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?

  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$ – Qmechanic Sep 30 '17 at 20:18
  • $\begingroup$ I can't make head nor tail of it. There was only silenlty mentioned that a hadron has a color singulett what is also not clear to me. $\endgroup$ – KarlPeter Sep 30 '17 at 20:49
  • $\begingroup$ My question is more elementary. It doesn't concern the coupling of quarks in arbitrary hadrons. It is just tend to what concretly is conserved as quantum number by color symmetry. The color charge? Apears it (mathematically) in the same way under r, , g, b base vectors as isospin number I under proton and neutron vectors it does? $\endgroup$ – KarlPeter Sep 30 '17 at 21:01
  • $\begingroup$ SU(3) has two, not just one, Casimir invariants, that is operators that have the same eigenvalue for every triplet, as you are considering, and analogously for every different representation. $\endgroup$ – Cosmas Zachos Sep 30 '17 at 21:02

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.