Lecture 1

Preface to Lecture Notes

A lot of my colleagues have written pompous, high-minded, self-indulgent lecture notes to accompany their lectures, this is another such set. However unlike these nameless, faceless colleagues, I'm hoping to generate some lecture notes that you might find useful. That's why I use big type and lots of paper.

Two texts that go well with this section of the course are Chapters 4 and 5 of "Introduction to Elementary Particles" by D. Griffiths and Chapters 8, 9, and 10 in "Nuclear and Particle Physics" by Burcham and Jobes. I would highly recommend the book by D. Griffiths in general for its no-nonsense writing style and an attitude that mathematics is a tool neither to be feared nor revered.

The things books seem to fail to explicitly point out, except in perhaps one summary page or paragraph, is that much of bound state hadronic particle physics is actually ad hoc. It works because some clever people, with no clue as to what is really going on, noticed certain similarities between the various particles. They would then postulate some sort of principle or symmetry that explained these observations. Sometimes that same principle would allow them to predict other things. In many cases this prediction would then hold.

The reason for this abysmal state of affairs is for two very fundamental reasons, which you will learn in this course, but will not be pointed out to you either in bold type or big letters except here:

1. The strong force coupling constant is ~1 compared to the inertial part of the wave equation.

2. We can only exactly solve for sinusoids and Gaussians. Everything else must use perturbation theory.

Because of these two facts it is impossible to calculate anything about bound states of quarks, which are completely non-perturbative, in some closed form. The only hope of doing these calculations is using a technique called 'lattice QCD' which requires such enormous computing power that it has only been quite recently that this technique has been able to get the mass of the proton from first principles.

Fortunately, for both of us, lattice QCD is outside the bounds of this course. Unfortunately, this leaves us with some fairly disjointed methods particularly around Isospin and the flavour SU(3) group which are useful, but one must be careful of context. I'll make an effort to point out where I think the context applies, but because this too is a pompous, high-minded, self-indulgent set of lecture notes; I'm afraid there are no guarantees.

Quark Flavour and Related Quantum Numbers:

Flavor	U	D	S	C	Charge	I	I3	B
u	1				2/3	1/2	1/2	1/3
d		1			-1/3	1/2	-1/2	1/3
c				1	2/3			1/3
s			-1		-1/3			1/3
t					2/3			1/3
b					-1/3			1/3
u-bar	-1				-2/3	1/2	-1/2	-1/3
d-bar		-1			1/3	1/2	1/2	-1/3
c-bar				-1	-2/3			-1/3
s-bar			1		1/3			-1/3
t-bar					2/3			1/3
b-bar					-1/3			1/3

Note: All blank entries have quantum number 0. I've also left off top and bottom quarks with "topness" and "bottomness" quantum numbers. All quarks are spin 1/2 particles and so they are also fermions.

Names of the weird quantum numbers:

U = "upness" D = "downness" S = "Strangeness" C = "Charm"

I and I₃ are the Isospin magnitude and the 'z' component of Isospin respectively.

B = Baryon Number

Notice that for all the quarks, B = (U + D + C - S)/3 also I₃ = (U - D)/2

This was chosen by design. With this choice you get the correct baryon number for the baryons and mesons. Example L_C (udc):

Baryon ” 3 quarks (U + D + C - S)/3 = (1 + 1 + 1 - 0)/3 = 1

Meson ” quark - antiquark ( p⁺ (u dbar)) B = (1 - 1)/3 = 0

Personally, when I was first learning this, it all seemed a bit disjointed. But there is one simple fact that cannot be denied from experiment. ALL of these seemingly ad hoc quantum numbers are, in fact, conserved in the interactions of the strong nuclear force.

Example in collisions of pions and protons:

Lecture 1

Combinations of the quark soup:

Mesons: Pions ” up and down quarks (u d-bar) ” p⁺

(u u-bar) ” p⁰

(d ubar) ” p^-

Kaons ” strange mesons (u s-bar) ” K+

(d s-bar) ” K⁰

(d-bar s) ” K⁰-bar

(u-bar s) ” K^-

Notice how the negatively charged mesons contain the strange quark while the positive mesons contain the anti-strange quark. This is another one of those historical accidents, which is like electric charge. About a century passed after the discovery that there were two kinds of electric charge to realize that the seemingly arbitrary choice as to which was 'positive' and which was 'negative' was made the wrong way… as far as electric current is concerned.

The same thing happened with the 'strange' particles, which were discovered about 15 years before the quark model was developed. Consequently, 'strangeness' is opposite in sign to that of the quark.

Baryons: Nucleons ” up and down quarks (u d d) ” n

(u u d) ” p

(d d d) ” D^-

(u u u) ” D⁺⁺

strange baryons (u d s) ” L⁰

(d d s) ” S^-

(u u s) ” S⁺

(u s s) ” X⁰

Now we see why there is no D^{- -}, it is impossible to combine three quarks with the electric charges that we have given them into these combinations. The D^{- -} is actually the antiparticle of the D⁺⁺ and forms a mirrored multiplet of antibaryons.

Spin Revision:

We start with a spin operator J and we assume that the Hamiltonian commutes with this operator. This makes J a constant of motion in classical mechanics, in quantum mechanics and this means that J will have discrete values that completely characterize the system.

Every text explains how you get to the algebra associated with J = J_xi+ J_y j+ J_z k I have found Burcham and Jobes Chapter 8 a good reference for this. In any event that algebra is:

More compactly written as:

Where e_ijk = 0 if any of the indices are equal

e_ijk = 1 if ijk = xyz and also if you re-arrange the letters by at least two jumps (i.e. e_xyz = e_zxy = e_yzx = 1)

e_ijk = -1 if ijk = xzy and also if you re-arrange the letters by at least two jumps (i.e. e_xzy = e_zyx = e_yxz = -1)

J² = J_x² + J_y² + J_z² has the property:

[J²,J_i] = 0 for i = x, y, or z

An 'eigenstate' of a given operator, 'S' say, is just that state (call the state |y>) which follows the following relationship:

S|y> = a|y> where 'a' is just some number, the eigenvalue, of
the eigenstate, 'S'.

Now our 'J', which will eventually be a combined momentum state, has the exact same algebra as the 'L' angular momentum operator which you have encountered in solving the Schrodinger equation for a spherically symmetric potential. This means that eigenstates of 'J' operators will produce the same answers as the eigenstates of 'L' operators. We can just go nuts with this and DEFINE a state which is an eigenstate of J_z. Which we will call |j m>. And thus we get:

This state is now not an eigenstate of J_x or J_y because of the algebra of how they behave. The only way |j m> could be an eigenstate of J_x,y is if J_z commuted with them…it doesn’t so it's no good hoping.

Lecture 2

BUT, we already have shown that J² does commute with J_z. So this means we can create a state that is an eigenstate of both J² and J_z.

Same algebra as 'L' so same eigenvalues.

Let's combine J_x and J_y into two combinations and see what we get:

J₊ = J_x + i J_y J_- = J_x - i J_y

[J_z, J±] = J_z J± - J± J_z

= J_zJ_x ± i J_zJ_y - J_xJ_z ± i J_yJ_z

= J_zJ_x - J_xJ_z ± i(J_zJ_y - J_yJ_z)

now use the commutation relations shown previously:

= iJ_y ± i(-iJ_x) = iJ_y ± J_x = ±J_±

suppose we operate on one of our J, J_z eigenstates with J_-:

J_zJ_- |j m > = (J_-J_z - J_-)|j m> from the above commutator relationship.

but 'm' and '1' are just numbers, which do commute with the 'J' operators.

But we know that by definition that so the J- operator must have had the effect of 'lowering' the z component of the total spin. The analogous result applies for J₊ but this operator will 'raise' the value of the z component of spin by one unit of Planck's constant.

OK, so can we find out what a, b is in this equation?

J_- |j m> = a |j m-1> or in J₊ |j m> = b |j m+1>

If we do both a 'raise' and then a 'lower' we will have an eigenstate

J_-J₊ |j m> = ab |j m> Using the commutation relations we see that

J_-J₊ = J² - J_z(J_z + 1)

So

Now remember that in this example, when J_- hits the state it is only after J⁺ has already raised it to (m+1). So a is the constant produced in going from (m+1) to m, while b is produced in going from m to (m+1). If a ¹ b then we mess up the relative normalization of the states, so a = b.

Therefore:

Spin 1/2 systems:

Now we shall look at the specific case of SU(2), which describes the quantum states of spin 1/2 particles.

Everyone else now goes about introducing the 'spinor' state and the Pauli spin matrices which have the exact same algebra as the 'J' operators if you use simple matrix multiplication. The key point is that the commutation algebra is the same.

Not wanting to break with this august and profound tradition:

The fundamental SU(2) state is the doublet

'Spin down' ™

'Spin up' ™

We can define 'raising' and 'lowering' operators on this doublets as 2x2 matrices:

This makes up a fundamental representation of the SU(2) group. Note that it is NOT THE ONLY fundamental representation of the SU(2) group. What this means is that you can model the group by these matrix manipulations, but there are other ways you can work with the group that do not explicitly use matrices. You can confirm that the algebra of these matrices is identical to the algebra of the 'J' operators that we established before.

At the moment we are looking to combine quantities that behave like the particle spin (integer or half-integer) or particle doublet properties (like Isospin). So we will want to combine two or more fundamental SU(2) representations to higher order representations of SU(2) that might mimic the states of baryons and mesons, which contain more than one of these quarks, each with it's own spin.

This is easiest to look at graphically. Graphs will tell you how many states will be in the multiplets. They will also give you some clue about whether the resulting final state wave functions will be symmetric or antisymetric.

Graphs will not easily give you any clue about what the coefficients in front of the old representation will be in the new one though, for that we really need to either use the raising and lowering operators. Or do what I do and look it up in the table of Clebsh-Gordon coefficients (see Appendix G in Burcham and Jobes).

Lecture 2

The REASON:

Now we finally get to the reason that we went through all of this.

The up and down quarks can be interpreted as a fundamental Isospin doublet.

Because:

1. Up and down have almost the same mass

2. 'upness' and 'downness' are conserved in the strong interactions.

If either of these conditions were not true then up and down would not form a doublet and the SU(2) symmetry would not be a good one.

So with a good SU(2) doublet established we can DEFINE a quantity and give it a name. Isospin was the unfortunate name given. Unfortunate because it seems to imply something is 'spinning' but it isn't.

We now define the quark doublets (and some different ways of writing the same thing):

The minus sign in the d-bar spinor allows us to treat quarks and antiquarks within the same formalism, and use the same Clebsh-Gordon tables.

Just as with angular momentum 'spin', Isospin will have a 'z' component and a magnitude which are conserved constants (i.e. commute with the strong interaction Hamiltonian).

I've ditched the h-bar factors because 'upness' and 'downess' don't have any units; they certainly don't have units of angular momentum. So it's crazy to keep factors of Planck's constant floating around.

This Isospin makes a few useful predictions. First there is the fact that you get groups of Baryons and mesons with almost the same mass. It explains why there is no D^-- states for example. The second and more important thing is it actually permits one to calculate the relative rates of decays of some of these particles, but ONLY if they decay via the strong or Electromagnetic interactions, both of which conserve Isospin.

Example: Decays of the D baryons:

First apply energy conservation, the mass of the Delta is 1232 MeV/c². So this rules out any decays to the Cascade (X) and the Sigma (S) particles since this wouldn't even leave enough room for even another pion.

Next start applying conservation rules:

Baryon number is conserved (remember, this is just built in, we don't know why this must be true actually, simply that it is) so we can't have a decay of the Delta to all pions or kaons. We need one baryon, the lightest of these is the proton or neutron. If you subtract off the mass of the proton from the mass of the D you are left only with enough Energy to give you the pions, the muon, and the electron and the photon; everything else is too heavy.

Lepton number conservation requires that you'd need two muons, or electrons. This rules out the muons on mass grounds again since you'd need two.

Now decays to electron-positron pairs or a photon are not forbidden, but must proceed through electromagnetic interactions, not the strong interactions. Since EM interactions proceed about 100,000 times more slowly than strong interactions, any process that is allowed by both forces would happen via the strong force first, with a very small probability of happening through the EM force. So we've effectively ruled out decays to electrons nearly all the time.

Furthermore the D is a baryon and baryon number is conserved, so there will have to be at least one baryon in the final state. Conservation of Energy implies that it has to be a nucleon and that there can only be one nucleon. Two nucleons and one anti-nucleon are not allowed.

Prediction 1 (but not using Isospin) -> The Delta will decay to nucleon+pion.

But there are two nucleons (p, n) and three pions (p⁺, p⁰, p^-)

Here we can use electric charge conservation to determine that some of the possible combinations are not allowed, but we don't actually need it at this stage. We now know that the D, an isospin 3/2 quadruplet, decays into an Isospin doublet (nucleons) and an Isospin triplet (pions). We can go to our favorite table of Clebsh-Gordon coefficients and read off the combinations of the doublet+triplet which make up the quadruplet. From the 1 x 1/2 table:

So Branching Ratio of:

Another example: If we have a proton target (like hydrogen) and a charged pion beam (why can we only get beams from the charged pions?) we can predict what the relative rates of interactions would be when the beam-target center of mass energy is equal to the D rest mass energy.

This also has usefulness in nuclear physics which is only a collection of up and down quarks, so Isospin is the only symmetry around.

The caveats involved in using Isospin symmetry to calculate quantities in particle physics are numerous. First you MUST guarantee that you have a strong or electromagnetic interaction. Then you MUST chose initial and final states that are pure Isospin eigenstates. (Well you don't have to, but then you end up with sums of the squares of different isospin amplitudes that don't cancel in the ratios. See Burcham and Jobes, Chapter 8.7.3) In the end this only really works well with the decays of the Delta Resonance and with the production of the different baryons in proton-pion or proton-nucleon scattering. It also works with strong decays when you get to the excited states of the baryons and mesons, where the mass is high enough to permit decays into some of the lower Isospin eigenstates.

Last lecture I promised that I'd show you how this Octet and Decuplet structure arises in this model. So far we've said nothing about the strange quark, charm, bottom, or top quarks. We have seen how the lucky accident that u and d quarks have the same mass causes certain symmetries. The strange quark has more mass than the up or down, but not that much more. So we should add it into our basic fundamental representation.

But if we do that we have added in another conserved quantum number, the strangeness (S = Y-1/3 for the quarks where Y is 'hypercharge'). It is fundamentally different than Isospin but also conserved by the strong interactions (and electromagnetic). So we have shifted away from SU(2) in a fundamental way. This requires that we delve into SU(3). This is the structure that explains most all Baryonic states, and shows why it is not worth carrying this symmetry any further to SU(4), for example, when we add in the charm quarks.

Lecture 3

Flavour SU(3)

We now wish to include strange quarks with the up and down quarks. We begin by using the same graphical method that we used previously for SU(2), but now we have a new quantum number 'strangeness' that will be orthogonal to the third component of Isospin. This means that our graphs, which were along a line in SU(2), will be in an Isospin-strangeness plane in SU(3).

A fundamental representation of SU(3) involves the use of 3-component column vectors to describe the wave functions (unlike the 2-component functions for Isospin).

Just like SU(2), SU(3) has a set of generators. Since we have two quantum numbers there are two of these generators that work like I_3. Here are the generators, F₃ and F₈ are the two which, when multiplied by the state vector, give you the Isospin and Strangeness.

Lecture 3

We see that the first and second column vectors are eigenstates of F₃ (same as I₃). So F₃ º I₃.

But the last state is not an eigenstate of Isospin…because it has the strange quantum number only. However, if we look at the equivalent of the 'strange' operator:

We would like the difference between F₈|y_1,2> and F₈|y₃> to be one unit, like isospin. But we can't get F₈|y_1,2> = 0 unless we make the F₈ generator have a non-zero trace (which destroys its properties as an SU(3) generator…very bad). So we DEFINE something called a hypercharge 'Y º S + B' (B = Baryon number). Also such that the hypercharge operator is:

There are also 3 'raising' or 'lowering' operators to take you around the triangle.

Now all of these properties are in place for both quarks and antiquarks in SU(3). And we are set up to look at what the graphical method can tell us about what happens when we combine quark - antiquark pairs and also groups of 3 quarks.

Lecture 3

Baryon Woes:

It is easy to see from the transparencies why the J=3/2 spin wave functions, which are all symmetric, must only couple to the symmetric flavor SU(3) wave function. What isn't so easy is to get the wave function for the octet.

Here we take the J=1/2 spin wave function and the octet from flavor SU(3). From the combinations of three spin 1/2 particles we had two doublets 2_ms and 2_ma. 2_ms is symmetric under the interchange of the first two quarks and 2_ma is antisymmetric under this interchange. This also happens with the octect where we get 8_ms and 8_ma, symmetric and antisymmetric under the interchange of the first two quarks. Here are the wave functions for these configurations in the case where we have a net spin 'up' proton. Notice that the flavour wave function follows the form of the spin wave function. But the flavour wave functions are supposed to be part of SU(3) whereas the spin wave functions are SU(2)! Do you have any ideas as to why this is?:

If we take :

If you work this through you will get a fully symmetri
c wave function in the exchange of two quarks which is the wave function of the spin-up proton. Does the constant in front properly normalize the proton wave function when we take |<p|p>|?

Turns out it doesn't, given what I have shown you here. In fact you will find that the normalization is off by a factor of root-2. Why did that happen? Similar wave functions can be calculated for other baryons, the easiest ones are the D baryons in the decuplet because they are completely symmetric and you don't have to fish around for the proper mixed-symmetric and mixed-anti-symmetric wave functions.

Meson and Baryon Identification Convention:

In the Particle Data Book you will see particles listed with their quantum numbers. The convention used is:

I(J^PC)

I = Isospin of the multiplet in which the particle sits.

J = Total orbital + intrinsic angular momentum

P = Parity quantum number of the particle

C = Charge conjugation quantum number of the particle.

The C quantum number is only a good quantum number for neutral particles, there will be more about C and P later. For now it is sufficient to say that they can only take on values of +1 or -1 and that C is not always a good quantum number for every hadron.

So the charged pions, for example have I(J^P) = 1(1^-) and the neutral pion is I(J^PC) = 1(1^-+). Because Parity for the mesons is P = (-1)^l+1 you can use this to puzzle out whether a given particle is believed to have a relative orbital angular momentum between the quark and antiquark. For example, because the pions all have parity -1, it is clear from P = (-1)^l+1 that l must be 0 or even. We already have a total spin J=1.

Now we do some guessing (if we aren't told already from other information). If L=0 this is very nice since then it's easy to see how two spin 1/2 quarks can combine to give J = S₁ + S₂ = 1. So L=0 is quite likely.

What if L = 2? Well if L = 2 and the two quarks spins are anti-aligned then that doesn't work. So if L = 2 the two quark spins would have to be aligned so that the spin part is S₁ + S₂ = 1. This would actually work as J could then take on values of 1, 2, and 3. Clearly L=4 is just too high to work. So by parity we are able to narrow our choices for the spin configuration of pions down to two choices.

Can you think of anything that might help narrow this down further? A hint is in the mass structure of the pions and how the J=1 pions group together. One might expect that if L=2 there would have clusters of pions with fine mass seperation corresponding to the J=1, J=2, and J=3 states…this is not seen for the lowest mass pions.

Also if L=2 then it would be reflected in the angular decay distribution of the pion decay products, but you can't tell that from just looking in the Particle Data Book. In general, the lowest mass mesons and baryons will have the simplest momentum/spin structure that describes all the quantum numbers.

Meson Mass Spectum Baryon Masses and Magnetic moments:

See the transparencies for Lectures 3&4