Isospin for Antiparticles

In Quarks & Leptons: An Introductory Course in Modern Particle Physics by Halzen and Martin page 42 reads:

The construction of antiparticle isospin multiplets requires care. It is well illustrated by a simple example. Consider a particular isospin transformation of the nucleon doublet, a rotation through $$\pi$$ about the 2-axis. We obtain $$\begin{pmatrix} p' \\ n' \end{pmatrix} = e^{-i \pi (\tau_2 /2)} \begin{pmatrix} p \\ n \end{pmatrix} = -i \tau_2 \begin{pmatrix} p \\ n \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1& 0 \end{pmatrix} \begin{pmatrix} p \\ n \end{pmatrix}.$$ We define antinuclear states using the particle-antiparticle conjugation operator C, $$Cp=\bar{p}, Cn = \bar{n}$$ Applying $$C$$ therefore gives $$\begin{pmatrix} \bar{p}' \\ \bar{n}' \end{pmatrix}=\begin{pmatrix} 0 & -1 \\ 1& 0 \end{pmatrix} \begin{pmatrix} \bar{p} \\ \bar{n} \end{pmatrix}.$$ However, we want the antiparticle doublet to transform in exactly the same way as the particle doublet, so that we can combine particle and antiparticle states using the same Clebsch-Gordan coefficients, and so on. We must therefore make two changes...

I do not understand what the issue is, however. What do they mean by "we want... to transform in exactly the same way"?

Didn't they just show that they do transform in exactly the same way?

• – rob
Commented Sep 19, 2021 at 22:47
• Near duplicate. Commented Mar 5, 2022 at 22:49

I do not understand what the issue is, however. What do they mean by "we want... to transform in exactly the same way"? Didn't they just show that they do transform in exactly the same way?

Yes and no; they really didn't! The antibaryon spinors you are looking at are upended! They certainly do not transform well under $$T_3$$.

Remember, the Gell-Mann Nishijima formula $$Q=T_3+ B/2$$ rules supreme here! It impels the highest-charged member of the doublet to be upstairs, in the upper position of the two-spinor, $$\begin{pmatrix} \bar{n} \\ \bar{p} \end{pmatrix}.$$ But now you have a problem: the rotation matrix you just computed is wrong by a minus sign. If you look at how the specific components transform, you see that $$\begin{pmatrix} \bar{n}' \\ \bar{p}' \end{pmatrix}=\begin{pmatrix} 0 & 1 \\ -1& 0 \end{pmatrix} \begin{pmatrix} \bar{n} \\ \bar{p} \end{pmatrix},$$ disastrously. But, as they explained, you need to be handling baryons and antibaryons even-handedly, isospinwise, as though they were spins.

But all is not lost. If you, in addition, flip the sign of the upper component, you do get the correct transformation law you had before,
$$\begin{pmatrix}- \bar{n}' \\ \bar{p}' \end{pmatrix}=\begin{pmatrix} 0 & -1 \\ 1& 0 \end{pmatrix} \begin{pmatrix} -\bar{n} \\ \bar{p} \end{pmatrix},\tag{2.41}$$ and a rephasing (the - sign) will not affect the GM-N formula in any way. Note the action of $$T_3$$ is now conventional, transformation in "exactly the same way", unlike the ugly monster you started with.

So now you may swing these two spinors, of baryons and antibaryons, as though they were spin 1/2s, and, e.g. compose two of them like (2.42).

You see then that the antisymmetric product of the spin composition, the singlet, has a counter-intuitive + sign for this symmetric isospin composition; and the antisymmetric isotriplet has a - sign, the $$\pi^0$$, unlike the symmetric spin triplet: a virtually ritual traditional point of puzzlement to novices.

This is all in a sly attempt to avoid formal mathematical language and the systematic definition of the conjugate representation of su(2). All the rigmarole that puzzled you is just a low-rent outline of the conjugate representation, geared to instant-gratification correct formulas. (NB. Remember it well if/when you learn about fermion masses in the SM via coupling to the Higgs field, which makes masses possible for both upstairs and downstairs fermions.)

You might choose the more mainstream path of the conjugate representation, like most intros to the SM, like Li& Cheng, a clearly superior reference, in comparison. Check that an equivalent way of appreciating (2.42) is to simply transpose your last equation, (2.40), to $$\begin{pmatrix} \bar{p}' & \bar{n}' \end{pmatrix}=\begin{pmatrix} \bar{p} & \bar{n} \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1& 0 \end{pmatrix} .$$

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As concerns to isospin we consider the up and down quarks $$\;\bl u, \bl d\;$$ as two states of a single particle, the quark $$\;q\in\mathsf Q$$, and so the anti-up and anti-down quarks $$\;\ol{\bl u\Vp{\bl d}},\ol{\bl d} ,\;$$ as two states of the antiquark $$\;\ol q\in\ol{\mathsf Q}$$. To examine the transformation of combinations $$\;q\ol q$$ ($$\bl\equiv\:\texttt{mesons}$$) under $$\mr{SU}\plr{2}$$ we apply a special unitary transformation $$\,^2U_{\bl q}\in \mr{SU}\plr{2}\;$$ on the Hilbert space $$\;\mathsf Q\;$$ of quarks which necessarily implies application of its complex conjugate $$\;^2\ol U_{\bl q}\;$$ on the Hilbert space $$\;\ol{\mathsf Q}\;$$ of antiquarks. But we must apply on the Hilbert space $$\;\ol{\mathsf Q}\;$$ an identical special unitary transformation $$\,^2U_{\ol{\bl q}}\bl\equiv{^2U} \bl\equiv {^2U_{\bl q}}\;$$ since, as in the spin$$\m1/2$$ case, we want the isospin 3-vector $$\;\mb I^{\ol{\bl q}}\e\plr{\mr I^{\ol q}_1,\mr I^{\ol q}_2,\mr I^{\ol q}_3}\;$$ of the antiquark to transform (rotate) in exactly the same way as the isospin 3-vector $$\;\mb I^{\bl q}\e\plr{\mr I^{q}_1,\mr I^{q}_2,\mr I^{q}_3}\;$$ of quark $$\;q$$. And this demand is due to the fact that these two isospin 3-vectors live in the same artificial 3-dimensional space $$\;\mathbb R^3\;$$ and so we must not rotate the former 3-vector differently from the latter. This incompatibility is resolved by a proper change of the basis $$\;\bl\lbrace\ol{\bl u\Vp{\bl d}},\ol{\bl d}\bl\rbrace\;$$ of the antiquark Hilbert space $$\;\ol{\mathsf Q}\;$$ as we'll see in the following.

So, consider a pair of quark-antiquark $$\;\bl\xi,\ol{\bl\zeta} \;$$ \begin{align} \bl\xi&\e\xi_1\bl u\p \xi_2\bl d \tl{01a}\\ \bl\zeta&\e\zeta_1\bl u\p \zeta_2\bl d\quad\bl\implies\quad\ol{\bl\zeta}\e\ol{\zeta_1}\,\ol{\bl u\Vp{\bl d}}\p \ol{\zeta_2}\,\ol{\bl d} \tl{01b} \end{align} Their representation by matrices is \begin{align} \bl\xi&\e\underbrace{ \begin{bmatrix} \xi_1 \vp\\ \xi_2 \vp \end{bmatrix}}_{{\lbrace\bl u,\bl d\rbrace}} \tl{02a}\\ \bl\zeta&\e\underbrace{ \begin{bmatrix} \zeta_1 \vp\\ \zeta_2 \vp \end{bmatrix}}_{{\lbrace\bl u,\bl d\rbrace}}\quad\bl\implies\quad \ol{\bl\zeta}\e\underbrace{ \begin{bmatrix} \ol{\zeta_1} \vp\\ \ol{\zeta_2} \vp \end{bmatrix}}_{{\lbrace\ol{\bl u\Vp{\bl d}},\ol{\bl d}\rbrace}} \tl{02b} \end{align} A matrix standing alone has no sense. If so, it is implied that it represents something (vector, linear operator etc) with respect to some basis of a linear space. For our case here it's important to know with respect to which basis a matrix represents a vector or a transformation. That's why we declare the basis in underbraces.

Consider now that we apply a special unitary transformation $$\;U \bl\in\mr{SU}\plr{2}\;$$ \begin{align} \bl\xi'&\e U\bl\xi \tl{03a}\\ \bl\zeta'&\e U\bl\zeta\quad\bl\implies\quad\ol{\bl\zeta'}\e\ol{U \Vp {\zeta'}}\,\ol{\bl\zeta\Vp '} \tl{03b} \end{align} with $$$$U\e\underbrace{ \begin{bmatrix} \hp\m g&\:\;h\:\:\vp\\ \m\ol{h}&\:\:\ol{g\Vp h}\:\:\vp \end{bmatrix}}_{{\lbrace\bl u,\bl d\rbrace}}\,, \quad\texttt{where}\quad g\ol{g\Vp h}\p h\ol{h}\e 1 \tl{04}$$$$ The matrix representation of equations \eqref{03a},\eqref{03b} gives \begin{align} \underbrace{ \begin{bmatrix} \xi'_1 \vp\\ \xi'_2 \vp \end{bmatrix}}_{{\lbrace\bl u,\bl d\rbrace}}&\e \underbrace{ \begin{bmatrix} \hp\m g&\:\;h\:\:\vp\\ \m\ol{h}&\:\:\ol{g\Vp h}\:\:\vp \end{bmatrix}}_{{\lbrace\bl u,\bl d\rbrace}} \underbrace{ \begin{bmatrix} \xi_1 \vp\\ \xi_2 \vp \end{bmatrix}}_{{\lbrace\bl u,\bl d\rbrace}} \tl{05a}\\ \underbrace{ \begin{bmatrix} \zeta'_1 \vp\\ \zeta'_2 \vp \end{bmatrix}}_{{\lbrace\bl u,\bl d\rbrace}}&\e \underbrace{ \begin{bmatrix} \hp\m g&\:\;h\:\:\vp\\ \m\ol{h}&\:\:\ol{g\Vp h}\:\:\vp \end{bmatrix}}_{{\lbrace\bl u,\bl d\rbrace}} \underbrace{ \begin{bmatrix} \zeta_1 \vp\\ \zeta_2 \vp \end{bmatrix}}_{{\lbrace\bl u,\bl d\rbrace}}\quad\bl\implies\quad \underbrace{ \begin{bmatrix} \ol{\zeta'_1} \vp\\ \ol{\zeta'_2} \vp \end{bmatrix}}_{{\lbrace\ol{\bl u\Vp{\bl d}},\ol{\bl d}\rbrace}}\e \underbrace{ \begin{bmatrix} \hp\m \ol{g\Vp h}&\:\:\ol{h}\:\:\vp\\ \m h&\:\:g\:\:\vp \end{bmatrix}}_{{\lbrace\ol{\bl u\Vp{\bl d}},\ol{\bl d}\rbrace}} \underbrace{ \begin{bmatrix} \ol{\zeta_1} \vp\\ \ol{\zeta_2} \vp \end{bmatrix}}_{{\lbrace\ol{\bl u\Vp{\bl d}},\ol{\bl d}\rbrace}} \tl{05b} \end{align} The complication is clear in above equations : while on the quark we apply the special unitary transformation $$\;U$$, on the antiquark we apply the special unitary transformation $$\ol{U \Vp {\zeta'}}\bl\ne U\;$$.

We override this complication expressing the relation to the right of equation \eqref{05b} as follows $$$$\left. \begin{cases} \ol{\zeta'_1}\e\hp\m\ol{g\Vp h}\,\ol{\zeta_1}\p\ol{h}\,\ol{\zeta_2} \\ \ol{\zeta'_2}\e\m h\,\ol{\zeta_1}\p g\,\ol{\zeta_2} \end{cases}\right\}\quad\bl\implies\quad \left. \begin{cases} \ol{\plr{\m\zeta'_2}}\e \hp\m g\,\ol{\plr{\m\zeta_2}}\p h\,\ol{\zeta_1} \\ \:\:\:\:\,\ol{\zeta'_1}\:\:\:\e\m\ol{h}\,\ol{\plr{\m\zeta_2}}\p\ol{g\Vp h}\,\ol{\zeta_1} \\ \end{cases}\right\} \tl{06}$$$$ so $$$$\underbrace{ \begin{bmatrix} \ol{\plr{\m\zeta'_2}} \vp\\ \ol{\zeta'_1}\vp \end{bmatrix}}_{{\lbrace \bl\m\ol{\bl d},\ol{\bl u\Vp{\bl d}}\rbrace}}\e \underbrace{ \begin{bmatrix} \hp\m g&\:\;h\:\:\vp\\ \m\ol{h}&\:\:\ol{g\Vp h}\:\:\vp \end{bmatrix}}_{{\lbrace \bl\m\ol{\bl d},\ol{\bl u\Vp{\bl d}}\rbrace}} \underbrace{ \begin{bmatrix} \ol{\plr{\m\zeta_2}} \vp\\ \ol{\zeta_1} \vp \end{bmatrix}}_{{\lbrace \bl\m\ol{\bl d},\ol{\bl u\Vp{\bl d}}\rbrace}} \tl{07}$$$$ We conclude that if in the space $$\;\ol{\mathsf Q}\;$$ of the antiquarks we use as up eigenstate of the isospin component $$\;\mr I^{\ol q}_3\;$$ the state $$\;\plr{\bl\m\ol{\bl d}}\;$$ instead of $$\;\ol{\bl d}\;$$ while keeping $$\;\ol{\bl u\Vp{\bl d}}\;$$ as the downstate then the transformation applied on the antiquark is identical to that of the quark.

To make this more clear the relation to the right of equation \eqref{01b} yields $$$$\ol{\bl\zeta}\e\ol{\zeta_1}\,\ol{\bl u\Vp{\bl d}}\p \ol{\zeta_2}\,\ol{\bl d}\quad\bl\implies\quad \ol{\bl\zeta}\e \ol{\plr{\m\zeta_2}}\plr{\bl\m\ol{\bl d}}\p\ol{\zeta_1}\,\ol{\bl u\Vp{\bl d}}\bl\equiv\underbrace{ \begin{bmatrix} \ol{\plr{\m\zeta_2}} \vp\\ \ol{\zeta_1} \vp \end{bmatrix}}_{{\lbrace \bl\m\ol{\bl d},\ol{\bl u\Vp{\bl d}}\rbrace}} \tl{08}$$$$ and $$$$\ol{\bl\zeta'}\e\ol{\zeta'_1}\,\ol{\bl u\Vp{\bl d}}\p \ol{\zeta'_2}\,\ol{\bl d}\quad\bl\implies\quad \ol{\bl\zeta'}\e \ol{\plr{\m\zeta'_2}}\plr{\bl\m\ol{\bl d}}\p\ol{\zeta'_1}\,\ol{\bl u\Vp{\bl d}}\bl\equiv\underbrace{ \begin{bmatrix} \ol{\plr{\m\zeta'_2}} \vp\\ \ol{\zeta'_1} \vp \end{bmatrix}}_{{\lbrace \bl\m\ol{\bl d},\ol{\bl u\Vp{\bl d}}\rbrace}} \tl{09}$$$$

$$\hebl$$

For the quark model of mesons consisting of two quarks $$\;\bl u\;$$ and $$\;\bl d$$, see my answer here

What is the symmetry of the pion triplet (π−,π0,π+)?, $$\color{blue}{\textbf{Example C}}$$.

For convenience we repeat the main equations of the mesons $$$$\begin{array}{ccccccccc} &\boldsymbol{\lbrace}\boldsymbol{u},\boldsymbol{d}\boldsymbol{\rbrace} \!\!\!\!\!&\boldsymbol{\otimes}& \!\!\!\!\boldsymbol{\lbrace}\OSB{\boldsymbol{u}},\overline{\boldsymbol{d}}\boldsymbol{\rbrace} & \!\!\boldsymbol{=}\!\! & \boldsymbol{\lbrace}\boldsymbol{\omega}\boldsymbol{\rbrace}& \!\!\!\!\boldsymbol{\oplus}\!\!&\boldsymbol{\lbrace}\BoldExp{\boldsymbol{\pi}}{-},\BoldExp{\boldsymbol{\pi}}{0},\BoldExp{\boldsymbol{\pi}}{+}\boldsymbol{\rbrace} & \\ & \boldsymbol{2}\!\!\!\!\! & \boldsymbol{\otimes} & \!\!\!\!\OSB{\boldsymbol{2}} & \!\!\boldsymbol{=}\!\!&\boldsymbol{1}&\!\!\!\!\boldsymbol{\oplus}\!\!&\boldsymbol{3}& \end{array} \tl{A-01}$$$$

\begin{align} &\left\{ \boldsymbol{\omega} = \sqrt{\tfrac{1}{2}}\left(\boldsymbol{u}\OSB{\boldsymbol{u}}+\boldsymbol{d}\overline{\boldsymbol{d}} \right)\hphantom{=\,}\right\} \quad \,\text{the singlet }\boldsymbol{1} \tl{A-02.1}\\ &\left. \begin{cases} \BoldExp{\boldsymbol{\pi}}{-} =\boldsymbol{d}\OSB{\boldsymbol{u}} \\ \BoldExp{\boldsymbol{\pi}}{0} =\sqrt{\tfrac{1}{2}}\left(\boldsymbol{u}\OSB{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}} \right)\\ \BoldExp{\boldsymbol{\pi}}{+} =\boldsymbol{u}\overline{\boldsymbol{d}} \end{cases}\right\}\quad \text{the triplet }\boldsymbol{3} \tl{A-02.2} \end{align}

In the above link we could see the details of the application of a special unitary transformation $$\;^{\bl 2}U\in \mr{SU}\plr{2}\;$$ on the Hilbert space of the quarks $$\;\mathsf Q$$
$$$$^{\bl 2}U \equiv \begin{bmatrix} \:\:\:g\hphantom{\boldsymbol{-}}\vphantom{\dfrac12} & h \vphantom{\dfrac12}\:\:\:\\ \!\!\!\!\!\boldsymbol{-}\overline{h} & \OSB{g}\vphantom{\dfrac12}\:\: \end{bmatrix}\,, \qquad g\OSB{g}+h\overline{h}=\vert g \vert ^{2} + \vert h \vert ^{2} = 1 \tl{A-03}$$$$ the special unitary transformation $$\;^{\bl 4}U\in \mr{SU}\plr{4}\;$$ on the product space $$\;\mathsf Q\ox\ol{\mathsf Q}$$, product itself of $$\;^{\bl 2}U\;$$ and $$\;^{\bl 2}\ol U$$ $$$$\begin{split} ^{\bl 4}U & \e \biggl(\,^{\bl 2}U\biggr)\ox\biggl(\,^{\bl 2}\ol U\biggr) \e\underbrace{\begin{bmatrix} \:\:\:g\hphantom{\boldsymbol{-}}\vphantom{\dfrac12} & h \vphantom{\dfrac12}\:\:\:\\ \!\!\!\!\!\boldsymbol{-}\overline{h} & \OSB{g}\vphantom{\dfrac12}\:\: \end{bmatrix}}_{{\lbrace\bl u,\bl d\rbrace}} \boldsymbol{\otimes} \underbrace{\begin{bmatrix} \:\:\:\OSB{g}\hphantom{\boldsymbol{-}}\vphantom{\dfrac12} & \overline{h} \vphantom{\dfrac12}\:\:\:\\ \!\!\!\!\!\boldsymbol{-}h & g\vphantom{\dfrac12}\:\: \end{bmatrix}}_{\lbrace\ol{\bl u\Vp{\bl d}},\ol{\bl d}\rbrace}\\ &\e \underbrace{\begin{bmatrix} \hp\m g\,\ol{g\Vp h} & \hp\m g\,\ol h & \hp\m h\,\ol{g\Vp h} & \!\!\!h\,\ol h \\ \bl\m g\,h & \hp\m g^{2} & \m h^{2} & hg\\ \m\ol h\,\ol{g\Vp h} & \,\m\ol h^{2} & \hp\m\ol{g\Vp h}^{2} & \OSB{g}\overline{h} \\ \hphantom{\boldsymbol{-}}\overline{h}h & \:\:\boldsymbol{-}\overline{h}g & \:-\OSB{g}h & \:\OSB{g}g \end{bmatrix}}_{\lbrace\bl u\ol{\bl u\Vp{\bl d}},\bl u\ol{\bl d},\bl d\ol{\bl u\Vp{\bl d}},\bl d\ol{\bl d}\rbrace}\\ \end {split} \tl{A-04}$$$$ which after a proper transformation of the basis in the product space $$$$\bl\lbrace\bl u\ol{\bl u\Vp{\bl d}},\bl u\ol{\bl d},\bl d\ol{\bl u\Vp{\bl d}},\bl d\ol{\bl d}\bl\rbrace \quad \bl{-\!\!\!-\!\!\!\longrightarrow}\quad \boldsymbol{\lbrace}\bl\omega,\BoldExp{\boldsymbol{\pi}}{-},\BoldExp{\boldsymbol{\pi}}{0},\BoldExp{\boldsymbol{\pi}}{+}\boldsymbol{\rbrace} \tl{A-05}$$$$ is transformed to the irreducible direct sum $$\;^{\bl 1}U_{\bl{\blr{1}}}\bl\oplus\,^{\bl 3}U_{\bl{\blr{2}}}\;$$ $$$$^{\bf 4}\OSS{U}\!\e \begin{bmatrix} \begin{array}{c|ccc} \:\: 1 \:\: &\rule [0ex]{20pt}{0.0ex}&\rule [-2.5ex]{0pt}{6.0ex} \rule [0ex]{16pt}{0ex}& \rule [0ex]{16pt}{0ex}\\ \hline \rule [-3ex]{0pt}{6ex}&g^{2} & \boldsymbol{-}\sqrt{2}gh &\boldsymbol{-}h^{2} \\ \rule [-3ex]{0pt}{6ex}& \sqrt{2}g\overline{h} & \left(g\OSB{g}\boldsymbol{-}h\overline{h}\right) & \sqrt{2}\OSB{g}h \\ \rule [-3ex]{0pt}{6ex}& \boldsymbol{-}\overline{h}^{2} & \boldsymbol{-}\sqrt{2}\OSB{g}\,\overline{h} & \OSB{g}^{2} \end{array} \end{bmatrix} = \begin{bmatrix} \begin{matrix} \begin{array}{c|ccc} ^{\bl 1}U_{\bl{\blr{1}}} \:\: &\rule [0ex]{20pt}{0.0ex}&\rule [-2.5ex]{0pt}{6.0ex} \rule [0ex]{16pt}{0ex}& \rule [0ex]{16pt}{0ex}\\ \hline \rule [-3ex]{0pt}{6ex}&\hp{g^{2}} & \hp{\m\sqrt{2}gh} &\hp{\m h^{2}} \\ \rule [-3ex]{0pt}{6ex}& \hp{\sqrt{2}g\overline{h}} & ^{\bl 3}U_{\bl{\blr{2}}} & \hp{\sqrt{2}\OSB{g}h} \\ \rule [-3ex]{0pt}{6ex}& \hp{\m\overline{h}^{2}} & \hp{\m\sqrt{2}\OSB{g}\overline{h} }& \hp{\OSB{g}^{2}} \end{array} \end{matrix} \end{bmatrix} \tl{A-06}$$$$

This is a less formal and more heuristic approach. What is confusing in Halzen and Martin's narrative is that the doublet $$\begin{pmatrix} \bar{p} \cr \bar{n} \end{pmatrix}$$ does formally transform correctly as a rotation $$\vec{\theta}=\theta\hat{n}$$ about the axis $$\hat{n}$$ by angle $$\theta$$ in the right handed sense in isospin configuration space (1-2-3 space). What they allude to with language like "transform exactly the same way as a particle doublet" is that the particle/antiparticle representations are characterized by both isospin and a hypercharge term $$Y=B$$ where $$B$$ is the baryon number. The Gell-Mann Nishijima (GMN) formula $$Q=T_3+B/2$$ links everything physically. So, for antiparticles, in addition to just trivially changing labels ($$p\rightarrow\bar{p}$$ and $$n\rightarrow\bar{n}$$), we are also inverting the baryon number so $$B=-1$$ (versus $$B=+1$$ for the particle states). This GMN convention tells us to place the highest $$Q$$ state so it is in the upper slot so $$\begin{pmatrix} 1 \cr 0 \end{pmatrix}$$ corresponds to a $$T_3=+1$$ eigenstate of the operator $$\tau_3=\begin{pmatrix}1 & 0\\\ 0 & -1\end{pmatrix}.$$

The $$\begin{pmatrix} \bar{p} \cr \bar{n} \end{pmatrix}$$ doublet, although it transforms in the right rotational right-handed sense with respect to the 1-2-3 basis, clearly violates this GMN convention since the charges are in the wrong order with $$|\bar{p}\rangle=\begin{pmatrix} 1 \cr 0 \end{pmatrix}.$$

However, when you try and naively fix this GMN issue by just swapping the doublet roles so the larger charge, $$\bar{n}$$, is in the upper slot, $$\begin{pmatrix} \bar{n} \cr \bar{p} \end{pmatrix},$$ now with $$|\bar{n}\rangle=\begin{pmatrix} 1 \cr 0 \end{pmatrix},$$ this flipping of the basis vectors puts the Hilbert space in an effective left-handed coordinate system.

This is heuristic, but just sketch the 2D Hilbert space starting from the original $$\begin{pmatrix} \bar{p} \cr \bar{n} \end{pmatrix}$$ with $$|\bar{p}\rangle$$ as the +horizontal axis and $$|\bar{n}\rangle$$ as the +vertical axis. This is the right-handed system (which just changed the label names of the original axes) matches the handedness 1-2-3 isospin space and all the transformation conventions. A swap of the axes roles creates a left-handed coordinate system in Hilbert space. Unfortunately, the isospin configuration space and all the unitary transformations were set up in a right-handed way (1-2-3 space and all the angle conventions as well as the $$\vec{\tau}$$ representation conventions were created with this right-handed sense). Mixing the right and left handed transformation senses of the different spaces is very confusing and nothing will quite work the same way.

So, if you want all of your usual unitary transformations to work the same way, particularly when you start combining representations in product spaces and expanding product spaces in superposition, you have to force the Hilbert space to be right-handed again. This is done by just defining $$-|\bar{n}\rangle=\begin{pmatrix} 1 \cr 0 \end{pmatrix}$$ so the antiparticle doublet looks like $$\begin{pmatrix} -\bar{n} \cr \bar{p} \end{pmatrix},$$ which is back to a right-handed coordinate system in Hilbert space while also respecting the GMN convention. Just sketch it and convince yourself that defining the +horizontal axis this way has the same right-handedness of the original coordinate system before you flipped $$\bar{n}$$ and $$\bar{p}$$.

A more conceptually familiar way of appreciating why this works is by analogy with ordinary spin SU(2). When you set up your 3D configuration space (x, y, z), you set up the coordinates in a right-handed sense. Your 2D Hilbert space in the fundamental representation is made of the usual orthonormal $$|\uparrow\rangle=\begin{pmatrix} 1 \cr 0 \end{pmatrix} \text{ and } |\downarrow\rangle=\begin{pmatrix} 0 \cr 1 \end{pmatrix}.$$

As column vectors, these are selected, for their $$\sigma_z$$ eigenvalues of $$\pm 1$$ when $$\sigma_z=\begin{pmatrix}1 & 0\\\ 0 & -1\end{pmatrix}$$ is in the familiar $$z$$-basis. The column vectors will conventionally represent the +horizontal and +vertical unit basis vectors in Hilbert space, so also represent the "right-handed" sense of these coordinates. From this setup, all the unitary transformations on states in Hilbert space then assume this right-handed convention for how states are transformed. That is, a transformation like $$U(\theta)=e^{-i\vec{\theta}\cdot\vec{\sigma}/2}$$ on any state on this Hilbert space, along with the associated rotation matrix in configuration space, assumes this right handed convention; it assumes this right-handed convention for $$\vec{\theta}$$ in configuration space and also for the specific form of the representation of all the $$\vec{\sigma}$$.

With this in mind, after setting all this up, we might naively decide to switch to a new basis convention in Hilbert space like $$|\downarrow\rangle=\begin{pmatrix} 1 \cr 0 \end{pmatrix} \text{ and } |\uparrow\rangle=\begin{pmatrix} 0 \cr 1 \end{pmatrix}.$$ These are, after all, still orthnormal and superficially obey the correct quantum projection rules. The $$\sigma_z$$ eigenvalues are also now reversed from what you are used to, but you might be willing to live with this as a mere convention (but would be confusing). However, even worse, this is now a left-handed coordinate system in Hilbert space with respect to all the transformation rules you worked hard to develop based on the right-handed conventions (e.g. x-y-z coordinate orientation in configuration space, $$\vec{\theta}$$, $$\vec{\sigma}$$, and thus unitary transformation forms in Hilbert space).

To see the left-handed character of this new basis in Hilbert space, just sketch the two axes and compare them to the conventional ones. You can't get from the old one to the new ones by a rotation alone. You have to use a discrete reflection, which messes up all the transformation rules.

But if you wanted to use all the right-handed transformation rules you know and love while still switching the roles of $$|\downarrow\rangle$$ and $$|\uparrow\rangle$$, you could do a trick where you define $$-|\downarrow\rangle=\begin{pmatrix} 1 \cr 0 \end{pmatrix} \text{ and } |\uparrow\rangle=\begin{pmatrix} 0 \cr 1 \end{pmatrix}.$$

This system is back to being right-handed and so respects all the transformation rules again.