Physics 557 – Lecture 9
Particles and their quantum numbers : Let us turn now to (re)introducing the various particles of the Standard Model. (See Chapters 1 and 4 in Griffiths and 6 and 15 in Rolnick.)
Leptons –
At the beginning of the last century we already knew about the electron as one of the basic building blocks, along with the proton. If was called a lepton to signify that it is light (compared to the proton). The electron (and the other charged leptons) do not participate in the strong interactions but do participate in the electromagnetic, weak, gravitational and, presumably, the Higgs interactions. The anti-electron or positron was observed in a cloud chamber by C. D. Anderson in 1933 and helped to confirm the prediction from the Dirac theory that all spin ½ particles had antiparticles. The electron provides the conventional unit of electric charge
e 1.602176 565 35 1019 C. (9.1)
Typically we write Qe 1 and Qe 1. The shared mass is
me m e 0.510 998 928 11 MeV. (9.2)
The close “friend” of the electron is the electron neutrino first suggested in 1930 by W. Pauli to explain the peculiar behavior of neutron decay. One observed the decay of the neutron into a proton and an electron, n p+ e, which conserved electric charge but little else. With n, p and e all fermions, how could angular momentum be conserved? Also a 2-body decay, as we have noted in the HW, should yield a definite energy for both the proton and the electron. But instead a spectrum of energies was observed. In fact, the observed event structure was consistent with a 3-body decay as discussed in the homework with 3 angles and 2 energies allowed to vary by energy- momentum conservation. Hence the existence of a spin ½, electrically neutral and essentially massless particle, which participated in this weak decay, was postulated. This was the first instance of what would later become standard procedure. When confronted with a mystery, predict a new particle! The current upper limit on the electron (anti-)neutrino mass is
m 2 eV (9.3) e
Lecture 9 1 Physics 557 Autumn 2012 from observations of the final state in tritium beta decay.
The next addition to the lepton family was the negatively charged muon (and the positively charged anti-muon). C. D. Anderson and Seth Neddermeyer found it in a cloud chamber looking at cosmic ray debris. (Neddermeyer founded the UW’s own cosmic ray program). It was observed to be much more penetrating in matter than an electron with comparable momentum. This suggested that the muon had a lower velocity than an electron (for the same momentum – see the discussion in the PDG article about energy loss in matter, for > 3 energy loss is an increasing function of ), 2 2 dE ln . (9.4) dx 2
The mass of the muon is now known to be
m 105.6583715 0.0000029 MeV, (9.5) some 200 times heavier than the electron! This discovery was doubly confusing. Yukawa had already predicted a particle with mass of approximately this magnitude in 1937 to explain the observed range of the strong interaction, i.e., the Yukawa potential looks like emr V r ~ , (9.6) Y r where the observed range of 10-15 m suggests m ~ 200 MeV. Thus one expected that the negative muon would be readily absorbed in nuclei (the positive anti-muon would electrically repelled) and thus not very penetrating. Instead the muon was observed to be penetrating and, when finally slowed enough to be electrically captured, it would happily orbit the nucleus with an orbit inside of the apparent size of the nucleus. It was observed to stay in this orbit without apparent strong interactions until it decayed via the weak interaction. This was clearly not the behavior of the fundamental carrier of the strong interaction! The confusion persisted until the discovery of the pion in the 1940’s, which clearly was the particle suggested by Yukawa.
Lecture 9 2 Physics 557 Autumn 2012 The muon was observed to have interactions just like the electron except that it was much more massive and, in fact, was observed to decay into the electron. A second, distinct neutrino was presumed to be at play in the process,
e e . (9.7)
Again an energy spectrum characteristic of a 3-body decay (although only one particle is easily detected) was expected and seen. The “direct” verification of the distinct nature of the two neutrinos was confirmed by a nifty experiment at BNL in 1962 by M. Schwartz, L. Lederman and J. Steinberger, which eventually was honored with a Nobel Prize. (It is interesting that in those [good olde] days a Nobel Prize winning experiment could be conceived, mounted and completed in a matter of days!) A neutrino beam was obtained from a beam of pions and kaons, which decay primarily into muons and muon neutrinos. The decay channel with electrons and electron neutrinos is disfavored by a factor of about 10-4! (We will return shortly to the question of why electrons and electron neutrinos play so little role here even though they are favored by phase space, i.e., they are so much lighter). When the neutrino beam struck a nuclear target, the experimenters saw
n p , (9.8) p n and not n pe , (9.9) p ne.
This clearly suggested not only that there is an overall additive lepton number L that is conserved, but also that there is a conserved lepton number for each generation.
e e e e
Le +1 -1 0 0 +1 -1 0 0 L 0 0 +1 -1 0 0 +1 -1
The total lepton number is the sum L = Le + L. Shortly we will add another lepton number for the tau generation. Until the recent discovery of neutrino oscillations, these leptons numbers were taken to be separately conserved. For example, the decay of the muon (mean lifetime = 2.1969811 0.000022 x 10-6 s) looks like
Lecture 9 3 Physics 557 Autumn 2012 e e
Le 0 1 1 0 (9.10) L 1 0 0 1 L 1 1 1 1
This process confirms the equality of the electric charges of the electron and muon. Their weak charges are also equal within experimental errors (a result often called lepton universality). Note that although the muon does decay via the weak interactions, L is not violated due to the muon neutrino in the final state. At the level of 1 part in 1011 we do not observe the decay e + , which would violate the separate lepton numbers. The mass of the neutrino is currently bounded above by 0.19 MeV.
Before the development of the gauge theory of the electro-weak interactions, both neutron decay and muon decay were treated as a point interaction involving 4 fermions with strength given by the dimensionfull Fermi constant
5 -2 GF 1.166 3787 6 10 GeV . (9.11)
While quite adequate for phenomenological applications at low energies, such an interaction was known to be non-renormalizable when iterated to higher orders. Another way to see that there is a problem is that, since the neutrino – proton scattering cross section must be quadratic in GF, its energy dependence must look like
2 p G F s (9.12) simply on dimensional grounds. While this form may be acceptable at low energy, this behavior cannot be true asymptotically as s without eventually violating the bounds of unitarity. This was a major stumbling block for understanding the weak interaction until the advent of the broken-symmetry theory of the electro-weak 2 2 interactions. We now understand GF g /MW for energies less than MW. At larger energies, i.e., at the current colliders, the more complex structure of the full gauge interaction is apparent and unitarity is respected.
Lecture 9 4 Physics 557 Autumn 2012 During the “glory” days of 1974-76 both the charm quark, to be discussed shortly, and the last known generation of leptons was found, the tau lepton. Again we have a lepton identical to the electron except for its mass, which is clearly not “light”,
m 1776.82 0.16 MeV. (9.13)
This (so-called light) lepton is heavier than a proton! The tau has decay channels analogous to those for the muon, e.g.,
, (9.14) e e .
Recall that the associated tau neutrino was confirmed just at the beginning of this century at Fermilab (the DONUT collaboration). The tau neutrino mass limit is much less restrictive than that on the other neutrinos, being an upper bound of 18.2 MeV. Note that the above leptonic decay channels amount to only a fraction of the total tau decay modes (17.410.04 % for the muon mode and 17.830.04 % for the electron mode) while the electron mode is essentially 100 % for muon decay. This is simply the result of competition from the hadronic channels that are now allowed by phase - - space, e.g., . If we think in the language of quarks, i.e., the pion is a quark- antiquark pair, we can still analyze all of these channels via the Fermi model of the weak interactions. Again by dimensional reasoning we have for the electron mode
2 5 e GF m 5 m (9.15) e . m
This simple expression assumes that me , m m , which is certainly a reasonable first pass to estimate this decay rate. Recall that the muon mean lifetime 2.2 x 10-6 s, while the observed lifetime of the tau is 2.9060.01 x 10-13 s. Using the formula above we estimate that the branching ratio to the electron channel is
5 2.9 1013 s 1777 MeV BR e 6 17.7%, (9.16) e 2.2 10 s 105.7 MeV
Lecture 9 5 Physics 557 Autumn 2012 in very good agreement with the observed value of 17.830.04 % noted earlier. In the same approximation the muon channel should exhibit an identical branching ratio as it (nearly) does (17.410.04 – we would expect a slightly reduced branching ratio for the muon channel as the available phase space is slightly smaller). A somewhat cruder model arises from looking at the tau decay in terms of the various 4-fermion channels of the form
f f , (9.17) where the fermion pairs f f are all those allowed by the weak interactions, i.e., the “weak” isospin doublets (we will discuss these doublets in more detail later) of sufficiently low mass
e u , , ,i . e ., e e : : d C u . (9.18) e d C
Recall that the “down” quark in the weak doublet with the up quark is really a mixture of the down and strange quarks (i.e., the mass eigenstates) specified by the mixing angle first defined by N. Cabibbo,
d(CCC ) d cos s sin . (9.19)
There are two sets of isospin basis vectors, one defined by the strong interactions and the mass (and flavor) eigenstates and one defined by the weak interactions. These two sets need not be and are not the same. This leads to the mixing which allows the weak interactions to violate the separate flavor quantum numbers.
To continue our discussion of the decay of the tau, we ignore the masses of the final state fermions, i.e., ignore the details of the 3-body phase space. One might expect that there are three distinct channels (electron, muon and quark) and that each would have a branching ration of 1/3. But wait! To calculate the contribution of the quark- antiquark channel to tau decay, we need to sum over all possible states. This sum includes a factor of 3 for the possible “color” states, e.g., red-anti-red, blue-anti-blue and green-anti-green. Thus the fact that the quarks have an extra quantum number, the color of the SU(3) symmetry of the strong interactions, has directly observable consequences. In this simple model calculation we now have 5 distinct f f channels and we expect the following branching ratios
Lecture 9 6 Physics 557 Autumn 2012 BR 20%, e e BR 20%, (9.20) BR 60%. + hadrons
These naïve results are in reasonable agreement with observation,
BR 17.8%, e e BR 17.4%, (9.21) BR 63%. + hadrons
The last channel includes a variety of final states (see the PDG summary). Specific - - -3 - examples include + (10.830.06 %), + K (7.000.10 x 10 %) and + + 0 (25.520.09 %). In this last case the 2-pion configuration is dominated by the resonance. Clearly the simple model of counting lepton and quarks seems to work for the global lepton versus hadronic branching ratios at the 10 % level but it is hopeless for providing finer details of the hadronic states. This is a general and useful result. To estimate the relative contributions of unspecified hadronic channels, i.e., the inclusive sum of the hadronic channels, relative to leptonic channels, it suffices to simply count the relevant quark and lepton states. The (presently incalculable – but lattice calculations are getting there) final state confining strong interactions will not generate large corrections to this naïve estimate, i.e., they cannot grossly change the relative probability that the short distance interaction (the Fermi weak interaction vertex in this case) specifies for quarks versus leptons. The final state strong interactions simply redistribute the probability within the various possible hadronic channels. For this reason the quark counting technique has little to say about the details of the hadronic final state. We will meet this situation again whenever we discuss electro-weak couplings.
In summary, we observe 3 generations of leptons with a wide range of masses and, at least until quite recently, a separately conserved lepton number for each generation (electron, muon, tau) of lepton. The coupling of these three generations to the weak and electromagnetic interactions is identical within experimental error, i.e., the couplings (charges) are universal. This universality of couplings is built into the Standard Model since the generations are simple replications of each other with only the masses changing.
Lecture 9 7 Physics 557 Autumn 2012 Hadrons –
As already noted, we see two basic types of hadrons or strongly interacting particles: Fermions (1/2 integer spin), which are called baryons (meaning heavy); Bosons (integer spin), called mesons.
We now understand that these two possibilities represent the two possible ways to construct a “color singlet” state from colored quarks. From the standpoint of the Quantum Chromo Dynamics (QCD) SU(3) gauge theory of the strong interactions, when we combine a 3 and a 3 of color (we will work out the details of this when we talk in more detail about the representations of SU(3)), we find 9 (=32) states that break into a singlet 1 and an octet 8 , 3 3 = 1 8 . The former yields the color singlet mesons and the latter corresponds to the coupling of quarks to gluons, which are in the 8 adjoint representation (this is the analog of the electron – positron – photon vertex of QED). Likewise when we combine 3 color triplets as in baryons, we find 27 (=33) states appearing: a 10 , 2 8 ’s and a 1 representations of SU(3). This last representation is the totally antisymmetric color singlet state defined by our old friend the antisymmetric 3-tensor, jkl qj qk ql, where the qj are the color wave functions of the quarks. The 10 is a symmetric state while the 8 ’s have mixed symmetry under interchange of the quarks, 3 3 3 = 10S 8MMA 8 1 .
In the 1930’s 2 baryons, the proton (Qp = -Qe) and the neutron (Qn = 0) were known, and no mesons. Like the electron, both were presumed to have antiparticles but it was not until 1955 that the antiproton was produced and detected. Due to the conservation of baryon number, the threshold process is
p p p p p p. (9.22)
By convention both the proton and the neutron have positive parity (JP = ½+). As already discussed, the strong interactions of the two are essentially identical and we describe this by putting the nucleons in an isodoublet (I = ½) of strong isospin,
p N . (9.23) n
As mentioned in lecture 7 and above, the first new particle (after the muon) was the pion, observed first in two charged states, 1, with mass
Lecture 9 8 Physics 557 Autumn 2012 m 139.57018 0.00035 MeV. (9.24)
They were observed to decay with a characteristically weak interaction lifetime
2.6033 0.0005 108 s. (9.25)
Note that this decay is essentially the inverse of (formally called the “crossed” version of) the tau to pion decay channel with the muon replacing the tau,
. (9.26)
Recall the basic coupling of the weak interaction is the same for all of the lepton generations, only the masses change. We can think of the decay process as occurring as illustrated in the figure, similar to that for the neutral pion. The truly interesting feature is that this channel has a branching ratio of 99.98770 0.00004 %, while the branching ratio of energetically favored decay into the electron channel is only (1.230 0.004) x 10-4. This result is so peculiar and so informative that it is a typical question on general and final exams! While we need the full structure of the “V-A” weak interaction theory, to get all the details correct, it is possible to motivate the result much more simply. The central point, which we have already mentioned, is that the charged current of the weak interaction couples to left-handed particles (i.e., the coupling is proportional to [1-5]) and right-handed antiparticles. Thus the weak decay of the pion wants to produce the changed lepton and neutrino with spins aligned as indicated in the figure to the right. This is clearly a state with Jz = -1 (if the z-axis is up) and cannot arise if the initial state is a spin zero pion. In the actually produced state one of the leptons must have its spin aligned opposite to the figure so that the spins of the two leptons can sum to 0. This can happen only to the extent that the charged lepton has nonzero mass (for the moment we are treating the neutrino as massless) because, as discussed in the Appendix, a massive propagating fermion is not an eigenstate of the operator [1-5]. In fact, the weak amplitude to produce the charged lepton with positive helicity (i.e., right-handed), if the coupling to the weak charged current is proportional to [1-5] (sometimes called a left-handed coupling), is proportional to the mass of the charged lepton. Thus we expect (ignoring the enhancement of the electron rate due to the difference in phase space) that
Lecture 9 9 Physics 557 Autumn 2012 2 2 m 105.658 4 ~ 2 10 . (9.27) me 0.511 e e
Argued the other way, if we know that the weak coupling is V-A, as was known from the nuclear decay process, this behavior of the branching ratios in pion decay tells us that the pion is spin zero.
Next came the observation of a neutral meson of similar, but not identical mass, the 0, m 134.9766 0.0006 MeV, 0 (9.28) with a much shorter mean lifetime
8.52 0.18 1017 s. 0 (9.29)
The much shorter lifetime (shorter than the charged pion) is explained by the stronger electromagnetic interactions that mediate this decay as we have observed earlier. The 2-photon channel provides 98.823 0.034 % of the decays of the neutral pion. As we discussed earlier the explicit polarization structure of the 2-photon final state (and the fact that EM conserves parity) tells us that the 0 is a JP = 0- particle.
To get a handle on the parity of the charged pions we can look at their strong interactions, in which parity is a conserved quantum number. Consider the process
D n n (9.30) near threshold, where a low energy pion is absorbed by the spin 1 deuteron producing a pair of neutrons. Since we are at low energies, only the l = 0 partial wave will be excited and the total spin is just 1, i.e., the spin of the deuteron. Recall that parity is a multiplicative quantum number. The intrinsic parity of the deuteron is the product of the intrinsic parity of the neutron (+1), intrinsic parity of the proton (+1) and the parity of their spatial wave function (l = 0, so P = +1),
D p n l0 1. (9.31)
This is analogous to our analysis of fermion-antifermion pairs, except that the intrinsic parities are identical instead of opposite. [Note that the overall odd behavior
Lecture9 10 Physics557Autumn2012 of the deuteron wave function under interchange of the two fermions required by Fermi statistics is provided by the I = 0, odd isospin wave function of the deuteron.] The parity of the initial state is the product of the intrinsic parity of the deuteron (+1), times the parity of the l = 0 partial wave (+1), times the desired intrinsic parity of the charged pion, . The final state is composed of two identical neutrons and thus must be antisymmetric under the interchange of the neutrons. Further it must have total spin 1. The cases (l = 0, S = 1) and (l = 1, S = 0) both produce wave functions that are symmetric under the interchange. Hence the minimal allowed state is (l = 1, S = 3 1) coupled to overall J = 1, the P1 state. Thus the parity of the 2-neutron state is -1 and hence the charged pions have = -1.
The three charged states of the pion have essentially identical strong interactions and form an isovector under strong isospin. The pions are also eigenstates of G-parity, as already described, with eigenvalue -1. Thus for the charged pions we have IG (JP) = 1- (0-), while for the neutral pion, which is also an eigenstate of C, we have IG (JPC) = 1- (0-+). The C operator operating on the charged pions interchanges them.
Our knowledge of the isospin of the nucleons (the neutron and the proton) and the pions, and the fact that the strong interactions respect isospin, allows us to write down the general form of the pion-nucleon coupling. We will focus first on just the isospin part of the interaction. The appropriate term in the Lagrangian describes the vertex where a nucleon emits a pion as indicated in the figure. To preserve isospin, we want the interaction to have the form of an isoscalar. In detail this means that, if we are careful about indices, the final expression must have no indices “left over”, all must participate in an appropriate multiplication. The vertex must involve an isospin wave function for both the incoming and outgoing nucleons and the corresponding SU(2) (matrix) representation of the of pion in the 2-D basis provided by the nucleons. From our previous discussion of SU(2) (in the context of ordinary spin) we know that the 2-D representation of a 3-component isovector multiplet, 3, is provided by the fundamental representation of the SU(2) generators themselves, i.e., the Pauli matrices (modulo a factor of 2). Thus we define a vector of matrices and write the representation of the pion acting in the 2-D basis of the nucleons as
Lecture9 11 Physics557Autumn2012 0 1 0i 1 0 1 2 3 1 0 i 0 0 1 (9.32) 3 1 i 2 . 1 i 2 3
Note that the natural basis here is the 1,2,3 or x,y,z basis. To make connection to the physical basis of the eigenstates of Iz, the charge eigenstates, we must identify the properly normalized creation operators. These are
1 †† i : 0 ; 2 1 2 1 †† i : 0 ; 2 1 2 (9.33) 0† 0† 0 3 : 0 , similar to the ladder operators. Formally, the creation operator for the + also destroys a - and vice versa. In this basis we have
0†2 † . (9.34) † 0† 2
The desired isoscalar form of the interaction, with all indices explicit and repeated indices implying summation, looks like (a,b = 1,2 ; k = 1,2,3)
~ gN† N L N, I a k kab b 0†2 † p (9.35) g p†† n , † 0† 2 n where g in this expression is the -N coupling constant. We can think of this expression as implying 4 processes (and their crossed versions) with relative strengths (in the amplitude)
Lecture9 12 Physics557Autumn2012 p 0 p:, g n p: 2 g , (9.36) p n: 2 g , n 0 n:. g
Thus, by knowing that isospin is a symmetry of the strong interactions and by knowing the isospin content of the particles, we can predict the relative strengths of these couplings, i.e., the 2 in the charged pion emission. We can verify this factor by looking at the corresponding Clebsch-Gordan coefficients:
1 12,121,012,12 , 3 2 12,121,112,12 , 3 2 (9.37) 12, 121,112,12 , 3 1 12,121,012, 12 , 3 which yield the same relative strengths in the different channels as the matrix above.
We now introduce a bit more notation and specify the full coupling (see Chapter 7 in Griffiths and Chapter 4 in Rolnick). Recall that we describe relativistic spin ½ particles in terms of 4-component Dirac spinors, x , which include both the spatial (or momentum) dependence and the spin wave function. If the fermion is non- interacting, it satisfies the free Dirac Equation (see the Appendix to this lecture for further discussion)
p m i m x i m1 x 0, (9.38) where in the last step the indices have been made explicit ( = 0,1,2,3 – the usual Lorentz index; , = 1,2,3,4 – the spinor index). The 4x4 Dirac matrices satisfy the anticommutation relations
Lecture9 13 Physics557Autumn2012 , 2g 2Diag 1,-1,-1,-1 . (9.39)
Recall this property is required so that a solution of the (linear) Dirac equation is also a solution of the (quadratic) Klein-Gordon equation. In particular, we can multiply the Dirac equation on the left by p m and use the above property of the Dirac matrices to prove that
2 2 2 p m 0 p m m 0. (9.40)
There is also a fifth matrix, which we have already used in our discussion of the weak interactions, conventionally defined by (with some variation in the literature),
5 i 0 1 2 3 , (9.41) that anticommutes with the others,
5 , 0. (9.42)
An explicit (but not unique) representation for the Dirac matrices, which is especially useful in the nonrelativistic limit, is given by [note that the above relations specify 2 2 2 2 2 that 0 5 1, 1 2 3 1 ]
1000 0001 000 i 0100 0010 00i 0 0 ,,, 1 2 0010 0100 0i 00 0001 1000 i 000 0 0 1 0 0 0 1 0 (9.43) 0 0 0 1 0 0 0 1 3 ,. 5 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0
(It is an informative exercise to verify that these matrices obey the above anticommutation relations and the reader is encouraged to do so.)
Lecture9 14 Physics557Autumn2012 As we have discussed before, the price of the Dirac equation is that we are necessarily discussing 4-component spinor fields. We understand these fields as describing 2 positive energy components and 2 negative energy components, corresponding to a spin ½ particle and a spin ½ antiparticle. The sign of the energy is identified from the components of 0. In the nonrelativistic limit the Hamiltonian is 0 just HNR ~ m.
The scalar product of two spinors has the form
† 0 , (9.44) i.e., it is this form that is invariant under Lorentz transformations. It is straightforward to verify that the following expressions transform as noted under Lorentz transformations,
:scalar, : vector, 5 : pseudoscalar, 5 : pseudovector.
Note from the second relation that the simple form † † 0 0 is not a scalar but rather a scalar density, i.e., it transforms as the zeroth component of a 4-vector.
Aside: Recall that to motivate the fact that fermions and antifermions have opposite parity we noted that under parity
x t,,, r P t r (9.45) 0 t,,, r P t r where the minus signs in the lower components of 0 tell us that the lower components of the spinor (the antifermion) have opposite parity from the upper components (the fermion). Let us verify that this is the appropriate way for parity to act on the spin wave function by considering the Dirac equation,
Lecture9 15 Physics557Autumn2012 0 i m t, r i t i m t , r 0 i m 0 t, r P (9.46) 0 0 i t i m t, r 0.
If we have a solution of the Dirac equation initially, then the transformed wave function is a solution in the parity-transformed frame due to the anticommutation relations of the Dirac matrices. Now we can confirm the parity properties of the forms above
0† 0 0 0 0 P :scalar, 0 0 0 0 P , , : vector 5 0 5 0 5 (9.47) P : pseudoscalar, 5 0 0 5 0 0 5 P , , : pseudovector.
With these results the simplest (but not unique) form describing the coupling of the pseudoscalar, isovector pions to the nucleons is
~g 5 , L N a k kab b (9.48) where again the indices are explicit.
Before returning to the historical story of new particle discovery let us consider the quark content of the pions and the nucleons (although the quark language arose much later in history). We have a doublet of quarks (the analog of the nucleon doublet), corresponding to the mass and flavor eigenstates,
u q . (9.49) d
We also need the corresponding isodoublet of antiquarks. However, we need to be careful about the phases in order for the iso-rotations to work out. As noted earlier i the charge conjugation process, qC e q , can produce an arbitrary phase. As long as no physical process mixes quarks with antiquarks (or nucleon with antinucleons) this phase is unobservable. It is conventional to include a relative minus sign, which Lecture9 16 Physics557Autumn2012 facilitates some issues about iso-rotations (see later), although the literature is not uniform on where to put it. We will use the choice made in Rolnick (but this is not unique in the literature),
d q . (9.50) u
[There is a similar definition of the antinucleon doublet.]
Since we know that 3 quarks will make a baryon, we have the following quantum numbers for the quarks:
Quark IIz Baryon # =Y Q = Iz + Y/2 u 1/2 1/2 1/3 2/3 d 1/2 -1/2 1/3 -1/3 u 1/2 -1/2 -1/3 -2/3 d 1/2 1/2 -1/3 1/3
The baryon number and electric charge are a bit strange at first sight but, as we will see, everything works out. Still acceptance of quarks as the correct algebraic and dynamic degrees of freedom was a long time coming.
As we have noted earlier we can easily construct a pseudoscalar, isovector state from a quark-antiquark pair. The spatial wave function is symmetric with l = 0, which we would naively expect to be the state with lowest energy (we will return to the issue of why the pion mass is so low). We can represent the isospin and spin wave functions as u d uu d d 0 . (9.51) 2 2 du
Note that, due to the charge conjugation convention, there is (typically) a minus sign in the Iz = 0 state even though it is a symmetric state. As discussed earlier the SU(3)
Lecture9 17 Physics557Autumn2012 “color” part of the wave function is the color singlet representation that appears in the decomposition of 3 3 8 1 .
The nucleons are composed of 3 quarks with the color part of the wave function given by the color singlet in the decomposition 3 3 3 = 10S 8MMA 8 1 . As noted earlier this wave function is antisymmetric under the interchange of any pair of quarks. Thus Fermi statistics tells us that the rest of the wave function must be symmetric under interchange. In the early days, before QCD, this was a major conundrum. As we will see, especially for the ++ particle, it was clear that the natural, lowest energy state for the baryons had a symmetric spatial wave function, i.e., l = 0, and the spin and isospin wave function together were also symmetric under interchange. How could this be if the underlying quarks were fermions? This situation generated many crazy ideas before QCD saved the day by adding an extra (antisymmetric) factor to the wave function! As we will see in more detail shortly, the spin and isospin wave functions of the nucleons are individually of mixed symmetry. Together they are overall symmetric. For the spin up case we can express the combined spin and isospin wave function as
2u u d 2 u d u 2 d u u u u d u u d u d u u d u d u u d u u p 1 N , n 18 (9.52) 2d d u 2 d u d 2 u d d d d u d d u d u d d u d u d d u d d with a corresponding form for spin down. [The interested student should verify the claimed symmetry.]
After thinking a bit about this simple quark model for the hadrons, the following questions immediately arise
Where are the spin 1 and isospin 0 states for the mesons?
Where are the spin 3/2 and isospin 3/2 states for the baryons?
Lecture9 18 Physics557Autumn2012 We will turn to these questions along with discussing the 3rd quark flavor in the next lecture. You should also consider how the historically earlier observation of (approximate) isospin symmetry arises in QCD.
Lecture9 19 Physics557Autumn2012