The conjugate representation in $\mathfrak{su}(2)$

Question

Cheng & Li gives the following problem:

Let $\psi_1$ and $\psi_2$ be the bases for the spin-1/2 representation of $\mathfrak{su}(2)$ and that for the diagonal operator $T_3$, \begin{align} T_3\psi_1 &= \frac{1}{2} \psi_1 \\ T_3\psi_2 &= -\frac{1}{2}\psi_2 \end{align} What are the eigenvalues of $T_3$ acting on $\psi_1^*$ and $\psi_2^*$ in the conjugate representation?

I originally thought this problem was trivial, just take the complex conjugate of both sides and use the fact that $T_3$ is real valued to get that $T_3\psi_1^* = \frac{1}{2}\psi_1^*$, but this is wrong.

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If we start from the arbitrary transformation $\psi'_i = U_{ij}\psi_j$ and complex conjugate both sides, we get ${\psi'}_i^* = U_{ij}^* \psi_j^*$. But for traceless Hermitian matrices such as $U$, there exists an $S \in \mathfrak{su}(2)$ such that $S^{-1}US = U^*$, and so, writing the previous equation in matrix form: \begin{align} \psi'^* = (S^{-1}US)\psi^* \implies S\psi'^* = U(S\psi^*) \end{align} So $S\psi^*$ transforms as $\psi$. It turns out that in the Pauli representation that $S = i\sigma^2$, and so: \begin{align} T_3\left(\begin{matrix} \psi_2^* \\ -\psi_1^*\end{matrix}\right) = \left(\begin{matrix} 1/2 & 0 \\ 0 & -1/2\end{matrix}\right) \left(\begin{matrix} \psi_2^* \\ -\psi_1^* \end{matrix}\right) \end{align}

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What I don't understand is why we couldn't just take the complex conjugate of both sides? Is this quantity $\psi^*$ not the traditional "algebraic" complex conjugate of $\psi$? If so, why could we complex conjugate $\psi'_i = U_{ij}\psi_j$ to get $\psi^*$? I feel like I thought I understood the conjugate representation but I clearly don't and I would appreciate any help understanding it.

Answer 1

What I don't understand is why we couldn't just take the complex conjugate of both sides?

Look at the Lie algebra which all representations need to satisfy, $$ [T_j,T_k]=i\epsilon_{jkm}T_m . $$ The generators are all Hermitean and the structure constants real, so this algebra is invariant under Hermitean conjugation. It is also invariant under similarity transformations $T_j\mapsto S^{-1}T_jS$, which provide useful changes of basis.

Now skip the transposition, and simply complex conjugate instead, $$ [T^*_j,T^*_k]=-i\epsilon_{jkm}T^*_m . $$
Do you have a representation of the algebra? Not really, since the r.h.s. sign difference spoils the broth--it is not quite the same algebra.

But, wait, $-T^*_j$ do provide a representation of the algebra. Moreover, luckily, $-T^*_j=S^{-1}T_j S$, so this turns out to be just the original rep in a different basis! The eigenvectors have moved and mutated, so the same eigenvalues are merely interchanged. I assume you have learned how to find S for the fundamental rep, since you already used it in upending your doublet ψ and slipping in a preferential sign--this is what $\sigma_2$ does.

Now consider the eigenvalues. The eigenvalues of $T_3$ are always paired, $\pm$, for all representations (all spins); and, moreover, all generators may be similarity-rotated to $T_3$. So S always exists, and merely scrambles the eigenvalues: all reps are real.

• A small point of notation: You might possibly be alarmed that a $-a^* \sim a$ situation would be called "real", when it is purely imaginary. But pure imaginary is just i times real. This is but an artifact of the "physics" choice of the Lie algebra convention, with an i in front of the real structure constant in a realization with Hermitean, not real, generators. (The adjoint representation consists of i multiplying real structure constants, so $S=1\!\!1$. In the undergraduate classical mechanics "Cartesian basis", one normalizes away the is to obtain real antisymmetric generators.) So a measly minus sign does not really matter.

This is a "good thing". If you looked at the anticommutator of two generators as above and complex conjugated again, if there were a non vanishing so-called d-coefficient on the right-hand-side beyond the identity, Hermiticity would require the i be missing, and so $-T^*_j$ would not satisfy the same anticommutation relation... there wouldn't be such an S preserving it.

So, for these real representations, d vanishes (and the anomaly coefficients based on these ds also vanish, for all representations of SU(2)).

This does not quite happen for the bigger SU(N)s, since not all of their representations are real. (You can illustrate this by looking at the eigenvalues of, e.g. the fundamental rep generators of SU(3), the Gell-Mann matrices. Hint: are the eigenvalues of $\lambda_8$ $\pm$-paired as above?) But, as you can see by inspection, the adjoint representation is always a real one (i times the real structure constants; and you may know how its eigenvalues are paired).

• An "academic aside": The conjugation rule for the doublet you illustrated, $(\psi_1, \psi_2)\mapsto (\psi_2^*, -\psi_1^*)$, is especially fortunate in the complex Higgs doublet of the EW SM. It allows you to write it compactly as $$ \Bigl((v+h)1\!\!1 + i \vec{\pi}\cdot \vec{\tau}\Bigr)\left(\begin{array}{c} 0\\ 1\end {array}\right) ~,$$ upon which its conjugate is but $$ \Bigl((v+h)1\!\!1 + i \vec{\pi}\cdot \vec{\tau}\Bigr)\left(\begin{array}{c} 1\\ 0\end {array}\right) ~,$$ of substantial utility in parsing out the custodial symmetries of the SM.