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

[binbagsss]

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.

 

[Orodruin]

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 $I=1/2$?

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

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 $I_{2}, I_{1}$?

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

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

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.

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?

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.)

 

[binbagsss]

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

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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?

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

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 $1$.

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.

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?

 

[Orodruin]

The definition of a doublet, quadruplet etc. is particles with the same II but different I3I_{3}, is this correct?

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

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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 $1$.

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 $\ell \geq m$.

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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?

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.