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   <title>Is There a Third Companion of Lambda and Sigma-0?</title>
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<center><b><font size=+2>Is There a Third Companion of <font face="Symbol">L</font>
and <font face="Symbol">S</font><sup>0</sup></font><sup>?</sup></b>
<br><font size=-1>A.V.Kopylov (Kopylov@al20.inr.troitsk.ru)</font>
<br><font size=-1>Institute for Nuclear Research of RAS, Moscow, Russia,</font>
<br><font size=-1>117312, Prospect of 60<sup>th</sup> Anniversary of October
Revolution 7a</font></center>

<p>The idea was exploited in Ref.1 that nucleon can be considered as quark-diquark
system following the proposal of Ida and Kobayashi [2] and Lingenfeld and
Tassie [3]. If to assume that magnetic moment of a nucleon can be expressed
as the sum of Dirac magnetic moments of different quarks
<center><b><font face="Symbol">m</font></b> = <font face="Symbol">m</font><sub>N.</sub><b><font face="Symbol">S</font></b>e<sub>i</sub>g<sub>i</sub><b>J</b><sub>i</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
(1)</center>
here e<sub>i</sub>, g<sub>i</sub> and <b>J</b><sub>i</sub> are the charge,
factor of Lande and moment of a constituent quark, <font face="Symbol">m</font><sub>N</sub><font face="Symbol">-</font>
is a nuclear magneton, then one can write the system of two equations:
for magnetic.moments of neutron and proton as different members of one
isospin multiplet. The expression (1) assumes that the only representative
mass parameter is a mass of a nucleon which determines the overall scale.
This supposition is quite natural for confined quarks. The nucleon is a
good object to compose this system because the masses of neutron and proton
are very close (the mass difference is about 0.1% of a mass of a nucleon
what is not observed for other baryons).
<p>So if to take into account that diquark can exist as a pair of like
quarks and as a pair of different quarks two systems of equations can be
written:
<p>Case I: n = u + dd p = d + uu
<center><img SRC="comp1.gif" height=87 width=174></center>

<p>From here it follows that: g<sub>1</sub>J<sub>1</sub><font face="Symbol">=</font>
-1.033 and g<sub>2</sub>J<sub>2</sub> <font face="Symbol">=1.836</font>
<p>The fact that g<sub>1</sub>J<sub>1</sub> turned out to be very close
to 1 means that the single quark acts as elementary particle placed at
the center of nucleon, some extra of 0.033 may be assumed here as a correction
similar to the Schwinger correction to magnetic moment of electron.
<p>Case II. The similar equations can be written for the case when diquark
<b>ud</b>
is in a state <sup>3</sup>S<sub>1</sub>, only here g<sub>1</sub>J<sub>1</sub>
is the value assigned to the single quarks, constituents of a diquark <b>ud</b>:
<p>n <font face="Symbol">=</font> ud + d p <font face="Symbol">=</font>
ud + u
<center><img SRC="comp2.gif" height=88 width=178></center>

<p>To satisfy this system one should take: g<sub>1</sub>J<sub>1</sub><font face="Symbol">=</font>
-1.033 and g<sub>2</sub>J<sub>2</sub> <font face="Symbol">=4.7</font>.
The fact that g<sub>1</sub>J<sub>1</sub> turned out to be close to 1 (with
the same correction) agrees perfectly with the assumption <sup>3</sup>S<sub>1</sub>.
Generally we have two more states in this case: one with a spin of a peripheral
quark up, and another one - with a spin down.
<p>The argument in favor that the model presented here is realistic is
that one finds agreement with the measured magnetic moments of both nucleons
if quarks at the center of nucleon act as elementary particles, this is
observed for all three states. Qualitatively this conforms the parton model
of the nucleon and <i>x</i>-scaling behavior in deep-inelastic lepton scattering.
A notable thing is also that if to consider a center of nucleon as the
place of the highest priority, the couple of mixed quarks is stronger
then a single quark which is stronger then a couple of like quarks. Figure
1 of Ref.1 illustrates these states. The question why these states can
be the stationary states is discussed in Ref. 4. Of course the arguments
presented here are very speculative but if these states are real and will
be found [5], this can be important in view of their possible application
in practice.
<p>For the quark content uds (or udc, udb) one can expect three possible
configurations: ud <font face="Symbol">+</font> s, us <font face="Symbol">+</font>
d, ds <font face="Symbol">+</font> u. Two baryons of this quark content
were found: <font face="Symbol">L</font> and <font face="Symbol">S</font><sup>0</sup>.
In the framework of our consideration one more baryon of this type is missing.
In fact it well may be that the third companion was really observed. The
last update of Particle Data Group [6] tells about <font face="Symbol">S</font>(1670)
and <font face="Symbol">S</font>(1670)B that Probably There are two states
at the same mass with the same quantum numbers, one decaying to <font face="Symbol">Sp
</font>and<font face="Symbol">
Lp, </font>the other to <font face="Symbol">L</font>(1405)<font face="Symbol">
p</font>. This can be the manifestation that the scheme discussed here
is valid.
<p><b>References</b>.
<ol>
<li>
A.V. Kopylov , .</li>

<li>
Ida M. and Kobayashi R. <i>Prog.Theor.Phys.</i> <b>36</b>, 846 (1966).</li>

<li>
D.B.Lichtenberg and L.J.Tassie, <i>Phys.Rev</i>. <b>155</b>, 1601 (1967)</li>

<li>
A.V. Kopylov , </li>

<li>
A.V. Kopylov , .</li>

<li>
C.Caso et al (Particle Data Group) URL:http://pdg.lbl.gov</li>
</ol>

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