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The group $SU(2)$ consists of $2\times2$ unitary matrices with determinant $1$ can be put in the form: $$U=\begin{pmatrix}a &b \\ c & d\end{pmatrix}$$ By invoking the conditions: $det(U)=1$ and $UU^\dagger=1$, we reduce the matrix $U$ to $$U=\begin{pmatrix}a &b \\ -b^* & a^*\end{pmatrix}$$ This is a 2 dimensional representation of $SU(2)$ group. Now, consider a three dimensional representation: $$U=\begin{pmatrix}a &b&c \\ d & e &f \\ g&h&i\end{pmatrix}$$ and then invoke the same conditions $det(U)=1$ and $UU^\dagger=1$. Isn't this just the same as the 3 dimensional representation for $SU(3)$ group?

I know this in't correct, but I don't know what's wrong.

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    $\begingroup$ Do you know what a representation of a group is? $\endgroup$
    – Moishe Kohan
    Mar 18, 2017 at 13:06
  • $\begingroup$ I'm very confused by the word representation. I think you are asking how a general matrix of $SU(3)$ looks like. This has nothing to do with representation theory. $\endgroup$
    – Mathematician 42
    Mar 18, 2017 at 13:07
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    $\begingroup$ That's very vague and not entirely correct. A group representation is a group morphism $\phi:G\rightarrow \text{GL}(V)$ where $V$ is a vector space. If $V$ is finite-dimensional, we say that $\phi$ is a $\dim(V)$-representation of $G$. $\endgroup$
    – Mathematician 42
    Mar 18, 2017 at 13:11
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    $\begingroup$ So suppose $\dim(V)=3$, what is the easiest group morphism you can think of? If you interested in a faithful representation you'll have think some more. Anyways, I would revise the basics on representations, I think you have not understood the theory. $\endgroup$
    – Mathematician 42
    Mar 18, 2017 at 13:13
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    $\begingroup$ Here's an example of a group representation. The group $G= \Bbb Z / 3\Bbb Z$ is represented by the three matrices $\begin{bmatrix} \cos t & -\sin t \\ \sin t & \cos t\end{bmatrix}$, where $t = 0, 2\pi/3, 4\pi/3$. The group operation in $G$ corresponds to multiplication of matrices. This is a (real) 2-dimensional representation of $G$ because it uses $2 \times 2$ real matrices. $\endgroup$
    – John Hughes
    Mar 18, 2017 at 13:41

1 Answer 1

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If you are a "physics lover" as your username suggests, then I suggest you think about it like this:

Any $SU(2)$ matrix can be written in the form $$ U = \exp \left(i \theta_x J_x + i \theta_y J_y + i \theta_z J_z \right),$$ where $\theta_x, \theta_y, \theta_z$ are real numbers and $$J_x = \frac 1 2 \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right), \ \ \ J_y = \frac 1 2 \left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right) \ \ \ J_z = \frac 1 2 \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right).$$ Hopefully you recognise $J_x, J_y, J_z$ as the angular momentum operators for spin-$1/2$ particles.

To get the three-dimensional representation, you still write $U$ in the form$$ U = \exp \left(i \theta_x J_x + i \theta_y J_y + i \theta_z J_z \right),$$ using the same $\theta_x, \theta_y, \theta_z$. But now, you change $J_x, J_y, J_z$ to $$J_x = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \end{array} \right), \ \ \ J_y = \left( \begin{array}{ccc} 0 & 0 & i \\ 0 & 0 & 0 \\ -i & 0 & 0 \end{array} \right), \ \ \ J_z = \left( \begin{array}{ccc} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{array} \right).$$ These are the angular momentum operators for spin-$1$ particles.

To construct the $n$-dimensional irreducible representation of $SU(2)$, use the spin $j = (n-1)/2$ angular momentum matrices as $J_x, J_y, J_z$.

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    $\begingroup$ I get it. Thanks!! Also,I'd like to ask what is an $SU(2)$ triplet (like the Higgs triplet)? How does it transform under $SU(2)$?Does it has anything to do with the three dimensional representation? All the particle physics lectures and textbooks I can find just use this terminology directly. $\endgroup$
    – Quantumania
    Mar 18, 2017 at 16:15
  • $\begingroup$ Particle physics has more than one $SU(2)$ symmetry group. The one that I was talking about is the group of rotational symmetries (well, sort of, up to an important minus sign...). This $SU(2)$ group that you're talking about in connection to the Higgs boson is called a gauge symmetry: it is associated with the electroweak force. And by the way, the Higgs is in a two-dimensional representation of this electroweak $SU(2)$. $\endgroup$
    – Kenny Wong
    Mar 18, 2017 at 18:03
  • $\begingroup$ But yeah, if your $SU(2)$ is the spin/rotation group, then an $SU(2)$ triplet is the same thing as a particle with spin one. People call it a triplet because you write the quantum wavefunction is really a column vector with three entries, and the angular momentum operators act on the wavefunction as 3x3 matrices. $\endgroup$
    – Kenny Wong
    Mar 18, 2017 at 18:05
  • $\begingroup$ Shouldn't the $J_\alpha$ matrices obey the same algebra as the Pauli matrices? When I try to do that explicitly they don't. $\endgroup$
    – twisted manifold
    Dec 12, 2019 at 9:58
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    $\begingroup$ These are not the usual angular momentum matrices for spin-1 systems (which has dimension 3). For example, $J_z$ is (usually) a diagonal matrix with diagonal elements (1,0,-1). How do we get the matrices in the answer (for arbitrary dimension), then? $\endgroup$
    – Archisman Panigrahi
    Sep 2, 2021 at 11:54

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