Introduc0on to Par0cle Physics - Chapter 2 · PDF fileClaudio Luci – Introduc0on to Par0cle Physics – Chapter 2 1 ... • Some reac0ons are observed while other ones are not

SymmetriesandConserva0onLaws

ClaudioLuci–Introduc0ontoPar0clePhysics–Chapter2 1

Introduc0ontoPar0clePhysics-Chapter2-

ClaudioLuci

lastupdate:070117

•  Conserva0onlaws•  Discreteandcon0nuoussymmetries•  Barionicandleptonicnumber•  parity•  Chargeconjuga0on•  TimeReversal,CPTandIden0calpar0cles•  Isospin•  Gell-mann-Nishijima’sformula

2ClaudioLuci–Introduc0ontoPar0clePhysics–Chapter2

ClicktoeditMaster/tlestyle

•  Ifsomethingdoesnothappen,itmeansthatisviolatedsomeconserva0onlaw


Conserva/onLawsWhatever is not forbidden, it happens

ν µ γ+→ + → + + → +/ / /- ; p ; n p e n e e

1.  “Sacred” conservation laws: they derive from a symmetry of the Lagrangian (Noether’s theorem)

•  Time translation → energy

•  Spatial translation→ momentum

•  Rotation invariance → angular momentum

•  gauge invariance (U(1)) → elettrical charge

•  etc… etc…

2.  “empirical” conservation laws: they are not “protected” by a symmetry of the Lagrangian and could be violated in some process •  for instance: barionic number or leptonic number


•  Con/nuoussymmetries:forinstancespa0altransla0on

Ø  Theycanbederivedfromtheiden0tybydoinginfinitesimaltransforma0on;Ø  itexistsageneratorofthetransforma0on:

Ø Fisahermi0anoperatorthatcommuteswiththeHamiltonianàitisanobservableØ Addi/vequantumnumbers

•  Discretesymmetries:forinstanceparityØ  youcannotgetitbydoing“liQlesteps”Ø Mul/plica/vequantumnumbers


Twokindofsymmetriesψ ψ ψ ψ ψ ψ〉 = 〉〈〉 = 〈〉| ' U| ; '| ' |

U = unitary operator of the symmetry

αα i FU( )=e α = real parameter; F = generator of the transformation



BarionicNumber• Experimentallyitwasobservedinnuclearreac0onsthatthenumberofnucleonswasalwaysconserved;thiscon0nuedtobetruealsotakingintoaccountthe“strange”fermionslikeΛeΣ.

•  Stuchelbergmadethehypothesisthatthebarionicnumbermustbeconserved

•  Nowadayswesaythatinareac0onmustbeconservedthenumberofquarksminusthenumberofan0quarks:

•  Howeverwehavenotfound(yet)asymmetryoftheHamiltonianthat“protects”thisconserva0onlaw,thereforetherearetheories(likeGUTforinstance)thatviolatetheconserva0onlawofbarionicnumber.

•  Theviola0onofbarionicnumberisoneofthe“base”ingredientstoexplainthe“disappearance”ofthean0maQer.

B=1

3Nq-Nq( )



LeptonicNumber• Experimentallyitisobservedthattheleptonicnumberisconservedtoo•  Moreoverwehave(wehad)threeindependentconserva0onlawsforthethreedifferentleptons(electron,muonandtau)becauseitdoesnotexistthedecay:

•  Howeverin2002wehadthefinalexperimentalevidenceofneutrinooscilla0ons(thatismasseigenstatesarenotequaltoflavoreigenstates),thereforetheleptonicnumberisnotconservedseparatelyanylonger

•  NewtheoriesbeyondSM(likeSuSyforinstance)foresee:

µ → e + γ but µ → e + νe + νµ

B.R.(µ → e + γ ) < 10-11 experimental limit before MEG exp. (2008)⎡⎣ ⎤⎦

B.R.(µ → e + γ ) ≈ 10-55 SM value ⎡⎣ ⎤⎦

B.R.(µ → e + γ ) ≈ 10-11÷-15

B.R. ≈ 4 i10-14 sensitivity expected by MEG-II⎡⎣ ⎤⎦

B.R.(µ+ → e+ + γ ) < 4.2 i10-13 MEG result (2016)⎡⎣ ⎤⎦



Parity

•  TheeigenvalueofPare:+1e-1•  Awavefunc0oncanorcannothaveagivenparity.Incaseithasaparity,itcanbeeven(eigenvalue+1)orodd(eigenvalue-1).Forinstancesphericalharmonics.

•  Theparityofsphericalharmonicsis:

•  Thestrongande.m.interac0onsdoconservetheparity

•  Somereac0onsareobservedwhileotheronesarenotobserved,likeforinstance

•  Thiscanbeexplainedbyassigningtothepionanintrinsicparity

→ − − −, , , , x y z x y z

( )ϑ ϕ π ϑ π ϕ ϑ ϕ−→ + =Y ( , ) Y ( - , ) Y (1 , ) ml

m ml l

l

π π π→ →- - 0 + d n + n ; + d n + n +

N.B. the parity operator does not act on the spin wave function

!r ⇒ −

!r

t ⇒ t!p ⇒ −

!p

!r ×!p ⇒

!r ×!p

!s ⇒

!s



Pionintrinsicparity:ηπ

• Theconserva0onofparityrequires:

It has been measured in the absorption of slow pions in deuterium

π →- + d n + n

deuteron: Sd = 1 ⇒ ηd = 1 η = −1( )L+S+1⎡

⎣⎢

⎤

⎦⎥

( ) ( )πη η η η⋅ ⋅ − = ⋅ ⋅ −d 1 1i fL Ln n

( )πη η ηη= ⋅ = −= ⇒ =d 1, 1, 0 1 fn n i

LL

• The two neutrons are identical fermion, therefore the total wave function must be antisymmetric. The initial state has J=1 because Sπ = 0, Sd=1 and Li = 0.

|ψnn(1)〉 =| J = 1,S = 1,Lf = 0,2〉 ⇒ψ symmetric

|ψnn(2)〉 =| J = 1,S = 1,Lf = 1〉 ⇒ψ antisymmetric

|ψnn(3)〉 =| J = 1,S = 0,Lf = 0,2〉 ⇒ψ symmetric

( )πη = − = −1 1 1



Chargeconjuga/on• Thechargeconjuga0onoperatortransformapar0cleintoitsan0par0cle,whohasallinternalquantumnumberofoppositesign(electricalcharge,strangeness,magne0cmoment,barionic/leptonicnumber,etc…).

•  Onlythe“neutral”par0clescanbeeigenstateofthechargeconjuga0onoperatorC;forinstanceπ0andphotonareeigenstatesofC…butneutronorneutrinoarenot!

•  TheeigenvalueCαare+1and-1

•  EigenstatesofCcanbeconstructedbyusingpar0cle-an0par0clepair,wheretheoperatorCexchangethetwopar0cles;

•  Ifthestateissymmetricorantysimmetricduetoexchange,wehave:

C|α,ψ 〉 = Cα|α,ψ 〉 eigenstate of C⎡⎣ ⎤⎦C|a,ψ 〉 = |a,ψ 〉 a=antiparticle of a⎡⎣ ⎤⎦

C|a,ψ1;a,ψ2〉 =|a,ψ1;a,ψ2〉 = ±|a,ψ1;a,ψ2〉

in this case |a,ψ1;a,ψ2〉 is an eigenstate of C



π+π-state• Let’stakeaπ+π-pairinastateofangularmomentumL:

becauseexchangingthetwopionsisequivalenttoinverttheirspa0alposi0on•  N.B.let’srecallthatthespinofthepioniszero,sowedonothavetotakeintoaccountthespinintheoverallstatewavefunc0on

( )π π π π〉〉L+ - + -C| ,L = -1 | ,L

π+ π- L

X1 X2

π+ π- L

X1 X2

Π- Π+ L

X1 X2

C



ffstate• fermionpairofspin½.•  Tobekeptinmindthefollowingproper0esofthespinwavefunc0on:

•  Thetwofermionsexchangeintroduce,becauseofthespin,afactor(-1)S+1

•  Moreover,becauseoftheopposedintrinsicparityofthefermion-an0fermionpair(seeDiraceq.),wehavetointroduceanotherfactor-1.

•  Thereforetheeigenvalueoftheoperatorchargeconjuga0onforthisstateis:

•  Thisresultimposessomeconstraintstothequarkcontentofthemesons

|↑↑〉 ; 12

|↑↓〉+|↓↑〉( ) ; |↓↓〉 Symmetric⎡⎣ ⎤⎦

S=1, SZ=1 S=1, S

Z=0 S=1, S

Z=-1

12

|↑↓〉 − |↓↑〉( ) S=0, SZ=0 ⎡⎣ ⎤⎦ antisymmetric⎡⎣ ⎤⎦

C|ff,J,L,S〉 = -1( )L ⋅ -1( )S+1

−1( )|ff,J,L,S〉 = -1( )L+S|ff,J,L,S〉

f f L

X1 X2



π0chargeconjuga/on

• ThesumofSandLofthequark-an0quarkpairgivesthespinofthepion.

•  Experimentallywefindthatthedominantdecaychannelis:

•  Sincethee.m.interac0onsareinvariantperchargeconjuga0on,wemusthaveaccordingtothequarkmodel.

π 0S =0

S = spin of qq pair forming the π0

L = angular orbital momentum of qq pair

q q L

X1 X2

( ) ( )π⇒ ⇒ =0L+S 0 C = L + S = -1 = -1 0 1

π γγ→0

C |π0〉 = Cπ 0 |π0〉 ;

C |γγ 〉 = Cγ ⋅Cγ |γγ 〉 = 1⋅|γγ 〉 since Cγ2 = 1

π 0=1 C

N.B. : Cγ = -1 since (q→-q ; E,B→ -E, -B) ⇒ C-parity of 3 photons is: Cγ( )3=-1

( )( )π γ

π γ−

Γ →= < ⋅Γ →

08

0

33 10

2R Naively one would expect something

of the order α=1/137



ques/ons

1)  Fromanexperimentalpointofview,whywecannotsimplysayR=0?2)  Howcanwebesurethattheπ0decaysintwophotons?Maybewecould

havejust“lost”andaddi0onalphoton.

3)  Doesthephasespaceplayaroleinconsideringthethirdphoton?Inotherwords,couldbethesuppressionaneffectofthephasespace?

( )( )π γ

π γ−

Γ →= < ⋅Γ →

08

0

33 10

2R



TimeReversal(tà-t)•  TheoperatorsPandCarehermi0anandunitaryoperators;theyhaveeigenstatesandgiverisetomul0plica0vequantumnumbers;

•  The time reversal operator T is antiunitary and does not have eigenstates; •  The correct definition of T in quantum mechanics was given by Wigner: now let’s take the complex conjugate of the equation and we go back to the original Schrödinger equation: PROVIDED THAT THE HAMILTONIAN H IS REAL !!! •  IfHcontainssomeimaginarytermsthenTimeReversalsymmetryisviolated!!

Tψ (t) = ψ *(−t) ⇒ -idψ

*(t ')dt '

= Hψ *(t ')

idψ (t ')

dt '= Hψ (t ')

N.B. Time Reversal violation can not be searched in some particle decays since there are no quantum numbers associated to T



CPTtheorem•  ThereisnofundamentalreasoninNaturetoenforcetheP,CandTsymmetryseparately;

•  howeveraquantumfieldtheorythatisinvariantforaLorentztransforma0on,mustbealsoinvariantforaCPTsymmetrytransforma0on;

•  this brings some consequences: for instance a particle and its antiparticle must have the same mass and the same life time;

•  possible violations of CPT are looked for in the mass differences between particle and antiparticle.

•  for instance the ASACUSA experiment at CERN established in 2003 the following limit for proton and antiproton mass difference:

mp − mp

mp≤10−8



Iden/calpar/cles•  TheQuantumMechanicsaffirmsthatinasystemcomposedbyiden0calpar0cles(alsocalledindis/nguishablepar0cles),wecannotperformanymeasurementabletoiden0fytheswapofanytwopar0clesofthesystem:

•  ifwedoasecondswapofthesamepar0cles:

•  BOSONS:eiα=1(symmetricpar0cles)•  FERMIONS:eiα=-1(an0symmetricpar0cles)

⇒ ζ2,ζ1 = eiα ζ1,ζ2

ζ1,ζ2 = eiα ζ2,ζ1 = ei2α ζ1,ζ2 ⇒ ei2α = 1 ⇒ eiα = ±1

N.B. the total wave function can be the product of various parts:

ζ = ψ (spatial) iσ(spin) iϕ(flavour) i τ(colour)

N.B. it is the total w.f. that must respect the symmetry and not the individual parts



Afewconservedquantumnumbers• N.B.notallinterac0onsdorespectthevariousconserva0onlaws.Forinstance:



Interac/onsrela/vestrength

• Duetointerac0onsrela0vestrengthandtoconserva0onlaws,theinterac0onstakeplaceinthefollowingorder:1.  Stronginterac0on(itactsonlyonquarksandgluons);2.  e.m.interac0on(forinstanceiftherearephotonsorleptonsas

interac0ngpar0cles;theydonotinteractstronglyande.m.comesfirst);3.  weakinterac0on(forinstanceifthereareneutrinosthatinteractonly

weaklyorifthereisaviola0onofaconserva0onlawfollowedbystrongande.m.interac0ons,theweakinterac0oncomefirst).

4.  (Gravityisnotconsideredatallinpar0clephysics)


•  weakinterac0on:forbidden(seeleptonnumber)

•  weakinterac0on:allowed,butwatch

outthethresholdenergy.

• weakinterac0on:forbidden(seebarionnumber)

• weakinterac0on:allowed.(whataboutstrangeness?)

•  ithappens(B.R.=20%).Whichinterac0on?(mindthestrangeness).


Afewexampleofreac/ons

µν µ→ ++ + p n

ν π→ + +- + + p ee p

π νΛ → + +- + e e

µµ π ν→ + ++ - 0K

π π→ ++ + 0K



Whythesereac/onsdonotoccur?It is not always straightforward to find the reason

K p π π+ + +→ Λstrangeness violation. It becomes evident by looking at the quark content.

φ

φ π π π+

→ ≈

→ ≈-

. .( (1020) ) 83%

. .( (1020) ) 15%o

BR KK

BROZI rule

( ) 195 mb 8.86 922 mb( )

p pp p

σ π πσ π π

+ +

− −→ = = ≈→

isospin

4( ) 1.2 10( )

e xe

µ

π νπ µ ν

− −−

− −Γ →

=Γ →

helicity



Isospin•  Heisembergproposedin1932thatprotonandneutronweretwodifferentstatesofthesamepar0cle:thenucleon.

•  Toimplementthisideawecanrepresentthenucleonasatwocomponentscolumnvector:

•  Themathema0calformalismisiden0caltotheoneusedbyPaulitodescribethespinoftheelectron.

•  TheprotonhasI3=½whiletheneutronhasI3=-½

•  Ifthestronginterac0onsareinvariantforrota0onintheisospinspace,thentheisospinmustbeconservedinallprocessesinvolvingstronginterac0on.

; 1 0

N= ; p= n= 0 1

αβ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠



Gell-Mann-Nishijima’sformula

The third component of the isospin distinguish the electrical charge within an isospin multiplet

( )31Q = I + B+S2

strangeness

Barionic number

charge

N.B. B+S = Y (hypercharge)

31 Q= I + Y 2

The electromagnetic interaction breaks the isospin symmetry; as a consequence the masses within a multiplet are different (mp different from mn)



Isospin

•  Let’ssupposetohavetwonucleons.Fromtheruleoftheaddi0onofangularmomentumweknowthatthetotalisospincanbe1or0.

•  Itexistsaboundstateproton-neutron(deuteron),butdonotexistboundstatesproton-protonorneutron-neutron,hencethedeuteronmustbeanisospinsinglet,otherwisetheyshouldexistalsotheothertwostatesthatdifferbyarota0onintheisospinspace.

Let’s see a dynamical consequence of the isospin conservation

( )

a) |1,1 = pp 1 b) |1,0 = pn+np 2

c) |1,-1 = nn

〉

〉

〉

Symmetric triplet; I = 1

( ) 1 |0,0 = pn - np 2

〉

Antisymmetric isosinglet; I = 0



nucleon-nucleonscaUering

•  SincethedeuteronhasI=0,fortherighthandprocesseswehave:whilefortheonesonthelevwehave:

•  SincethetotalisospinImustbeconserved,onlythestatewithI=1willcontribute.ThescaQeringamplitudeshavetobeinthera0o:

•  Theprocessesa)andb)havebeenmeasured,andoncewetakeintoaccountthee.m.interac0on,theyareinthepredictedra0o.

Let’s consider the following processes: +

0

-

a) p + p d +

b) p + n d +

c) n + n d +

π

π

π

→

→

→

the π has isospin 1 because it exists in three different states

+ - 0d + d + |1,1 ; ; d + |1,-1|1,0 πππ = =〉= 〉〉

( ) 1p + n = |1,0 |0,02

p + p = |1,1 ; ; n + n = |1,-1〉+ 〉〉〉

11: : 1 2

2 : 1: 2 and the σ



pion-nucleonscaUering

•  Let’sexpressthevariousstatesinthebaseofthetotalisospinbyusingtheClebsch-Gordancoefficients

Let’s consider the four reactions:

1 1 3 12 2 2

⊗ = ⊕

0

+ + a) + p + p- b) + p + n- - c) + p + p- - d) + n + n

π π

π π

π π

π π

→

→

→

→

The ini t ia l states are the composition of I=1 and I=1/2 that give I=1/2 and I=3/2

3 3 3 1 1 1 | , | , ,

2 2 2 2 2 2

3 3 3 1 1 10 | , | , ,2 2 2 2 2 2

1 2+ - ,p | ; ,p | - |3 3

2 1-,n | ; ,n | + |3 3

π π

π π

+ − −

− − −

〉 = 〉〉 = 〉〉

〉 = 〉〉 = 〉〉

1/3



pion-nucleonscaUering•  Let’swritethefourprocessesinthenewbase:

•  tocomputetheprobabilityamplitude,wehavetoperformthescalarproduct

2/3

+ +

− − − −

− − − −

− → −

〉 → 〉

〉〉 → 〉〉

〉〉 → 〉〉

〉〉

3 3 3 3 , ,

2 2 2 2

3 1 1 1 3 1 1 1 , , , ,

2 2 2 2 2 2 2 2

3 1 1 1 3 1 1 1, , , ,

2 2 2 2 2 2 2 2

3 3 3 3 , ,

2 2 2 2

a) | |

1 2 2 1 b) | - | | + |3 3 3 31 2 1 2 c) | - | | - |3 3 3 3

d) | |

〈〉 f|S|i

〈〉〈〉 =3 3 3 3

3 3 1 1, | , , | ,3 12 2 2 2 22

S| = A ; S| A I I I I ≠1 32 2

N.B. A A

tot 3 2

tot 3 2 1 2

tot 3 2 1 2

tot 3 2

a) A =A

2 2 b) A = A A3 3

1 2 c) A = A A3 3

d) A =A

⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞+⎜ ⎟⎝ ⎠

N.B. strong interactions do not mix states with different total isospin



pion-nucleonscaUering•  Thecross-sec0onsofthefourprocessesarepropor0onal,bymeansofafactorKequalforall4processes(thattakesintoaccountthephasespace,2πfactors,etc…),to:

•  Fromtheserela0onsweinferthattheprocessesa)andd)musthavethesamecross-sec0onatthesameenergy.Thishasbeenverifiedexperimentally.

•  FortheotherprocessesweneedtoknowA1/2andtherela0vephasebetweentheamplitudes.

3/3

( )( )( )( )

2 3

22

3 2 1 2

2

3 2 1 2

2 3

2

+ + a) + p + p

2 2- 0 b) + p + n A A3 3

1 2- - c) + p + p A A3 3

- - d) + n + n

K A

K

K

K Aσ

σ π π

σ π π

σ π π

π π

→ =

→ = −

→ = +

→ =



TheΔ resonance•  TheΔ resonance has isospin 3/2 (it exists in 4 states of different charge), therefore all

processes in which it appears as formation resonance can proceed only through the channel with I=3/2. As a consequence:

•  from these relations we can infer now: that we know they are true.

( )( )( )( )

2 3

22 2

3 2 32

2 2

3 2 32

2 3

2

+ + a) + p + p

2 2- 0 b) + p + n A3 9

1 1- - c) + p + p A3 9

- - d) + n + n

K A

K A

K A

K A

σ π π

σ π π

σ π π

σ π π

→ =

→ = =

→ = =

→ =

( )( )

( )( )

0

+ + - + p + p + p + n 9 ; 2

- - - - + p + p + p + p

σ π π σ π π

σ π π σ π π

→ →= =

→ →



Clebsch-Gordancoefficients



exercize•  WehaveasystemcomposedbyaΣ-andaproton.Writethetotalwaveinthebaseofthetotalisospinofthesystemandcomputetheprobabilitytofindthesysteminastateoftotalisospin½.

•  TheΣ-hasI=1eI3=-1(seeGell-mann-Nishijimaformula),whiletheprotonhasI=1/2eI3=+1/2.Combiningtogetherthetwostateswecanhavehastotalisospin½or3/2withthethirdcomponentI3equalto-1/2.

•  Theprobabilitytofindthesysteminastateoftotalisospin½is2/3.

- 1 3 1 2 1 1| | , | ,3 2 2 3 2

2

Σ 〉 = − 〉 − − 〉p



exercize•  ThebarionΛdecaysinproton–π-orneutron-π0.Inthedecayas-quarkoftheΛtransformsintoau-quarkofthenucleon,thereforeitsstrongisospinchangesby½.AssumingthatintheΛdecaysthisselec0onruleisretained,andignoringothersmallcorrec0on,deducethera0ooftheB.R.ofthep-π-decaywithrespecttothen-π0

decay.•  The nucleone has isospin ½ while the pion has isospin 1, therefore a nucleon plus a pion can give

total isospin equal to ½ or 3/2. The Λhas isospin zero, so in the total wave function of the system nucleon-pion we need to take into account only the component with isospin ½ , due to the selection rule ΔI = ½.

•  The transition probability is equal to the square of the w.f.:

•  The experimental values are:

•  (mostlikelythereisahigherordercontribu0onwithΔI=3/2)

p + π − = 1

2;12

+ 1;−1 = − 23

12

;− 12

+ 13

32

;− 12

0 1 1 1 1 1 2 3 1; 1;0 ; ;2 2 3 2 2 3 2 2

n π+ = − + = − + −

( )( )

21 1;2 2

20 1 10 ;2 2

2|. . 3 21. . | 3

pBR p

BR n n

ππ

π π

−− −

−

〈 + 〉Λ → += = =

Λ → + 〈 + 〉

( ) ( )0. . 63.9% ; . . 35.8%BR p BR nπ π−Λ → + = Λ → + =

( )( )0

. . 63.9 1.7835.8. .

B R p

B R n

π

π

−Λ → += =

Λ → +



exercize Il K0

S può decadere in due pioni carichi oppure in due pioni neutri. Trovare il rapporto tra il B.R. del decadimento in pioni neutri rispetto a quello in pioni carichi. Si ricorda che per ragioni di simmetria lo stato finale deve avere isospin totale zero

Nei decadimento deboli con DS=1 si ha DI=1/2, quindi dato che il K ha I=1/2, lo stato finale dei due pioni deve avere I=0 oppure I=1. La funzione d’onda dei due pioni deve essere simmetrica rispetto allo scambio delle due particelle, quindi dato che essi hanno spin zero e si trovano in uno stato di momento angolare l=0, anche la parte di isospin deve essere simmetrica, quindi I=0.

Utilizzando i coefficienti di Clebsh-Gordan si ha:

1 1 10;0 1, 1;1 1 1,0;1,0 1, 1;1 13 3 3

= + + − − + − + = 0 01 1 13 3 3π π π π π π+ − − += + − +

Di conseguenza abbiamo:

( )( )

20 0 0 0 0 ;

20;

. . |0 0 12. . |0 0

S

S

BR K

BR K

π π π π

π π π π+ − + −

→ + 〈〉= =

→ + 〈〉

I valori sperimentali sono:

( ) ( )0 0 0 0. . 30.7% ; . . 69.2%S SBR K BR Kπ π π π+ −→ + = → + =

( )( )0 0 0

0

. . 30.7 0.4469.2. .

S

S

B R K

B R K

π π

π π+ −

→ += =

→ +

Probabilmente vi è un contributo di ordine superiore con DI=3/2



exercizeDedurre attraverso quali canali di isospin possono avvenire le seguenti due reazioni:

Nel caso in cui il canale dominante sia quello con isospin 0 per entrambe le reazioni, trovare il rapporto tra le sezioni d’urto σa/σb

- 0 0 - a) K ; b) K p pπ π+ −+ → Σ + + → Σ +

Ricordiamo l’isospin totale e la terza componente delle particelle coinvolte nella reazione e scriviamo lo stato iniziale ed i due stati finali in termini degli autostati di isospin utilizzando i coefficienti di Clebsh-Gordan.

1 11;0 0;02 2

K p− + = + −1 1 1 1

3 32 2 2 2; ; ;K I I p I I− = = = − = = =

0 03 31; 0 ; 1; 0I I I IπΣ = = = = = = 0 0 2 12;0 0;0

3 3πΣ + = + −

3 31; 1 ; 1; 1I I I Iπ+ −Σ = = = = = = −1 1 12;0 1;0 0;06 2 3

π+ −Σ + = + + +

Di conseguenza la reazione a) può avvenire soltanto attraverso il canale di isospin totale 0, mentre la reazione b) può avvenire attraverso il canale con isospin 0 ed anche con isospin 1.

Nel caso in cui il canale dominante sia quello con isospin 0 per entrambe le reazioni, allora il rapporto tra le sezioni d’urto è pari al rapporto dei quadrati dei coefficienti di C.G. dell’autostato di isospin 0 nei due stati finali:

220 0 1

0;0 3 110;03

a

b

πσσ π+ −

−Σ += = =

Σ +


Endofchapter2


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Introduc0on to Par0cle Physics - Chapter 2 · PDF fileClaudio Luci – Introduc0on to Par0cle Physics – Chapter 2 1 ... • Some reac0ons are observed while other ones are not