Transcript Elementary Particle Physics

Lecture 5 – Symmetries and Isospin
●
●
●
What is a symmetry ?
Noether’s theorem and conserved quantities
Isospin
Obs! Symmetry lectures largely covers sections
of the material in chapters 5, 6 and 10 in Martin
and Shaw though not always in the order used
in the book.
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Symmetry – an intuitive example and a definition
An experiment measures the reaction rate R for    p        n at x  0.
If the experiment was moved to x  a the same rate R would be observed.
This is equivalent to not moving the original experiment but redefining the x-axis through
a translation: x '  x  a instead.
The laws of physics are invariant to a transformation of a translation in space - symmetry.
To put it another way, we can't know if our experiment took place at a certain place just
by looking at the measured reaction rate.
Definition of a symmetry in particle physics: under a transformation one or more observables
will be unchanged/"invariant to the transformation".
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Types of symmetry
Continuous symmetries
●

Depend on some continuous variable, eg linear translation
which can be built up out of infinitesimal steps

Eg linear spatial translation x’=x+d x, translation in time
t’=t+d t
Discrete symmetries (subsequent lectures)
●

Two possibilities (all or nothing)

Take a process, eg A+B
●
C and measure it!
Does this process happen at the same rate if


Eg. The mirror image is considered (parity – move from left-handed to
right-handed co-ordinate system)
Eg. All particles are transformed to their anti-particles: A+B
C
(charge conjugation)
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Conserved quantities in quantum mechanics
Consider observable system described by time-independent Hamiltonian.
Wave function of the system ψ satisfies Schrödingers equation: i
Ψ
ˆ (1.18)
 HΨ
t
Consider operator for observable Qˆ and its expectation value's time-dependence.
 Q  ψ | Qˆ |ψ  (5.01)
d Q d
 ˆ
Qˆ

ˆ
  ψ | Q |ψ 
| Q    ψ |
ψ     | Qˆ

dt
dt
t
t
t
 ˆ


| Q     | Qˆ
 (5.02)
t
t
(assume no time dependence on the ope rator)
d Q
1 ˆ
1
ˆ ˆ   (5.03)
   HΨ
| Qˆ    Ψ | QH
dt
i
i
ˆ | Qˆ  Ψ | HQ
ˆ ˆ 
Hˆ is Hermitian  HΨ
d Q 1  ˆ ˆ

  Q, H   (5.04)
dt
i
If  Hˆ , Qˆ   0 (5.05) i.e. the operator for an observable commutes with the Ham iltonian
then that observable is a conserved quantity.



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Conserved quantities – why we need them
and how we find them
In an ideal world in which we can write down the equations for all of the
fundamental forces in all circumstances and calculate that, eg      e      0.
We don't have such a theory so we need conserved quantities/conservation laws
to provide important constraints on what can and can't occur. How do we find
such laws and how can we interpret them in the light of our incomplete knowledge
of nature ?
We don't know the form of Hˆ for every situation so we can't try all
commutator combinations: eg  Hˆ , pˆ  ,  Hˆ , Lepton number  etc. and
see which pairs of operators commute.
There is a way to find out if Hˆ commutes with another operator without knowing or asking
what Hˆ is.
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Translational invariance
All positions in space are physically indistinguishable.
If a closed system of particles (no external forces acting) is moved
from one position to another its properties are unaltered.
Consider shift: ri  ri '  ri  d r
Hamiltonian becomes: Hˆ  r   Hˆ  r '
i
x1 x2
i
This is unchanged by the translation, eg free particle
x



Hˆ  2  2  2 insensitive to x1'  x1  d x
x1 y1 z1
 Hˆ  r '  Hˆ  r  (5.06) - generally true.
2
i
2
2
i
x1+dx
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Consider single particle  ( x) and 1-dimension (x )
Displacement operator Dˆ  ( x)   ( x  d x) (5.07)
 ( x  d x)   ( x) 


ˆ
p


i
 x
x 
 Dˆ  1  id xpˆ
x
Small displacement d x 
  x 
  
d x   ( x)  id x  i    x   1  id xpˆ x   ( x)
x
 x 
(5.08)
ˆ ˆ  x   ( x)
Define  '( x)  Hˆ  x   ( x)  Dˆ  '( x)  DH
From (5.07): Dˆ  '( x)   '( x  d x) = Hˆ  x  d x   ( x  d x )= Hˆ  x   ( x  d x )  Hˆ  x  Dˆ  ( x )
ˆ ˆ  x   ( x)  Hˆ  x  Dˆ  ( x)
 DH
  Dˆ , Hˆ   0 (5.09)
 1  id xpˆ x , Hˆ   0   pˆ x , Hˆ   0 (5.10) Linear momentum conservation!!
Can generalise to many particles, 3-dimensions and finite displacements.
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What have we done ?
(1) Derived the quantum mechanical version of a conservation law:
for an observable Q :  Hˆ , Qˆ   0 (5.05)
(2) Realised it was impossible just to take a Hamiltonian and try out
various observables to see which commute.
(3) Took an example of a displacement in space. Saw that if the Hamiltonian of a system
is invariant to the displacement, then the transformation operator Dˆ commutes with the
Hamiltonian:  Dˆ , Hˆ   0. (5.09)
(4) Dˆ is related to the linear momentum operator: Dˆ  1  ipˆ xd x (5.08). This led to
 pˆ x , Hˆ   0 (5.10): linear momentum conservation.


The invariance of the Hamiltonian to a displacement in space leads to the conservation
of linear momentum. Symmetry!!
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Noether’s theorem
Previous slides showed a quantum mechanical ”version” of
Noether’s theorem.
If a system of particles shows a symmetry, eg its Hamiltonian is
invariant to a continuous transformation then there is a conserved
quantity.
Symmetry
Conservation Law
Translation in space
Translation in time
Linear momentum
Energy
Rotation in space
Angular momentum
Gauge transformations
Electric, weak and
colour charge
Discrete symmetries also lead to conserved quantities (to come)
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Charge conservation – (not for lecture/exam)
Electron in an electromagnetic field: . Is nature insensitive to a local phase shift (gauge transformation)
depending in position and time:   ei Q r ,t  ? (5.11)
Yes, but the solution is subtle. The el ectron is charged and interacts via the electromagnetic force.
E   (  scalar potential) ; B    A
 A  vector potential. (5.12)
Gauge freedom with potentials (from electrodynamics):
' 

t
A '  A    r , t 
;
  r , t   arbitrary function.
Electron in an electrostatic potential: Hˆ  Hˆ 0  e
Schrödinger's equation for electron: i
' e
i  r ,t Q


Hˆ 0  
2m
(5.13)
2
(5.14)

 Hˆ 0  e .
t


- for simplicity only consider a time variation.
Is it possible to have the same form for Schrödinger's equation: i
 '
 Hˆ 0  e  ' ? (5.15)
t


 '
 Hˆ 0  e '  ' should also work (gauge freedom).
t
 2
 i  r ,t Q
  i  r ,t Q
i e
  
 e  e

e
t
t 
 2m
i



 
 ei  r ,t Q  i
 Q
t
 t
2
 


 i  r ,t Q  

e

 e  e
e
- ok, since  ,  are abitrary.

  : set Q

2
m

t

t

t




   2
i
 
 e   (5.16)
From 5.15 and 5.16, and  ' would behave the same way!
t  2m

The long range electromagnetic field cancels the local phase shift. Charge must be conserved since the cancellation
would not work if the electron lost its charge for a period of time and couldn't interact with the electromagnetic field.
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Angular Momentum (no spin)
Transformation is an infinitesimal rotation about the z -axis with angle 
 x '   cos ε sin ε   x  1 ε   x 
 
 

 y
y
'
sin
ε
cos
ε
y
-ε
1
  
  
 
x  x  x   y
y  y  y   x
r  x, y , z   r  x, y , z   d r
y’
z  z   z (5.17)
d r    y,  x, 0  (5.18)
ˆ ( x, y, z )  Ψ ( x   y, y   x, z ) (5.19)
DΨ
y
xp
xp’ y
p
P
yp’
 Ψ
Ψ 
 Ψ ( x, y , z )    y
x

y 
 x


 1  i  xpˆ y  ypˆ x  Ψ
x’

x



ˆ
ˆ
p


i
p


i
y
 x
x
y 

 Dˆ  ( 1  iεLˆ3 ) (5.20) Lˆz  xpˆ y  ypˆ x  operator for z - component of orbital angular momentum. (5.21)
ˆ ˆ  r   (r )
Define  '(r )  Hˆ  r   (r )  Dˆ  '(r )  DH
From (5.19): Dˆ  '(r )   '(r  d r ) = Hˆ  r  d r   (r  d r )= Hˆ  r   (r  d r )  Hˆ  r  Dˆ  ( r )
ˆ ˆ  r   (r )  Hˆ  r  Dˆ  (r )
 DH
  Dˆ , Hˆ   0 (5.22)
 1  iεLˆ3 , Hˆ   0   Lˆ3 , Hˆ   0 (5.23) Orbital angular momentum conservation!!
Angular momentum conservation is demanded if we require the laws of physics are invariant to a rotation.
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Angular momentum
A rotation in space (around z-axis) was considered for a situation
involving no spin.
The Hamiltonian was invariant to it  symmetry.
The observable quantity which is conserved is orbital angular momentum: Lz
 Hˆ , Lˆz   0 (5.23) ; Lˆz  xpˆ y  ypˆ x (5.21)


Considering rotations around x and y axes would similarly give:
 Hˆ , Lˆ y   0 , Lˆ y  zpˆ x  xpˆ z ;  Hˆ , Lˆz   0




, Lˆz  xpˆ y  ypˆ x ;  Lˆi , Lˆ j  = ijk iLˆk (5.24)
Generalise to include spin S :
Total angular momentum: J  L  S ;  Sˆi , Sˆ j  = ijk iSˆk ;  Jˆi , Jˆ j  = ijk iJˆk (5.25)
Total angular momentum conserved:  Hˆ , Jˆ x    Hˆ , Jˆ y    Hˆ , Jˆ z   0 (5.26)
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Spin
• Simplest case for studying angular momentum.
• Algebraically a carbon copy of orbital angular momentum:
 Sˆi , Sˆ j   i ijk Sˆk
(5.25)


• Useful to use matrices to represent states and operators.
• A spin ½ particle can have a spin-up or spin-down projection
along an arbitary z-axis. 1 1 1  1 1  0 
|
  
22
 0
|    
2 2
1 
(5.27)
• General state - linear combination:
 
1 
 0




 
 
 
 
 0
1 
|  |2  |  |2  1 (5.28)
• Operators with 2x2 matrices.
0
1  0 1
1 0 i 
ˆ  1 1
Sˆx   
Sˆ y  
S
z


 (5.29)
2 1 0 
2 i 0
2  0 1 
1
Drop
and they become the Pauli spin matrices
2
0
 0 1
0 i 
1
x    y  


(5.30)
z



1
0
i
0
0

1
 




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Transformations
Intuitively easy to understand how the components of a vector
change under a rotation.
Consider how a two component spinor is affected by a
rotation of a co-ordinate system.
3


 ' 
 
 i ˆ  /2
A
A
A
(5.31) U is a matrix: e  1  
 ..
   U ( )   U ( )  e
2! 3
  '
 




ˆ
U ( )  cos I  i    sin (5.32)
2
2
 - direction angle of rotation (right-hand sense), | | angle magnitude.


I  identity matrix ,    x ,  y ,  z   Pauli matrices.
U ( ) is a 2  2 matrix belonging to the set of all such matrices: SU(2)
Unitary 2x2 matrices with determinant 1.
Unitary  U *U  I ; U 1  U * (5.33)
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Question
Find the matrix representing a rotation by  around the y-axis.
Show it converts a spin-up to spin-down particle.
   y (5.34)




U ( )  cos I  i ˆ   sin  U  cos I  i y sin  i y (5.35)
y
2
2
2
2


 0  i   0 1 
U  i 
 
 (5.36)
 i 0  1 0 
1 
Spin-up particle (in z - direction):     (5.37)
0
 0  1 1   0 
U  
     (5.38)
1 0  0  1 
spin-up
spin-down
z
Can transform between possible states – equivalent to rotating
co-ordinate axes.
Obs! – in an experiment a rotation would affect all states and
there wouldn’t be observable effect in the physics
measurement – nature doesn’t care where your axes are.
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Question
Repeat the procedure in the previous question and converts a
spin-down to a spin-up particle.
0
Spin-up particle (in z - direction):     (5.39)
1 
 0  1  0 
1 
1 1
U  




(5.40)
 
 
2 2
1 0 1 
0
Obs! - sign. Unimportant (for the most part).
y
spin-up
spin-down
The amplitude squared is the important quantity.
z
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Addition of angular momentum
●
Two spin ½ particles can form multiplets
●
A triplet and a singlet can be formed (n=2S+1)
11 11
|

22 22
1  11 1 1
1 1 11 
| 10 
|

|



|
 |


2 2 22 
2 22 2 2
1 1 1 1
| 1  1 |  |  
2 2 2 2
1  11 1 1
1 1 11 
| 00 
|    |  |

|
2 2 22 
2 22 2 2
| 11 |
triplet
singlet
●
(5.41)
(5.42)
If nature chooses to use a multiplet, it must use
all ”members”.

Eg deuteron (np) – 3 x spin 1 states (SZ=-1,0,1)

No stable spin 0 np state
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SU(2) - isospin
●
Proton and neutron have similar masses:
mp  0.9383 GeV mn  0.9396 GeV
●
Heisenberg postulated that proton and neutron are two states
of the same particle – isospin doublet. Mass differences due to
electromagnetic effects.
11
1 1
p |
 n |  
(5.34)
22
2 2
| I I 3  I  isospin quantum number,
I 3  quantum number for 3rd component of isospin.
●
Proton and neutron have different projections in internal
”isospin space”.
●
Strong force invariant under rotations in isospin space.
●
Isospin conserved for strong interactions (Noether’s theorem).
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Testing isospin
●
●
Best way to understand anything is to look at physical situations
Two approaches to testing the hypothesis that isospin is a
symmetry of the strong force.

Isospin invariance
●
●

Isospin conservation
●
●
If a strong reaction/decay takes place then the reaction/decay of the isospinrotated particles must also happen at the same rate.
If a particle within an isospin multiplet is found in nature then the other
multiplet members must also exist since they correspond to different
projections in isospin space and the strong force is invariant to a rotation in
isospin space.
Isospin quantum numbers must ”add up” when considering
reactions/decays.
Both approaches are complementary (Noether’s theorem)
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Isospin multiplets
If the electromagnetic field could be turned off
the masses of the particles within the isospin
multiplets would be the same according to
isospin symmetry.
p (9383 MeV)
11
22
n (9396 MeV)
1 1

2 2
 140 MeV) 1 1
0 135 MeV) 1 0
 140 MeV) 1  1
I3
Multiplicity of states: N=2I+1 (5.43)
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(5.44)
(5.45)
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Rotation in isospin space
Slightly more interesting than rotations in real space.
A rotation by 180o around the ' y ' -axis in isospin space
converts a proton  neutron and      .
Can measure:
(a) p  p  d    and (b) n  n  d   
(a) and (b) are the 'same' reaction as seen
by the strong force if isospin is a good symmetry.
Strong reaction rates are measured to be the same.
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Question
Consider reactions (a), (b) and (c)
(a) p  p  d    and (b) n  n  d    (c) p  n  d   0 .
(i) Verify that I 3 is conserved for each reaction.
(ii) Estimate the ratio of the reaction rate (a):(b):(c) for identical experimental conditions
i.e. same energies, momenta etc, the only differences are the interacting particles.
1
1
1 1
1 1

 0  1 (b) - -  0  1 (c)   0  0
2
2
2 2
2 2
Ok, processes can happen but at what relative rates ?
(i) Sum I 3 : (a)
(ii ) (a)
(b)
1 1 1 1
1 1
2 2 2 2
1 1 1 1
-  1 -1
2 2 2 2
(LHS)
(LHS)
0 0 1 1  1 1 (RHS)
0 0 1 -1  1 -1 (RHS)
1 1 1 1
1
- 
 1 0  0 0  (LHS) 0 0 1 0  1 0 (RHS)
2 2 2 2
2
(When it gets complicated use Clebsch-Gordan co-efficients to do the above)
Only the I  1 piece is important since isospin must be conserved.
1
 Ratio of amplitudes: M a : M b : M c  1:1:
.
2
1
2
2
2
 Ratio of rates/cross section: M a : M b : M c  1:1: .
2
(c)
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Gell-Mann-Nishijima Formula
1
For all hadrons and quarks: Q  I 3  (B  S  C  B) (5.36)
2
Q  charge ; I 3  3rd component of isospin ; B  Baryon number
S  strangeness  ns  ns , C  charmness  nc  nc , B  "bottomness"  nb  nb (4.01)
(no need to consider "top" hadrons - they don't exist.)
Baryon octet (spin ½)
Meson nonet (spin 0)
-1
-½
½
+1
-1
I3
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½
+1
I3
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Question
The decay  c   c   0 takes place with a rate typical of the strong force.
The isosinglet  c has a quark content udc.
Deduce the quark content and values of the quantum numbers of  c .
Does it possess any isospin partners ?
If so, what is their quark content ?
Verify that isospin is conserved in the above decay.
Strong rate  flavour isn't violated!!
 c has content udc, B  1, C  1, B  S  0
1
From 5.36 Q  I 3  (B  S  C  B)
2
 c is an isosinglet  0 0 ,  0 : 1 0
1
 : I 3  1  1  1  0 ,I  1 1 0
2
Part of a Multiplet:  c  udc   1 0  ,  c0  ddc   1 -1  ,  c  uuc   1 1 
Mass  2520 (from tables)

c
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Conserved quantities/symmetries
Quantity
Strong
Weak
Electromagnetic
Energy



Linear momentum



Angular momentum



Baryon number



Lepton number



Isospin

-
-
Flavour (S,C,B)

-

Charges (em, strong
and weak forces)



Parity (P)

-

C-parity (C)

-

G-parity (G)

-
-
CP

-

T

-

CPT



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Summary
●
●
Symmetry: i.e. under a given transformation certain
observable aspects of a system are invariant to the
transformation.
If the Hamiltonian of a system is invariant to a
symmetry operation a conservation law is obtained.

●
●
Noether’s theorem
Isospin is an algebraic copy of spin but covering an
internal ”isospin space”
Isospin symmetry a powerful way to understand the
masses of particles and their reactions.

Isospin violation/conservation can be studied by considering
a rotation in isospin space or counting a conserved quantum
number.
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