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Why do proton and neutron form isospin doublet? I3 or I?

  1. May 5, 2015 #1
    As far as I understand, ##I_{3}##, the component of isospin in a certain direction is additive,

    but ##I## is to be treated as a vector sum, is this correct?

    So, ##I_{3}=1/2## for ##u## quark,
    ##I_{3}=-1/2 ## for ##d## quark.

    Adding ##I_{3}## then for a proton we find ##I_{3}=1/2##
    and for a neutron ##I_{3}=-1/2##

    Is it from this that we conclude both the proton and neutron form a isospin doublet with ##I=1/2##?

    What is the formulae for ##I##? I read somewhere that ##I## is greater than or equal to ##I_{3}##,
    So is ##I= |I_{3}| + |I_{2}| + |I_{1}|## ? But if this is the reasoning for the doublet, what about ##I_{2}, I_{1}##?


    On instead should the approach be a vector sum of the isospin of the quarks. So both the down and up quarks have ##I=1/2##,

    So doing a vector sum of the three quarks gives the three possible values: ##3/2,1/2,1/2##,

    So why then, for the doublet would you take the ##1/2## and not ##3/2##. Is there an observation side to this?


    Thanks in advance.
     
  2. jcsd
  3. May 5, 2015 #2

    Orodruin

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    No, this only tells us that they are either a part of the 3/2 isospin quadruplet or of the isospin 1/2 doublet.

    The mathematics of isospin are equivalent to that of spin, as both are based on an SU(2) symmetry.

    The correct thing to say here is that the u and d quark together form an isospin doublet, i.e., I = 1/2, where I3(u) = +1/2 and I3(d) = -1/2, much in the same way that a spin 1/2 particle can be in a spin up or spin down state.

    It is not a matter of "taking". Both the quadruplet and doublet states exist. The quadruplet isospin state is fully symmetric, which means you need to anti-symmetrise your state in some other way (the quarks are fermions after all), ultimately leading to a larger mass for the quadruplet. (The quadruplet are the ##\Delta(1232)## resonances.)
     
  4. May 6, 2015 #3
    .

    Sorry could you expand here more? So once ##I_{2}, I_{3}## are taken into account?

    The definition of a doublet, quadruplet etc. is particles with the same ##I## but different ##I_{3}##, is this correct?



    I know that ##S=S_{1}+S_{2}+S_{3}##, this is in operator form. But I can't think how this explains the statement that if ##S_{3} =1## , ##S## is greater than or equal to ##1##.



    So the up and down quark form a doublet, whereas a proton and neutron can form either a ##I=1/2## doublet or are part of the ##3/2## quadruplet?
     
  5. May 6, 2015 #4

    Orodruin

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    The isospin multiplets are collections of ##2I+1## particles, one for each of the ##2I+1## possible third component values.

    This is not true. What is true is ##S^2 = S_1^2+S_2^2+S_3^2##. S is the total spin, not the sum of the components in each direction. Alternatively, you need to write it as a vector.

    The representation ##\ell## of SU(2) is a ##2\ell+1##-plet. The maximal value of the third component is ##m = \ell##, which means that if the representation contains a state with third component m, then ##\ell \geq m##.

    No, the proton and neutron are the two states of a doublet. The quadruplet made out of three up-down doublets are the delta resonances. These are different particles.
     
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