# What are Quantum Numbers?

For quite some time I didn’t really understand what quantum numbers are. For example, why do we use the words “red”, “blue” and “green” for the charges of the strong interaction? Why does a gluon carry “red anti-green + green anti-red” color? From the group theoretical perspective these things actually make a lot of sense and maybe this post helps someone who is equally confused as I was a few years ago.

First, we recall that a Lie algebra representation is a map $R$ from the Lie algebra $\mathfrak{g}$ of a group $G$ to the linear operators $\mathrm{Lin}(\cdot)$ over some vector space $V$.

R: \ \mathfrak{g}  \rightarrow \mathrm{Lin}(V) \, .

The easiest example is the fundamental $2$-dimensional representation of $\mathfrak{su}(2)$, which is a map

R: \ \mathfrak{su}(2)  \rightarrow \mathrm{Lin}(\mathbb{C}^2) \, .

In words this means that this representation maps each element of $\mathfrak{su}(2)$ onto a $2 \times 2$ matrix that acts on $2$-dimensional vectors. A basis for this Lie algebra is given by

\begin{align}
T_1&=\frac{1}{2} \sigma_1 =  \frac{1}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \, , \notag \\
T_2&=\frac{1}{2} \sigma_2 =  \frac{1}{2} \begin{pmatrix} 0 & -\mathrm{i} \\ \mathrm{i} & 0 \end{pmatrix} \, , \notag  \\
T_3&=\frac{1}{2} \sigma_3 =  \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \, ,
\end{align}

where $\sigma_i$ denote the usual Pauli matrices.

From linear algebra we know that the eigenvectors of a linear operator always form a basis for the vector space in question. In addition, for any Lie group, one or more of the generators can be simultaneously diagonalized using similarity transformations. The set of generators that can be diagonalized simultaneously are called Cartan generators. Thus, a suggestive and particularly easy basis for the vector space of each representation is given by the eigenvectors of the Cartan generators. An easy way to label these basis vectors is to use the corresponding eigenvalues.

In particle physics the fundamental particles are (among others) labelled by their color and weak-isospin. These quantum numbers correspond to eigenvalues of the Cartan generators of the corresponding gauge groups. For $SU(2)_L$, the gauge group of weak-interactions, there is only one Cartan generator and in the fundamental two-dimensional representation it is given by

I_3 = \frac{1}{2} \sigma_3 = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \, .

The factor $\frac{1}{2}$ is there, because we usually normalize our generators such that the Dynkin index Tr$(T_a T_a)$ for this generator in the fundamental representation of the Lie algebra is $\frac{1}{2}$.

The corresponding eigenvalues are $\frac{1}{2}$ and $-\frac{1}{2}$. This means particles that correspond to $SU(2)_L$ eigenstates in the fundamental $2$-dimensional representation are with respect to $SU(2)_L$

\begin{pmatrix} 1 \\ 0 \end{pmatrix} \ \text{ with isospin } \frac{1}{2} \quad , \quad \begin{pmatrix} 0 \\ 1 \end{pmatrix} \ \text{ with isospin } -\frac{1}{2} \, .

Less trivial is $SU(3)$, the gauge group of strong interactions, because there are two Cartan generators. In the representation that acts on the fundamental $3$-dimensional representation they can be written as

\label{eq:cartansu3}
H_1=
\frac{1}{2} \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 1
\end{array} \right) \, .

Again, we label the trivial eigenvectors using the the eigenvalues of the Cartan generators. However, now there are two eigenvalues for each eigenvector and therefore we define objects, called weights, that collect these numbers for each eigenvector. This means the $SU(3)_C$ quantum numbers for the basis vectors of the fundamental $3$-dimensional representation are

\left(\frac{1}{2} , \frac{1}{2 \sqrt{3}} \right) \mathrm{ \ for \ }  \begin{pmatrix}
1 \\ 0 \\ 0
\end{pmatrix} \quad , \quad \left(0 , \frac{-1}{ \sqrt{3}} \right) \mathrm{ \ for \ }  \begin{pmatrix}
0 \\ 1 \\ 0
\end{pmatrix}  \quad , \quad \left(-\frac{1}{2} , \frac{1}{2 \sqrt{3}} \right) \mathrm{ \ for \ }  \begin{pmatrix}
0 \\ 0 \\ 1
\end{pmatrix}  \,  .

It is conventional to replaces these weights with names: red $:=(\frac{1}{2} , \frac{1}{2 \sqrt{3}})$, blue $:=(0 , \frac{-1}{ \sqrt{3}})$, green $:=(\frac{-1}{2} , \frac{1}{2 \sqrt{3}})$. The labels for the conjugate fundamental representation $\bar 3$ are simply minus the labels of the fundamental $3$ and are called anti-red, anti-blue and anti-green.

We can use these basis vectors and the corresponding weights to derive the basis vectors and weights for product representations like

\label{eq:3x3bardecomposition}
3 \otimes \bar 3 = 1 \oplus 8 \, .

The $8$ is the adjoint representation of $SU(3)$ and Eq. \ref{eq:3x3bardecomposition} tells us that we can write each element of the adjoint as $3\times 3$ matrix. For the basis vectors we use

e_{ij} = e_i \otimes e_j \, .

For the quantum numbers of the product representations we use

QN \big ( (a \otimes b)_{ij} \big) = QN(a_i) + QN( b_j) \, .

Formulated in terms of weights this means that we can compute the weight corresponding to the $ij$ element of the product representation $3 \otimes \bar{3}=1 \oplus 8$ by adding the weights of $3_i$ and $\bar{3}_j$

w\Big ( (3 \otimes 3)_{ij}\Big) = w(3_i) + w(3_j) \,.

The $3 \times 3$ matrix that correspond to this weights is given by the Kronecker product of the $i$-th and $j$-th basis vector. Thus we have

\left(
\begin{array}{cc}
\frac{1}{2} & \frac{\sqrt{3}}{2} \\
\end{array}
\right) = \left( \begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array} \right) \, , \quad  \left(
\begin{array}{cc}
1 & 0 \\
\end{array}
\right)=  \left( \begin{array}{ccc}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array} \right) \, , \quad \left(
\begin{array}{cc}
\frac{1}{2} & -\frac{\sqrt{3}}{2} \\
\end{array}
\right) = \left( \begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array} \right) \, ,   \left(
\begin{array}{cc}
-\frac{1}{2} & -\frac{\sqrt{3}}{2} \\
\end{array}
\right)=
\left( \begin{array}{ccc}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{array} \right) \, , \quad \left(
\begin{array}{cc}
-1 & 0 \\
\end{array}
\right) = \left( \begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{array} \right) \, , \quad \left(
\begin{array}{cc}
-\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{array}
\right) = \left( \begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 1 & 0
\end{array} \right) \,

and two zero weights $(0,0)$ that span a basis for the Cartan subalgebra

\left(
\begin{array}{cc}
0 &0\\
\end{array}
\right)_1=  \frac{1}{2} \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1
\end{array} \right)  \, , \quad   \left(
\begin{array}{cc}
0 & 0 \\
\end{array}
\right)_2 =  \frac{1}{2 \sqrt{3}} \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 1
\end{array} \right) \, .

These $8$ matrices are given in a basis, known as Cartan-Weyl basis for $\mathfrak{su}(3)$. The special thing about this basis is that each matrix here is an “eigenmatrix” of the Cartan generators $H_i$, which means

H_i \circ M = [H_i,M]=\lambda_i M \, ,

where $\lambda_i$ is the eigenvalue for the Cartan generator $H_i$. In physical terms, each of these $8$ matrices represents a different gluon. This is completely analogous to how the three basis vectors for $\mathbb{R}^3$ for the fundamental representation correspond to three different quarks: a red quark, a blue quark and a green quark.

There is another way to denote gluons, analogous to the color notation for quarks in the fundamental $3$. We can use the fact that each gluon corresponds to a product of a basis vector of the fundamental $3$ and basis vector of the anti-fundamental $\bar{3}$, to name each gluon in terms of color, too. For example, for $i=1$ and $j=2$

8_{12} \hat= w(3_1) + w( \bar{3}_2)  = \left(\frac{1}{2} , \frac{1}{2 \sqrt{3}} \right) +\left(0 , \frac{1}{ \sqrt{3}}\right) = \text{red+ anti-blue} =\left( \frac{1}{2},  \frac{\sqrt{3}}{2} \right)  \, .

Therefore just as we have red, blue and green quarks, we have $8$ gluons that we can label by color combinations of the form color-anticolor.

There is another basis for the $8$ basis matrices of the adjoint representation, more popular among physicists, called the Gell-Mann basis\footnote{The Gell-Mann basis is more popular due to the close connection to the Pauli matrices of $SU(2)$.}. In this basis the $8$ basis elements $T_a$ of $\mathfrak{su}(3)$ are given in terms of the $8$ Gell-Mann matrices $\lambda_a$ by $T_a= \frac{1}{2} \lambda_a$, where
\label{lambda1-3}
\lambda_1 =
\begin{array}{ccc}
\left(
\begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{array}
\right),
&
\lambda_2 =\left(
\begin{array}{ccc}
0 & -i & 0 \\
i & 0 & 0 \\
0 & 0 & 0
\end{array}
\right),
&
\lambda_3 =\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1
\end{array}
\right),
\end{array}

\label{lamdba4-6}
\begin{array}{ccc}
\lambda_4 =\left(
\begin{array}{ccc}
0 & 0 & 1 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{array}
\right),
&
\lambda_5 =\left(
\begin{array}{ccc}
0 & 0 & i \\
0 & 0 & 0 \\
-i & 0 & 0
\end{array}
\right),
&\lambda_6 =\left(
\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}
\right),
\end{array}

\label{lambda7-8}
\begin{array}{cc}
\lambda_7 =\left(
\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & -i \\
0 & i & 0
\end{array}
\right),
&
\lambda_8=\frac{1}{\sqrt{3}}\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 1
\end{array}
\right).
\end{array}

Again, we can give names in the form color-anticolor to the gluon states in this basis. For example,

\begin{align}
T_1 &= \frac{1}{2}\lambda_1 =
\left(
\begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{array}
\right)
= \left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}
\right) + \left(
\begin{array}{ccc}
0 &0 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{array}
\right)  \notag \\ &= \text{red anti-green} + \text{green anti-red}
\end{align}

and this is a popular way to label the gluons.

P.S. I wrote a textbook which is in some sense the book I wished had existed when I started my journey in physics. It's called "Physics from Symmetry" and you can buy it, for example, at Amazon. And I'm now on Twitter too if you'd like to get updates about what I'm recently up to.