# 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
Sep 19, 2021 at 22:47
• @CosmasZachos Thank you, I've fixed the typos. I do know how to evaluate the exponential, at least in theory. Sep 20, 2021 at 2:52
• Near duplicate. Mar 5 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 $$\begin{pmatrix} \bar{p}' & \bar{n}' \end{pmatrix}=\begin{pmatrix} \bar{p} & \bar{n} \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1& 0 \end{pmatrix} .$$


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}}}\;$$ $$$$^{\bf 4}\OSS{U}\!\e \begin{bmatrix} \begin{array}{c|ccc} \:\: 1 \:\: &\rule [0ex]{20pt}{0.0ex}&\rule [-2.5ex]{0pt}{6.0ex} \rule [0ex]{16pt}{0ex}& \rule [0ex]{16pt}{0ex}\\ \hline \rule [-3ex]{0pt}{6ex}&g^{2} & \boldsymbol{-}\sqrt{2}gh &\boldsymbol{-}h^{2} \\ \rule [-3ex]{0pt}{6ex}& \sqrt{2}g\overline{h} & \left(g\OSB{g}\boldsymbol{-}h\overline{h}\right) & \sqrt{2}\OSB{g}h \\ \rule [-3ex]{0pt}{6ex}& \boldsymbol{-}\overline{h}^{2} & \boldsymbol{-}\sqrt{2}\OSB{g}\,\overline{h} & \OSB{g}^{2} \end{array} \end{bmatrix} = \begin{bmatrix} \begin{matrix} \begin{array}{c|ccc} ^{\bl 1}U_{\bl{\blr{1}}} \:\: &\rule [0ex]{20pt}{0.0ex}&\rule [-2.5ex]{0pt}{6.0ex} \rule [0ex]{16pt}{0ex}& \rule [0ex]{16pt}{0ex}\\ \hline \rule [-3ex]{0pt}{6ex}&\hp{g^{2}} & \hp{\m\sqrt{2}gh} &\hp{\m h^{2}} \\ \rule [-3ex]{0pt}{6ex}& \hp{\sqrt{2}g\overline{h}} & ^{\bl 3}U_{\bl{\blr{2}}} & \hp{\sqrt{2}\OSB{g}h} \\ \rule [-3ex]{0pt}{6ex}& \hp{\m\overline{h}^{2}} & \hp{\m\sqrt{2}\OSB{g}\overline{h} }& \hp{\OSB{g}^{2}} \end{array} \end{matrix} \end{bmatrix} \tl{A-06}$$$$