Chapter 4 Symmetries And Conservation Laws
Chapter 4 4.1 Why Conservation And Symmetry 4.2 Angular Momentum 4.3 Flavour Symmetry: Isospin 4.4 Parity and Charge Conjugation 4.5 Interactive Exercise
4.4 Parity and charge Conjugation
Symmetry and Conservation Laws
Particle Physics
Discrete Symmetries: Parity Parity, P denotes inversion or parity operator. For a system which is a right hand, P turns it into an upsidedown and backward left hand. →
Applying parity to a vector, a , P produces a vector pointing in the opposite direction: →
→ P(a) = -a : called ordinary vectors or polar vectors →
→
→
axb=c
→ → Also for P(c) = -c :called 'pseudo' or 'axial' vectors.
Examples of pseudo vector is angular momentum, magnetic field. Rule: Do not add a vector to a pseudo vector, in a theory with parity invariance. →
→
i.e. F = q [E+ (v xB/c)]: → v is vector, B is pseudo vector, so →
→
→
→
→
we add v x B to E and not B to E. P(a.(b x c)) changes sign, as the product of vector and pseudo vector changes sign. Thus we define two kinds of scalars: ordinary scalar and pseudo scalar.
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SCALARS
ORDINARY SCALAR
Fig. 1 Scalars
PSEUDO SCALAR
Eigenvalues Scalar
P(s)= S
+1
Pseudo Scalar
P(p)= -p
-1
Vector (Polar vector)
P(v)= -v
-1
Pseudo vector (axial vector)
P(a)= a
+1
Table 1: Scalars and vectors under Parity
Application of parity operator twice gives P2=I and hence eigenvalues of P are ¹1. From quantum field theory: the parity of a fermion (½ integer spin) must be opposite to that of corresponding antiparticle; while the parity of a boson (integer spin) is the same as its antiparticle. Quarks have positive intrinsic parity, and anti-quarks have negative. Dayalbagh Educational Institute
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Parity of a composite system in its ground state is the product of the parities of its constituents. Thus the baryon octet, decuplet have positive parity (+1)3, whereas the pseudo-scalar, vector meson nonets have negative parity. Example Meson & baryon parity
P meson = (-1)L Pq1 Pq2 P baryon = (-1)L 12 (-1)L 3Pq1Pq2Pq3 where L is the orbital angular momentum (-1 comes from spherical harmonics) In the lowest mass mesons and baryons. L = 0 and
meson, antimeson -1,-1 baryon. antibaryon +1,-1 As with charge conjugation, parity must be conserved in strong and EM interactions As we will see, this is not true for weak interactions For many years It was believed that parity is also conserved In the weak interactions. Historical note: The first theory of weak interactions was formulated in 1934 by Enrico Fermi specifically for nuclear beta decays. This theory was modeled after the theory of electromagnetic interactions. It was a parity conserving theory. Dayalbagh Educational Institute
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For a long time the only known weak interaction processes were beta decays of atomic nuclei. There were no experimental tests of conservation in weak interactions until 1956
parity
http://www.din.uniroma1.itimagefermi.gif
Fig. 2 E. Fermi. Nobel Prize in Physics 1938.
Soon after the discovery of the first few elementary particles it was understood that many of their decays were also weak interaction processes. In particular, pion and muon decays are weak processes: π+ → μ+υμ ;
μ+ → e+υe υμ
In order to maintain parity conservation it was necessary to ascribe to each particle its parity as an intrinsic property like mass and charge. Dayalbagh Educational Institute
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The parity of a particle can be either positive (+1) or negative (-1). But intrinsic parity is a relative property: one assigns a positive parity to one particle, for instance to the proton. and then one finds the parities of other particles by analyzing processes between these and protons (this is no different from charge!). Thus it was found that for instance the pion has negative parity relative to the proton, which is conventionally given a positive parity. The relation between the parities of particles and antiparticles is different for spin 1/2 particles (fermions) and spin 0 or spin 1 particles (bosons): An antifermion has the opposite parity of the fermion An antiboson has the same parity as the boson. Therefore the antiproton has negative parity since we arbitrarily give the proton a positive parity. (Proton and antiproton are spin ½ fermions.) Since it was found that the π- had negative parity, therefore the π+ also had to have negative parity. (Pions are spin 0 bosons.) Parity is multiplicative: the parity of a system of two particles is the product of their parities and of the parity of their relative motion. Then it follows rigorously that a two-pion system has positive parity and a three-pion system has negative parity, provided their relative angular momentum is zero. Dayalbagh Educational Institute
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Then, if we discover a particle that decays into a pair of pions, then we know that it has positive parity, and if it decays into three pions, then it has negative parity. By 1956 there arose a difficulty, the ‘tau-theta paradox”: there were two particles that looked much the same but one of them decayed by weak interaction into two pions and the other into three pions: τ → π+ π+ π - ;
θ→π+ π 0
(this tau not to be confused with today’s tau lepton!). At first the experimental data were not very accurate, so the tau and theta could be different. But with improved techniques they looked more and more similar, so it became hard to maintain that they were different particles. In particular, from their production in strong interaction processes one had to conclude that they both had negative parity. T.D. Lee and C.N. Yang proposed that they were the same particle, and therefore parity had to be violated in weak interactions. Today this particle is called K +. Parity violation in weak interactions was soon demonstrated in several experiments: in nuclear beta decay and in muon decay. Dayalbagh Educational Institute
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Today parity violation is understood to be a general property of the weak interactions. Fig.3 For their discovery of parity violation. T.D. Lee (left) and C.N. Yang (right) received the Nobel Prize for Physics in 1957
Fig 4
http://lappweb06.in2p3.fr/neutrinos/neutimg/ nacteurs/leeyang.jpg
Parity conservation was considered obvious In 1956, Lee & Yang (Nobel 1957!) questioned this for the weak interaction Wu looked at β-decay of 60 Co
http://physics.nist.gov/GenInt/Parity/parity.ht ml
Parity conservation would resuit in isotropic emission Supercooled Co showed nonisotropic β emission The opposite was seen with β+ emission (58Co)
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Actually, it is not quite correct to say that Lee and Yang did discover the violation of parity. What they did was to show that there had been no experiments to test the hypothesis of parity conservation in the weak interactions, and they suggested ways of setting up experiments to carry out such tests. Parity violation in processes involving neutrinos can be understood in the following way. The neutrino is a zero-mass spin 1/2 particle. Therefore its spin must be pointing on general grounds either along its direction of motion or in the opposite direction, but never at an angle to its direction of motion. In the former case it is said to be right-handed, and in the latter case it is left-handed. There are therefore three possibilities: i. All neutrinos are right-handed. ii. All neutrinos are left-handed: iii.Neutrinos can be left-handed and right-handed. There is no theory that can tell which of these is realised in nature. The empirical evidence is that all neutrinos are Lefthanded and all antineutrinos are right-handed.
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Now if we look at the mirror image of a neutrino, then that is still a neutrino. But a left-handed screw (or a left-handed helix) seen in a mirror is a right- handed screw, and a left-handed neutrino seen in a mirror is a right-handed neutrino. But that does not exist. And that is why parity is violated. The argument is not so simple when we consider the weak interactions of particles of nonzero mass. Here we take the evidence of parity violation to construct the theory. But first some jargon. The projection of spin on the momentum vector is called helicity. The helicity of a massive particle can be positive (righthand screw!) or negative (left-hand screw). The weak interactions are carried by the intermediate vector bosons. In order to account for the empirically established parity violation one is forced to accept the following result: The W boson couples only to the left-handed components of fermions; The Z boson couples to the left- and tight-handed components but with different strength.
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Parity Parity transformation reflection:
is
the
transformation
by (1)
A system is invariant under parity transformation if Parity is not an exact symmetry: it is violated in weak interaction! A parity operator
is defined as (2)
Since two consecutive reflections must result in the identical to initial system, (3) From equations (2) and (3), Pa =+1, -1 Considering then an eigenfunction of momentum: it is straightforward that The latter is always true for , i.e., a particle at rest is an eigenstate of the parity operator with eigenvalue Pa. Different particles have different values of parity Pa. For a system of particles,
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In polar coordinates, the parity transformation is: and a wave function can be written as (4) In Equation (4), Rnl is a function of the radius only, and
Ylm are spherical harmonics, which describe angular dependence. Under the parity transformation, Rnl does not change, while spherical harmonics change as
which means that a particle with a definite orbital angular momentum is also an eigenstate of parity with an eigenvalue Pa(-1)l . Considering only electromagnetic and strong interactions, and using the usual argumentation, one can prove that parity is conserved:
ˆ =0 [P,H] Recall: the Dirac equation (4) suggests a four-component wavefunction to describe both electrons and positrons Intrinsic parities of e- and e+ are related, namely: Pe + Pe- = –1
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This is true for all fermions (spin-1/2 particles), i.e., Pf Pf = –1
(5)
Experimentally this can be confirmed by studying the reaction e+e-→ γγ where initial state has zero orbital momentum and parity of Pe- Pe+ . If the final state has relative orbital angular momentum lγ, its parity is Pγ2 (–1)lγ. Since Pγ2=1 , from the parity conservation law stems that Experimental measurements of lγ confirm (5) While (5) can be proved in experiments, it is impossible to determine Pe- or Pe+ , since these particles are created or destroyed only in pairs. Conventionally defined parities of leptons are: (6) And consequently, parities of anti-leptons have opposite sign. Since quarks and anti-quarks are also produced only in pairs, their parities are defined also by convention: (7) with parities of anti-quarks being -1. For a meson M=(ab), parity is then calculated as (8)
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For the low-lying mesons (L=0) that means parity of -1, which is confirmed by observations For a baryon B=(abc), parity is given as (9) and for antibaryon PB = -PB similarly to the case of leptons. For the low-lying baryons (9) predicts positive parities, which is also confirmed by experiment. Parity of the photon can be deduced from the classical field theory, considering Poisson’s equation:
Under a parity transformation, charge density changes as and ∇ changes its sign, so that to keep the equation invariant, the electric field must transform as (10) On the other hand, the electromagnetic field is described by the vector and scalar potentials: (11) For the photon, only the vector part corresponds to the wavefunction: Under the parity transformation,
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and from (11) stems that (12) Comparing (12) and (10), one concludes that parity of photon is Pγ = –1
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Charge Conjugation An operator which changes the sign of the charge: called charge conjugation, C, and it coverts each particle into its anti particle. C operator can also be applied to a neutral particle, such as neutron, yielding anti-neutron, which changes the sign of all internal quantum numbers, while leaving mass, energy, momentum, spin untouched. Also C2=I, hence eigenvalues of C are +1. Charge conjugation involves replacing all charges with negative charges & vice versa.
positive
e- → e+ The operator also inverse other quantum numbers, including lepton, strangeness & baryon number, but leaves mass, energy, momentum & spin unchanged Charge conjugation also inverts electric and magnetic fields EM interactions are invariant under charge conjugation Motion of a charge particle in an EM field Spectrum from anti-hydrogen Charge conjugation involves replacing all charges with negative charges & vice versa. e- → e+ Dayalbagh Educational Institute
positive
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The operator also inverse other quantum numbers, including lepton, strangeness & baryon number, but leaves mass, energy, momentum & spin unchanged Charge conjugation also inverts electric and magnetic fields EM interactions are invariant under charge conjugation Motion of a charge particle in an EM field Spectrum from anti-hydrogen Charge conjugation flips EM fields, and so we expect the EM field of the photon to flip Ĉψγ= -1ψγ We observe the decay πo → γ + γ and in terms of charge conjugation εc (πo)= εc (γ)εc (γ) = (-1)2 = +1 But the EM interaction πo → γ + γ + γ is forbidden as εc (3γ)= -1 and charge conjugation would be violated For |p⟩ to be an eigen state of C, we have C |p⟩ = ±|p⟩ = |p⟩ The equation implies that |p⟩ and |p⟩ differ at most by a sign, which implies that they represent the same physical state.
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Examples are πο, γ, η, η1, ρο, Φ, ω, Ψ. The proton γ is the quantum of EM field, which changes sign under C, thus C |γ⟩ = (-1) |γ⟩ Hence only those particles which are their own antiparticles can be eigen states of C. A system consisting of a spin ½ particle and its antiparticle, is a configuration with orbital angular momentum ℓ, total spins,constitute an eigen state of C with eigenvalue (-1)ℓ+s. e.g. mesons: for pseudo scalars, ℓ=0=s⇒ C= +1. for vectors, ℓ=0, s=1 ⇒ C= -1. Charge conjugation is a multiplicative quantum number and like Parity, it is conserved in strong and electromagnetic interactions. e.g. πο → γ + γ But πο → γ + γ + γ is not allowed. C= (-1)n for n photon decay, so C= +1 before and after in earlier equation, but latter has C= +1, C= -1. Apply to a neutrino (left-handed), C gives a left handed anti-neutrino which does not occur. Charge conjugation is not a symmetry of the weak interactions. Dayalbagh Educational Institute
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Fig. 5: P, C and CP on a neutrino
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Symmetry: Charge Conjugation Charge conjugation involves replacing all charges with negative charges & vice versa.
positive
e- → e + The operator also inverse other quantum numbers, including lepton, strangeness & baryon number, but leaves mass, energy, momentum & spin unchanged Charge conjugation also inverts electric and magnetic fields EM interactions are invariant under charge conjugation Motion of a charge particle in an EM field Spectrum from anti-hydrogen Charge conjugation flips EM fields, and so we expect the EM field of the photon to flip Ĉψγ = -1ψγ We observe the decay π0 → γ + γ and in terms of charge conjugation εC (π0) = εC(γ)εC(γ) =(-1)2 = +1 But the EM interaction π0 → γ + γ + γ is forbidden as εC(3γ) = -1 and charge conjugation would be violated
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Charge Conjugation Charge conjugation replaces particles by their antiparticles, reversing charges and magnetic moments Charge conjugation is violated by the weak interaction For the strong and electromagnetic interactions, charge conjugation is a symmetry: It is convenient now to denote a state in a compact notation, using Dirac's “ket” representation: denotes a pion having momentum , or, in general case, (13) Next, we denote particles which have antiparticles by “a”, and otherwise - by “α”
distinct
In these notation, we describe the action of the charge conjugation operator as: (14) meaning that the final state acquires a phase factor Cα, and otherwise (15) meaning that the from the particle in the initial state we came to the antiparticle in the final state. Since the second transformation turns antiparticles back to particles, and hence (16)
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For multiparticle states the transformation is: (17)
From (14) it is clear that particles α = γ,π0,... etc., are eigenstates of with eigenvalues Cα= ± 1. Other eigenstates can be constructed from particleantiparticle pairs: For a state of definite orbital angular momentum, interchanging between particle and antiparticle reverses their relative position vector, for example: (18) For fermion-antifermion pairs theory predicts (19) This implies that π0, being a 1S0 state of uū and dd, must have C-parity of 1.
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Tests Of C-Invariance Prediction of can be confirmed experimentally by studying the decay π0→γγ. The final state has C=1, and from the relations
it stems that
.
Cγ can be inferred from the classical field theory: under the charge conjugation, and since all electric charges swap, electric field and scalar potential also change sign: which upon substitution into (2) gives Cγ = –1 . To check predictions of the C-invariance and of the value of Cγ, one can try to look for the decay π0 → γ + γ + γ If both predictions are true, this mode should be forbidden: which contradicts all previous observations. Experimentally, this 3γ mode have never been observed. Another confirmation of C-invariance comes from observation of η-meson decays: η→γ+γ
;
η → π 0 + π0 + π 0
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η → π + + π - + π0 25
Symmetry and Conservation Laws
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They are electromagnetic decays, and first two clearly indicate that CΡ=1. Identical charged pions momenta distribution in third confirm C-invariance.
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Charge Conjugation; combined inversion CP
Charge conjugation is the name of the operation that takes a particle to its antiparticle without affecting space and time. If we denote this operation by C, then we have for example νL
C
νL
where the subscript L reminds us of the handedness of the neutrino and antineutrino. But left-handed antineutrinos do not exist. Therefore the charge conjugation symmetry is also violated. However, if we combine the two operations C and P (parity), then we have the following: νL
C
νL
P
νR
This Is shown In the following figure: The simple arrows represent the momenta: the broad arrows represent helicity; the parity transformation P takes the neutrino to an (non existing) right-handed neutrino: charge conjugation C takes the neutrino to an (non existing) left-handed antineutrino: the combined inversion CP takes the neutrino to the right-handed antineutrino. Dayalbagh Educational Institute
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νL νL νL
P
Particle Physics
νR νL
C CP
νR
Fig 9
CP Conservation Parity violation came as a shock to physicists Wu’s experiment also violates charge conjugation Applying both charge conjugation and a parity transformation results in a physically acceptable process, observable in the universe While the weak force individually does not conserve C and P. it was thought that the combined property (CP) was conserved This was essentially the argument of Landau (1957) by which he showed that the combined inversion CP is a good conservation law. But the argument is not convincing if the weak process does not involve neutrinos. Therefore experimentalists were quick to start checking CP conservation. The experiment was difficult and the first attempts did not have the necessary sensitivity. Dayalbagh Educational Institute
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The experiment was difficult and the first attempts did not have the necessary sensitivity. But in 1964 a team of four physicists succeeded in showing that the CP symmetry was broken by a small amount. The four physicists were J.H. Christenson, J.W. Cronin, V.L. Fitch and R. Turlay. For this discovery Cronin and Fitch received the Nobel Prize of Physics in 1980.
CP violation means that there is in nature an asymmetry between matter and antimatter. In fact, we know that there is this asymmetry because everything we observe in the universe is matter and for all we know there is no antimatter to any significant amount. On the other hand, It is a fair assumption that at the Big Bang as much antimatter was created as matter. Then, under complete symmetry. matter should have annihilated with antimatter, leaving a lot of photons but no quarks and leptons (except neutrinos). So the existence of the universe as we know it is evidence of CP violation. This argument was first put forward by A.D. Sakharov.
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A.D. Sakharov
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K° Meson K° and K° are strange particles produced in strong interactions π- + p → Λ + K° π+ + p → K++ K° + p they are a particle/antiparticle pair with a mass of 0.498 GeV and quark make up K° (ds)
K° (ds)
These particles can mix via the weak force K° ↔ K°
Fig 6
Mixing is allowed because Charge (Q = 0) is conserved Baryon number (B = 0) is conserved Weak interaction allows change of strangeness (S = ±1)
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If we make a pure beam of K° particles, it soon becomes a mix of K° and K°
Fig 7
The particles in the beam can decay into pions K → 2π
K → 3π
τS = 9 x 10 -11 τL = 5 x 10 -8s Initially there are both 2π and 3π decays But the 2π decays end rapidly, leaving only the 3π decays Does this reflect different behaviour of the K° and K°? i.e K° → 2π K° → 3π
Fig 8
K° → π+ + π-
K° → π+ + π-
Both K° and K° can decay into 2π or 3π states Dayalbagh Educational Institute
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So how can be explain the difference in lifetimes? (why doesn’t everything quickly decay into 2π?) The decay mode (2π or 3π) depends on the relative amounts of K° and K° This suggests the beam of particles consists of a quantum superposition of K° and K° We can define two superpositions KS = 1/√2 (K° + K°) KL = 1/√2 (K° - K°) If CP is not violated, what are the CP eigen values of KS and KL? Applying both charge conjugation and parity operations CP K° = K° CP K° = K° so CP KS (CP = +1) CP KL (CP = -1) What CP values do the 2π and 3π decays have?
Fig 9
The 2π state has CP = +1 While the argument is a little more complex, the 3π state has CP = -1 Dayalbagh Educational Institute
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So we can identify KS → 2π (CP = +1 ) KL → 3π (CP = -1 ) KL and KS act like they are particles What does this mean for your understanding of particles? KS Regeneration: After many KS lifetimes, the beam will be purely KL The beam can then interact with matter, with differing reactions for K° and K° (due to S = ±1) This can regenerate the KS → 2π component of the beam
CP Violation In 1964, Christenson et al studied KL decays One in a thousand KL decayed into 2π KL(CP = -1 ) → 2π (CP = +1 ) which should be forbidden if CP is perfectly conserved This implies that the weak interaction imprints a fundamental asymmetry on nature As we will see later, this appears to be the reason there is any matter in the universe!
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Talking to Aliens While KL decays into 3π, the dominant decay (66%) is into KL → π ± + I + + υI (υI) where I is a lepton (e, μ, but not τ – why?) While KL is electrically neutral etc, it is found that rate (KL → e+ + υe + π-) rate (KL → e- + υe + π+)
= 1.00648 ± 0.00035
So, all we need to do is ask the aliens to set up the K°/K° experiment and tell them that the dominant charged lepton emitted is what we call positive How do we agree on what is right and what is left?
Violations: Summary Individually parity and charge conjugation are strongly violated by the weak force Combined, however, the weak force mildly violated CP conservation It is thought that C, P & T are absolutely conserved when applied together (CPT Symmetry) CP violation therefore reveals that the weak interaction must have some time asymmetry!
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Symmetry: Time Reversal The operation of time reversal reverses momentum and angular momentum, while leaving the positions of particles unchanged. Basically this means we can run any interaction backwards Applying time reversal means that A+B→C+D C+D→A+B should have the same rate. This is true for the strong and EM interaction Again, this appears not to be true for the weak interaction!
Discrete symmetries: The CPT-theorem Charge conjugation, parity, time reversal Despite of CP violation, up to now, no violation of the combined version of all three discrete symmetries has been found. So CPT seems to be a true symmetry of the world. Ultimately, this allows for causal structures of the theory as realized up to now.
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Problem 1 The deuteron is a bound state of neutron and proton and has spin 1 and positive parity. Prove that it can exist only in the 3S1 and 3D1 states.
Solution 1 The deuteron which is the nucleus of deuterium (heavy hydrogen) consists of one proton and one neutron. Since the parity of neutron and proton are +1, that of deuteron is also +1. Spin of deuteron is 1, and l = 0 mostly, with 4% admixture of l = 2 so that the parity determined by (−1)l = +1 for l = 0 or 2. The deuteron is in a state of total angular momentum
J = 1. Thus J p = 1+ Using the spectroscopic notation
2s+1
LJ, deuteron
state is described by 3S1 and 3D1.
Problem 2 Conventionally nucleon is given positive parity. What does one say about deuteron’s parity and the intrinsic parities of u and d-quarks?
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Solution 2 Parity of deuteron πd = πp πn.(−1)l As πp = πn = +1 for s or d state l = 0 or 2.
Πd = +1 The intrinsic parity of quarks is assumed to be positive because the intrinsic parity of a nucleon (+) comes from the parities of three quarks and l = 0.
Problem 3 Explain how the parity of K meson has been determined. −
Solution 3 Since strange particles are always produced in pairs, as in the reaction
π− + p → Λ+K0 the intrinsic parity of a strange particle can only be determined relative to that of another. Thus, for example, one can determine the kaon parity relative to that of Λ, which by convention is assigned a positive parity. Consider the reaction
K − + He4 → ΛHe4 + π − → ΛH4 + π Dayalbagh Educational Institute
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These reactions are known to occur in a helium bubble chamber. Now, the Λ is bound in an s-state relative to the nuclear core He3 or H3 which have positive parity. Furthermore, all the participants in the reaction are spinless.
Linitial = Lfinal. The orbital angular momentum does not contribute to the parity because of s-state. The only relevant parities in the above reaction are
PK = PΛ. Pπ − = −PΛ, as Pπ − = −1 The validity of the argument obviously hinges on the hyper-nuclei having zero spin. If the spin were 1, for example, Angular momentum conservation would require l = 1 in the final state, thus reversing the conclusion. The spin of ΛH4 has also been experimentally determined to be indeed zero. It is concluded that the relative parity of K− is negative.
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