So, we will continue our discussion on the
Quark Model. So, yesterday we discussed the
grouping of mesons, like pions, under what
is called the isospin. So, we said that we
can consider pions as made of quarks and anti
quarks and think about an isospin symmetry
or isospin grouping under which a u quark
and a d quark belongs to isospin equal to
half. So, I equal to half. With u having I
3 equal to plus half and I 3 of u half u is
plus half, and I 3 of d quark is minus half.
And similarly, we can think about the anti
quark isospin half group d bar and u bar and
we said we will put a minus sign in whenever
the wave function of u bar is considered.
This will give you I 3 equal to plus half
for the d bar quark and I 3 equal to minus
1 by 2 for the 
u bar quark. With this we worked out that
pi plus is actually a combination of u and
d bar, and pi 0 when you actually get from
the pion wave function thought of as isospin
1, isospin 1 and isospin 3, by plus 1. Then
we can actually use step down operator to
get the wave function corresponding to pi
0, and then that is what we did yesterday;
and we got 1 over 2 d d bar minus u u bar.
The minus sign is because of the minus sign
associated with u bar in the I 3 equal to
half multiplet, and pi minus is again going
to be d and u bar. Only one such combination
possible.
So, this first one is our I 3 equal to plus
1, second one is I 3 equal to 0, and third
one is I 3 equal to minus 1; all belonging
to I equal to 1. This is what we discussed
yesterday and now we said we can also think
about other mesons like K-ons.
In that case what we do is we will have to
consider the fact that u d, let me write that
again, belongs to an I equal to half group
and s, the strange quark s quark belongs to
an I equal to 0 group. And similarly, d bar
u bar minus belongs to an I equal to half
and s bar belongs to an I equal to 0. So,
with this u d and s bar for example, we can
consider u s bar combination.
This will have an isospin half, because S
has and S bar as isospin equal to 0. u has
isospin equal to half. So, the only possible
result and isospin of the combination is half
and I 3 similarly is plus half. So, we have
I equal to half, I 3 equal to plus half for
the combination, combined u s bar system.
We associate this or we identified this as
the K plus meson.
And from the other member of the quark doublet
u isospin doublet, u d be how d s bar which
is again I equal to half, but I 3 equal to
minus half, because d has isospin third component
equal to minus half. And this, if you look
at the electric charge, it is a neutral particle,
we denote that as K 0. If you look at the
mass of this you see that experimentally they
all have the same must. We will come to some
mass relations as we progress in quark model.
So, this is one thing. But then you also have
an antiquark d bar and u bar combination and
the s quark. So, with that again you can make
u bar s or let me take since we are talking
about the isospin half I 3 equal to half first
plus half first; let me take the d bar s.
So, now, electric charge of d bar s combination
is also neutral. So, I should be denoting
this also as a neutral particle with a 0 there
in the place of charge. But then this is different
from K 0. It is actually the antiparticle
of k 0. So, if you look at the quark content
it is clear that d goes to its antiparticle
s bar goes to its antiparticle. We said that
the antiparticle is the relative concept in
some sense. So, together d s bar goes to its
antiparticle, so k 0 goes to its antiparticle;
if you take this change. And this is again
I equal to I 3 equal to plus half system.
And similarly, you can have a K minus with
u bar s. We can put a minus sign, because
u bar is always there is a minus sign associated
with u bar wave function. So, I 3 equal to
minus half belonging to an isospin doublet.
So, if you look at again K minus, it is the
antiparticle of K plus, because u has gone
to u bar; s bar has going to s. So, in a similar
fashion we can actually think about other
mesons also. So, there are many mesons which
are discovered over the years, and all of
them can be understood in terms of their quark
contents. And we can apply the isospin symmetry
properties the way we have done to all of
them. Only thing is now you have to consider
the isospin along with this one you have to
also consider the c quark, b quark, c quark
and b quark, essentially. t quark top quark
does not really form bound states; we discussed
that.
So, basically considering the isospin of c
quark equal to 0, b quark equal to 0 and s
quark equal to 0 and an isospin of u quark
as plus half, with along with the d quark
we can work out the isospin of any of the
mesons and also different combinations which
will combine together to give isospin multiplets
of any mesons.
In exactly similar, way we can also look at
the baryons, which are now made of only quarks
not anti quarks. So, for example, if you look
at proton you can see that if you look at
u u u, then sorry u quark and a u quark and
a d quark then you will have a proton. And
if you work out the isospin then you will
see that u quark has an isospin third component
is equal to plus half. Another u quark, so
together they will have an isospin combination
of plus 1 and then you have a d quark which
has an isospin third component minus half.
So, together they have an isospin third component
plus half. And you will see that you can actually
work it out to get both proton and neutron
belonging to one isospin group with I equal
to plus half and I 3 equal to plus half for
proton and I 3 equal to minus half for the
neutron.
Similarly, for other baryons we can look at
the isospin grouping. And in fact, if you
are interested a complete list of all the
known mesons up to today is available at the
particle data group which actually compiles
all this information from various different
experiments and keeps it in one place for
our easy reference. The URL of the particle
data group or pdg is given here. So, you can
take a look at that then you will see that
different mesons what their other properties,
charges are; what the quark content is, what
their isospin is, what their other quantum
numbers and other properties are. When they
decay what they do decay into, etcetera. So,
a lot of information including what we have
discussed here and many more information that
we could not discuss in here also there. So,
please take a look at it.
Underlying these isospin, we kept calling
saying that the isospin symmetry at some point
I said that there is some isospin symmetry
and then we can group these particles under
this symmetry. What is the symmetry? The symmetry
is basically this; and in fact, the whole
thing originated from the fact that, from
the observation that neutrons and protons,
although are electrically different and; they
will be behaving differently under electromagnetic
interactions because proton has charge plus
one and neutron has no charge electrical charge.
Although they behave differently under electromagnet
interactions they behave the same way under
the strong interactions. This was one observation
that people made.
Now, this then said that in some sense they
are; they can be thought of as 2 different
states of a single kind of an entity called
nucleon; as far as strong interactions are
concerned. So, we can actually think about
proton and neutron belonging to a nucleon
the 2 states of a nucleon. In a sense this
is similar to the spin up and spin down states
of an electron.
%%%%
If you forget about the spin part; spin splitting,
etcetera or the magnetic interactions including
the spin, then electron whether it has an
up type, it is, it has a spin orientation
in the up direction with respect to some quantization
direction or down direction; opposite or along
to the quantization direction. It behaves
the same way for all purposes except some
special cases; except in some special cases
like magnetic interactions, etcetera.
So, now it is this analogy that we can take
again here. And then this additional quantum
number that will distinguish the nucleons;
called the isospin. And we can make a statement
that the strong interaction, since it does
not distinguish between p and n, it is symmetric
under isospin transformations. In the sense
that if you make a transformation from one
state to another which differs in isospin
states. It is not going to make any difference
as far as the strong interactions are concerned.
This observation led to is the underlying
theme that brings out the concept of isospin
symmetry.
And in the same way we can actually talk about
u quark and the d quark, and then say that
the strong interaction does not really distinguish
between the u quark and d quark. And whenever
there is a symmetry in particle physics or
in physics in general, in physics we try to
associate that symmetry with some group. It
is an easy mathematical way of understanding
symmetry.
For example, rotation can be understood through
what is called a special orthogonal group.
Rotation in 3 dimensions can be understood
with the help of special orthogonal group
of dimensions 3. And the symmetry that is
associated with the isospin is, the symmetry
group of SU 2. What is SU 2? SU 2 is basically
the following.
If you take the element, an element of SU
2 any member of SU 2 this member is unitary.
If you take that member and Hermitian conjugate
of that member. Say for example, in a matrix
representation you take a matrix, complex
matrix; the Hermitian conjugate of the same
matrix, and then that is going to be equal
to the multiplication, the product of these
two is going to be equal to 1. They are called
unitary member, unitary matrices.
So, the members of SU 2 are unitary. In matrix
representation it is a unitary matrix, they
are unitary matrices. And special unitary
is what the SU stand for. And the special
here means that the determinant of the member,
element is equal to plus 1. You can show that
the determinant of any unitary matrix is going
to be plus or minus 1. And if you take only
those with determinant equal to plus 1, then
you will get the special unitary group. So,
this is the symmetry group of isospin symmetry.
So, as we said the, if you take p n or u d
they belong to a doublet of p n or u and d
they belong to the doublet of SU 2 isospin.
And the strong interaction does not distinguish
between the members of a particular multiplet.
Whether it is p or n does not know, u or d
you cannot distinguish and the isospin in
a strong interaction. So, let me write that
here for you.
Strong interactions do not distinguish p from
n 
or u from d. In fact, if you take a particular
multiplet it cannot distinguish one member
from the other under SU 3, I mean under the
strong interaction. This is actually similar
to the case as we said that the spin case
or the other way of saying is that, another
example is if you take charged particle, electromagnetic
interaction of charged particles, if a particular
particle say A has charge plus 1 and some
other particle B has charge plus 1 their interaction,
electromagnetic interaction will be exactly
the same with something else.
If you take some background electric field
put a charged particle of charge plus 1 and
another particle with charge plus 1; say under
electromagnetic interaction a particle a with
charge plus 1 unit I mean, and another particle
with charge plus 1 unit behave 
in a similar fashion. Which means that if
you take A, put it in an electric field and
take B and then put it in an electric which
we put it in the same electric field they
will experience exactly the same kind of force,
ok.
So, this is not a strange thing and then this
means that there is a particular quality or
quantity of quantum mechanical, or I mean
property which called charge electric charge
and then that is the only thing which will
distinguish the particle. Nothing else as
such. Similarly, there has to be something
which actually is associated with the strong
interaction we call this the charge strong
charge or the strong color charge and that
is the only thing to distinguish this particle.
And then it is not the electric charge say
there that will distinguish this. So, as far
as that is concerned, u quark and the d quark
behave in a similar fashion. We will come
to the color charges later.
Now, there is another way of grouping these
quarks which is called a flavor symmetry,
ok.
Which says that if you take a u quark a d
quark and an s quark, you can put them in
a particular group ok, and this flavour symmetry,
I mean the transformation within this it is
not going to make any difference as far as
the strong interactions are concerned. With
this we can associate the symmetry group of
SU 3.
Now, in a similar way we can also take u bar,
d bar and s bar and then put it under SU 3.
And this is called a triplet under SU 3. Similarly,
this is a, u bar, s bar, d bar is a triplet
under SU 3. Now, when we combine this to make
quarks if you take a 3 triplet let me denote
the triplet by 3; and an antiquark triplet
by 3 bar. Under the SU 3 we can understand
this, as a combination of an 8 under one.
The objects that you will get; the result
of this combination will actually can be classified
under a group of 8 and 1 standing alone. What
do I mean by this? This is something which
we just now, earlier said. Think about the
spin half object which is a doublet; in a
spin half multiplet, you have spin up, or
spin third component of the spin as plus half
or minus half there are 2 members. So, that
is why it is a doublet.
And take another spin half combine them you
will get one spin one particle which belongs
to which is actually a triplet. So, this is
spin half, cross half another way of writing
it has spin here in the first case we are
writing it as 2 cross 2 equal to 3 which means
that ;3 plus 1 ok; which means we are denoting
this by the multiple’s; number of members
in a particular group. And second way of writing
it is basically denoting it by the spins.
Spin half as a doublet for spin half cross
spin half will give you a spin 1 and then
spin 0. So, spin 1 has 3 projections plus
1, minus 1 and 0, there are 3 values plus
1, minus 1 and 0. In the case of spin 0 there
is only 1 such that things. So, that belongs
to a singlet, one multiplet of member one.
And spin 1 belongs to multiplet of 3 which
is triplet. In a similar fashion if you do
the algebra and consider the SU 3 grouping
then 3 cross 3 bar is going to give you an
8. Remember a group or a multiplet of 8 members
and 1 singlet. This should add up 3 into 3
is 9, so that is equal to 8 plus 1. Similarly,
for the spin case 2 into 2 is 4 and 3 plus
1 is also equal to 4, ok.
So, now, let us take u d s find out what are
the different types of mesons you can make
with a u bar and d bar and s bar under the
flavor symmetry.
So, this tells you that you can have a uu
bar you can have a ud bar and you can have
a us bar, you can have a dd bar, you can have
a ds bar, you can have an s s bar, you can
have d bar u, sorry du bar du bar and you
can have an su bar and an sd bar. There are
9 such possibilities of course, so 9 combinations.
Now, how do you make it an 8 and a 1? We saw
that ud bar is nothing but a pi plus and the
pi 0 is 1 over root 2 uu bar, dd bar minus
uu bar and pi minus is, pi minus is a u bar
d ok. So, these are taken up. Actually, then
you see that ud bar is taken over ud bar is
taken. Let me circle that. And then u bar
d is also taken by pi minus dd bar and a combination
of dd bar and uu bar is also considered in
that. So, there must be another combination
of uu bar and dd bar or along with something
else which will come up. That is another one.
And you have a k plus which we had considered,
which is us bar, us bar and k 0 was ds bar
then you had k 0 bar was or k 0; k minus 
u bar s k 0 bar is equal to d bar s ok.
So, these are already identified, you know
this thing. Now we already have one particle
which has electric charge equal to 0, which
is one when particle pi 0 which is taking
up the combination dd bar and uu bar. There
is one singlet that we had to make as per
our split of 3 cross 3 equal to 8 plus 1 using
uu bar, dd bar and ss bar so that singlet
should be treating u, s and d in a similar
fashion. So, let me call that singlet us eta
bar and write it as uu bar plus dd bar plus
ss bar, taking it in exactly the same footing
compared to each other. Normalization constant
is 1 over root 3, so that eta bar eta bar
dot product, or be normalized. Eta bar is
normalized, the wave function is normalized
to 1 over 3; sorry 1.
Now, an orthogonal combination of this is
given here pi 0. If you look at pi 0 and eta
prime they are orthogonal to each other ok.
Now, if you take you have to make one more
orthogonal combination and that can be written
as, so orthogonal combination which is orthogonal
to both eta prime and pi 0. You have to have
a u bar u; uu bar and dd bar plus, with a
plus sign because that will make sure that
it is orthogonal to pi 0.
And to make sure that it is orthogonal to
ss bar, now that tear across the u and d at
the same that will give you a 1 plus 1 which
is 2, here to cancel that by 2 ss bar. So,
that the whole thing is orthogonal to each
other uu bar that will give you a 1, dd bar
dd bar will give you another 1 and minus 2
ss bar into one ss bar will give you minus
2. So, that is orthogonal to the combination
eta prime.
So, thus we have 9 combinations all right.
So, 9 combinations out of uu bar, u d s and
u bar, d bar, s bar. And since we have put
them in different multiplets, let me collect
those things and then put it in a group of
8 and a group of 1. Conventionally what is
done is you actually put it in a plane of
in a plane of I 3 and S, where S is called
the strangeness.
So, since we did not make this especially,
explicitly earlier; strangeness, similar to
the isospin you will also assign what is called
a strangeness quantum number. So, the strangeness
is assigned so that u quark and d quark has
strangeness equal to 0 and s quark has strangeness
equal to plus; actually minus 1. So, with
this we will see that if you put in a plane
of I 3 and S the pions are made of u quark
and d quark alone. So, strangeness is 0, so
they should be coming in the s equal to 0
plane line and I 3 equal to plus 1 0 and minus
1.
If you take k plus, k plus is an; the quark
content of k plus is us bar. So, the strangeness
of this is plus 1 because strange quark has
strangeness is equal to minus 1 and s bar
the strange anti quark has strangeness equal
to plus 1. So, k plus will have a strangeness
plus 1 and k 0 will have a strangeness plus
1. And they have isospin third component of
the isospin equal to plus half and minus half,
respectively. So, we have k plus and k 0 and
similarly you have a k minus and k 0 bar with
strangeness is equal to minus 1 because the
quark content has an S in it. So, they have
an S in it. So, let me put a dotted line to
join these.
So, you if you have 4 plus 3, 7 members here
and you put the eta which is not a singlet,
along with it. And this makes a group of 8
and then this is called the meson octet. And
the singlet eta prime in the I 3 s plane stays
alone at I equal to 0, I 3 equal to 0, s equal
to 0 position, because if you look at eta
prime it has uu bar, dd bar strangeness equal
to 0 and a term with ss bar again strangeness
is 0 there because s quark a strangeness minus
1 and s bar has strangeness plus 1. So, this
is how this grouping goes.
So, we have a meson octet which we have somehow
supported the way we put it in an octet in
some fashion. We have some kind of a support
from the quark model where we can think of
the s, u and d quarks belonging to a triplet
of SU 3. And then similarly for the anti quarks.
And then make a combination of these quark
and anti quark and then see that they actually
can be thought of as a group of 8 and a group
of 1.
Baryons, we can actually think about the other
mesons also in a similar fashion put them
in different groups, multiplets; different
multiplets under the SU 3 grouping. But now
coming to the baryons we will be able to;
we have; they are made of 3 quarks. So, we
should be taking under SU 3 of flavor you
should be taking a member from a triplet,
and a member from a triplet, and a member
another member from another triplet.
Then see that group theoretically this 3 cross
3 cross 3 will go to a 10 and 8 and then 8
and then one. So, you will have a group of
10, a group of 8, a group of another group
of 8 and a singlet ok. So, with u d s groupings
and we can actually work it out in exactly
the same way as we have done earlier and we
will see that.
So, I will write only the quark contents here.
Not exactly the wave functions. So, quark
content of p the proton is uud and quark content
of neutron is udd. So, again in a plane of
I 3 and strangeness group, so they will come
at ok. So, let me, so this is your I 3. So,
proton and neutron are spin, isospin half
and I 3 equal to plus half for proton; I 3
equal to minus half for neutron; strangeness
is equal to 0 for this thing. And then you
can actually; you have another set of these
baryons which are called sigma baryons. There
are 3 different types of them; sigma, sigma
minus, sigma 0 uds and sigma plus, which is
uus.
All of them has one strange quark in it and
therefore, the strangeness is minus 1. So,
strangeness minus 1, isospin half isospin
minus half is isospin 3 third component of
this is minus half (read as minus one) a sigma
minus and sigma 0 has isospin 0 and sigma
plus has isospin third component of isospin
equal to plus 1.
And then there are 2 other baryons which are
called cascade 0 which can be thought of as
made of a u quark and the s quark, ss u; uss
and cascade with a charge minus 1 unit which
is dss. And they have isospin their component
plus half which is cascade 0 and minus 1 unit,
which is cascade minus. And along with that
you have a another this one called with I
equal to 0 which is u d s the lambda quark.
So, these things together ok, so lambda will
also come there, can be thought of as a multiplet
of 8. ok, and when you put it in an I 3 s
plane they will be arranged in this fashion.
This is called the baryon octet.
We indeed have another baryon octet and the
10 is also there we will discuss that in the
next class.
