Download Quantum numbers II - Sezione di Fisica

Quantum numbers II
M. Cobal, PIF 2006/7
Where does isospin comes from?
In quark model: proton = uud, neutron = udd
One is obtained by the other by changing a u for a d
®The closeness in mass of p and n reflects the fact that Mu and Md
must be very similar
p and n have finite size: rms of charge distribution of order≅1fm.
The Coulomb energy term (needed to put charge on p) is then ≅
Since Mn-Mp = 1.3 MeV → Md >Mu by 2-3 MeV
The costituent mass of each quark is of order 300 MeV (one third
of the nucleon mass), Mu and Md are equal within 1%
M. Cobal, PIF 2006/7
This near equality in the mass of the lightest quarks is the
Lo spin isotopico
Dà corpo all’idea che adroni aventi masse prossime, e gli stessi
spin e parità intrinseca ma cariche elettriche diverse, sono la stessa
cosa per quanto riguarda l’interazione forte (indipendenza dalla
carica).
Questo fu formalizzato assegnando a ciascuno dei gruppi di adroni
di cui sopra uno stesso numero quantico I, analogo a quello di spin
(isospin o spin isotopico).
Come per lo spin, la terza componente dell’isospin si supponeva
quantizzata, e i suoi diversi valori distinguevano fra i membri di
uno stesso gruppo (multipletto).
M. Cobal, PIF 2006/7
Sarebbero la stessa cosa, per quanto riguarda le interazioni forti,
per cominciare, il protone e il neutrone. Essi divengono quindi
stati di una stessa particella. Lo stato di un nucleone può allora
essere scritto come una matrice colonna a due elementi, che
danno rispettivamente l’ampiezza di probabilità che si tratti
di un protone o di un neutrone.
Gli stati di protone e neutrone si scriveranno allora come:
e
Essi sono autostati dell’operatore matriciale
corrispondenti agli autovalori 1 e -1
M. Cobal, PIF 2006/7
Introdotti anche gli operatori matriciali
e
si verifica che le tre matrici soddisfano alle regole di commutazione:
Inoltre
e
trasformano un neutrone in protone e viceversa.
M. Cobal, PIF 2006/7
Una generica trasformazione nello spazio degli stati si scrive:
La trasformazione deve trasformare una base ortonormale in una
base ortonormale, la matrice U deve essere unitaria.
Le matrici unitarie hanno determinante di modulo 1.
Se il determinante vale 1, si chiamano unimodulari.
Le matrici in questione formano gruppo: il gruppo SU(2), gruppo
delle trasformazioni unitarie unimodulari.
Le
M. Cobal, PIF 2006/7
individuano l’algebra di Lie del gruppo.
L’algebra dei generatori può essere generalizzata in termini di
operatori astratti:
Le matrici 2x2 date sopra che soddisfano a quest’algebra
agiscono sulla rappresentazione bi-dimensionale (fondamentale)
del gruppo. Si possono trovare via via matrici NxN che soddisfano
all’algebra. I multipletti che sono trasformati da esse costituiscono
rappresentazioni N-dimensionali di SU(2). La rappresentazione
tridimensionale è nota come la regolare. Essa descrive particelle
di isospin 1, dunque un tripletto (es.: i pioni, nei loro tre stati di
carica).
M. Cobal, PIF 2006/7
Ora, l’indipendenza dalla carica si può tradurre nell’invarianza
dell’interazione forte per rotazioni nello spazio dell’isospin, cui
corrisponde la conservazione dello spin isotopico.
L’isospin si supponeva rigorosamente conservato nelle interazioni
forti. Le piccole differenze di massa all’interno di un multipletto si
attribuivano alle interazioni elettromagnetiche che distinguevano fra
di esse (esempio: protone e neutrone, doppietto di isospin ½.)
M. Cobal, PIF 2006/7
Isospin I
M. Cobal, PIF 2006/7
M. Cobal, PIF 2006/7
Nuovi numeri quantici: la stranezza
Negli anni 50, i kaoni e le Λ erano noti come particelle
strane. La stranezza del loro comportamento stava nel fatto
che erano prodotte con sezioni d’urto tipiche delle interazioni
forti, ma decadevano con vite medie proprie delle deboli.
Nel 1952 Abraham Pais avanzò la sua ipotesi della “produzione
associata”: un kaone e una Λ interagivano “forte” soli in coppie,
e come tali potevano essere prodotti da interazioni forti; ma,
lasciate a se stesse, potevano decadere solo via interazione debole.
M. Cobal, PIF 2006/7
Particles with I=0 are isosinglets :
Λ(116)=uds, I(J)P = 0(1/2)+
-By introducing isospin, new criteria for non-exotic particles:
In all observed interactions these criteria are satisfied, confirming
once
M. Cobal, again
PIF 2006/7the quark model
I due nucleoni non si distinguerebbero nel caso di una simmetria
esatta. L’esatta simmetria è rotta dall’interazione elettromagnetica.
Anche i barioni ordinari, nucleoni,e iperoni, per un totale di 8, si
assomigliano abbastanza. Si può pensare a una simmetria rotta, che
di per sé li porrebbe in un unico multipletto.
D’altra parte, le interazioni forti possiedono altre leggi di
conservazione oltre a quella dello spin isotopico: quella del
numero barionico e quella della stranezza.
La simmetria, una volta rotta, dovrebbe lasciare inviolate la
conservazione dell’isospin, del numero barionico e della stranezza.
M. Cobal, PIF 2006/7
SU(3). Ritrovare un ordine.
Furono Murray Gell-Mann e Yuval Ne’eman (1961) a rendersi
conto, indipendentemente, che il gruppo di simmetria che garantiva
tutto ciò era la più immediata generalizzazione di SU(2), cioè il
gruppo SU(3) delle matrici unitarie unimodulari 3x3.
M.Gell-Mann, Y. Ne’eman, The Eightfold Way, Benjamin, 1964
M. Cobal, PIF 2006/7
Sembra presentarsi subito una difficoltà: la rappresentazione
fondamentale è tridimensionale, può quindi albergare tre
particelle e non otto. Bisognerà dunque saltarla, questa
rapresentazione fondamentale, e partire dalla regolare, che
otto-dimensionale. Va osservato che il contenuto in multipletti
di isospin e l’attribuzione ad essi di un valore definito della
stranezza è prescritto.
La figura mostra
l’ottetto barionico.
M. Cobal, PIF 2006/7
Ma non è proibita l’esistenza di altri ottetti – non barionici:
l’ottetto dei mesoni pseudoscalari
M. Cobal, PIF 2006/7
e quello dei mesoni vettoriali
N.B.: c’è qualche
problema di
sovrabbondanza,
per il quale rimando
a Gell-Mann e
Ne’eman, op. cit.
M. Cobal, PIF 2006/7
Si passa poi alle rappresentazioni di maggiore dimensionalità. La
prima è il decupletto. Vi trovano posto stati eccitati degli iperoni,
nonché la Δ, ex “risonanza 33”. Ma il vertice, la Ω, è mancante …
M. Cobal, PIF 2006/7
Deve trattarsi di un singoletto isotopico con spin 3/2, parità +,
massa di circa of 1.680 MeV, carica negativa, numero barionico
+1, stranezza = -3, stabile rispetto a decadimenti forti.
Un barione con queste caratteristiche
fu scoperto nel 1964 da un gruppo di
fisici di Brookhaven e delle Università
di Rochester e Syracuse, condotto da
N. Samios, usando la camera a bolle
da 80 pollici.
M. Cobal, PIF 2006/7
Strangeness S
M. Cobal, PIF 2006/7
M. Cobal, PIF 2006/7
M. Cobal, PIF 2006/7
-Suppose we observe production of the Σ+ baryon in a strong
interaction:
Which then decays weakly:
It follows that Σ+ baryon quantum numbers are: B=1, Q=1, S=-1, and
hence Y=0 and I3=1
Since I3> 0 ⇒there are more multiplet members!
If a baryon has I3=1, the only possibility for isospin is I=1 and we
have a triplet:
Σ+, Σ0, Σ-
M. Cobal, PIF 2006/7
All
these particles have been observed:
Λ+γ
π- + n
Masses and quark composition of Σ-baryons:
Σ+ (1189)=uus, Σ0(1193)=uds, Σ-(1197) = dds
It clearly indicates that d-quark is heavier than u-quark
Combining different methods, it has been found:
2≤ md-mu ≤ 4 (MeV/c2)
which is negligible comparing to hadron masses
M. Cobal, PIF 2006/7
Isospin in two-nucleon
Isospin of two-nucleons, each with I = ½ :
(in analogy with combination of two spin ½)
triplet with I=1
symmetric under label
inter-charge
Other notation:
M. Cobal, PIF 2006/7
singlet with I= 0
antysimmetric for label
inter-charge
• Example: In each case: final state has I = 1
p+p → d + π+
I= 1
0 1
p+n → d + π0
I= 0 o 1 0 1
This is a pure I= 1
In this case: 50%
I=0
reaction
and 50% I=1
Conservation of isospin → reactions can proceed ONLY through the
I=1 channel.
• Important application of I conservation: in strong interactions of
non-identical particles (mixture of different isospin states)
e.g.: pion-nucleon scattering Iπ = 1, IN = ½ → IToT = ½ or
M. Cobal, PIF 2006/7
If strong interactions depend only on I and not on I3, the 3x2 π-n
scattering processes can be described in terms of two isospin
amplitudes
These 2 decays have I3 = ± 3/2:
described by I= 3/2 amplitude
These 4 decays have I3 = ± 1/2:
described by I= 3/2 or ½
amplitudes
The weights of the two amplitudes in the mixture are given by the
Clebsh-Gordan coefficients (see Appendix C1 of Perkins)
M. Cobal, PIF 2006/7
M. Cobal, PIF 2006/7
• For pion-nucleon scattering:
I = 3/2
Pion
Nucleon
I3 = 3/2
π+
p
1
π+
n
π0
p
π0
n
π-
p
π-
n
1/2
• Calculate cross-section for:
M. Cobal, PIF 2006/7
-1/2
I = 1/2
-3/2
1/2
-1/2
1
Where H is an isospin operator: H1 if it operates on initial/final
states with I = ½, H3 for states with I = 3/2
By conservation of isospin: no operator connecting I = ½ with I = 3/2.
• Reaction ( π+ p → π+ p) is a pure I = 3/2, I3 = +3/2→
• For reaction ( π- p → π. p) we may write:
M. Cobal, PIF 2006/7
• For reaction ( π- p → π0 n) we may write:
• Then:
• Limiting situations are:
Experimentally: (σπ+p/ σπ−p)tot = 3, proving that the I = 3/2
amplitude dominates this region
M. Cobal, PIF 2006/7
Other symmetries
Finally, there are 2 discrete symmetries associated with
reversing the direction of some quantity. These are:
1. Charge conjugation – particles into anti-particles.
2. Time reversal – changing the direction of time.
Interesting because it is not obvious whether the laws of nature
should look the same for any of these changes, and the answer
was surprising when these symmetries were first tested.
Example of a neutron and its decay to illustrate each of the two
symmetries. Neutrons have spin angular momentum of ½ and
decay as:
M. Cobal, PIF 2006/7
n→p e ν
Charge conjugation
M. Cobal, PIF 2006/7
M. Cobal, PIF 2006/7
Let’s denote particle which have distinct anti-particles by “a”, and
otherwise by “α”
Then:
That is, the final state acquires a phase factor
Otherwise:
That is, from the particle in the initial state,
antiparticle in the final state
we arrive to the
The 2nd transformation turns antiparticles back to particles,
and hence:
For multiparticle states the transformation is:
It is clear that particles α = γ, π0,..etc, are eigenstates of C with
M. Cobal, PIF 2006/7
eigenvalues
-Other eigenstates can be constructed from particle-antiparticle
pairs:
For a state of definite L, interchanging between particle and antiparticle reverses their relative position vector. For example:
For fermion-antifermion pairs theory predicts:
This implies that π0, being a 1S0 state of
C-parity of 1.
M. Cobal, PIF 2006/7
and
must have
Tests of C-invariance
Cγ can be inferred from classical field theory:
and since all electric charges swap, electric field and scalar potential
also change sign:
Which upon substitution into:
gives Cγ = -1
• Another confirmation of C-invariance comes from observation of
η-meson decays:
These are em decays, and first two clearly indicate that C =1.
Identical charged pions momenta distribution in third, confirms
M. Cobal, PIF 2006/7
C-invariance.
Charge conjugation (C) simply means to change each particle into its antiparticle. This changes the sign of each of the charge-like numbers. The neutron
is neutral, nonetheless it has charge-like quantum numbers. It is made of three
quarks, and charge conjugation change them into three anti-quarks. Charge
conjugation leaves spin and momentum unchanged.
The interesting question is, does a world composed completely of anti-matter
have the same behavior. For example, in neutron decay, there is a correlation
between the spin of the neutron and the direction of the electron that is
emitted when the neutron decays. The electron spin is also directed opposite to
its direction of motion.
←
→
momentum direction
n
ν p e
⇐
⇐ ⇒ ⇐
spin direction
Charge conjugated:
n
⇐
←
→
ν p e
⇐ ⇒ ⇐
momentum direction
spin direction
This is not what an anti-neutron decay looks like! The laws of physics responsible for
neutron decay are not invariant with respect to charge conjugation.
M. Cobal, PIF 2006/7
The parity operation (P) changes the direction (sign) of each of the spatial
coordinates. Hence, it changes the sign of momentum. Since spin is like angular
momentum (the cross product of a vector direction and a vector momentum,
both of which change sign under the parity operation), spin does not change
direction under the parity operation.
momentum direction
n
⇐
←
→
ν p e
⇐ ⇒ ⇐
momentum direction
n
⇐
→
←
ν p e
⇐ ⇒ ⇐
spin direction
Parity operation:
spin direction
The world would look different under the parity operation, since now the
electron’s spin would be in the same direction as its momentum.
The world is not symmetric under the parity operation!
M. Cobal, PIF 2006/7
Parity violation occurs only in the weak interaction.
The lack of symmetry under the parity operation was
discovered in the fifties following the suggestion of Lee and
Yang that this symmetry was not well tested experimentally. It
is now known that parity is violated in the weak interaction, but
not in strong and electromagnetic interactions.
The situation with charge conjugation symmetry is similar; the
lack of symmetry under charge conjugation exists only in the
weak interaction.
The Standard Model incorporates parity violation and charge
conjugation symmetry violation in the structure of the weak
interaction properties of the quarks and leptons and in the
form of the weak interaction itself.
M. Cobal, PIF 2006/7
In weak interactions C invariance is broken:
LH neutrino ν → LH antineutrino
(which dos not exists)
However under combined CP:
LH neutrino ν → RH antineutrino
Weak interactions are eigenstates of CP
This statement is not completely true:
CP violation in weak interactions does occur at the 10-4 level
M. Cobal, PIF 2006/7
Time Reversal
Time reversal means to reverse the direction of time. Here we need to be a bit
more careful. There are a number of ways in which we can consider time
reversal. For example, if we look at collisions on a billiard table when the cue
ball strikes the colored balls on the break, it would clearly violate our sense of
how things work if time were reversed. It is very unlikely that we would have a
set of billiard balls moving in just the directions and speeds necessary for
them to collect and form a perfect triangle at rest, with the cue ball moving
away. However, if we look at any individual collision, reversing time results in a
perfectly normal looking collision (if we ignore the small loss in kinetic energy
due to inelasticity in the collision). The former lack of time reversal invariance
has to do with the laws of thermodynamics; we here are interested in
individual processes for which the laws of thermodynamics are not important.
Time reversal reverses momenta and also spin, since the latter is the
cross product of a momentum (which changes sign) and a coordinate, which
does
not.
M. Cobal, PIF 2006/7
Now let’s consider what happens when we apply time reversal (T) to the case of
the neutron decay.
←
→
momentum direction
n
ν p e
⇐
⇐ ⇒ ⇐
spin direction
Time reversal:
n
⇒
→
←
ν p e
⇒ ⇐ ⇒
momentum direction
spin direction
This looks just fine, the electron spin is opposite to its momentum and the electron
direction is opposite to the neutron’s spin.
So, at least for neutron decay, the laws of physics appear to be symmetric under
time reversal invariance.
M. Cobal, PIF 2006/7
Now let’s consider what happens when we apply all three symmetry operations to
the case of neutron decay.
←
n
⇐
Time reversal:
p
⇐ ⇒ ⇐
spin direction
→
momentum direction
←
ν
⇒
⇒ ⇐ ⇒
p
←
e
→
n
ν
⇒
⇒ ⇐ ⇒
T, P and Charge conjugation:
momentum direction
e
n
Parity plus time reversal:
M. Cobal, PIF 2006/7
ν
→
p
←
momentum direction
e
→
p
spin direction
n
ν
⇒
⇒ ⇐ ⇒
spin direction
momentum direction
e
spin direction
The result that applying C, P, and T leaves the physical laws unchanged is not
surprising. Since CP leaves things unchanged (for neutron decay) and T also does,
applying all three should also work fine.
In fact, there is a theorem that says that under rather general conditions, any set of
physical laws that can be described by a field theory will be unchanged under the
CPT operation. There are many consequences to this theorem, for example that
the total lifetime and mass of a particle is identical to that of its anti-particle.
There are some considerations of conditions under which physical laws are not
invariant under CPT, but sensitive experimental tests of CPT invariance have not
shown any evidence for its breakdown.
M. Cobal, PIF 2006/7
Could there be evidence of violation of one or more of these symmetries in
neutrons?
Consider the case if neutrons had an electric dipole moment (edm). The neutron
has no charge, but it does have charged quarks inside it. If the charge is distributed
such that the negative and positive charge is separated by some distance (within
the neutron), then it would have a dipole moment. The value of the dipole moment
is the value of the positive charge times the distance between the positive and
negative charges. The direction of the dipole moment must be aligned with the
spin. Assume the neutron dipole moment points in the same direction as the spin:
.
T
P
C CP CPT
←
←
→ → ←
← dipole moment direction
n
n
n
n
n
n
⇐
⇒ ⇐ ⇐ ⇐
⇒ spin direction
Charge conjugation correctly turns a neutron into an anti-neutron, with the spin and
electric dipole moment in opposite directions. CPT does the same. However, both
T and CP produce non-physical particles, with the relative direction of spin and edm
incorrect. The existence of a neutron edm explicitly violates CP and T symmetries.
No such evidence for a neutron edm is seen.
M. Cobal, PIF 2006/7
There is evidence for violations of CP symmetry and hence of T symmetry. Until
recently, that evidence existed solely in the decays of neutral kaons. A neutral kaon
is a meson consisting of a strange quark and a down quark. The physical particles
with definite mass and lifetime are combinations of a kaon and an anti-kaon, much
the way that circularly polarized light is a combination of vertically and horizontally
polarized light.
Now, one combination of K0 and K0 that makes a physical particle with definite
mass and lifetime is mostly a CP eigenstate with eigenvalue +1 (K0S) and another
combination is a CP eigenstate with eigenvalue –1 (K0L) . It was a surprising result
found in 1964 that the K0L decayed into a pair of pions that were in a CP eigenstate
with eigenvalue +1. This implied violation of CP symmetry in kaon decays.
The manifestation of CP violation is restricted to the weak interaction. Hence
processes that involve only the electromagnetic and strong interactions appear to
be CP conserving.
M. Cobal, PIF 2006/7
• Until 1964, it was believed that CP simmetry was respected. That
year Christensons discovered that:
Normally:
(CP = -1)
But also:
(CP= 1, prob = 2x10-3)
• Origin of this CP violation is not known, but it is a necessary prerequisite to the development of a baryon-antibaryon asimmetry
• CP implies T violation through the CPT theorem
• In weak interactions limits found on T-violating contributions are
below the 10-3 level.
• In strong interactions, studied with forward-backward reactions:
if the interactions are invariant under T and P → matrix elements
-4
for
the
forward
and
backward
reactions
must
be
the
same
(<
5x10
)
M. Cobal, PIF 2006/7
CPT conservation
• CP violation will be mentioned, linked to violation under time
reversal T
• CPT Theorem: All interactions are invariant under the successive
operation of C(=charge conjugation) P(=parity) and T=time reversal)
taken in any order
Consequences of the CPT theorem that may be tested with
experiments relate to properties of particles and antiparticles:
Must have:
- same mass
- same lifetime
- equal and opposite electric charge and magnetic moments
M. Cobal, PIFresults
2006/7
(these
would follow from C invariance alone, but weak
interactions violate C and CP simmetry)
Summary
M. Cobal, PIF 2006/7
Quantity
Under C
Under T
Under P
position
r
r
-r
momentum
Spin
p
σ
-p
−σ
-p
σ
E field
E
E
-E
B field
B
-B
B
Magnetic dipole
momentum
σ⋅B
σ⋅B
σ⋅B
Electric dipole
momentum
σ⋅E
−σ ⋅ E
−σ ⋅ E
Long. polarisation
σ⋅p
σ⋅p
−σ ⋅ p
Transverse
σ ⋅(p1xp2)
−σ ⋅(p1xp2)
σ ⋅(p1xp2)
polarisation
M. Cobal, PIF 2006/7
Hadron Quantum Numbers
M. Cobal, PIF 2006/7
Hadron quantum numbers
• Characteristics of a hadron:
1) Mass
2) Quantum numbers arising from simmetries: J,P,C. Common
notation:
- JP (e.g. for proton: ½+), or
- JPC if a particle is also an eigenstate of C
(e.g. For π0: 0+)
3) Internal quantum number’s: Q and B (always conserved), S, C, ,
T (conserved in em and strong interactions)
How it is known the quantum numbers of a new hadron?
M. Cobal,it
PIF 2006/7
How
is known that mesons are done of 2 and barions of 3
quarks?
Some a priori knowledge is needed:
Considering the lightest 3 quarks (u,d,s), the possible 3- and 2quarks states will be
M. Cobal, PIF 2006/7
• Particles which fall out of above
restrictions are called exotic
particles (
)
• From observations of strong
interactions processes, quantum
numbers of any particles can be
deduced.
•Observation of pions
confirm these predictions
ensuring that pions are nonexotic
M. Cobal, PIF 2006/7
Assuming that K- is a strange meson, one can predict quantum
numbers of Λ-baryon:
And further, for K+ meson:
 All of the more than 200 observed hadrons satisfy this kind of
predictions, and NO exotic particles have been found so far,
M. Cobal,
PIF 2006/7 validity of the quark model which suggests that only

Confirms
quark-antiquark and 3-quarks (or antiquark) states exist
Even more quantum numbers..
It is convenient to introduce another quantum number which are
conserved in strong and em interactions:
-Sum of all internal quantum numbers, except of Q
hypercharge Y = B+S+C+ +T
- Instead of Q:
I3 = Q – Y/2
which is to be treated as a projection of a new vector
Isospin I = (I3)max
So that I3 takes 2I+1 values from –I to I
M. Cobal, PIF 2006/7
-Hypercharge Y, isospin I and its projection I3 are additive quantum
numbers, so that the quantum numbers for hadrons can be deduced
from those of quarks:
-Proton and neutron both have isospin of ½, and also very close
M. Cobal, PIF 2006/7
masses.
They are said to belong to an isospin doublet
•
• Heisemberg (1932): neutron and proton might be treated as
different charge sub-states of one particle: the neutron
• For every nucleon: new quantum number – isospin- with value I = ½
• There are 2 substates, with Iz, or I3 = ± ½
• Charge is given by Q/e = ½ + I3 if we assign I3 = + ½ to the proton
and I3 = - ½ to the neutron
Isospin ‘up’ en isospin ‘down’
• Complete analogy with the description of a particle of ordinary
M. Cobal, PIF 2006/7
spin ½, with substates Jz = ± ½
Other examples:
• Useful concept, because it is quantum number conserved in strong
interactions ⇒ these depends on I, not on I3.
• In strong interactions we do not distinguish between n and p:
I3 = +1
they are degenerate states
• Em interactions do not conserve I.
They couple to electric charge Q
and then “single-out” the I3-axis
in the isospin space
I3 = -1
• Isospin simmetry apply between all baryons and mesons which
M. Cobal, PIF 2006/7
transform one to another by change of u and d quarks.