,
[/tex]
which we call the Baryon Number.

ii) The invariance of the theory under iso-spin transformations, [itex]SU( 2 )[/itex],

[tex]\delta \Psi = - i \vec{ \beta } \cdot \frac{ \vec{ \tau } }{ 2 } \Psi ,[/tex]

leads to the conserved isotopic spin current

[tex]\vec{ J }_{ \mu } ( x ) = \bar{ \Psi } ( x ) \gamma_{ \mu } \frac{ \vec{ \tau } }{ 2 } \Psi ( x ) .[/tex]

Its associated constant charges (iso-vector) are given by

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

iii) The invariance under [itex]U_{ em } (1)[/itex] transformation, given by,

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

gives us the conserved electromagnetic current

[tex]J_{ em }^{ \mu } = \bar{ \Psi } \gamma^{ \mu } \left( \frac{ I }{ 6 } + \frac{ \tau^{ 3 } }{ 2 } \right) \Psi .[/tex]

This can be written as

[tex]J_{ em }^{ \mu } ( x ) = \frac{ 1 }{ 6 } S^{ \mu } ( x ) + J^{ \mu }_{ 3 } ( x )[/tex]

Integrating the time component of this current leads to the well-known relation

[tex]Q_{ em } = \frac{ B }{ 6 } + T_{ 3 } .[/tex]

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 [itex]\langle 0 | J_{ em }^{ \mu } ( 0 ) | n \rangle[/itex], where [itex]| n \rangle[/itex] is a set of ( small mass) intermediate states. From the properties of the vacuum, [itex]| 0 \rangle[/itex], and [itex]J_{em}^{ \mu }[/itex] (under the above symmetries), we can prove the following properties of [itex]| n \rangle[/itex]:

1) [itex]| n \rangle[/itex] has zero electric charge and zero Baryon number.

2) [itex]| n \rangle[/itex] is an eigenstate of the charge conjugation operator [itex]C[/itex], with eigenvalue [itex]C = - 1[/itex].

3) The total angular momentum and parity ([itex]J^{ P }[/itex]) of [itex]| n \rangle[/itex] is [itex]1^{ - 1 }[/itex].

4) The total isotopic spin of [itex]| n \rangle[/itex], [itex]T[/itex], is either [itex]0[/itex] or [itex]1[/itex].

Try to prove these properties yourself. If you got stuck on any one, you can ask me for help.

Regards

Sam
 
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  • #5
thanks.
 

1. What is an electromagnetic transition?

An electromagnetic transition is a process in which a nucleus undergoes a change in its energy level by emitting or absorbing a photon, which is a particle of electromagnetic radiation.

2. What does ΔI mean in terms of electromagnetic transitions?

The symbol ΔI represents the change in the nuclear spin of a nucleus during an electromagnetic transition. In this case, |ΔI|=1 means that the nuclear spin changes by one unit, and ΔIz=0 indicates that the change is only in the z-direction.

3. How are electromagnetic transitions classified?

Electromagnetic transitions are classified based on the type of radiation emitted or absorbed. This includes gamma rays, X-rays, ultraviolet, visible light, infrared, and radio waves.

4. What is the significance of |ΔI|=1, ΔIz=0 in electromagnetic transitions?

This specific type of electromagnetic transition is known as an E1 transition and is important in nuclear spectroscopy. It is also used in studies of nuclear structure and in determining the properties of excited nuclear states.

5. Can you provide an example of an electromagnetic transition with |ΔI|=1, ΔIz=0?

An example of such a transition is the decay of an excited state of the nucleus 60Co to its ground state. During this process, a gamma ray of energy 1.17 MeV is emitted, and the nuclear spin changes from I=2 to I=1, with no change in the z-component of the spin.

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