SYMMETRIES in PHYSICS

SYMMETRIES in PHYSICS 

Jerzy DUDEK 

Institute for Subatomic Research 

University of Strasbourg I 

2nd October 2007 

Jerzy DUDEK 

SYMMETRIES in PHYSICS

Symmetry and Groups: Historical Aspects 

Part I 


Jerzy DUDEK 



Symmetry in Physics: A Short History 

Group Theory: First Concepts 

The Origin and the Meaning of Word: Symmetry 

The word symmetry (συµµɛτρια) originates from the Greek 

language: συµ (’together’) and µɛτρων (’measure’) 

The implied meaning: ’measured together’, well proportioned 

The meaning has evolved in time into: beauty, unity, harmony 

In sciences the first meaning of the word symmetry was that 

related to proportions 

Todays meaning as the equality of elements under geometrical 

transformations (translations, rotations, reflexions) arrived only 

towards the end of the Renaissance 

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The implied meaning: ’measured together’, well proportioned 







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The implied meaning: 

’measured together’, well proportioned 







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Symmetry: Plato (428 - 347 BC) and even earlier ... 

The symbol of ’beauty in symmetry’ are five Platonic Figures 

Platonic Figures: Three-dimensional polyhedra whose faces are 

identical planar regular convex polygons 

The allowed polygons are either equilateral triangles, or squares 

or regular pentagons 

There exist only five regular convex (=platonic) polyhedra: 

tetrahedron, cube, octahedron, icosahedron & dodecahedron 

As it seems, neolithic people from Scotland have developed the 

five Platonic solids about 1000-3000 years before Plato (stone 

models in Ashmolean Museum, Oxford) 

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Models of Platonic Figures 



Recall: Platonic Figures = Polyhedra whose 

faces are identical regular convex polygons 

Allowed polygons are equilateral triangles, or 

squares or regular pentagons 

Plato 

Tetrahedron, Cube, Octahedron, Icosahedron, Dodecahedron 

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Plato 


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Plato 


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Plato 


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Fascination by Symmetry of Platonic Figures: Kepler 

A. Many natural scientists and philosophers were 

fascinated by symmetry of Platonic figures 

B. One of the first ones in the modern times was 

Johannes Kepler 

Kepler’s Planetary System 


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Kepler ”Theory” of Planetary Constellation [Failure] 

The Earth orbit is the measure of all orbits 

Around it we circumscribe the dodecahedron 

The Mars orbit is circumscribed around the dodecahedron 

Around the Mars orbit we circumscribe the tetrahedron 

The Jupiter orbit is circumscribed around the tetrahedron 

Around of the Jupiter orbit we circumscribe the cube 

The Saturn orbit lies in the sphere surrounding the cube 

In the Earth orbit the regular icosahedron is inserted 

The orbit entered in it is the Venus orbit 

In the Venus orbit the octahedron is inserted 

The orbit entered in it is the Mercury orbit 

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Paper on Snow-Crystal Symmetry (N o 1 in History) 

In the winter of 1609 Kepler got caught by the snow-storm on 

one of the bridges in Prague 

In 1611 he published an article ’Strena seu de nive sexangula’ 

examining the planar hexagonal symmetry 1 

Examples of Hexagonal Symmetry 2 

1 ’Strena seu de nive sexangula’ - ’On the Six-Cornered Snowflakes’; 

2 From Kenneth K. Librecht, Caltech 

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Multiple Examples of the Hexagonal Symmetry 

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What Kepler Could Not Know ... [1] 

The Secret of Hexagonal Symmetry of Snow Crystals? 

Figure: The water molecules in an ice crystal form a hexagonal lattice. 

Each red ball represents an oxygen atom, the grey sticks represent two 

hydrogen atoms. 

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Other Forms of Symmetry: 12-Fold and Cylindrical 

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Empirical Systematics of Nakaya 

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Crystal Growing, Humidity and Temperature 

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Observations from a Certain Time-Perspective: 

Despite all (human) efforts the understanding of nature is a 

long process 

There were many hypotheses and scientific quarrels about this 

particular and many other puzzles of nature in the past ... 

It is a good advice to remeber the question: 

Are we sure? 

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long process 




Are we sure? 

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long process 




Are we sure? 

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Symmetry: How Did It All Begin? 

Kepler (1611) - observes hexagonal symmetry of snow crystals 

Steno (1669) - observes that inclination angles of faces of the 

quartz crystals are the same → independent of the crystal size 

De Lisle (1783) - angles between crystal faces identify crystals 

Haüy (1784) - studies the symmetries: rotations & reflections 

Wollaston (1809) - first goniometer to measure crystal angles 

Hausmann (1821) - introduces the spherical trigonometry 

Hessel (1830) - shows existence of 32 polyhedral symmetries 

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The Idea and Some Properties of Permutations 

Joseph-Louis Lagrange (1736-1813) 

J. L. Lagrange 

A. Lagrange studied (among many other things) 

algebraic equations 

B. He introduced the notion of permutations in 

relation to solutions of these equations ... 

C. ... however, he did not develop it further - this 

will come only later ... 

During the years 1772-1785 Lagrange created a series of memoirs about 

’Differential Equations’, he developed calculus of variations; in 1788 he 

published his ’Mécanique analytique’ including the ’Multiplier method’ 

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The Concept of Permutation Group 



Evariste Galois (1811-1832) 

E. Galois 

A. One of the most adventurous figures in the 

world of mathematics of XIX century 

B. His works were judged incomprehensible by the 

contemporary - among others by Poisson ... 

C. ... they were correctly interpreted by Liouville, 

published in 1846, 24 years after his death ... 

Figure: A fragment of the original text written the night before the duel 

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E. Galois 








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E. Galois 








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E. Galois 








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Another Young Genius 



Niels Abel (1802-1829) 

A. Another tragic figure; he left behind important 

mathematical ideas despite his very short life 

B. Today we find in literature Abel’s integral 

equations, Abelian functions, Abel theorem 

about convergence; commutative groups are 

called Abelian 

N. Abel 

Abel died in the age of 26 of tuberculosis and malnutrition; only two 

days after his death the letter offering him a teaching position in Berlin 

arrived in Norway ... 

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Niels Abel (1802-1829) 







N. Abel 




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Niels Abel (1802-1829) 







N. Abel 




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Niels Abel (1802-1829) 







N. Abel 




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Further Development in Studies of Permutations 

Augustin Louis CAUCHY (1789-1857) 

A. L. Cauchy 

A. One of the most successful mathematicians of 

his times, pioneered studies of infinite series, 

differential equations, probability ... 

B. He undertook the studies of the properties of 

permutations essential for the development of 

group theory 

It was Arthur CAYLEY who, after CAUCHY, follows the study of groups 

of permutations - the only groups known at that time ... 

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A. L. Cauchy 






group theory 



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A. L. Cauchy 






group theory 



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A. L. Cauchy 






group theory 



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Arrival of the Definition of Abstract Groups 

Arthur CAYLEY (1821-1895) 

A. Defines for the first time the concept of an 

abstract group 

B. Introduces the idea of Group Multiplication 

Table 

C. Observes that matrices (and quaternions) 

form groupes 

A. Cayley 

A. CAYLEY, graduated from Cambridge, spent several year as lawyer, 

later became professor of pure mathematics at Cambridge. With over 

900 publications one of the most active mathematicians of his times. 

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Table 


form groupes 

A. Cayley 




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Table 


form groupes 

A. Cayley 




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Table 


form groupes 

A. Cayley 




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Table 


form groupes 

A. Cayley 




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Abstract Groups - Today’s Formulation 

One of the most powerful tools to study transformation properties 

and symmetries in physics is the theory of groups. 

Definition (Group, CAYLEY) 

Abstract elements g ∈ G form a group under the operation ”◦” if: 

1 o For any g 1 , g 2 ∈ G the ’product’ g 1 ◦ g 2 ≡ g ∈ G 

2 o For any g 1 , g 2 , g 3 ∈ G we have (g 1 ◦ g 2 ) ◦ g 3 = g 1 ◦ (g 2 ◦ g 3 ) 

3 o There exists a neutral element e ∈ G: ∀ g ∈ G : e ◦ g = g 

4 o For any g ∈ G we find inverse g −1 ∈ G such that g ◦ g −1 = e 

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Very Special Role of the Group of Permutations 

One of the most powerful tools to study finite group properties is 

the Theorem of Cayley. 

Theorem (CAYLEY) 

Every group G composed of n elements is necessarily isomorphic 

with an n-element sub-group of the group S n 

Conclusion: 

It is sufficient to examine the group structure of the permutation 

group - the structure of all other ones is the same !!! 

Example: Group S 3 

j„ 

df . 1, 2, 3 

S 3 = 

1, 2, 3 

« „ 

1, 2, 3 

, 

2, 1, 3 

« „ 

1, 2, 3 

, 

1, 3, 2 

« „ 

1, 2, 3 

, 

3, 2, 1 

« „ 

1, 2, 3 

, 

2, 3, 1 

« „ 

1, 2, 3 

, 

3, 1, 2 

«ff 

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Conclusion: 




j„ 

df . 1, 2, 3 

S 3 = 

1, 2, 3 

« „ 

1, 2, 3 

, 

2, 1, 3 

« „ 

1, 2, 3 

, 

1, 3, 2 

« „ 

1, 2, 3 

, 

3, 2, 1 

« „ 

1, 2, 3 

, 

2, 3, 1 

« „ 

1, 2, 3 

, 

3, 1, 2 

«ff 

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Conclusion: 




j„ 

df . 1, 2, 3 

S 3 = 

1, 2, 3 

« „ 

1, 2, 3 

, 

2, 1, 3 

« „ 

1, 2, 3 

, 

1, 3, 2 

« „ 

1, 2, 3 

, 

3, 2, 1 

« „ 

1, 2, 3 

, 

2, 3, 1 

« „ 

1, 2, 3 

, 

3, 1, 2 

«ff 

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Conclusion: 




j„ 

df . 1, 2, 3 

S 3 = 

1, 2, 3 

« „ 

1, 2, 3 

, 

2, 1, 3 

« „ 

1, 2, 3 

, 

1, 3, 2 

« „ 

1, 2, 3 

, 

3, 2, 1 

« „ 

1, 2, 3 

, 

2, 3, 1 

« „ 

1, 2, 3 

, 

3, 1, 2 

«ff 

Jerzy DUDEK 


Symmetry and Spontaneous Symmetry Breaking 

Part II 


Jerzy DUDEK 



Selected Groups and Symmetries 

Spontaneous Symmetry Breaking 

Identical Particles in 3D-Spaces and Permutations 

Given system with n identical particles ↔ H = H (x 1 , x 2 , . . . x n ) 

Particles being identical - Hamiltonian must be totally symmetric 

P ij H(x 1 , . . . x i , . . . x j , . . . x n ) P −1 

ij 

= H(x 1 , . . . x j , . . . x i , . . . x n ) 

and H(x 1 , . . . x i , . . . x j , . . . x n ) = H(x 1 , . . . x j , . . . x i , . . . x n ) 

so that 

P ij H P −1 

ij 

= H → [P ij , H] = 0, ∀ i ≠ j ≤ n 

Implications: 

1 o Operators H and P can be diagonalised simultaneously 

2 o H |Ψ〉 = E |Ψ〉 → P ij Ĥ P −1 

ij 

(|P ij Ψ〉) = E (|P ij Ψ〉) 

... consequently, if |Ψ〉 is an eigenstate of H so is |P ij Ψ〉 

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P ij H(x 1 , . . . x i , . . . x j , . . . x n ) P −1 

ij 

= H(x 1 , . . . x j , . . . x i , . . . x n ) 


so that 

P ij H P −1 

ij 

= H → [P ij , H] = 0, ∀ i ≠ j ≤ n 

Implications: 


2 o H |Ψ〉 = E |Ψ〉 → P ij Ĥ P −1 

ij 

(|P ij Ψ〉) = E (|P ij Ψ〉) 


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P ij H(x 1 , . . . x i , . . . x j , . . . x n ) P −1 

ij 

= H(x 1 , . . . x j , . . . x i , . . . x n ) 


so that 

P ij H P −1 

ij 

= H → [P ij , H] = 0, ∀ i ≠ j ≤ n 

Implications: 


2 o H |Ψ〉 = E |Ψ〉 → P ij Ĥ P −1 

ij 

(|P ij Ψ〉) = E (|P ij Ψ〉) 


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P ij H(x 1 , . . . x i , . . . x j , . . . x n ) P −1 

ij 

= H(x 1 , . . . x j , . . . x i , . . . x n ) 


so that 

P ij H P −1 

ij 

= H → [P ij , H] = 0, ∀ i ≠ j ≤ n 

Implications: 


2 o H |Ψ〉 = E |Ψ〉 → P ij Ĥ P −1 

ij 

(|P ij Ψ〉) = E (|P ij Ψ〉) 


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P ij H(x 1 , . . . x i , . . . x j , . . . x n ) P −1 

ij 

= H(x 1 , . . . x j , . . . x i , . . . x n ) 


so that 

P ij H P −1 

ij 

= H → [P ij , H] = 0, ∀ i ≠ j ≤ n 

Implications: 


2 o H |Ψ〉 = E |Ψ〉 → P ij Ĥ P −1 

ij 

(|P ij Ψ〉) = E (|P ij Ψ〉) 


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P ij H(x 1 , . . . x i , . . . x j , . . . x n ) P −1 

ij 

= H(x 1 , . . . x j , . . . x i , . . . x n ) 


so that 

P ij H P −1 

ij 

= H → [P ij , H] = 0, ∀ i ≠ j ≤ n 

Implications: 


2 o H |Ψ〉 = E |Ψ〉 → P ij Ĥ P −1 

ij 

(|P ij Ψ〉) = E (|P ij Ψ〉) 


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P ij H(x 1 , . . . x i , . . . x j , . . . x n ) P −1 

ij 

= H(x 1 , . . . x j , . . . x i , . . . x n ) 


so that 

P ij H P −1 

ij 

= H → [P ij , H] = 0, ∀ i ≠ j ≤ n 

Implications: 


2 o H |Ψ〉 = E |Ψ〉 → P ij Ĥ P −1 

ij 

(|P ij Ψ〉) = E (|P ij Ψ〉) 


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P ij H(x 1 , . . . x i , . . . x j , . . . x n ) P −1 

ij 

= H(x 1 , . . . x j , . . . x i , . . . x n ) 


so that 

P ij H P −1 

ij 

= H → [P ij , H] = 0, ∀ i ≠ j ≤ n 

Implications: 


2 o H |Ψ〉 = E |Ψ〉 → P ij Ĥ P −1 

ij 

(|P ij Ψ〉) = E (|P ij Ψ〉) 


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Identical Particles Must Be either Fermions or Bosons! 

From definition of the permutation operator it follows that P 2 

ij 

= 1 

while P ij Ψ = p ij Ψ → P 2 

ij Ψ = p 2 

ij Ψ ↔ p 2 

ij = 1 → p ij = ±1 

In the particle-number representation we may write for short 

df . 

Ψ 1,2, ... n = Ψ(x 1 , x 2 , . . . x n ) 

Conclusion: We thus Discover the Pauli Principle ! 

1 o P ij Φ n1 , ... n i , ...n j , ... n n 

= +Φ n1 , ... n j , ...n i , ... n n 

→ Bosons 

2 o P ij Ψ n1 , ... n i , ...n j , ... n n 

= −Ψ n1 , ... n j , ...n i , ... n n 

→ Fermions 

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ij 

= 1 


ij Ψ = p 2 

ij Ψ ↔ p 2 

ij = 1 → p ij = ±1 


df . 

Ψ 1,2, ... n = Ψ(x 1 , x 2 , . . . x n ) 


1 o P ij Φ n1 , ... n i , ...n j , ... n n 

= +Φ n1 , ... n j , ...n i , ... n n 

→ Bosons 

2 o P ij Ψ n1 , ... n i , ...n j , ... n n 

= −Ψ n1 , ... n j , ...n i , ... n n 

→ Fermions 

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ij 

= 1 


ij Ψ = p 2 

ij Ψ ↔ p 2 

ij = 1 → p ij = ±1 


df . 

Ψ 1,2, ... n = Ψ(x 1 , x 2 , . . . x n ) 


1 o P ij Φ n1 , ... n i , ...n j , ... n n 

= +Φ n1 , ... n j , ...n i , ... n n 

→ Bosons 

2 o P ij Ψ n1 , ... n i , ...n j , ... n n 

= −Ψ n1 , ... n j , ...n i , ... n n 

→ Fermions 

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ij 

= 1 


ij Ψ = p 2 

ij Ψ ↔ p 2 

ij = 1 → p ij = ±1 


df . 

Ψ 1,2, ... n = Ψ(x 1 , x 2 , . . . x n ) 


1 o P ij Φ n1 , ... n i , ...n j , ... n n 

= +Φ n1 , ... n j , ...n i , ... n n 

→ Bosons 

2 o P ij Ψ n1 , ... n i , ...n j , ... n n 

= −Ψ n1 , ... n j , ...n i , ... n n 

→ Fermions 

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ij 

= 1 


ij Ψ = p 2 

ij Ψ ↔ p 2 

ij = 1 → p ij = ±1 


df . 

Ψ 1,2, ... n = Ψ(x 1 , x 2 , . . . x n ) 


1 o P ij Φ n1 , ... n i , ...n j , ... n n 

= +Φ n1 , ... n j , ...n i , ... n n 

→ Bosons 

2 o P ij Ψ n1 , ... n i , ...n j , ... n n 

= −Ψ n1 , ... n j , ...n i , ... n n 

→ Fermions 

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ij 

= 1 


ij Ψ = p 2 

ij Ψ ↔ p 2 

ij = 1 → p ij = ±1 


df . 

Ψ 1,2, ... n = Ψ(x 1 , x 2 , . . . x n ) 


1 o P ij Φ n1 , ... n i , ...n j , ... n n 

= +Φ n1 , ... n j , ...n i , ... n n 

→ Bosons 

2 o P ij Ψ n1 , ... n i , ...n j , ... n n 

= −Ψ n1 , ... n j , ...n i , ... n n 

→ Fermions 

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Arrival of the Theory of Continuous (Lie) Groups 

Sophus LIE (1842-1899) 

A. Introduces new type of groups whose elements 

are continuous functions of some parameter(s) 

S. LIE 

B. Most important for physics applications are 

groups of matrices 

C. Examples: GL n , SL n , U n , SU n , O n , Sp 2n ... 

Distinguished role in physics play the space-time Lie groups: Rotation, 

Lorentz and Poincaré, dimensions 3, 6 and 10, respectively, as well as 

the unitary and special unitary groups together with all their sub-groups 

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Sophus LIE (1842-1899) 

S. LIE 









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Sophus LIE (1842-1899) 

S. LIE 









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Sophus LIE (1842-1899) 

S. LIE 









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About Special Unitary Groups: SU n 

Groups SU n ↔ n × n unitary matrices U ↔ [det(U) = 1] 

It turns out that these matrices can be expressed as 

U = exp [ ∑ ] 

p αβ ĝ αβ , pαβ ↔ real parameters 

αβ 

Constant operators (generators) ĝ αβ satisfy 

[ ĝ αβ , ĝ γδ ] = δ βγ ĝ αδ − δ αδ ĝ γβ 

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U = exp [ ∑ ] 


αβ 



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U = exp [ ∑ ] 


αβ 



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N-Body Systems with Two-Body Interactions 

Consider an N-particle system with a two-body Hamiltonian 

Ĥ = ∑ αβ 

h αβ c α + c β + 1 ∑ ∑ 

v αβ;γδ c α + c + 

2 

β c δ c γ 

αβ 

γδ 

Introduce operators 

ˆN αβ 

df . 

= c + α c β 

It is easy to verify that 

[ ˆN αβ , ˆN γδ ] = δ βγ ˆN αδ − δ αδ ˆN γβ thus ˆN αβ ↔ ĝ αβ 

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Ĥ = ∑ αβ 

h αβ c α + c β + 1 ∑ ∑ 


2 

β c δ c γ 

αβ 

γδ 


ˆN αβ 

df . 

= c + α c β 



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Ĥ = ∑ αβ 

h αβ c α + c β + 1 ∑ ∑ 


2 

β c δ c γ 

αβ 

γδ 


ˆN αβ 

df . 

= c + α c β 



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N-Body Hamiltonians and the SU n -Generators 

N-Body Hamiltonians are functions of SU n -group generators 

Two-body interactions lead to quadratic forms of ˆN αβ = c + α c β 

, 

three-body interactions to the cubic forms of ˆN αβ , etc. 

Hamiltonians of the N-body systems can be diagonalised within 

bases of the irreducible representations 

Solutions can be constructed that transform as the SU n -group 

representations thus establishing a link H ↔ SU n 

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N-Body Hamiltonians and More General Observations 

Consequences in Terms of Chains of Sub-Groups 

Group SU n has numerous sub-groups: U m and SU m with m < n, 

similarly O m , SO m and in particular R(3) and all the point groups 

More Generall Intellectual Challenges 

If a group is a symmetry group of a physical system - so are its 

sub-groups! Do we know their physical meaning? ... 

Consequences?... Interpretation? 

... but even if not ... 

The group theory offers a guaranteed guidance through the jungle 

of possibilities! 

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Subgroup Structure Can Be Very, Very Rich ... 

32 Point Groups: Subgroups 

. 

D 4h 

T h 

O T d 

D 6h 

C 4h 

D 4 

D 2d 

C 4v 

D 2h 

T D 6 

C 6h 

C 6v 

D 3d 

D 

3h 

S 4 

C 4 

D 2 

C 2h 

C 2v 

C 6 

C 3i 

D 3 

C 3v 

C 3h 

O h 

C C C C 

i 2 s 3 

Figure: Richness of the sub-group 

structures at the end of chain... 

C 

1 

. 

Dashed lines indicate that the 

subgroup marked is not invariant 

Trivial groups are denoted here 

C 1 ≡ {1I}, C s ≡ {1I, ˆσ}, 

C i ≡ {1I, ˆπ} 

Here we show the structure only 

at the very end of the SU n chain 

- it helps imagining how rich the 

full group structure is ... 

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Point Groups - Examples from Molecular Physics 

Group C 3v - Molecule: [NH 3 ] Group C 3h - Molecule: [B(OH) 3 ] 

Group D 3d - Molecule: [C 2 H 6 ] Group D 3h - Molecule: [C 2 H 6 ] 

From J. Goss, University of Newcastle 

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The Highest Symmetries in Molecular Physics 

Group T d - Molecule: [CH 4 ] Group O h - Molecule: [SF 6 ] 

Group D 6d - Mol.: [Cr(C 6 H 6 ) 2 ] Group I h - Molecule: [C 60 ] 

From J. Goss, University of Newcastle 

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Strong Interactions and Sub-Atomic Physics 

Except for the stellar objects (for which N → ∞) the subatomic 

objects are mesoscopic: N max ∼ 10 2 ≪ Avogadro Number ∼ 10 23 

Standard Models of Nuclear Interactions Symmetries 

1 o Mean-Field Methods SU n , SO n , point groups 

2 o Nuclear Shell-Model SU n , SO n , point groups 

3 o Nuclear Cluster Models point groups 

4 o Compound (Interacting) Bosons group algebras 

5 o Collective Nuclear Models point groups 

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Symmetry vs. Asymmetry - Historical Perspective [1] 

Principle of Sufficient Reason (LEIBNIZ) 

If there is no sufficient reason for things to happen - then the 

original situation does not change 

The underlying idea: the presence of symmetry guarantees that 

one choice is as good as any other - thus nothing happens 

Pierre CURIE (’Sur la symétrie dans les phénomènes physiques’) 

For a phenomenon to take place - an absence of symmetry is 

needed - the asymmetry gives the origin to phenomena 

If one phenomenon is a ’cause’ and another ’result’ - the symmetry 

of cause must be higher than the symmetry of the result 

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Caricature - A Classical (Macro) Example: Buridan’s Ass 

Figure: Buridan’s ass, having two strictly identical bundles of carrots on 

both sides has no reason to select one of the two - and dies of starvation 

Jerzy DUDEK 






... already Aristotle (384-322 BC) 

In his ’De Caelo’ Aristotle asks how a dog faced with the choice 

of two equally tempting meals could rationally choose one ... 

French philosopher Jean BURIDAN, XIV century: an ass faces 

starvation when it is unable to choose between two equally 

appetising piles of hay ... 

Remarks: 

The ideas of Leibniz and Curie do not necessarily correspond to the 

present day view - today the spontaneous symmetry breaking effects 

are given much more attention 

Jerzy DUDEK 






... already Aristotle (384-322 BC) 

In his ’De Caelo’ Aristotle asks how a dog faced with the choice 

of two equally tempting meals could rationally choose one ... 

French philosopher Jean BURIDAN, XIV century: an ass faces 

starvation when it is unable to choose between two equally 

appetising piles of hay ... 

Remarks: 

The ideas of Leibniz and Curie do not necessarily correspond to the 

present day view - today the spontaneous symmetry breaking effects 

are given much more attention 

Jerzy DUDEK 





An Alternative: Spontaneous Symmetry Breaking 

Aren’t there ”Some Miracles” Hidden Somewhere? 

1 There exist phenomena not involving any obvious symmetry 

2 There exist transitions from less symmetric to more symmetric 

situations 

3 Let us turn to a somewhat misleading picture of s. symmetry 

breaking 

Weinberg’s Caricature of the Spontaneous Symmetry Breaking 

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situations 




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Symmetry: Exact, Approximate, Spontaneously Broken 

Consider an exact symmetry defined by transformations ĝ ∈ {G}: 

ĝ H ĝ −1 = H ↔ Hψ = Eψ → ĝHĝ −1 (ĝψ) = E(ĝψ) 

Conclusion: ψ and ĝψ are solutions. There are two possibilities: 

1 o An n-fold degenerate spectrum: ĝψ p α = ∑ k≤n ψk α D α kp (ĝ) 

Matrices Dkp α (ĝ) form an n-dimensional representation of {G} 

2 o Non-degenerate spectrum: ĝψ p α = D α (ĝ) ψ k q 

Numbers D α (ĝ) form one-dimensional representations of {G} 

In both cases the index α enumerating irreducible representations 

plays a role of the label (quantum number) characterising symmetry 

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Example: An Exactly Spherically-Symmetric Nucleus 

Consider 132 Sn nucleus with 50 protons and 82 neutrons 

Are the wave-functions obeying the spherical symmetry? 

How the Pauli principle can be seen from the 3D perspective? 

Hint: 

To answer these questions we solve the corresponding realistsic problem 

of motion and obtain the energies and wave functions 

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Spatial Structure of Orbitals (Sperical 132 Sn) (|ψ(⃗r )| 2 ) 

Limit 80% Limit ??% Limit ??% Limit ??% Limit ??% 

Density distribution |ψ π (⃗r )| 2 ≥ Limit, for π = [2, 0, 2]1/2 orbital 

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Limit 80% Limit 50% Limit ??% Limit ??% Limit ??% 


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Limit 80% Limit 50% Limit 10% Limit ??% Limit ??% 


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Limit 80% Limit 50% Limit 10% Limit 3% Limit ??% 


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Limit 80% Limit 50% Limit 10% Limit 3% Limit 1% 


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Limit 20% Limit ??% Limit ??% Limit ??% Limit ??% 

Bottom: N=3 shell b-[303]7/2, w-[312]5/2, y-[321]3/2, p-[310]1/2 

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Limit 20% Limit 15% Limit ??% Limit ??% Limit ??% 


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Limit 20% Limit 15% Limit 12% Limit ??% Limit ??% 


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Limit 20% Limit 15% Limit 12% Limit 10% Limit ??% 


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Spatial Structure of N=3 Spherical Shell (|ψ ν (⃗r )| 2 ) 

132 Sn: Distributions |ψ ν (⃗r )| 2 for single proton orbitals. Top O xz , 

bottom O yz . Proton e ν ↔ [ν=30, 32, ... 38] for spherical shell 

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132 Sn: Distributions |ψ ν (⃗r )| 2 for single proton orbitals. Top O xz , 

bottom O yz . Proton e ν ↔ [ν=40, 42, ... 48] for spherical shell 

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132 Sn: distributions |ψ ν (⃗r )| 2 for consecutive pairs of orbitals. Top 

O xz , bottom O yz . Proton e ν ↔ [n=30:32, ... 38:40], spherical shell 

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132 Sn: distributions |ψ ν (⃗r )| 2 for consecutive pairs of orbitals. Top 

O xz , bottom O yz . Proton e ν ↔ [n=40:42, ... 48:50], spherical shell 

Jerzy DUDEK 



What Did We Learn? 



In the 132 Sn nucleus with 50 protons and 82 neutrons: 

... the wave functions do not obey the spherical symmetry ... 

They transform under irreducible representations of the SO 3 

Calculation shows that the energy eigen-values are degenerate: 

(degeneracy equal to 2j+1) 

Question: What happens if we continue adding nucleons to 132 Sn? 

Experiments will show that the resulting systems are NOT spherically 

symmetric! Strange! - since the Hamiltonian of the system obeys 

the rotational symmetry exactly!! Let us have closer look... 

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Fundamental Symmetries of Nuclear Forces [1] 

Denote ˆx df . = {⃗r,⃗p,⃗s,⃗t}. Nuclear interactions have the form 

̂V (ˆx 1 , ˆx 2 ) ≡ ̂V C (ˆx 1 , ˆx 2 ) + ̂V T (ˆx 1 , ˆx 2 ) + ̂V LS (ˆx 1 , ˆx 2 ) + ̂V LL 2(ˆx 1 , ˆx 2 ) 

where: C-central, T -tensor, LS-spin-orbit and LL 2 -quadratic LS 

Central Interaction (r 12 ≡ |⃗r 1 −⃗r 2 |) 

̂V C (ˆx 1 , ˆx 2 ) = V 0 (r 12 ) + V s (r 12 ) [⃗s (1) · ⃗s (2) ] 

+ V t (r 12 ) [⃗t (1) ·⃗t (2) ] 

+ V s−t (r 12 ) [⃗s (1) · ⃗s (2) ] [⃗t (1) ·⃗t (2) ] 

Invariant under rotations, translations, inversion and time-reversal 

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Central Interaction (r 12 ≡ |⃗r 1 −⃗r 2 |) 

̂V C (ˆx 1 , ˆx 2 ) = V 0 (r 12 ) + V s (r 12 ) [⃗s (1) · ⃗s (2) ] 

+ V t (r 12 ) [⃗t (1) ·⃗t (2) ] 

+ V s−t (r 12 ) [⃗s (1) · ⃗s (2) ] [⃗t (1) ·⃗t (2) ] 


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Tensor Interaction [Non-Central] 

⃗ S 

(12) df . 

= 3 (⃗s 1 ·⃗r 12 )(⃗s 2 ·⃗r 12 ) − (⃗s 1 · ⃗s 2 ) r12 

2 

r12 

2 

and r 12 

df . 

= |⃗r 1 −⃗r 2 | 

̂V T (ˆx 1 , ˆx 2 ) = [V t0 (r 12 ) + V t1 (r 12 )⃗t 1 ·⃗t 2 ] ⃗ S (12) 


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Spin-Orbit Interaction [Non-Local] 

⃗ L 

df . 

= 1 2 (⃗r 1 −⃗r 2 ) ∧ (⃗p 1 − ⃗p 2 ), r 12 

df . 

= |⃗r 1 −⃗r 2 | and ⃗ S df . = ⃗s 1 + ⃗s 2 

̂V LS (ˆx 1 , ˆx 2 ) = V LS (r 12 ) ⃗ L · ⃗S 


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Quadratic Spin-Orbit Interaction [Non-Local] 

⃗ L 

df . 

= 1 2 (⃗r 1 −⃗r 2 ) ∧ (⃗p 1 − ⃗p 2 ) and r 12 

df . 

= |⃗r 1 −⃗r 2 | 

̂V LL (ˆx 1 , ˆx 2 ) = V LL (r 12 ){(⃗s 1 · ⃗s 2 ) ⃗ L 2 − 1 2 [(⃗s 1 · ⃗L)(⃗s 2 · ⃗L) + (⃗s 2 · ⃗L)(⃗s 1 · ⃗L)]} 


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Nuclear Forces and the Mean Field Theory 

1 o From the above discussion it follows that the N-N interaction 


is invariant under rotations, translations, inversion and time-reversal 

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Nuclear Forces and the Mean Field Theory 

1 o From the above discussion it follows that the N-N interaction 


is invariant under rotations, translations, inversion and time-reversal 

Spont. Sym. Breaking 

Z=fixed 

E 

N N ’ N" 

The Nuclear Mean Field Theory ... 

... is usually very successful. It is based on 

∫ 

̂V mf (ˆx) = ψ ∗ (x ′ ) ̂V (ˆx, ˆx ′ )ψ(x ′ ) dx ′ 

spherical 

deformed 

Def. 

Some or all of the above symmetries will be 

broken by the mean-field Hamiltonian 

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Spontaneous Symmetry Breaking - An Example 

Given a system with symmetry {G} i.e. [H(β), G] = 0. Here β is a 

parameter. Often a critical value, β crit. , exists such that: 

1 For β < β crit. symmetry of solution is compatible with {G} 

2 For β > β crit. symmetry of solution is not compatible with {G} 

A Classical (Macro) Example 

Original system ↔ axial symmetry 

For F > F crit. we find infinitely 

many solutions at the same energy 

Yet: The original axial symmetry 

will be spontaneously broken and 

only one among many directions - 

privileged !!! 

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Spontaneous Symmetry Breaking - Classical Physics 

A small bead of mass m threaded on a rotating circle of radius r; 

constant frequency ω. We use the Lagrangian formalism: 

1 L = 1 2 m˙⃗r 2 −V (⃗r ) = 1 2 mr 2 ˙ϑ 2 + 1 2 mω2 r 2 sin 2 ϑ−mgr(1−cos ϑ) 

df . 

2 L = T ϑ − U ϑ ↔ U ϑ = mgr[ (1 − cos ϑ) − 1 

} {{ } 

2 (mω2 /g) sin 2 ϑ ] 

} {{ } 

Potential Centrifugal 

Bead on a Circle 

z 

ω 

π _ υ 

y 

r 

m 

x 

f = −mg 

U=mgr(1−cos υ) 

The lowest energy solutions ↔ ˙ϑ = 0 ↔ dU 

dϑ = 0 

Result: sin ϑ s = 0 and cos ϑ o = 1/β; β df . = rω 2 /g 

Conclusion. Two types of lowest-energy solutions: 

1. ϑ s = 0 - with the axial symmetry of the problem 

2. ϑ o ≠ 0 - breaking the original axial symmetry !! 

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df . 


} {{ } 


} {{ } 



z 

ω 

π _ υ 

y 

r 

m 

x 

f = −mg 



dϑ = 0 





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df . 


} {{ } 


} {{ } 



z 

ω 

π _ υ 

y 

r 

m 

x 

f = −mg 



dϑ = 0 





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df . 


} {{ } 


} {{ } 



z 

ω 

π _ υ 

y 

r 

m 

x 

f = −mg 



dϑ = 0 





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Spontaneous Symmetry Breaking - Discussion 

1 o Under which conditions are the so obtained solutions stable? 

2 o Are the derivatives d2 U 

dϑ 2 ≡ U ′′ = mgr(cos ϑ − β cos 2ϑ) > 0? 

1 Solution ϑ s = 0 → U ′′ (ϑ s ) = 1 − β → ϑ s stable for β < 1 

2 Solution ϑ o ≠ 0 → U ′′ (ϑ o ) = β − 1/β → ϑ o stable for β > 1 

U is Symmetric when ϑ → −ϑ 

U ( υ) 

Define frequency: ω crit. ≡ √ g/r 

ω < ω crit. : bead rests at ϑ s = 0 

β1 

ω > ω crit. : climbs up ϑ o : 0 → π 2 

_ π /2 

β crit. = 1 

π /2 

υ 

ϑ s - sub-critical ϑ o - over-critical 

... but why should there be any 

spontaneous symmetry breaking? 

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U ( υ) 



β1 


_ π /2 

β crit. = 1 

π /2 

υ 




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U ( υ) 



β1 


_ π /2 

β crit. = 1 

π /2 

υ 




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U ( υ) 



β1 


_ π /2 

β crit. = 1 

π /2 

υ 




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Spontaneous Symmetry Breaking - Interpretation [1] 

Essential Facts about our Physical System: 

1 Considered system has 1 degree of freedom and axial symmetry 

2 ... but in the generalised coordinates only 1 discrete symmetry: 

U(ϑ) = 2mgr sin 2 (ϑ/2)[1 − β cos 2 (ϑ/2)] ↔ U(+ϑ) = U(−ϑ) 

Essential Conclusions from Our Analysis: 

1 This symmetry holds for β < β crit. and β > β crit. potentials 

2 This symmetry holds only for one i.e. the β < β crit. solution 

3 For the β > β crit. solutions we have either ϑ o > 0 or ϑ o < 0 

4 For the β > β crit. solutions the original symmetry is violated 

5 We have considered only the lowest energy (’ground’) states 

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This transition is called spontaneous since the ϑ o -sign chosen by the 

system depends on ’irrelevant details’: fluctuations, non-uniformities 


1 We used no information about those ’irrelevant details’ yet ... 

2 We were able to predict the absolute value of ϑ o in experiment 

3 Is spontaneous symmetry breaking something so very unusual? 

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Jerzy DUDEK 





Symmetry Groups and Degeneracies of Levels 

Hamiltonian and a Group of Transformations 

Given Hamiltonian H and a point-group G = {O 1 , O 2 , . . . O f } 

Assume that G is a symmetry group of H i.e. 

[H, O k ] = 0 with k = 1, 2, . . . f 

Spectra and Group’s Irreducible Representations 

Let irreducible representations of G be {R 1 , R 2 , . . . R r } 

Let their dimensions be {d 1 , d 2 , . . . d r }, respectively 

Then the eigenvalues {ε ν } of the problem 

Hψ ν = ε ν ψ ν 

appear in multiplets d 1 -fold, d 2 -fold ... degenerate 

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[H, O k ] = 0 with k = 1, 2, . . . f 







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[H, O k ] = 0 with k = 1, 2, . . . f 







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[H, O k ] = 0 with k = 1, 2, . . . f 







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[H, O k ] = 0 with k = 1, 2, . . . f 







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[H, O k ] = 0 with k = 1, 2, . . . f 







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Point Groups - ... and What We Need to Remember 

Consider a group of symmetry of an equilateral triangle (group C 3v ) 

Two Types of Transformations - Observe Existence of a Fixed Point 

a 

120 o 

b 

c 

C^ 3 

240 o 

(C 

^ 2 

3) 

b 

a 

c 

Figure: This group contains rotations Ĉ3 and Ĉ 2 3 (120o and 240 o ), left, 

and three vertical reflections ˆσ va , ˆσ vb and ˆσ vc , right, in addition to 1I. 

Six Element Point Group: C 3v = {1I, Ĉ 3 , Ĉ 2 3 , ˆσ va , ˆσ vb , ˆσ vc } 

Jerzy DUDEK 





Point Groups - ... and What We Need to Remember 

Consider a group of symmetry of an equilateral triangle (group C 3v ) 

Two Types of Transformations - Observe Existence of a Fixed Point 

a 

120 o 

b 

c 

C^ 3 

240 o 

(C 

^ 2 

3) 

b 

a 

c 

Figure: This group contains rotations Ĉ3 and Ĉ 2 3 (120o and 240 o ), left, 

and three vertical reflections ˆσ va , ˆσ vb and ˆσ vc , right, in addition to 1I. 

Six Element Point Group: C 3v = {1I, Ĉ 3 , Ĉ 2 3 , ˆσ va , ˆσ vb , ˆσ vc } 

Jerzy DUDEK 



Discrete Symmetries in Nuclei 



Let us recall one of the magic forms introduced long time by Plato. 

The implied symmetry leads to the tetrahedral group denoted T d 

A tetrahedron has four equal walls. 

Its shape is invariant with respect to 

24 symmetry elements. Tetrahedron 

is not invariant with respect to the 

inversion. Of course nuclei cannot be 

represented by a sharp-edge pyramid 

... but rather in a form of a regular spherical harmonic expansion: 

λ∑ 

max λ∑ 

R(ϑ, ϕ) = R 0 c({α})[1 + α λ,µ Y λ,µ (ϑ, ϕ)] 

λ 

µ=−λ 

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Discrete Symmetries in Nuclei 




The implied symmetry leads to the tetrahedral group denoted T d 

A tetrahedron has four equal walls. 

Its shape is invariant with respect to 

24 symmetry elements. Tetrahedron 

is not invariant with respect to the 

inversion. Of course nuclei cannot be 

represented by a sharp-edge pyramid 


λ∑ 

max λ∑ 


λ 

µ=−λ 

Jerzy DUDEK 



A Basis for Tetrahedral Symmetry 



Only special combinations 1 of spherical harmonics may form a 

basis for surfaces with tetrahedral symmetry: 

Three Lowest Orders: 

Rank ↔ Multipolarity λ 

λ = 3 : α 3,±2 ≡ t 3 

λ = 7 : α 7,±2 ≡ t 7 ; α 7,±6 ≡ − q 11 

13 · t 7 

λ = 9 : α 9,±2 ≡ t 9 ; α 9,±6 ≡ + q 28 

198 · t 9 

1 Collaboration with Andrzej GÓŹDŹ and Daniel ROS̷LY, UMCS Lublin, Poland 

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λ = 3 : α 3,±2 ≡ t 3 

λ = 7 : α 7,±2 ≡ t 7 ; α 7,±6 ≡ − q 11 

13 · t 7 

λ = 9 : α 9,±2 ≡ t 9 ; α 9,±6 ≡ + q 28 

198 · t 9 


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λ = 3 : α 3,±2 ≡ t 3 

λ = 7 : α 7,±2 ≡ t 7 ; α 7,±6 ≡ − q 11 

13 · t 7 

λ = 9 : α 9,±2 ≡ t 9 ; α 9,±6 ≡ + q 28 

198 · t 9 


Jerzy DUDEK 










λ = 3 : α 3,±2 ≡ t 3 

λ = 7 : α 7,±2 ≡ t 7 ; α 7,±6 ≡ − q 11 

13 · t 7 

λ = 9 : α 9,±2 ≡ t 9 ; α 9,±6 ≡ + q 28 

198 · t 9 


Jerzy DUDEK 





Nuclear Tetrahedral Shapes - 3D Examples (Part 1) 

Illustrations below show the tetrahedral-symmetric surfaces at three 

increasing values of rank λ = 3 deformations t 1 : 0.1, 0.2 and 0.3: 

Figure: t 3 = 0.1 Figure: t 3 = 0.2 Figure: t 3 = 0.3 

Observations: 

There are infinitely many tetrahedral-symmetric surfaces 

Nuclear ’pyramids’ do not resemble pyramids! 

Jerzy DUDEK 





Nuclear Tetrahedral Shapes - Proton Spectra 

Double group Td 

D has two 2-dimensional - and one 4-dimensional 

irreducible representations → three distinct families of levels 

Proton Energies [MeV] 

-2 

-3 

-4 

-5 

-6 

-7 

-8 

-9 

-10 

-11 

-12 

226 

90 Th 136 

{07}[5,4,1] 3/2 

{19}[5,1,4] 7/2 

{07}[5,0,5] 9/2 

{20}[5,0,5] 9/2 

{08}[5,2,3] 5/2 

{06}[6,1,5] 11/2 

{10}[6,2,4] 9/2 

{08}[5,0,3] 7/2 

{06}[6,3,3] 7/2 

{04}[4,4,0] 1/2 

{12}[5,1,4] 9/2 

{09}[5,2,3] 7/2 

{16}[5,0,5] 11/2 

{08}[5,2,1] 1/2 

{11}[5,0,5] 11/2 

{09}[4,0,0] 1/2 

{08}[4,4,0] 1/2 

{16}[4,0,2] 3/2 

{13}[4,1,3] 5/2 

{10}[4,0,4] 7/2 

{15}[4,0,4] 7/2 

{09}[4,2,2] 3/2 

{07}[4,3,1] 1/2 

{09}[5,1,4] 9/2 

100 

64 

82 

58 

100 

64 

90 

70 

94 

-.2 -.1 .0 .1 .2 .3 .4 

Tetrahedral Deformation 

{04}[4,1,1] 3/2 

{06}[4,2,0] 1/2 

{06}[5,0,5] 9/2 

{05}[6,1,5] 9/2 

{07}[4,1,3] 7/2 

{07}[4,2,2] 5/2 

{11}[5,0,5] 11/2 

{06}[5,4,1] 1/2 

{05}[4,0,0] 1/2 

{04}[3,0,1] 1/2 

{04}[4,1,3] 7/2 

{03}[4,1,3] 7/2 

{04}[5,0,3] 7/2 

{10}[3,1,2] 3/2 

{05}[6,1,5] 11/2 

{05}[4,3,1] 1/2 

{04}[5,0,5] 11/2 

{06}[4,0,4] 7/2 

{06}[4,0,4] 7/2 

{06}[3,0,1] 1/2 

{04}[3,1,0] 1/2 

{03}[3,1,0] 1/2 

{10}[3,1,2] 5/2 

{06}[3,1,2] 3/2 

Strasbourg, August 2002 Woods-Saxon Universal Params. 

α 32(min)=-.200, α32(max)=.400 

Figure: Full lines ↔ 4-dimensional irreducible representations - marked 

with double Nilsson labels. Observe huge gaps at N=64, 70, 90-94, 100. 

Jerzy DUDEK 





Nuclear Tetrahedral Shapes - Neutron Spectra 

Double group Td 

D has two 2-dimensional - and one 4-dimensional 

irreducible representations → three distinct families of levels 

Neutron Energies [MeV] 

-4 

-5 

-6 

-7 

-8 

-9 

-10 

-11 

226 

90 Th 136 

{13}[6,2,4] 7/2 

{05}[6,5,1] 3/2 

{03}[6,6,0] 1/2 

{07}[6,0,2] 5/2 

{04}[6,4,0] 1/2 

{11}[6,0,6] 11/2 

{07}[6,0,6] 11/2 

{06}[6,0,4] 9/2 

{05}[7,2,5] 11/2 

{06}[7,2,5] 11/2 

{08}[6,0,4] 9/2 

{04}[7,3,4] 9/2 

{08}[6,1,5] 11/2 

{09}[6,2,4] 9/2 

{05}[6,1,5] 11/2 

{04}[6,1,5] 11/2 

{08}[6,3,1] 1/2 

{19}[6,0,6] 13/2 

{09}[6,0,6] 13/2 

{08}[5,0,1] 1/2 

{07}[5,5,0] 1/2 

{13}[5,0,3] 5/2 

{06}[5,1,2] 3/2 

{07}[4,1,1] 1/2 

{04}[5,5,0] 1/2 

148 

136 136 

126 

124 

142 

112 

-.2 -.1 .0 .1 .2 .3 .4 


{04}[6,0,6] 11/2 

{02}[5,2,1] 3/2 

{03}[5,2,1] 3/2 

{04}[5,3,2] 5/2 

{06}[6,5,1] 1/2 

{10}[6,0,6] 13/2 

{06}[5,2,3] 7/2 

{04}[5,0,5] 9/2 

{05}[5,3,2] 5/2 

{08}[5,1,4] 9/2 

{03}[5,0,1] 3/2 

{03}[5,0,1] 3/2 

{06}[6,2,4] 9/2 

{04}[7,1,6] 13/2 

{04}[4,1,3] 5/2 

{04}[5,3,0] 1/2 

{05}[7,2,5] 11/2 

{04}[5,1,2] 5/2 

{03}[5,4,1] 1/2 

{02}[4,1,3] 5/2 

{02}[3,0,1] 3/2 

{06}[6,0,6] 13/2 

{03}[5,0,3] 5/2 

{03}[5,0,3] 5/2 

{05}[4,1,1] 3/2 

Strasbourg, August 2002 Woods-Saxon Universal Params. 


Figure: Full lines ↔ 4-dimensional irreducible representations - marked 

with double Nilsson labels. Observe huge gaps at N=112, 136. 

Jerzy DUDEK 





Introducing Nuclear Octahedral Symmetry 


The implied symmetry leads to the octahedral group denoted O h 

An octahedron has 8 equal walls. Its 

shape is invariant with respect to 48 

symmetry elements that include inversion. 

However, the nuclear surface 

cannot be represented in the form of 

¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ 

§ § § § § § § § § § § § § § § § § § § § § § § § § § § 

 

¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ 

§ § § § § § § § § § § § § § § § § § § § § § § § § § § 

 

¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ 

§ § § § § § § § § § § § § § § § § § § § § § § § § § § 

 

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£ £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ 

¡ 

£ £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ 


λ∑ 

max λ∑ 


λ 

µ=−λ 

Jerzy DUDEK 



A Basis for Octahedral Symmetry 




basis for surfaces with octahedral symmetry: 



λ = 4 : α 40 ≡ o 4 ; α 4,±4 ≡ ± q 5 

14 · o 4 

λ = 6 : α 60 ≡ o 6 ; α 6,±4 ≡ − q 7 

2 · o 6 

λ = 8 : α 80 ≡ o 8 ; α 8,±4 ≡ q 28 

198 · o 8; α 8,±8 ≡ q 65 

198 · o 8 

1 Collaboration with Andrzej G ÓŹDŹ and Daniel ROS̷LY, UMCS Lublin, Poland 

Jerzy DUDEK 










λ = 4 : α 40 ≡ o 4 ; α 4,±4 ≡ ± q 5 

14 · o 4 

λ = 6 : α 60 ≡ o 6 ; α 6,±4 ≡ − q 7 

2 · o 6 

λ = 8 : α 80 ≡ o 8 ; α 8,±4 ≡ q 28 

198 · o 8; α 8,±8 ≡ q 65 

198 · o 8 


Jerzy DUDEK 










λ = 4 : α 40 ≡ o 4 ; α 4,±4 ≡ ± q 5 

14 · o 4 

λ = 6 : α 60 ≡ o 6 ; α 6,±4 ≡ − q 7 

2 · o 6 

λ = 8 : α 80 ≡ o 8 ; α 8,±4 ≡ q 28 

198 · o 8; α 8,±8 ≡ q 65 

198 · o 8 


Jerzy DUDEK 










λ = 4 : α 40 ≡ o 4 ; α 4,±4 ≡ ± q 5 

14 · o 4 

λ = 6 : α 60 ≡ o 6 ; α 6,±4 ≡ − q 7 

2 · o 6 

λ = 8 : α 80 ≡ o 8 ; α 8,±4 ≡ q 28 

198 · o 8; α 8,±8 ≡ q 65 

198 · o 8 


Jerzy DUDEK 





Nuclear Octahedral Shapes - 3D Examples 

Illustrations below show the octahedral-symmetric surfaces at three 

increasing values of rank λ = 4 deformations o 4 : 0.1, 0.2 and 0.3: 

Figure: o 4 = 0.1 Figure: o 4 = 0.2 Figure: o 4 = 0.2 

Recall: α 40 ≡ o 4 ; α 4,±4 ≡ ± q 5 

14 · o 4 

Jerzy DUDEK 





Nuclear Octahedral Shapes - Neutron Spectra 

Double group Oh 

D has four 2-dimensional and two 4-dimensional 

irreducible representations → six distinct families of levels 

Neutron Energies [MeV] 

-2 

-4 

-6 

-8 

-10 

-12 

160 

70 Yb 90 

{17}[6,1,3] 7/2 

{09}[6,4,2] 5/2 

{08}[5,0,5] 9/2 

{18}[6,1,5] 11/2 

{21}[5,0,5] 9/2 

{10}[5,4,1] 3/2 

{19}[6,3,3] 7/2 

{20}[6,1,5] 11/2 

{07}[5,4,1] 3/2 

{16}[6,2,4] 9/2 

{11}[5,0,1] 3/2 

{16}[5,4,1] 1/2 

{09}[5,0,3] 7/2 

{10}[5,3,0] 1/2 

{07}[5,0,5] 11/2 

{13}[5,2,1] 1/2 

{12}[5,0,5] 11/2 

{13}[5,0,5] 11/2 

{21}[5,1,4] 7/2 

{10}[5,2,3] 5/2 

{12}[4,0,0] 1/2 

{22}[4,4,0] 1/2 

{21}[4,0,2] 3/2 

118 

114 116 

94 94 

88 

100 

-.35 -.25 -.15 -.05 .05 .15 .25 .35 

Octahedral Deformation 

82 

126 

110 

86 

88 

{08}[6,0,6] 11/2 

{07}[6,4,0] 1/2 

{06}[8,0,2] 5/2 

{10}[6,1,3] 7/2 

{11}[6,5,1] 3/2 

{17}[6,0,6] 13/2 

{07}[5,1,4] 7/2 

{12}[5,3,2] 3/2 

{11}[5,2,1] 3/2 

{11}[5,3,0] 1/2 

{07}[8,8,0] 1/2 

{08}[4,0,2] 5/2 

{09}[6,0,4] 9/2 

{08}[5,3,2] 5/2 

{21}[5,2,3] 7/2 

{14}[5,3,2] 5/2 

{16}[5,1,4] 9/2 

{08}[5,0,5] 9/2 

{21}[4,1,3] 5/2 

{13}[4,1,1] 1/2 

{23}[4,2,2] 3/2 

{08}[5,0,5] 11/2 

{08}[5,4,1] 1/2 

Strasbourg, August 2002 Dirac-Woods-Saxon 



Figure: Full lines correspond to 4-dimensional irreducible representations - 

they are marked with double Nilsson labels. Observe huge gap at N=114. 

Jerzy DUDEK 





Nuclear Octahedral Shapes - Proton Spectra 

Double group Oh 

D has four 2-dimensional and two 4-dimensional 

irreducible representations → six distinct families of levels 

Proton Energies [MeV] 

0 

2 

-2 

-4 

-6 

-8 

-10 

-12 

160 

70 Yb 90 

{10}[5,0,5] 11/2 

{08}[5,0,3] 7/2 

{10}[5,0,3] 7/2 

{10}[5,4,1] 3/2 

{13}[5,0,5] 11/2 

{19}[4,4,0] 1/2 

{18}[4,0,2] 3/2 

{11}[4,0,0] 1/2 

{15}[5,1,4] 7/2 

{07}[5,2,3] 5/2 

{11}[5,1,2] 5/2 

{10}[5,3,2] 5/2 

{23}[5,1,4] 9/2 

{13}[4,0,4] 7/2 

{08}[4,2,2] 3/2 

{08}[4,3,1] 1/2 

{17}[5,2,3] 7/2 

{15}[4,3,1] 1/2 

{12}[4,0,2] 5/2 

{11}[4,1,3] 5/2 

{09}[4,2,0] 1/2 

{13}[4,0,4] 9/2 

{09}[4,3,1] 3/2 

72 

58 

70 

88 

52 

64 

-.35 -.25 -.15 -.05 .05 .15 .25 .35 


82 

94 

56 

88 

52 

{09}[4,0,2] 5/2 

{16}[5,3,2] 5/2 

{18}[5,1,4] 9/2 

{08}[5,0,5] 9/2 

{21}[4,1,3] 5/2 

{17}[4,1,3] 7/2 

{24}[4,2,2] 3/2 

{09}[5,1,2] 5/2 

{10}[5,0,5] 9/2 

{07}[5,1,0] 1/2 

{11}[5,4,1] 3/2 

{21}[4,2,0] 1/2 

{10}[5,0,5] 11/2 

{10}[4,3,1] 1/2 

{09}[4,2,2] 3/2 

{08}[3,3,0] 1/2 

{13}[3,0,1] 3/2 

{08}[5,0,3] 7/2 

{25}[4,2,2] 5/2 

{16}[4,1,3] 7/2 

{09}[4,3,1] 3/2 

{24}[3,1,2] 3/2 

{10}[4,3,1] 1/2 

Strasbourg, August 2002 Dirac-Woods-Saxon 



Figure: Full lines correspond to 4-dimensional irreducible representations 

- they are marked with double Nilsson labels. Observe huge gap at Z=70. 

Jerzy DUDEK 





Nuclear High-Rank Symmetries and Challenges 

There are several new-physics aspects related to the nuclear highrank 

symmetries - tetrahedral and octahedral ones! 

Properties High Symmetries ’Usual’ symmetries 

or features Tetrahedral Octahedral Ellipsoid 

No. Sym. Elemts. 48 96 4 + . . . 

Parity NO YES YES 

New Degeneracies 4, 2, 2 4, 2, 2 

} {{ } 

π = + 

4, 2, 2 

} {{ } 

π = − 

}{{} 

2 

π = + 

2 

}{{} 

π = − 

New Q. Numbers 3 3 + 3 2: π = ±1 

We call these new quantum numbers τρ ι − τιµυκoσ (tri-timeric) 

’possessing three values’ 

Jerzy DUDEK 



Point Groups: Simple and Double 



Vector Fields Ψ vs Spinor Fields Φ 

If the objects Ψ do not change under the rotation ˆR(2π) 

ˆR(2π) Ψ = +Ψ → bosons ↔ simple groups: Example T d 

If the objects Φ change the sign under the rotation ˆR(2π) 

ˆR(2π) Φ = −Φ → fermions ↔ double groups: Example T D d 

Remarks - Comments 

Objects in classical mechanics, integer rank tensors like boson 

wave-functions transform under simple point-groups: D 2h , T d 

Spinors - the non-collective wave functions of fermions - transform 

under double point-groups: D D 2h , T D d , OD h ... 

Jerzy DUDEK 
















Jerzy DUDEK 
















Jerzy DUDEK 
















Jerzy DUDEK 
















Jerzy DUDEK 





Coexistence of T d - and O h -Symmetries in Nuclei 

Octahedral Def. (Rank 1) 

0.30 

0.20 

0.10 

0.00 

-0.10 

-0.20 

Tetrahedral Symmetry / Instability 

1 

1 

1 

-0.30 

-0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 

158 Tetrahedral Deformation (Rank 1) 

70 

Yb 88 

1 

1 

1 

E [MeV] 

6.00 

5.75 

5.50 

5.25 

5.00 

4.75 

4.50 

4.25 

4.00 

3.75 

3.50 

3.25 

3.00 

2.75 

2.50 

2.25 

2.00 

1.75 

1.50 

1.25 

1.00 

0.75 

0.50 

0.25 

UNIVERS_COMPACT (D=3, 23) 


Emin=-2.34, Eo=-1.62 

Jerzy DUDEK 







0.30 

0.20 

0.10 

0.00 

-0.10 

-0.20 


1 

2 

2 

3 

3 

-0.30 

-0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 

160 Tetrahedral Deformation 

70 

Yb 90 

3 

3 

2 

2 

1 

E [MeV] 

6.00 

5.75 

5.50 

5.25 

5.00 

4.75 

4.50 

4.25 

4.00 

3.75 

3.50 

3.25 

3.00 

2.75 

2.50 

2.25 

2.00 

1.75 

1.50 

1.25 

1.00 

0.75 

0.50 

0.25 



Emin=-2.18, Eo= 0.40 

Jerzy DUDEK 







0.30 

0.20 

0.10 

0.00 

-0.10 

-0.20 


1 

2 

2 

3 

3 

-0.30 

-0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 


70 

Yb 92 

3 

3 

2 

2 

1 

E [MeV] 

6.00 

5.75 

5.50 

5.25 

5.00 

4.75 

4.50 

4.25 

4.00 

3.75 

3.50 

3.25 

3.00 

2.75 

2.50 

2.25 

2.00 

1.75 

1.50 

1.25 

1.00 

0.75 

0.50 

0.25 




Jerzy DUDEK 







0.30 

0.20 

0.10 

0.00 

-0.10 

-0.20 

1 


1 

2 

3 

2 

-0.30 

-0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 


70 

Yb 94 

3 

2 

3 

2 

1 

E [MeV] 

6.00 

5.75 

5.50 

5.25 

5.00 

4.75 

4.50 

4.25 

4.00 

3.75 

3.50 

3.25 

3.00 

2.75 

2.50 

2.25 

2.00 

1.75 

1.50 

1.25 

1.00 

0.75 

0.50 

0.25 



Emin= 0.98, Eo= 4.20 

Jerzy DUDEK 







0.30 

0.20 

0.10 

0.00 

-0.10 

-0.20 

1 


1 

2 

2 

2 

3 

-0.30 

-0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 


70 

Yb 96 

2 

3 

2 

2 

3 

2 

1 

E [MeV] 

6.00 

5.75 

5.50 

5.25 

5.00 

4.75 

4.50 

4.25 

4.00 

3.75 

3.50 

3.25 

3.00 

2.75 

2.50 

2.25 

2.00 

1.75 

1.50 

1.25 

1.00 

0.75 

0.50 

0.25 




Jerzy DUDEK 


3 






0.30 

0.20 

0.10 

0.00 

-0.10 

-0.20 

1 


1 

2 

2 

2 

-0.30 

-0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 


70 

Yb 98 

2 

3 

2 

3 

2 

1 

E [MeV] 

6.00 

5.75 

5.50 

5.25 

5.00 

4.75 

4.50 

4.25 

4.00 

3.75 

3.50 

3.25 

3.00 

2.75 

2.50 

2.25 

2.00 

1.75 

1.50 

1.25 

1.00 

0.75 

0.50 

0.25 




Jerzy DUDEK 







0.30 

0.20 

0.10 

0.00 

-0.10 

-0.20 

1 


2 

1 

2 

3 

2 

-0.30 

-0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 


70 

Yb 100 

3 

2 

2 

3 

2 

1 

1 

E [MeV] 

6.00 

5.75 

5.50 

5.25 

5.00 

4.75 

4.50 

4.25 

4.00 

3.75 

3.50 

3.25 

3.00 

2.75 

2.50 

2.25 

2.00 

1.75 

1.50 

1.25 

1.00 

0.75 

0.50 

0.25 




Jerzy DUDEK 







0.30 

0.20 

0.10 

0.00 

-0.10 

-0.20 

1 


1 

1 

2 

3 

2 

-0.30 

-0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 


70 

Yb 102 

3 

2 

2 

3 

2 

1 

1 

E [MeV] 

6.00 

5.75 

5.50 

5.25 

5.00 

4.75 

4.50 

4.25 

4.00 

3.75 

3.50 

3.25 

3.00 

2.75 

2.50 

2.25 

2.00 

1.75 

1.50 

1.25 

1.00 

0.75 

0.50 

0.25 




Jerzy DUDEK 







0.30 

0.20 

0.10 

0.00 

-0.10 

-0.20 

1 


2 

1 

2 

3 

2 

-0.30 

-0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 


70 

Yb 104 

1 

3 

1 

2 

2 

3 

2 

1 

1 

E [MeV] 

6.00 

5.75 

5.50 

5.25 

5.00 

4.75 

4.50 

4.25 

4.00 

3.75 

3.50 

3.25 

3.00 

2.75 

2.50 

2.25 

2.00 

1.75 

1.50 

1.25 

1.00 

0.75 

0.50 

0.25 




Jerzy DUDEK 




Nuclear Tetrahedral-Symmetry Effects 


A32 

0.30 

0.20 

0.10 

0.00 

-0.10 

-0.20 

1 


1 

1 

1 

1 

-0.30 

0.0 0.1 0.2 0.3 0.4 

156 

A20 

70 

Yb 86 

2 

3 

3 

4 

4 

5 

5 

E [MeV] 

6.00 

5.75 

5.50 

5.25 

5.00 

4.75 

4.50 

4.25 

4.00 

3.75 

3.50 

3.25 

3.00 

2.75 

2.50 

2.25 

2.00 

1.75 

1.50 

1.25 

1.00 

0.75 

0.50 

0.25 




Jerzy DUDEK 






A32 

0.30 

0.20 

0.10 

0.00 

-0.10 

-0.20 

1 

1 


1 

1 

1 

1 

-0.30 

0.0 0.1 0.2 0.3 0.4 

158 

A20 

70 

Yb 88 

1 

1 

2 

2 

3 

3 

4 

4 

E [MeV] 

6.00 

5.75 

5.50 

5.25 

5.00 

4.75 

4.50 

4.25 

4.00 

3.75 

3.50 

3.25 

3.00 

2.75 

2.50 

2.25 

2.00 

1.75 

1.50 

1.25 

1.00 

0.75 

0.50 

0.25 




Jerzy DUDEK 


2 






0.30 

0.20 

0.10 

0.00 

-0.10 

-0.20 

1 

2 

2 

1 


2 

2 

1 

1 

-0.30 

0.0 0.1 0.2 0.3 0.4 

160 Quadrupole Deformation 

70 

Yb 90 

1 

1 

1 

2 

2 

3 

3 

E [MeV] 

6.00 

5.75 

5.50 

5.25 

5.00 

4.75 

4.50 

4.25 

4.00 

3.75 

3.50 

3.25 

3.00 

2.75 

2.50 

2.25 

2.00 

1.75 

1.50 

1.25 

1.00 

0.75 

0.50 

0.25 




Jerzy DUDEK 


2 





A32 

0.30 

0.20 

0.10 

0.00 

-0.10 

-0.20 

2 

3 

4 

2 


4 

3 

3 

3 

3 

2 

1 

-0.30 

0.0 0.1 0.2 0.3 0.4 

162 

A20 

70 

Yb 92 

1 

1 

1 

1 

2 

2 

3 

E [MeV] 

6.00 

5.75 

5.50 

5.25 

5.00 

4.75 

4.50 

4.25 

4.00 

3.75 

3.50 

3.25 

3.00 

2.75 

2.50 

2.25 

2.00 

1.75 

1.50 

1.25 

1.00 

0.75 

0.50 

0.25 




Jerzy DUDEK 






A32 

0.30 

0.20 

0.10 

0.00 

-0.10 

-0.20 

4 

5 

6 

4 


6 

5 

5 

5 

5 

4 

4 

3 

-0.30 

0.0 0.1 0.2 0.3 0.4 

164 

A20 

70 

Yb 94 

3 

2 

2 

1 

1 

1 

1 

2 

3 

E [MeV] 

6.00 

5.75 

5.50 

5.25 

5.00 

4.75 

4.50 

4.25 

4.00 

3.75 

3.50 

3.25 

3.00 

2.75 

2.50 

2.25 

2.00 

1.75 

1.50 

1.25 

1.00 

0.75 

0.50 

0.25 




Jerzy DUDEK

SYMMETRIES in PHYSICS

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