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129A Lecture Notes
Strong Interactions I
1 Four Forces
Becquerel discovered radioactivity at the end of 19th century, and Rutherford classified different types according to the charge of the emitted particles. α- ray has charge 2, β-ray −1, and γ-ray zero. In today’s terminology, the α-ray is the emission of 4 He nucleus (two protons and two nucleons), β-ray is an electron, and γ-ray a very energetic photon. This was the first point in the history when three types of interactions manifested themselves: strong, weak, and electomagnetic. The γ-ray is emitted then the nucleus is in an excited state, and decays into a lower state. This process is very similar to the emission of photons from excited states of an atom. The α-ray is emitted from large nuclei when it is energetically favored to release α-particle to reduce its Coulomb repulsion within the nucleus. The parent nucleus of atomic number Z and mass number A changes to Z −2 and A − 4. The dynamics of this decay is quite complicated, but the lifetime is determined primarily by the Gamov factor, the tunneling probability for the α-particle to go through the Coulomb barrier. The fact that the positively charged particles can be packed inside small nuclei is already quite puzzling; that was the original puzzle about the nuclear strong interactions. The β-ray is the only interaction in nuclei that changes the number of protons and neutrons. The parent nucleus of atomic number Z and mass number A becomes Z + 1 and A. The net effect is to turn one neutron into a proton, by the emission of an electron. This is the manifestation of the weak interactions. Of course there is another important force in nature: gravity. It plays lit- tle role in the world of microscopic particles, as its effect is always suppressed by the Newton’s constant GN = (10^19 GeV)−^2 in the natural unit. But it is important at macroscopic distances and governs the motion of heavenly bodies as you know well. In the rest of the lecture notes, we will discuss the strong interactions.
2 Proton statistics
How do we know that protons obey Fermi-Dirac statistics? We of course know that because of the spin-statistic theorem, but this theorem needed to be established experimentally anyway. We need to know that the proton is a fermion independent of its spin and the spin-statistics theorem. For that purpose, we consider molecular band spectrum. A molecular band spectrum is what appears in the emission lines from a gas of molecules mostly from vibrational spectra (infrared), but the “lines” appear to be a “band”, i.e. a thick line. Looking more closely, the thick line actually consists of many many fine lines, which come from rotatinal deexcitations. Generally, a diatomic molecule has a rotational spectrum due to the rigid body Hamiltonian
H =
L~^2
2 I
¯h^2 l(l + 1) 2 I
Here it is assumed that the molecule has a dumb-bell shape and can rotate in two possible modes. In the case of hydrogen molecules H 2 , two atoms are bosons because they consist of two fermions (one electron and one proton). Therefore the total wave function must be symmetric under the exchange of two hydrogen atoms. A part of the wave function comes from the spin degrees of freedom of two protons. Depending on S = 0 or S = 1 for two proton spins, the spin part of the wave function is either anti-symmetric or symmetric. Everything else being the same between two hydrogen atoms, the anti-symmetry of the S = 0 spin wave function must be compensated by the rotational wave function. Using the relative coordinate ~r = ~x 1 − ~x 2 between two protons, the interchange of two protons will flip the sign of ~r → −~r. The rotational wave function is nothing but spherical harmonics Y (^) lm (~r), which satisfies the property Y (^) lm (−~r) = (−1)l(~r). Therefore the interchange of two protons would result in a sign factor (−1)l^ from the rotational wave function. In order to compensate the unwanted minus sign for S = 0 case, we need to take l odd. On ther other hand, for S = 1 case, we need to take l even to keep the wave function symmetric. Transitions among rotational levels take place only between the same S, because the nuclear magneton is too small to cause spin flips in the tran- sitions. Therefore rotational spectra appear from transitions among even l states or odd l states, but not among odd and even l’s. Because the tran- sitions are most frequent between two nearest states, the S = 0 case causes
4 Nuclei
Nuclei sit at the center of any atoms. Therefore, understanding them is of central importance to any discussions of microscopic physics. Due to some reason, however, the nuclear physics had not been taught so much in the standard physics curriculum. I try to briefly review nuclear physics in a few lectures. Obviously I can’t go into much details, but hope to give you at least a rough idea on nuclear physics. As you know, nuclei are composed of protons and neutrons. The number of protons is the atomic number Z, and the mass number A is approximately the total number of nucleons, a collective name for protons and neutrons. Therefore A = N + Z (4)
where N is the number of neutrons. We know that nuclei are very small. An empirical formula for the size of the nuclei, which can be measured using the form factor in elastic electron-nuclei scattering, is
R = r 0 A^1 /^3 , r 0 = 1.12 fm. (5)
This is a good approximation practically for all nuclei with A > ∼ 12. Here, fm = 10−^13 cm, or sometimes called also “Fermi” rather than femto-meter, and the nuclei are smaller by five orders of magnitude than the atoms. What the formula means is that the nuclear density is more-or-less constant for any nuclei, ρ = 1. 72 × 1038 nucleons/cm^3 = 0.172 nucleons/fm^3. Of course, the nuclear density does not drop to zero abruptly. The form factor measurement is often fitted to the size and the “surface thickness,” within which the density smoothly falls from the constant to zero. The result is that the surface thickness is about t ' 2 .4 fm.
4.1 Empirical Mass Formula
Gross properties of nuclei are manifested in the empirical (or Weizs¨acker) mass formula. Recall Einstein’s relation E = mc^2 , which tells us that the to- tal mass of nuclei has information on its composition as well as its interaction energies. The empirical mass formula is
mnucleus(Z, N ) = Zmp + N mn −
B
c^2
Figure 1: From “Theoretical Nuclear Physics,” by Amos deShalit and Her- man Feshbach, New York, Wiley, 1974.
Figure 3: A more realistic shape of nuclei. From “Subatomic Zoo,” by Hans Frauenfelder and Ernest M. Henley, Prentice-Hall, Inc., NJ, 1974.
Figure 4: Nuclear binding energy is more-or-less independent of its size, roughly about 8.5 MeV/nucleon. The first few peaks are for 4 He, 12 C, 16 O. The maximum is for 56 Fe. From “Theoretical Nuclear Physics,” by Amos deShalit and Herman Feshbach, New York, Wiley, 1974.
where the last term is the “mass deficit” due to the binding energy B, and is given by
B = avA − asA^2 /^3 − asym
(Z − N )^2
A
− aC
Z^2
A^1 /^3
Among all these terms, the first terms is the most important one, giving roughly constant binding energy per nucleon. If you neglect all the other terms, the binding energy is roughly 8.5 MeV/nucleon. But if you fit the data with all other terms, the number of course comes out differently. We discuss each of the terms below. The first term is called the volume term with av = 15.68 MeV, represent- ing that the total binding energy is roughly proportional to the number of
gives a preferred fraction of protons for a given A, which becomes smaller and smaller as A increases, consistent with the observed band of stable isotopes. Finally the last term is called the pairing term. There is a tendency that nucleons want to be paired between a given state and its time-reversed state, i.e., the opposite orbital and spin angular momenta. Because of this property, even-even nuclei (nuclei with even number of protons and even number of neutrons) have all 0+^ ground state. There is a sizable difference in the binding energies between nuclei with all nucleons paired (even-even ones) and those with some nuclei unpaired (even-odd, odd-even, and odd- odd). The pairing term represents the energy difference among them, given by
δ(A) =
34 A−^3 /^4 MeV for odd-odd nuclei 0 MeV for odd-even nuclei − 34 A−^3 /^4 MeV for even-even nuclei
Looking at the plot Fig. (4), there are a few anomalously high binding energies for low A. They are 4 He, 12 C, 16 O. The maximum is for 56 Fe. The presence of a maximum means that any thermonuclear fusion process, such as stellar burning, cannot produce nuclei beyond iron. We will come back to the question how different nuclei and hence elements were born in our Universe later. Fig. 5 shows observed nuclides. For small numbers of nucleons, the band is nearly diagonal, i.e., Z ≈ N. As the size grows, the band bends and is below the diagonal, Z < N. Using the empirical mass formula, the existence of the band is easy to understand. As you go away from the band, the symmetry term becomes important and the mass of the nucleus grows. What it means it that such a nuclide, if exists, decays immediately by ejecting excess neutrons or protons until the symmetry term becomes small enough to make it energetically impossible to eject free neutrons or protons. For large nuclei, Coulomb term is important and smaller number of protons is preferred. That is why the band bends downwards. Even within the band, the number of stable nuclides is not so large. All the colored ones decay either by β-decay (N, Z) → (N − 1 , Z + 1)e−^ ¯νe or anti-β-decay (N, Z) → (N + 1, Z − 1)e+νe to approach the narrow band of stability moving along − 45 ◦^ line. Unstable nuclei can also emit an α-particle, a unusually tightly bound 4 He nucleus, to lower the mass number, approaching the maximum binding energy of A = 56.
Figure 5: Table of nuclear isotopes, from http://www2.bnl.gov/CoN/. The horizontal axis is for the number of neutrons N , while the vertical axis the protons (i.e., atomic number) Z. The black squares represent stable isotopes, while the others decay either by α- or β-decays to more stable nuclei. Double lines are for magic numbers.
- Even though the nuclear force is attractive to bind nucleons, there is a repulsive core when they approach too closely, around 0.5 fm. They basically cannot go closer.
- The nuclear force has “charge symmetry,” which means that we can make an overall switch between protons and neutrons without chang- ing forces among them. For instance, nn and pp scattering are the same (except for the obvious difference due to the electric charge). For example, “mirror nuclei,” which are related by switching protons and neutrons, have very similar excitation spectra. Examples include 13 C and 13 N, 17 O and 17 F, etc.
- A stronger version of the charge symmetry is “charge independence.” Not only nn and pp scattering are the same, but also np scattering is also the same under the “same configuration” which I specify below using the concept of isospin.
The last item needs some more explanations. There is a new symmetry in the nuclear force called isospin, proposed originally by Heisenberg. The idea is very simple: regard protons and neutrons as identical particles. But of course, you can’t; they are different particles, right? They even have different masses! Well, the trick is to introduce a new quantum number, isospin, which takes values +1/2 and − 1 /2 just like the ordinary spin. We say a proton is a nucleon with Iz = +1/2, while a neutron with Iz = − 1 /2. At this point, it is just semantics. But the important statement is this: the nuclear force is invariant under the isospin rotation, just like the Hamiltonian of a ferromagnet is invariant under the rotation of spin. Then you can classify states according to the isospin quantum numbers because the nuclear force preserves isospin. But what about the mass difference, then? The point is that their masses are actually quite similar: mp = 938.3 MeV/c^2 and mn = 939 .6 MeV/c^2. To the extent that we ignore the small mass difference, we can treat them identical. Another question is the obvious difference in their electric charges +|e| and 0. Again, the Coulomb force is not the dominant force in nuclei, as we have seen in the empirical mass formula. We can ignore the difference in the electric charge and put it back in as a “small” perturbation. The charge symmetry is a limited example of the isospin invariance. It corresponds to the overall reversal of all isospins. If you reverse all spins sz , that is basically the 180◦^ rotation around the y-axis, and you obtain another
-150 0 0. 5 1 1. 5 2 2. 5 3
0
150
300
450
600
750
Radius [fm]
Potential [MeV]
(^1) S 0 (np)
(^3) P 0 (np)
(^1) P 1 (np)
(^3) P 1 (np)
-150 0 0. 5 1 1. 5 2 2. 5 3
0
150
300
450
600
750
Radius [fm]
Potential [MeV] (^1) D 2 (np) (^3) D 2 (np)
(^1) F 3 (np)
(^3) F 3 (np)
Figure 6: A recent analysis of nucleon-nucleon scattering data to obtain the nucleon-nucleon potential. Taken from A. Funk, H. V. von Geramb, K. A. Amos, nucl-th/0105011.
state with degenerate energy. Likewise, if you reverse all isospins, by rotat- ing the isospin around the “isospin y-axis” by 180◦, you interchange protons with neutrons, just like interchanging spin up and spin down states. If the nuclear force is indeed invariant under the isospin rotation, it must also be invariant under the isospin reversal. Fig. 7) shows that indeed the nuclear spectra approximately respect this invariance. Of course, isospin is not an exact symmetry because protons and neutrons have different electric charges. But the isospin invariance goes even further (“charge independence”). It says that the not only the interaction between pp and nn are the same (“charge symmetry”), also np is, except that you have to carefully select the config- uration. Here is what is required. Because proton and neutron both carry I = 1/2 (and opposite Iz = ± 1 /2), two nucleon states would have both I = 1 and I = 0 components. Both pp and nn states are said to be in the I = 1 state. On the other hand, the np state can either be in the I = 1 or I = 0 states. But the fermion wave function must be anti-symmetric while I = 1 (I = 0) isospin wave function is symmetric (anti-symmetric). Therefore, if the space and spin wave function of a np state is symmetric (anti-symmetric), it selects I = 0 (I = 1) isospin wave function. This way, you can separate purely I = 1 part of the np wave function, and compare the interaction to that of the nn and pp states. And they are indeed the same up to corrections from Coulomb interaction. On the other hand, the force in the I = 0 state can be different. For instance, the only two-nucleon bound state is the deu- terium, an np state. What is suggests is that the bound state is in the I = 0 state, and anti-symmetric isospin wave function. Then the rest of the wave function must be symmetric. For a given potential, the S-wave is always more binding than the P -wave just because it lacks the centrifugal barrier. Therefore the deuterium is likely to be in the S-wave, a symmetric spatial wave function. Then the spin wave function must be symmetric, S = 1. Indeed deuterium does have spin one. A more quantitative test can be seen in Fig. 8. 21 F, 21 Ar, 21 Na, and 21 Mg all have the mass number 21. Assuming (^18) F is in the I = 0 state, all four nuclei can be obtained by adding three
neutrons to it, which can be in either I = 3/2 or I = 1/2 states. The nuclear excitation spectra show states common only between 21 Ar and 21 Na, which are in the I = 1/2 state, or states common to all four of them, which are in the I = 3/2 state. Similarly check can be done among 14 C, 14 N, 14 O, which show states common to all of them (I = 1) or states special to 14 N (I = 0).
Figure 8: Comparison of excitation spectrum of four nuclei with the same mass number, showing states with I = 1/2 and I = 3/2 multiplet struc- ture. From “Theoretical Nuclear Physics,” by Amos deShalit and Herman Feshbach, New York, Wiley, 1974.
immediate consequence of the finite mass. The presence of the charge exchange reaction suggests that the force carrier is (or at least can be) electrically charged. This particle is called charged pion π−^ or π+^ in the modern terminology. The charge exchange reaction, producing the backward peak in the np scattering is caused by the following process. When the neutron comes close to the proton, the neutron emits the force carrier π−, and it becomes a proton (!). Even though (from the neutron point of view) she is still going pretty much straight ahead, we see the proton coming along the original direction of the neutron, namely the “backscattered proton.” The emitted π−^ is then absorbed within the time allowed by the uncertainty principle and the proton becomes a neutron. By 40’s there was discovered a particle that weighs 200 times electron in cosmic rays (or more precisely, 105.7 MeV/c^2 ). This of course raised hope that the discovered particle may be the force carrier for the nuclear force. After intensive research, however, especially that carried out by Italians hid- ing (literally) underground in Rome under Nazi’s occupation in 1945, it was shown that the new particle does not show any sign to feel the nuclear force. This particle is what is now called muon μ±. Indeed, underground is a good place to study muons! Later on people speculated that there may be two new particles weighing 200 times electron, and this is indeed what happened. By going to higher altitudes on the Andes in cosmic ray studies, people have found that the charged pions exist in cosmic rays, which quickly (within about 10−^8 sec) decay to muons which live longer (about 10−^6 sec) and reach the surface of the Earth. (Of course their life is stretched by the relativistic time dilation effect. Otherwise we didn’t have a chance to detect them even on the Andes.) Only at higher altitudes, pions had chance to enter the de- tector (photographic films). Later on, a neutral pion π^0 was also discovered that decays into two photons. They are later determined to have no spin and odd parity. Once found, it seemed to confirm Yukawa’s suggestion. Read Chapter 2 of Cahn–Goldhaber for more details. The potential between nucleons caused by the exchange of a “virtual pion” was calculated to have the following form
V =
g^2 hc ¯
m^2 π 4 m^2 N
mπc^2 (~τ 1 · ~τ 2 )
[ (~σ 1 · ~σ 2 ) +
( 1 +
μr
(μr)^2
) S 12
] e−μr μr
Here μ = mπc/¯h with mπ with the small difference between mπ± = 139.6 MeV/c^2 and mπ^0 = 135.0 MeV ignored in the same spirit as we ignore the proton-
neutron mass difference and call it mN. The factor
S 12 =
r^2
[3(~σ 1 · ~r)(~σ 2 · ~r) − (~σ 1 · ~σ 2 )r^2 ] (10)
is the form for the phenomenologically required tensor force. The matrices ~τ = 2I~ are the analogs of Pauli matrices for the isospin. The important point with the potential is that it is indeed invariant under the rotation of the isospin space because of the form (~τ 1 · ~τ 2 ). The OPE (one-pion-exchange) exchange Eq. (9) works well in the two- nucleon system. We have seen that there is only one bound state in two- nucleon system, namely deuteron, with I = 0, L = 0, S = 1. Let us see if this is consistent with the OPE potential. We focus on the s-wave (L = 0) which doesn’t have the centrifugal barrier and presumably binds the most. When I = 1 ((~τ 1 · ~τ 2 ) = +1), the Fermi statistics requires S = 0 (~σ 2 = −~σ 1 and hence (~σ 1 · ~σ 2 ) = −3). Then the tensor force is proportional to
S 12 =
r^2
[3(~σ 1 · ~r)(~σ 2 · ~r) − (~σ 1 · ~σ 2 )r^2 ] =
r^2
[−3(~σ 1 · ~r)(~σ 1 · ~r) + 3r^2 ]. (11)
At the lowest order in the potential in perturbation theory, using the fact that the s-wave is isotropic, we find 〈rirj^ 〉 = 13 〈r^2 〉, and hence the tensor force vanishes identically. Therefore, the OPE potential is
V = −
g^2 ¯hc
m^2 π 4 m^2 N
mπc^2
e−μr μr
This potential is attractive, of finite range, and may or may not have a bound state depending on the size of the coupling g^2 /¯hc and mπ. For the actual values, there is no bound state. On the other hand, for the I = 0, L = 0, S = 1 case, we have (~τ 1 ·~τ 2 ) = − 3 and (~σ 1 · ~σ 2 ) = +1. Let us take Sz = +1 state as an example. Then the tensor force does not vanish, and its expectation value is proportional to
〈S = 1, Sz = +1|
r^2
[3(~σ 1 · ~r)(~σ 2 · ~r) − (~σ 1 · ~σ 2 )r^2 ]|S = 1, Sz = +1〉
= 〈S = 1, Sz = +1|
r^2
[3(σ 1 z z)(σ 2 z z) − r^2 ]|S = 1, Sz = +1〉
2 z^2 − x^2 − y^2 r^2
