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I have following (maybe a bit general) question about the $SU(3)$-symmetry of color by quarks:

If I consider the analogy to the $SU(2)$-symmetry of isospin $I$ crucially it concers the conservation of the quantum number $I =1/2$ under $SU(2)$-rotations, because we have an arbitrary linear combination $\begin{pmatrix}p \\\ n \end{pmatrix} =p \begin{pmatrix}1 \\\ 0 \end{pmatrix} + n \begin{pmatrix}0 \\\ 1 \end{pmatrix}$ where $\begin{pmatrix}1 \\\ 0 \end{pmatrix} \equiv |1/2$ +$1/2> $ and

$\begin{pmatrix}0 \\\ 1 \end{pmatrix} \equiv |1/2$ -$1/2>$. Futhermore $I^2 |1/2$ +$1/2> = \hbar^2I(I+1)|1/2$ +$1/2> = \hbar^23/2|1/2$ +$1/2> $ and also $I^2 |1/2$ -$1/2> \hbar^23/2|1/2$ +$1/2> $. Knowing that $\begin{pmatrix}1 \\\ 0 \end{pmatrix} \equiv |1/2$ +$1/2> $ and

$\begin{pmatrix}0 \\\ 1 \end{pmatrix} \equiv |1/2$ -$1/2>$ span the whole $R^2$ whe conclude that every linear combination $\begin{pmatrix}p \\\ n \end{pmatrix} $ of "neuton and proton vectors" of unity length has the same $I$. That's what I understand under invariance of isospin $I$ (resp. $SU(2)$(ratation in 2D)-symmetry).

If we come back to my initial question about $SU(3)$-color symmetry how can I understand here the "conservation" (of what?)? I know that the 3D color space is spaned by $\begin{pmatrix}1 \\\ 0 \\\ 0 \end{pmatrix} \equiv r$, $\begin{pmatrix}0 \\\ 1 \\\ 0 \end{pmatrix} \equiv g$, $\begin{pmatrix}0 \\\ 0 \\\ 1 \end{pmatrix} \equiv b$ but what has every color vector $\begin{pmatrix}r \\\ g \\\ b \end{pmatrix} $ as it’s invariant? If I remind the analogy to isospin as above, can I interpret this invariant quantum number as $=:c \equiv 1$ with „basical“ colors r, g, b („=“ vector base) as a triplett with $\begin{pmatrix}1 \\\ 0 \\\ 0 \end{pmatrix} \equiv r \equiv |1$ +$1> \equiv |c$ +$c>$, $\begin{pmatrix}0 \\\ 1 \\\ 0 \end{pmatrix} \equiv b\equiv |1$ $0> \equiv |c$ $0>$ and $\begin{pmatrix}0 \\\ 0 \\\ 1 \end{pmatrix} \equiv b \equiv |1$ -$1> \equiv |c$ -$c>$ or is this interpretation wrong?

user267839
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1 Answers1

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Colour space is actually a 2 dimensional charge-space. We can parameterise it in terms of two quantum numbers.

Let's call them $X$ and $Y$. The pair $(X,Y)$ give all possible charges. $X$ is sort of the red-ness and $Y$ the green-blueness.

We could define the colours to be three points on an equilateral triangle centred at the origin:

$$red= (1,0)$$

$$green = (-1/2,\sqrt{3}/2)$$

$$blue = (-1/2,-\sqrt{3}/2)$$

It is easily confirmed that red+green=anti-blue and so on.

Therefor we can map the triplet $(R,G,B)$ onto: $$(R-G/2-B/2,\sqrt{3}/2(G-B))$$

So we could use the pair $(X,Y)$ as an equivalent of isospin. However this is less useful as all hadrons (i.e. mesons and baryons) would have total zero values for X and Y as they have to be colourless.

ersbygre1
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