Lecture 9: Quarks II

• Quarks and the Baryon Multiplets

• Colour and Gluons

• Confinement & Asymptotic Freedom

• Quark Flow Diagrams

Section 6.2, Section 6.3, Section 7.1

Useful Sections in Martin & Shaw:

Baryons: Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions

Absence of anti-particles suggests there is not substantial anti-quark content

So try building 3-quark states

Start with 2:

(note that m() ≠ m(+) so they are not anti-particles, and similarly for the * group)

So try building 3-quark states

Now add a 3rd:

The baryon decuplet !!

and the sealed the Nobel prize

ddd ddu duu uuu

dds uus

dss uss

sss

uds

Baryons: Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions

Absence of anti-particles suggests there is not substantial anti-quark content

(note that m() ≠ m(+) so they are not anti-particles, and similarly for the * group)

-1

1

I3

Y

p

(938)

(1321)

0

(1315)

n

(940)

(1197)

0 (1193)(1116)

(1189)

But what about the octet?

It must have something to do with spin... (in the decuplet they’re all parallel, here one quark points the other way)

We can ''chop off the corners" byartificially demanding that 3 identicalquarks must point in the same direction

But why 2 states in the middle? ddd ddu duu uuu

dds uus

dss uss

sss

uds

ways of getting spin 1/2:

u d s

u d s

u d s

these ''look" pretty much the same as far as the strong force is concerned (Isospin)

0

J=1/2

J=3/2

ddd ddu duu uuu

dds uus

dss uss

sss

uds

We can patch this up again by altering the previous artificial criterion to:

''Any pair of similar quarks must be in identical spin states"

What happened to the Pauli Exclusion Principle ???

Why are there no groupings suggestingqq, qqq, qqqq, etc. ??

What holds these things together anyway ??

Not so crazy lowest energy states of simple, 2-particle systems tend to be ''s-wave" (symmetric under exchange)( )

So having 2 states in the centre isn’t strange...but why there aren’t more states elsewhere ?!

J=1/2

J=3/2

u u s

u u s

i.e. why not and ???

-1

1

I3

Y

p

(938)

(1321)

0

(1315)

n

(940)

(1197)

0 (1193)(1116)

(1189)

The lowest energy state has

them

Pauli Exclusion Principle

Perhaps, like charge, it also helps hold things together !

We see states containing up to 3 similar quarks this ''charge" needs to have at least 3 values (unlike normal charge!)

Call this new charge ''colour," and label the possible values as Red, Green and Blue

We need a new mediating boson to carry the force between colours (like the photon mediated the EM force between charges)

call these ''gluons"

there must be another quantum number which further distinguishes the quarks (perhaps a sort of ''charge")

In p-n scattering, u and d quarks appear to swap places.But their colours must also swap (via gluon interactions). This suggests that an exchange-force is involved...

But then we run into trouble while trying to conserve chargeat an interaction vertex:

R G

(a quark-screw!)

the only way out is to attribute colour to the gluonsas well. For the above case, the gluon would have tocarry away RG quantum numbers

But we currently have no reason to exclude exchanges which do not change the quark color as well!

So, for example, a gluon composed of the superpostion

RR1/2 BB

would couple red and blue quarks without changing their colours

To handle all possible interchanges, we therefore needdifferent gluons with colour quantum numbers

RG, RB, GR, GB, BR, BG

To couple to green as well, we just need one more gluon:

RR1/ + BB GG

Allowing these 2 additional gluons results in a higherdegrees of symmetry since we are making use of allpossible pair combinations of RGB with the anti-colours

Maybe this is a good thing to do... let’s try it and see !

Since appropriate superpositions of this gluon with will yield the necessary red-green and blue-green couplings1/2 RR BB

IC

YC

BR

G

IC

YC

B

G

R

RG BG

RB BR

GB GR

RR BBGG

Another way:

...starting to look familiar ?!

CentralSU(3) states: 1/2 ( )

1/6( ) 1/3( )

RR BB

GGRR

RR

BB

BB GG

In ''SU(3)-speak", the last state is actually a separate (singlet) representation of the group which is not realized in nature, so we end up with 8 gluons.The reason we get 9 states for the mesons is that the symmetry there is not perfect, so there is mixing. But, for colour, the symmetry is assumed to be perfect.

We could explain only having the quark combinations seen if we only allowed ''colourless" quark states involving either colour-anticolour, all 3 colours (RGB) , or all 3 anticolours.

If the carriers of the force (the gluons) actually carrycolour themselves, the field lines emanating from asingle quark will interact:

q q q

''flux tube"

* formally still just a hypothesis(calculation is highly non-perturbative)

(hence the analogy with ''colour", since white light can be decomposed into either red, green & blue or their opposites - cyan, magenta & yellow)

For this configuration, the field strength (flux of lines passing through a surface) does not fall off as 1/r2 any more

Can be stopped by terminating field line on another colour charge

Ah! So only colourless states have finite energy !

''Confinement"

q q q q q q q q q q

PoP !

''fragmentation"

The field energy will thus scale with the length of the string and so

as L then E

it will remain constant.

Clearly we can’t allow this!!

Need Colourless States...

So what about qqqqqq states ?

Sure that’s basically the deuteron (np = uuuddd)

How about qqqqq ?

Pentaquarks

for q > 0.16e, the number offractionally charged particlesis less than 4x1022 per nucleon

(Halyo et al., PRL 2000)

Search for Free Fractional Charges (M. Perl et al.)

vx = qE/6r

vz = 2r2g/9

q q

q

Getting very close to a quark:

q

RG

q

RB

q

So, on average, thecolour is ''smeared" outinto a sort of ''fuzzy ball"

Thus, the closer you get, the less colour charge you see enclosedwithin a Gaussian surface. So, on distance scales of ~1 fm, quarksmove around each other freely ''Asymptotic Freedom"

the opposite of what happens with vacuum polarization in QED!)

Note that asymptotic freedom means that the running coupling will decrease with higher momentum transfer

This also means that perturbative QCD calculations will work at high energies!

Where Are The Coupling Constants Running ???

G

RB

B

u = qq annihilation

R

So how do we now interpret pion exchange??

duu

dud

udu

ddu

p

p

n

n

= qq creation

B

d

R

GB

G

(ud)or

(ud)

K + K+p + p p + n + +

ss

su

us

K

K+

udu

uud

udd

du

uud

p

p

n

p

ud

duu

ud

duu

+

p

+

p

++

p + + ++

p + +

Quark Flow Diagrams: