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Isospin current

  1. Mar 23, 2013 #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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  3. Mar 25, 2013 #2

    samalkhaiat

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    Science Advisor

    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
     
  4. Mar 25, 2013 #3
    Thanks for reply.

    Why I3 is zero? What type of transitions should we understand? Why is there Delta?
     
    Last edited: Mar 25, 2013
  5. Mar 28, 2013 #4

    samalkhaiat

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    Science Advisor

    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
     
    Last edited: Mar 28, 2013
  6. Mar 31, 2013 #5
    thanks.
     
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