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I SU(3) quark model and singlet states

  1. Apr 10, 2016 #1

    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!
     
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  3. Apr 10, 2016 #2

    Vanadium 50

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    The physical states have the phi being mostly ssbar, and the rho and omega being isotriplets and an isosinglet. Essentially, the broken SU(3) is not a good symmetry, and SU(2) + strangeness works better.
     
  4. Apr 10, 2016 #3

    CAF123

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    Hi Vanadium 50,
    I see. I was confused by the wording of the paragraph - it says 'there are two singlet vector states belonging to the octet and one pure singlet'. But what are these two singlet vector states? As far as I understand, there are 7/8 particles that can be easily found in the octet and the 8th one is found by saying it is linearly independent from the other 7. In using the I,U,V operators, we also see there is another state that transforms as a singlet (since such a state under permutation of u,d,s (i.e the approximate flavour symmetry) does not change). So where/what are these two singlet vector states and pure singlet?

    Also a few follow up questions if that's ok:
    1) How is the ##K^{*,+}##, say, related to the ##K^+##?
    2) In the decomposition ##3 \otimes \bar 3 = 8 \oplus 1##, what meson octet is this ##8## refering to? The one with the ##\pi##'s along the centre line or the one with the ##\rho##'s? And why?
    3) I see that indeed ##| \phi \rangle = | s \bar s \rangle## and ##| \omega \rangle \propto ( | u \bar u \rangle + |d \bar d \rangle)## but is there anyway I can infer that this is the correct content of the ##\phi## and ##\omega## states from the fact that the ##\phi## decays to kaons and the ##\omega## decays to three pions as given in the paragraph?

    Thanks!
     
  5. Apr 10, 2016 #4

    Vanadium 50

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    1) The K* and K have different spins, and so reside in different multiplets.
    2) The decomposition is just the group algebra. No physics goes into this (other than picking the correct group).
    3) If a state decays to kaons, it is likely there is ssbar in the wavefunction. The more it decays to kaons, the stronger this statement becomes.
     
  6. Apr 10, 2016 #5

    CAF123

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    Ok. In terms of the group algebra do you mean that for example we couldn't have had ##3 \otimes \bar 3 = 6 \oplus 3## because the lhs is always real and the rhs is not? So ##8 \oplus 1## was the only sensible combination? Then we say that this is consistent with the observation of mesons forming states that produce an octet and singlet? Analogously with the baryons:

    Baryon states transform under the ##3 \otimes 3 \otimes 3 ## representation of SU(3) (a reducible direct product of three copies of the fundamental representation) which may be written as a direct sum ##10 \oplus 8 \oplus 8 \oplus 1## involving just irreps of SU(3). So, if I understand right, the two copies of 8 doesn't mean there is two independent baryon octets (in fact there isn't), just that the presence of 8 in such a decomposition is consistent with the fact we see a baryon octet in nature?

    I've seen that this decomposition is sometimes written like ##10 \oplus 8_S \oplus 8_A \oplus 1##? Do you know what the A and S stand for here?
     
  7. Apr 10, 2016 #6

    ChrisVer

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    the A and S stand for Antisymmetric and Symmetric... so the two 8s are not so identical (it's not just 2 copies)...
    I'm trying to see how to get 6+3 from 3x3* ..

    What does the 8 mean? It means that there is an 8-dimensional vector... after you act with your symmetry, its individual base vectors may change and mix, but the overall result will describe the same vector space as before.
     
  8. Apr 11, 2016 #7

    CAF123

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    Hi ChrisVer,
    I see, where does the terminology antisymmetric and symmetric come from though?

    I don't think it's allowed - I think you could have 6+3 = 3x3 (two copies of fundamental on rhs) since one side is not real and the other complex for example. By similar reasoning I suppose 6+3*=3x3 would work too.

    Ok, so in the case of ##3 \otimes \bar 3##, we have a nine component vector whose basis components would be all possible permutations of the three quarks and three antiquarks? Then in this space can form all nine linearly independent states (8 transforming non trivially and one trivially) . Same goes for the baryon decompositions in which we have a 10 component vector whose basis vectors are |uds>, |uuu>, |ddd>, |uud> etc....

    Is it correct to say that in the OP, ##|\omega_8 \rangle## is transforming non trivially and ##|\omega_1\rangle## is transforming trivially? I'm just confused with the wording 'two singlet vector states and pure singlet'
     
  9. Apr 11, 2016 #8

    vanhees71

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    Have a look at the good old booklet by Lipkin

    H. Lipkin, Lie groups for Pedestrians, North-Holland Publ. (1965)

    The reason is clear when you look at it in terms of Young diagrams. The fundamental representation ##3## is drawn as one square and its complex conjugate ##\bar{3}## (which is truly different from the fundamental representation itself) is equivalent to the antisymmetrized states of two fundamental representations and thus notated as two vertically arranged squares. To reduce a tensor product of a fundamental and a antifundamental representation you can either put the single square of the ##3## representation underneith to the two squares of the ##\bar{3}## representation, leading to a singlet representation (since there's no 1 totally antisymmetric state) or you can attach it to the right-upper place near the ##\bar{3}## representation, which represents the octet representation.

    For ##3 \otimes 3## it's different. There you place one square either underneath or to the right of the other single square, i.e., you have
    ##3 \otimes 3=\bar{3} \oplus 6##.
     
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