In the book QFT by Ryder on the topic $\rm SU(2)$ and the rotation group, it is stated that,

The group $\rm SU(2)$ consists of $2\times2$ unitary matrices with unit determinant,

$$UU^\dagger = 1, \qquad \det U=1 \tag{1}\label{1}$$

Let $$U = \begin{bmatrix}a & b\\ c & d\end{bmatrix}; \tag{2}\label{2}$$ applying the unitary condition leads to,

$$U = \begin{bmatrix}a & b\\ -b^* & a^*\end{bmatrix}, \qquad |a|^2+|b|^2=1. \tag{3}\label{3}$$

This is regarded as the transformation matrix in a 2-d complex space with a basic spinor $\xi$ that transforms under $\rm SU(2)$ as,

$$\xi \rightarrow \xi' = U \xi, \quad \xi^\dagger \rightarrow \xi'^\dagger = \xi^\dagger U^\dagger, \qquad \xi = \begin{bmatrix}\xi_1\\ \xi_2\end{bmatrix} .\tag{4}\label{4}$$

It is clear that $$\xi^\dagger\xi = \vert \xi_1\vert^2+\vert \xi_2\vert^2 \tag{5}\label{5}$$ is invariant under \eqref{4}. The outer product is, $$\xi\xi^\dagger = \begin{bmatrix}\vert \xi_1\vert^2 & \xi_1\xi_2^*\\ \xi_1^*\xi_2 & \vert \xi_2\vert^2\end{bmatrix} \rightarrow U\xi\xi^\dagger U^\dagger, \tag{6}\label{6}$$ where $\xi\xi^\dagger$ is a Hermitian matrix.

We see that $\xi$ and $\xi^\dagger$ transform in different ways, but we may use the unitarity of $U$ to show that $$\begin{bmatrix}\xi_1\\ \xi_2\end{bmatrix}, \qquad \begin{bmatrix}-\xi_2^*\\ \xi_1^*\end{bmatrix}$$ transform in the same way under $\rm SU(2)$: Comparing \eqref{3} and \eqref{4}, $$\begin{align} \xi'_1 & = a \xi_1 + b \xi_2 \\ \xi'_2 & = -b^* \xi_1 + a^* \xi_2 , \end{align} \tag{7}\label{7}$$ hence $$\begin{align} -\xi'^*_2 & = a (-\xi^*_2) + b \xi^*_1 \\ \xi'^*_1 & = -b^* (-\xi^*_2) + a^* \xi^*_1 . \end{align} \tag{8}\label{8}$$

Now, $$\begin{bmatrix}-\xi^*_2\\ \xi_1^*\end{bmatrix} = \begin{bmatrix}0 & -1\\ 1 & 0\end{bmatrix} \begin{bmatrix}\xi^*_1\\ \xi_2^*\end{bmatrix} = \zeta \xi^*, \qquad \qquad \zeta = \begin{bmatrix}0 & -1\\ 1 & 0\end{bmatrix} \tag{9}\label{9}$$

so we have shown that $\xi$ and $\zeta \xi^*$ transform in the same way under $\rm SU(2)$. In symbols $\sim$ means transforms like $$\xi \sim \zeta \xi^* .\tag{10}\label{10}$$

Thus, $$\xi^\dagger \sim (\zeta \xi)^T = (-\xi_2, \xi_1) .\tag{11}\label{11}$$

$$\xi \xi^\dagger \sim \begin{bmatrix}\xi_1\\ \xi_2\end{bmatrix} \begin{bmatrix}-\xi_2, \xi_1\end{bmatrix} = \begin{bmatrix} -\xi_1\xi_2 & \xi_1^2\\ -\xi_2^2 & \xi_1\xi_2\end{bmatrix} \tag{12}\label{12}$$

Call this matrix $-H$, we see from \eqref{6} that under an $\rm SU(2)$ transformation, $$H \rightarrow UHU^\dagger, \tag{13}\label{13}$$ where $H$ is a traceless matrix.

We may now construct from the position vector $\vec{r}$, a traceless Hermitian $2\times2$ matrix transforming under $\rm SU(2)$ like $H$.

$$h = \vec{\sigma} \cdot \vec{r} = \begin{bmatrix}z & x-iy\\ x+iy & -z\end{bmatrix}, \tag{14}\label{14}$$

  1. What does he mean when he said "but we may use the unitarity of $U$ to show that...transform in the same way under $SU(2)$"? What is the purpose of this construction, i.e. two things that transform in the same way \eqref{8}, \eqref{9}?
  2. How did he get $\begin{bmatrix}-\xi_2^*\\ \xi_1^*\end{bmatrix}$ and knows that it will transform in the same way with $\begin{bmatrix}\xi_1\\ \xi_2\end{bmatrix}$ under SU(2)?
  3. I applied \eqref{4} on \eqref{12}, cannot get \eqref{13}.
  4. All the calculations lead to equation \eqref{12}, what is the point of this? Is it just to show something that is traceless? I don't get the point.
  5. Just to confirm, in the end he wants to construct a traceless Hermitian $2\times2$ matrix that transforms under $\rm SU(2)$, but I still cannot see the connection of this statement to his constructions/calculations. Can anyone explain more on this?
  • $\begingroup$ He proves to you, by explicit calculation, that $\zeta\xi^* \rightarrow {\zeta\xi^*} ' = U \zeta \xi^*$. He writes explicitly what the transformation means. Have you seen $-\zeta U \zeta =U^*$ ? $\endgroup$ Dec 12, 2020 at 14:38
  • 1
    $\begingroup$ @CosmasZachos So $\zeta \xi^*$ transforms in the same way as $\xi$ in such a way that $\zeta \xi^* \rightarrow U\zeta \xi^*$ which is the same as how $\xi$ transforms, $\xi \rightarrow U\xi$, and that is how $\rm{SU(2)}$ transformation would look like. Is that correct? $\endgroup$
    – mathemania
    Dec 12, 2020 at 15:10
  • $\begingroup$ Yes, of course; that is what the author says and it is evidently true. $\endgroup$ Dec 12, 2020 at 15:13
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    $\begingroup$ Is it not the case that the matrix on the RHS of (12) (Ryder's (2.47)), call it $-H$, is absolutely not Hermitian, and can't be, but in the book after the section quoted in the question he actually says he wants to identify $h$ (Hermitian) and $H$ (not Hermitian), but you absolutely cannot identify them if this is the case. If so some of his derivation is basically irrelevant/useless, there is no reason for him to construct (12). This point also seems to have been noticed here. $\endgroup$
    – bolbteppa
    Sep 15, 2021 at 16:34
  • 1
    $\begingroup$ These notes do a better job of this and a better job of Ryder's Lorentz section. $\endgroup$
    – bolbteppa
    Sep 15, 2021 at 16:36

2 Answers 2


I do not have that text, and I would rather not shadow-box with its logic. There are better ways to introduce the SU(2) conjugate doublet representation, $\tilde \xi\equiv \zeta \xi^*$, linked to, e.g. this or this.

  1. He is using the special form of U, (3), dictated by unitarity, to demonstrate what he claims, namely that $\xi \sim \tilde \xi$, (10). However, he is dangerously misleading, in that the "magic" you observe here, reality of representations, does not carry through for other SU(N)s. The conjugate representation is equivalent to the fundamental only for SU(2).

  2. It is due to the conjugacy isomorphism of the SU(2) Lie algebra, $[\sigma_j,\sigma_k]=2i\epsilon_{jkl}\sigma_l$, invariant under $\zeta^{-1} \sigma^k\zeta =- \sigma^{k~*}$, whence $\zeta^{-1} U\zeta = U^*$. This is reviewed in good texts (Li & Cheng, (4.71), etc..) and several questions on this site. Take it as a given if not obvious—the "magic" above.

  3. I'm not sure I can second-guess the author's pedagogy. He is demonstrating that $\xi \tilde{\xi}^\dagger$ transforms like (6), $\xi \xi^\dagger$, whence you really should be able to see the two dyadics transforming identically, (~), as claimed!

  4. This transformation is the form for the transformation of the adjoint, so, then, like, ~, (14). Personally, I would not introduce the adjoint this way, but I do not have that book to follow his design. Note this latter matrix H, is not hermitian, $(\xi \tilde{\xi}^\dagger)^\dagger= \tilde{\xi}\xi^\dagger $, unlike the pure dyadic $\xi \xi^\dagger$ and (14); but he does not claim anything of the sort. (It is evident that, for (12) to be hermitian, you need the phases of $\xi_1$ and $\xi_2$ to be opposite, so $\xi_1\xi_2$, $\xi_2^2-\xi_1^2$ and $i(\xi_2^2+\xi_1^2)$ to all be real.)

  5. Not apparent with the information you provide, indeed, undecidable. In sympathy with your bafflement, there are superior introductions to SU(2) around, even on this very site, and on MSE. Persistence pays, but if he is not a good match, why, then, do you perseverate?

  1. Extra credit. This has presumably nothing to do with your text, but now that one developed the elegant compact language, one would be remiss to not pursue it to the core of the Higgs sector of the SM, the full display of its custodial symmetries. Instead of your dyadic above, bilinear in the spinors $\xi$, one may define another 2×2 matrix, linear in them, whose first column is $\tilde \xi$, and the second $\xi$, $$M\equiv (\tilde\xi , \xi)=\begin{bmatrix}-\xi_2^* & \xi_1 \\ \xi_1^* & \xi_2 \end{bmatrix}~~~\leadsto \\ M^\dagger M = \xi^\dagger \cdot \xi ~~ {\mathbb I} , \hbox{hermitian}. $$ It is evident by above that $$ M'= UM,~~\implies ~~ (M^\dagger M)'= M^\dagger M . $$ since each of its columns transforms thusly. So $M^\dagger M$ is invariant under U action. However, imagine a different SU(2) transformation, completely unrelated and oblivious to the above U one, now acting on the right, $$ M\to MV, $$ for unitary V. The two SU(2)s commute with each other, by inspection, so the group structure is a Cartesian product, SU(2)×SU(2); and $M^\dagger M$ transforms in the adjoint of the new (global) SU(2), V. By suitable squaring and tracings of the hermitian matrix, one may construct the higher-symmetric Higgs potential.

For the benefit of the viewers, ever since this post was written, I think what helped me a lot to understand my questions are the references listed,

  1. From Spinors to Quantum Mechanics by Gerrit Coddens
  2. The Geometrical Meaning of Spinors Lights the Way to Make Sense of Quantum Mechanics
  3. Spinors in Physics by Jean Hladik

All my questions are directly pointing to the theory of spinors. A combination of references 1&2 is what helped me a lot. I now disagree with the above comment that it has something to do with the conjugate spinor. All the formalism shown by Ryder is done using the "regular" spinor, although the formalism could also be done using the conjugate spinor. For reference 3, ignore the bad reviews on Amazon since they are related to criticisms about the typos, I believe that they are not that vital to the understanding.


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