As stated in the question title, I am trying to prove the relation (pg. 144 of Griffiths' Introduction to Elementary Particle' or pg. 126 on Riazuddin's 'High Energy Physics') concerning the eigenvalues of a multiplet of isospin $I$ and charge conjugation number $C$: $$G=(-1)^IC,$$ My problem: I am guided to do it by proving first, an here is where I have my problem, that $$e^{-i\pi \hat{I_2} }|I,0\rangle=(-1)^I|I,0\rangle,$$ where $\hat{I_2}$ stands for the component $2$ of the isospin operator.

My attempt: I have attempted the proof by expanding the exponential factor in the series, by using $\hat I_2=\sigma_y$ without any success. I have also checked the original Yang, Lee paper where they describe the 4-matrix representation of $\hat I_2$: $$\hat{I_2}=\begin{bmatrix} 0 & -i&0 &0\\i&0&0&0\\0&0&0&-i\\0&0&i&0 \end{bmatrix},$$ but again achieving nothing.

Any help would be appreciated.


1 Answer 1


I will desist from doing your homework problem for you, so I won't do your isospin rotation as your text presumably wishes you to do it, by analogy to angular momentum rotation matrices $d^I_{00}(\pi)=\langle I,0 | \exp (i \pi I_2)|I,0\rangle$ . I'll just get the point across in a geometric way, as evoked by your text.

Observe that, for an isosinglet, your isorotation will have no effect, $I_2 |I(=0),0\rangle=0$; while for an isotriplet, I=1, just about the only case you'll encounter, you know from the Rodrigues' rotation formula for vectors that, for an antisymmetric generator of a rotation around unit axis k, $$ \mathbf{K}= \left (\begin{array}{ccc} 0 & -k_z & k_y \\ k_z & 0 & -k_x \\ -k_y & k_x & 0 \end{array}\right )~; $$ so that the orthogonal finite rotation matrix is just $$ \mathbf{R} (\theta)=e^{\theta \mathbf{K}} = 1\!\!1 + (\sin\theta) \mathbf{K} + (1-\cos\theta)\mathbf{K}^2 ~. $$ (Note your isospin generators are hermitean, not antisymmetric, so $K_2\mapsto iI_2$. You may find them in conventional isotriplet notation in WP and, since they are strictly equivalent, they would yield the same conclusion, with the neutral pion now in the middle of the triplet.).

Thus for $R_2$, an isorotation around the 2/y axis by π, a parity reflection in isospace flipping the charges of the charged pions, ($k_y=1$, $k_x=k_z=0$), the above formula collapses to $$ R_2=1\!\!1+2\mathbf{K}^2= \mathrm {diag} (-1,1,-1). $$

Recall that the eigenstate of $K_z$ with eigenvalue 0 in this specific representation is (0,0,1), hence its eigenvalue w.r.t. the action of $R_2$ so found is -1.

$R_2$ actually is the conjugacy operation on SU(2), so, combined with C, yields G commuting with isospin, as observed by Charlie Goebel in 1956.

It is actually not that hard to prove that for the isoquintet, I=2, Curtright, Fairlie, Zachos (2014), the eigenvalue is +1, as per the generic formula.

A note on your attempt: it could be pursued, but with blood, sweat, and tears. What you have is the 1/2 x 1/2 reducible quartet rep, in bad normalization. You may also work out the antidiagonal 4x4 C, take it out of their G, confirm it is R(2) proportional to their I(2), by coincidence, and reduce the rep into a triplet and a singlet, and Clebsch to observe the opposite signs for the $\pi^0$ versus the $\eta$. You might be able to climb the Eiger in slippers, but do you really want to?

  • $\begingroup$ Thanks for the edit and for the complete answer. My attempt was the only way I could think of solving it, so I don't really have any special crush in it, so, I prefer to climb the Eiger in an easier way. Clarified that, I'm working on your proposed way to solve it at the moment! $\endgroup$
    – mcejalvo
    Commented Jun 8, 2017 at 8:40
  • $\begingroup$ yes, given the definition, it amounts to a trivial symmetry of the Legendre polynomials. $\endgroup$ Commented Jun 8, 2017 at 16:10

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