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CAF123
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'In the SU(3) quark model there are two singlet vector states $$|\omega_8 \rangle = \frac{1}{\sqrt{6}} \left(|u \bar u \rangle + |d \bar d \rangle - 2 |s \bar s \rangle \right) $$ belonging to the octet and the pure singlet state $$|\omega_1 \rangle = \frac{1}{\sqrt{3}} \left(|u \bar u \rangle + | d \bar d \rangle + |s \bar s \rangle \right)$$ The two physical states are the ##\omega## and ##\phi## mesons. The ##\omega## decays primarily into three pions while the ##\phi## prefers to decay into kaons. This therefore suggests that the mass eigenstates ##| \omega \rangle## and ##\phi \rangle ## are mixtures of ##| \omega_8 \rangle## and ##| \omega_1 \rangle ## with a mixing angle of about ##\sqrt{2}##.' I am just trying to understand why this is the case.
I think that, in a similar vein as to how we build up one octet of SU(3) involving ##K^0, K^+, K^-, \bar K^0, \pi^-, \pi^+, \pi^0## and then seeing that the combination in the middle ## \propto (\langle u \bar u \rangle + |d \bar d \rangle - 2| s \bar s \rangle)## is a linearly independent combination transforming non trivially under SU(3) (flavour) transformations thereby producing the 8th member of the octet we also see that there is a combination ##\propto ( |u \bar{u} \rangle + |d \bar{d} \rangle + |s \bar s \rangle)## which does not transform thereby constitutes a singlet. These final two states combine in such a way to produce the physical states ##\eta## and ##\eta'## that we observe in the particle spectrum. It is my understanding that this octet is the 8 in ##3 \otimes \bar 3 = 8 \oplus 1##?
My notes give a picture where the ##\omega## and ##\phi## are part of another octet involving the ##K^{*,0}, K^{*,+} , \rho^-, \rho^0, \rho^+, K^{*,-}, \bar K^{*,0}... etc##. So either ##\omega## or ##\phi## is a singlet and the other makes up the remaining member of this octet. How do I tell which is the singlet based on what I have wrote above and how is the mixing angle defined in this context?
Thanks!
I think that, in a similar vein as to how we build up one octet of SU(3) involving ##K^0, K^+, K^-, \bar K^0, \pi^-, \pi^+, \pi^0## and then seeing that the combination in the middle ## \propto (\langle u \bar u \rangle + |d \bar d \rangle - 2| s \bar s \rangle)## is a linearly independent combination transforming non trivially under SU(3) (flavour) transformations thereby producing the 8th member of the octet we also see that there is a combination ##\propto ( |u \bar{u} \rangle + |d \bar{d} \rangle + |s \bar s \rangle)## which does not transform thereby constitutes a singlet. These final two states combine in such a way to produce the physical states ##\eta## and ##\eta'## that we observe in the particle spectrum. It is my understanding that this octet is the 8 in ##3 \otimes \bar 3 = 8 \oplus 1##?
My notes give a picture where the ##\omega## and ##\phi## are part of another octet involving the ##K^{*,0}, K^{*,+} , \rho^-, \rho^0, \rho^+, K^{*,-}, \bar K^{*,0}... etc##. So either ##\omega## or ##\phi## is a singlet and the other makes up the remaining member of this octet. How do I tell which is the singlet based on what I have wrote above and how is the mixing angle defined in this context?
Thanks!