Braibant S., Giacomelli G., Spurio M. Particles and Fundamental Interactions: An Introduction to Particle Physics
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118 6 Invariance and Conservation Principles
Rotations in space. Proceeding as in the case of translations, let us define an
infinitesimal rotation around the z axis, i.e., '
0
D 'Cd'; the operator dR
z
associated
with these rotations is
dR
z
D 1 C d' @=@': (6.16)
Recalling that L
z
D .„=i/@=@' corresponds to the z component of the orbital
angular momentum operator, one has
dR
z
D 1 C .i=„/L
z
d': (6.17)
A finite rotation ' can be obtained as a series of infinitesimal rotations (d' D
'=n):
R
z
D lim
n!1
1 C
i
„
L
z
d'
n
D exp
i
„
L
z
'
: (6.18)
It follows that the invariance of the Hamiltonian under a rotation around the z axis
implies that ŒL
z
;H D 0 and therefore corresponds to the conservation of the z
component of the orbital angular momentum L
z
.
6.3 Spin-Statistics Connection
The spin-statistics connection is one of the most important in the submicroscopic
world.Particleswithhalf-integerspin(1/2,3/2,...),inunitsof„, follow the Fermi–
Dirac statistics and are called fermions. Particles with integer spin (0, 1, 2, ...)
follow the Bose–Einstein statistics and are called bosons. Statistics fix the symmetry
properties of the wave function for a pair of identical particles with respect to their
exchange.
Consider a pair of identical particles, denoted by (1,2), and consider the operator
I which reverses the positions of two particles:
I.1; 2/ ! .2; 1/: (6.19a)
For the wave function describing two identical particles, one has
I .1;2/ D .2; 1/: (6.19b)
Applying twice the operator I , the initial situation is obtained, that is,
I
2
.1; 2/ D I Π.2; 1/ D .1; 2/: (6.19c)
An eigenvalue equation exists for I
2
, with eigenvalue C1. The possible eigenvalues
for the I operator are ˙1
I .1;2/ D˙ .1; 2/: (6.19d)
6.4 Parity 119
Comparing (6.19b) and (6.19d), we obtain .2; 1/ D˙ .1; 2/. It is assumed that
1. The wave function for two identical bosons must be symmetric under the
exchange 1 $ 2: .1; 2/ D .2; 1/
2. The wave function for two identical fermions must be antisymmetric under the
exchange 1 $ 2: .1; 2/ D .2; 1/
The total wave function can be expressed as the product of a spatial function, ˛
(space), and a spin function, ˇ (spin). The spatial part describes the orbital motion
of a particle with respect to the other, and it is given (as already known from atomic
physics) by spherical harmonics Y
m
`
.; '/ functions (see Sect. 6.4). The exchange
of two particles corresponds to an inversion of coordinates which introduces the
factor .1/
`
.If` is even (odd), the wave function ˛ is symmetric (antisymmetric)
under the exchange operation.
From the Dirac theory, the function describing the spin status ˇ is symmetric
if the two spins are parallel, and antisymmetric if the two spins are antiparallel.
Identical bosons must therefore have both ˛ and ˇ symmetric or antisymmetric,
while identical fermions must have ˛ symmetric and ˇ antisymmetric, or vice versa.
6.4 Parity
The inversion of spatial coordinates .x; y; z !x; y; z/,orr !r,or
“exchange of the right with the left” is an example of a discrete transformation.
The inversion is generated by the parity operator P , which reverses the spatial
coordinates (Fig.6.1), that is,
P r Dr: (6.20)
Its application on a wave function gives
P .r/ D .r/: (6.21)
Fig. 6.1 Angles , ' for the
vector r and angles . /,
.' C / for r
ϕ
π−θ
θ
z
r
y
x
–r
120 6 Invariance and Conservation Principles
The parity operator applied two times gives
P
2
.r/ D PP .r/ D P .r/ D .r/: (6.22)
This implies P
2
D 1. The eigenvalue equation
P D p D˙ (6.23)
has eigenvalues p D˙1 (assuming that it admits eigenvalues). The parity of the
system is positive or negative. An example of a wave function with positive (even)
parity is the function .x/ D cos x
P cos x D cos.x/ D cos x; that is p DC1; positive: (6.24)
An example of a wave function with negative (odd) parity is .x/ D sin x
P sin x D sin.x/ Dsin x; p D1; negative: (6.25)
A wave function with undefined parity is .x/ D sin x C cos x
P.sin x C cos x/ D sin.x/ C cos.x/ Dsin x C cos x: (6.26)
.x/ D sin x Ccos x is not a P eigenfunction and the parity is not defined.
In a physics process, the parity of the system is a conserved quantity if the parity
operator commutes with the Hamiltonian, ŒH; P D 0. For example, any potential
with spherical symmetry has the properties H.r/ D H.r/ D H.r/. In this case,
one has ŒH; P D 0 and the bound states of the system have defined parity. Familiar
examples are the atomic bound states described by (neglecting the spin)
.r;;'/D .r/Y
m
`
.; '/: (6.27)
The angular functions Y
m
`
.; '/ are the spherical harmonics
Y
m
`
.; '/ D
s
.2` C 1/.` m/Š
4.` C m/Š
P
m
`
.cos /e
im'
(6.28)
where P
m
`
.cos / are the Legendre polynomials. The space inversion r !r is
equivalent to ! , ' ! C ' (see Fig.6.1). The application of the parity
operator to the functions e
im'
and P
m
`
gives
Pe
im'
D e
im.'C/
D e
im
e
im'
D .1/
m
e
im'
PP
m
`
.cos / D .1/
`Cm
P
m
`
.cos /
(6.29a)
(`; m are the orbital and azimuthal quantum numbers); finally, one has
PY
m
`
.; '/ D .1/
`
Y
m
`
.; '/: (6.29b)
6.4 Parity 121
p
s
ν
e
ν
e
Fig. 6.2 Momentum p and spin s for an electron neutrino and an antineutrino. The momentum is
indicated by the simple arrow at the top and the spin by the arrow at the bottom
The parity of the spherical harmonic functions Y
m
`
is then .1/
`
; states with
` D 0; 2; ::: have parity p DC1, while those with ` D 1; 3, ... have parity p D1.
The so-called electric dipole transitions between two states are characterized by the
selection rule ` D 1. In an electric dipole transition, the parity of the atomic
(nuclear) state changes, while the parity of the entire system, which consists of
the final state photon C atom (nucleus) is conserved. The parity of the emitted
electromagneticradiation (the photon) must be p D1 in order to conserve the total
parity of the system. The spin-parity assignment for the photon is J
P
D Spin
parity
D
1
: the photon is a vector particle. All observed electromagnetic processes conserve
parity.
The conservation of parity gives rise to a multiplicative law and parity is a
multiplicative quantum number. The parity of a system described by D
1
2
is
given by P D P
1
P
2
.
Because the parity operator corresponds to a mirror reflection, it was believed
until 1956 that Nature laws are symmetric under parity transformations. Mathemat-
ically, it corresponds to the fact that the parity operator gives rise to eigenvalues that
are conserved in each transition. It is an experimental fact that parity is conserved
in all transitions due to electromagnetic and strong interactions. The wave functions
describing the particles have definite parity for (6.24)or(6.25). Parity is instead
violated in weak interaction transitions (Chap.8).
Parity violation in weak interactions. Let us anticipate an example of parity
violation in the weak interaction. The electron neutrino has spin s D 1=2 and it may
therefore have two polarization states, s
z
D˙1=2. It is experimentally found by
only the neutrino with polarization antiparallel to the momentum (s
z
D1=2): the
neutrino is said to be left-handed). Likewise, the antineutrino has only s
z
DC1=2,
and it is right-handed (see Fig. 6.2).
Consider an electron neutrino and apply the parity operator. This operation
changes the neutrino momentum, p !p, but not its spin s. The application of
the parity operator to a left-handed neutrino gives rise to a right-handed neutrino,
i.e., a state which is not observed (schematically: P.
!
(H
e
/ D .
(H
e
/). The
neutrino is subject only to the weak interaction: this interaction is thus not invariant
under spatial inversion, meaning that it does not conserve parity.
122 6 Invariance and Conservation Principles
6.5 Spin-Parity of the Meson
As in the case of the photon, most particles have defined spin and parity which must
be experimentally determined. In this section, the important case of the spin and
parity determination of the charged pion is reviewed. Pions are abundantly produced
in strong interactions, and their spin and parity values are used in other parts of the
book. The reader can skip this section during the first reading.
6.5.1 Spin of the Meson
The spin of the
C
meson was determined by applying the “principle of detailed
balance” to the reaction pp !
C
d and to its reverse reaction,
C
d ! pp.The
deuteron d is a bound pn state, its ground state is
3
S
1
(the notation is
2sC1
l
j
). The
orbital angular momentum between p and n is zero .l
pn
D 0; Swave/; therefore,
the total angular momentum corresponds to the spin of the deuteron, s
d
D 1.This
means that the spin states of n and p are aligned and that the state corresponds to a
spin triplet .2s
d
C 1 D 3/.
The cross-section for the direct reaction is given by (4.40):
.pp !
C
d/DjM
if
j
2
matrix element
.2s
C1/.2s
d
C1/
v
i
v
f
p
2
phase space and flux factors
(6.30)
where the flux term contains the dependence on the relative speed of the initial state
v
i
D v
p
v
p
0
, and that of the final state v
f
D v
d
v
C
.
The cross-section is averaged over initial spins, and it is summed over all orbital
angular moments `. The element M
if
contains the dependence on the interaction
dynamics, and is unknown for the strong interaction. The density of final states
(D phase space) gives the terms .2s
C1/.2s
d
C1/p
2
=v
f
,wherep
D is the pion
momentum in the c.m.
The cross-section for the reverse reaction is given by
.
C
d ! pp/ D
1
2
ˇ
ˇ
M
fi
ˇ
ˇ
2
.2s
p
C 1/
2
v
0
f
v
0
i
p
2
p
: (6.31)
The factor 1/2 is due to the fact that in the final state, there are two identical
fermions. Assuming that the strong interaction is invariant under time reversal
(Sect. 6.7) and parity, it can be formally assumed that
jM
if
j
2
DjM
fi
j
2
: (6.32)
This relation is called detailed balance principle. If the cross-section for the direct
and the reverse reaction are measured at the same c.m. energy, one has v
i
v
f
D v
0
f
v
0
i
.
6.5 Spin-Parity of the Meson 123
Using the ratio between the cross-sections of the direct and reverse reactions, one
obtains
.pp !
C
d/
.
C
d ! pp/
D 2
.2s
C 1/.2s
d
C 1/
.2s
p
C 1/
2
p
2
p
2
p
D
3
2
.s
C 1/
p
2
p
2
p
: (6.33)
The measurements of the direct (6.30) and of the inverse (6.31) cross-sections have
been carried out since the early 1950s, with proton and pion beams of different
momenta (typically a few tens of MeV in the c.m. system). Once the ratio (
p
2
p
2
p
)
between the momentum of the particles in the final state was calculated and the
ratio between the cross-sections was measured, it was possible to determine the
factor
3
2
.s
C 1/; it was found to be compatible with the value s
D 0.Itwas
therefore concluded that the spin of the
C
meson was zero.
The spin of the
and
0
mesons are determined with a simpler and qualitative
reasoning. In e
C
e
, pp and pp collider at high energy (E
c:m:
>10GeV)
C
;
;
0
are abundantly produced in equal numbers. This suggests that the three mesons
must have the same spin and isospin (see Sect. 7.2), otherwise they could not be
produced with the same abundance. Therefore, the spin of the
and
0
mesons
must be zero as the spin of the
C
.
6.5.2 Parity of the Meson
The intrinsic parity of the
meson was determined by observing low-energy
absorption in deuterium. The process is due to the strong reaction (capture):
d ! nn: (6.34)
Initial
d state. The deuterium nucleus (deuteron) has spin s
d
D 1,the
has
spin 0. The orbital angular momentum between the
and deuteron is zero (`
d
D
0/ because the pion is captured at very low energies. The total angular momentum
of the initial
d state is then jJ
i
jDjS
C S
d
C L
d
jD1:
Final nn state. The angular momentum jJ
f
jDjL
nn
CS
nn
j must be equal to that of
the initial state. L
nn
is the orbital angular momentum between the two neutrons; S
nn
is their total spin, respectively denoted here as ` and S.
Let us make some considerations on the symmetry of the system before consid-
ering the consequences of the conservation of J. The wave function describing the
system of two neutrons can be written as the product of a space coordinates term
and a spin term, that is,
tot
D ˛.space/ ˇ.spin/: (6.35)
124 6 Invariance and Conservation Principles
As the two neutrons are identical fermions, their wave function
tot
must be
antisymmetric under their exchange. The product of the symmetries ˛(space) and
ˇ(spin) is considered separately and must be antisymmetric.
˛(space) is described by spherical harmonic functions, Y
m
`
.; '/. The exchange of
the two neutrons is equivalent to the transformation ! , ' ! C ',for
which one has
Y
m
`
.; '/
1$2
! .1/
`
Y
m
`
.; '/: (6.36)
The factor .1/
`
gives the symmetry of the space function ˛ under the exchange of
the two particles.
The functions ˇ(spin) for the combination of two spin 1/2 particles are written as
the first neutron is described by ˇ
1
with S
1
D 1=2; S
1z
D˙1=2
the second neutron is described by ˇ
2
with S
2
D 1=2; S
2z
D˙1=2:
If S
1
and S
2
are parallel, we have S
nn
D S D S
1
C S
2
D 1, S
z
D 0; ˙1: If S
1
and S
2
are antiparallel one has S D 0; S
z
D 0. In the first case, the three states
with S D 1 and S
z
D 0, ˙1 represent a spin triplet, which is symmetric under the
exchange of neutron 1 with neutron 2. The wave functions ˇ for the three states
of the triplet (the second number in parentheses refers to the third component) are
written as
(
ˇ.1; 1/ D ˇ
1
.1=2; C1=2/ˇ
2
.1=2; C1=2/
ˇ.1; 0/ D .1=
p
2/Œˇ
1
.1=2; C1=2/ˇ
2
.1=2; 1=2/ C ˇ
2
.1=2; C1=2/ˇ
1
.1=2; 1=2/
ˇ.1; 1/ D ˇ
1
.1=2; 1=2/ˇ
2
.1=2; 1=2/:
(6.37)
The state with S D 0, S
z
D 0 is an antisymmetric singlet:
ˇ.0; 0/ D .1=
p
2/Œˇ
1
.1=2; C1=2/ˇ
2
.1=2; 1=2/ˇ
2
.1=2; C1=2/ˇ
1
.1=2; 1=2/:
(6.38)
The symmetry of the spin wave function is therefore .1/
SC1
. It follows that
the symmetry of the final state wave function
tot
under the exchange of the two n
is .1/
`CSC1
; this quantity should be equal to 1 because the two n are identical
fermions:
tot
1$2
! .1/
`CSC1
tot
D
tot
. In conclusion, (` C S C 1)mustbe
odd, and (` C S ) must be even. The final state must be such that
1. ` C S D even for symmetry arguments;
2. jJjD1 for angular momentum conservation.
The possible combinations of ` and S for jJ jD1 are
8
<
:
` D 0
S D 1
` C S D odd
8
<
:
` D 1
S D 0
` C S D odd
8
<
:
` D 1
S D 1
` C S D even
8
<
:
` D 2
S D 1
` C S D odd;
6.5 Spin-Parity of the Meson 125
among these, only the combination ` D 1, S D 1 has ` C S D even and one can
conclude that the two neutrons are in a state
3
P
1
. (The notation is as follows: P !
Pwave,` D 1; index D 1 ! J D 1; superscript D 3 ! 2S C 1 D 3.)
The parity of the final state is P.2n/ D P.n/P.n/ .1/
`
D1: The parity of
n and p is conventionally taken to be equal to C1. Note that in all interactions, the
baryon number is conserved. Accordingly, the absolute value of the nucleon parity
is irrelevant because it cancels out in each reaction.
Let us assume that parity is conserved in the strong interaction. No experimental
indication is reported so far to sustain the contrary. The parity of the initial state is
equal to that of the final state, i.e., P.
d/ D P.nn/ D1. On the other hand, one
has P.
d/ D P./P.d/.1/
`D0
;sinceP.d/ D P.p/P.n/.1/
0
DC1, one
has P.
d/ D P.
/. The reaction (6.34) can therefore occur through the strong
interaction only if P.
/ D1. It can be concluded that the
has negative
parity. One can show that P.
C
/ D P.
0
/ D P.
/. Systems with n pions have
parity P.n/ D .1/
n
.
The mesons have spin-parity J
P
D 0
, and are called pseudoscalar mesons.
Bosons with J
P
D 0
C
are called scalar, those with J
P
D 1
vector mesons and
those with J
P
D 1
C
are axial mesons.
6.5.3 Particle–Antiparticle Parity
The intrinsic parity of the proton is defined by convention. The Dirac theory predicts
opposite parity for fermion–antifermion, and the same intrinsic parity for boson–
antiboson (for example,
C
–
and K
C
– K
).
It is possible to make a precise experimental assignment of the intrinsic parity of
the because the can be individually produced. Similarly, a relative parity can be
assigned to the p–
p pair because it can also be produced individually, for example,
in the reaction
pp ! pp.p
p/: (6.39)
The parity of the particle–antiparticle system was found to be equal to 1.The
intrinsic parity of strange mesons that are produced in association, for example, in
p ! K
C
K
n, cannot be individually measured. For instance, in the reaction
pp ! K
C
p; (6.40)
only the relative parity of the K
C
pair (which is found to be odd) with respect to
the proton can be determined. Conventionally, the same even parity of the proton is
assigned to the and the same odd parity of the is assigned to the K
C
.
126 6 Invariance and Conservation Principles
6.6 Charge Conjugation
The charge conjugation operator was originally defined through its action on an
electric charge q,thatis,
Cq Dq (6.41a)
C .q/ D .q/: (6.41b)
As with parity, the charge conjugation is a discrete operator. Applying the charge
conjugation twice, one has
C
2
q D CCq D C.q/ D q (6.42a)
C
2
.q/ D C .q/ D .q/: (6.42b)
Equation 6.42b is an eigenvalue equation with c
2
D 1; it follows that the only
possible eigenvalues are c D˙1.
Generalizing, the operator C transforms a particle, even if it is electrically
neutral, in the corresponding antiparticle. After the application of this operator
to a particle wave function, the baryon and lepton numbers are not conserved
and no real physical process corresponds to this transformation. It follows that
after a transformation due to the operator C , an arbitrary phase ' appears in the
wave function, that is, C .q/ D .
q/exp.i'/. The phase can be chosen at our
convenience.
Let us analyze the effects of the C operator on some characteristic parameters of
a particle (see Table 6.1). Applying C to a proton, one obtains an antiproton, which
has electric charge, baryon number and magnetic dipole moment opposed to those
of the proton; instead, the spin remains unchanged. Note that a positive magnetic
dipole moment means that it has the same direction and orientation of the spin. This
occurs for the proton and the positron; spin and magnetic dipole are opposite for the
Table 6.1 Effect of the charge conjugation operation C
applied to a proton and to an electron. The magnetic moment
is in units [e„=.2mc/], the spin s in units „. and s
are parallel and in the same direction for the proton and the
positron
Electric Baryonic Magnetic
charge number moment Spin
qB .e„=2m
p
c/ („)
Proton p Ce C1 2.793 1/2
Cp D
p e 1 2:793 1/2
Electric Electronic Magnetic
charge number moment Spin
qL
e
.e„=2m
e
c/ („)
Electron e
e C1 1:0012 1/2
Ce
D e
C
Ce 1 C1:0012 1/2
6.6 Charge Conjugation 127
antiproton and the electron. A neutron is different from the antineutron because the
magnetic dipole moment of the neutron is directed in the opposite direction to that
of the spin, while it has the same direction as the spin for the antineutron.
6.6.1 Charge Conjugation in Electromagnetic Processes
Applying the C operator to a
C
meson, a
is obtained:
C j
C
iDj
i:
Note that the Dirac notation ket introduced in Eq. 6.3 is used for states such as j
C
i.
It is obvious that for
C
and
charged mesons, an eigenvalue equation cannot be
written since the application of the operator C transforms the particle in a different
one. Instead, the application of C to
0
does not change the state and an eigenvalue
equation can be written as
C j
0
iDj
0
i (6.43)
with DC1 or 1. To assign the correct sign, let us consider the
0
electromag-
netic decay into two photons; since the electromagnetic interaction conserves C ,the
effect of C on the photon must be examined.
Photons are emitted by accelerated charged particles. A charge changes sign
when the operator C is applied. The application of C to a photon must keep track
of this effect, suggesting that
C j iDji: (6.44)
Since C is a multiplicative operator, for a system of n photons, one therefore has
C jn iD.1/
n
jni: (6.45)
For the considered electromagnetic decay of the
0
into two photons,
0
! 2,the
application of C to the
0
is equivalent to the application of C to the system of the
two photons. Assuming that the EM interaction conserves C, one can write
C j
0
iDC j2iD.1/
2
j2iDj2iDCj
0
i: (6.46)
As the decay
0
! 2 exists, and represents 99% of all
0
decays, we can conclude
that the
0
is an eigenstate of C with even C-parity. The decay
0
! 3 would
lead to a state of odd C-parity; this decay must therefore be prohibited if the EM
interaction is invariant under C . Experimentally, the decay
0
! 3 has never been
observed. The upper limit .
0
! 3/=.
0
! 2/ < 3:1 10
8
represents one of
the strong confirmations of charge conjugation conservation in the electromagnetic
interaction. For neutral mesons, for example, the
0
, the notation J
PC
D 0
C
is
often used.