Symmetries and Group Theory - Lecture 2

1 Introduction

Physics attempts to look for the global aspects of a system, since much of system behaviorcan be understood from general principles without investigating the details. It is well knownthat conservation of energy and momentum completely describe the kinematic motion of asystem. Other conservation principles and symmetries are also inserted in the mathematicaldescription of interactions and physical laws. There are many types of symmetries, some ofthese are geometrical and some more mathmatical, but all symmetries lead to conservationprinciples - the Noether theorem.

There are discrete and continuous symmetries, but they must be coded into a mathematicalform, or applied in a way to represent a symmetry. Thus the symmetry of baryon conserva-tion requires that any reaction involving baryons must balance before and after a reaction.As an example, an equation for pion production in proton-proton collisions is written;

P + P = P + P + π

The number of baryons to the left and right of the equal sign is the same, as baryon numberis conserved. This demonstrates that a pion, π, is not a baryon but it is a hadron (stronglyinteracting particle like the ptoton). Obviously, hadron number is not conserved. However,the reaction must also conserve other quantities such as energy, momentum, and charge.Electric charge is a mathematical symmetry and energy and momentum are related to timeand sparial translations (geometric symmetries). There are 2 units of positive charge on theleft of the equation and there must be 2 units on the right. Thus the pion in this reactionmust be charge neutral.

The mathematical tool for investigating symmetry is group theory.

2 The Noether theorem

Although one does not need to invoke the Noether theorem to consider simple, particularlyclassical symmetries, it is a powerful tool in quantum theory, and for the physical interpre-tation of a symmetry. The theorem is stated in terms of the Lagrangian formulation of aninteraction.

Noether Theorem

For every continuous variation of the action which leaves it invarient, there is a con-served quantity (a currrent)

1

An example of a conserved quantity is the charge of a system. This is mathematically statedby the equation of continuity,

∂ρ∂t

= −~∇ · (~V ρ)

In this equation, ρ is the charge density and ~V the velocity of an incremental volume ofcharge. The 3-vector of the charge current is ~J = ρ~V . Now consider the relativistic 4-vectorin space-time.

j = ( ~J, ρ)

Operate on this vector by the 4-gradient, (~∇, ∂∂t

), to obtain;

~∇ · ~J +∂ρ∂t

= 0

This states that the 4-current does not change in space-time so it is conserved, representingof course, conservation of charge.

To demonstrate the Noether theorem, write a Lagrangian density as, L(φk(x),∂uφk(x)∂xu

). In

this expression, φk,∂uφu

∂xuare the fields, which for example, give the position and velocity

amplitudes. The Lagrangian of this system is;

L =∫

d3xL

The local action is S =t2∫

t1

dtL. The equations of motion follow from stationary action.

δ S = 0

This introduces the Euler-Lagrange equations by applying the calculus of variations. Thenlook at changes to these equations due to small changes in the fields. This produces a con-served current. Without developing a general proof which takes some effort, look at thesimple example of energy conservation. In this case there is a particle of mass, m movingunder the influence of a potential, V . The action is;

S =∫

dtL[x(t), x(t)] =∫

dt[(m/2)∑

x2i − V (x)]

Let Q = ∂∂t

. For example Q[x] = x. Then;

QL = m∑

xixi − ∑ ∂V∂xi

xi = ddt

[m/2∑

x2 − V ] = ddt

[F ]

2

L

V(x)

x

Figure 1: A figure showing the potential function, V (x), of a 1-D crystal lattice which has asymmetry upon translation by the lattice dimensions, L

Here, F = [m/2∑

x2 − V ]. Set

j =∑ ∂L

∂xQ[x] − f

j = m∑

x2 − F

j = m/2∑

x2 + V (x)

This is an expression for the energy of the system. Thus if the energy is conserved (current)then the system is invariant under a time translation.

3 Examples of discrete symmetries

As a simple example consider the potential well of a 1-D atomic crystal lattice. This has asymmetry about the position of the atoms in the lattice as shown on Figure 1. A transla-tion by the lattice dimensions produces the same potential. This is an example of a discrete,geometric symmetry.

In this example there is a translation operator Tn(L) which operates on a state vector of aquantum system changing the position vector from x to x + nL, where n is a positive ornegative integer. Therefore if |ψ(x)〉 is a state vector, then;

|ψ(x+ nL)〉 = Tn |ψ(x)〉

The state vector |ψ(x)〉 and |ψ(x + nL)〉 are equivalent descriptions of the system. There-fore the expectation values of an operatorA in this space is the same using either state vector.

〈ψ|A|ψ〉 = 〈Tnψ|A|Tnψ〉

This can be written;

3

K

π

K0

π0 π+

K0

K+

−1 0 1Charge

−2

−1

0

Hyp

erch

arge

Figure 2: A figure showing a discrete symmetry among different elementary particles in the2-D space of charge vs hypercharge

〈ψ|A|ψ〉 = 〈ψ|(T−1n ATn)|ψ〉

Then A = (T−1n ATn). Therefore the invarience of the potential operator is obvious;

〈ψ|V (x)|ψ〉 = 〈ψ|(T−1n V (x)Tn)|ψ〉 = V (X + nL)

The Hamiltonian H = p2/2m+V commutes with the operator Tn since a spatial translationdoes not change the value of p. The above clearly connects the symmetry to the mathemat-ical description of the state.

In another example, this time from particle physics, we have already observed a symmetryin charge conservation and baryon number. However, a pion is not a baryon. We introduceanother quantum number called hypercharge, Y , which is baryon number plus a strangenessquantum number, Y = B + S. Strangeness is to be associated with a strange quark whichproduces an additional hadronic meson called a kaon. The symmetry of the pion/kaon sys-tem is discrete in a 2-D plot of charge vs hypercharge, as shown in Figure 2. It is an exampleof a mathematical symmetry which underlies the strong interaction.

This figure suggests that there is an underlying symmetry between mesons under the stronginteraction. In fact the splitting along rows is due to the electromagnetic and not the stronginteraction. The above symmetry is due to a gauge transformation and is developed later,but a gauge transformation leaves a Lagrangian invarient as did the simple case demon-strated above.

4

4 Groups

We begin with the mathematical description of a group. This is introduced by a considera-tion of discrete, as opposed to continuous, group elements. To form a group, there is a set ofelements and a algebra for combining these elements (e.g. “multiplication”). The set formsa group if;

1. The algebra of combinng the elements of a group, ai, produces another element of thegroup. aiaj = ak.

2. There is an idenity element, I, such that, Iai = ai

3. There is an inverse element, a−1, such that aiai = I

4. All elements are mapped into all elements of the group by these operations

5. The algebra is associative. ai(bjck) = (aibj)ck

Any group may have different representations. For example the identity element is al-ways mapped into itself. Then we classify group representations. An important class arehomomorphic maps. These maps preserve the group algebra. Suppose the map takes thegroup F into the group G; F → G with an element F (i) → G(i), then apply the groupmultiplication.

F (kk) = F (fi)F (jk)

Applying the above map, if ;

F (fk) → G(gi)G(gj) = G(gk)

then the map is homomorphic. Mapping onto the identity is homomorphic but not alwaysisomorphic. For the map to be isomorphic, the two groups must have a one-to-one corre-spondance between elements which also preserves the multiplication.

fi → gi fj → gj fifj = fk gigj = gk

An example of a simple group is C2, represented by the multiplication Table 1

The numbers +1 and −1 form the elements of this group under normal multiplication. Theorder of a group is the number of group elements. Other representations of this group are thespatial inversion (parity) and a 2-element permutation. C2 is the lowest order representationof the groups generated by cyclic permutations, Cn. The cyclic group of order 3 has themultiplication table shown in Table 2. Geometrically, the rows and column represent angle

5

rotations of 2π/3, ie the verticies of an equilateral triangle. This group is referenced as thepermutation or symmetric group.

In classical physics, symmetry equations are applied to the solutions of a pde representingthe physics. These solutions usually are eigenfunctions which can be represented as vec-tors in a linear, N-dimensional vector space. Group theory can then be realized as lineartransformations on vectors. The multiplication of vectors and vector transformations is in-corporated into matrix algebra. Thus we can use matrix representations for group elements.Then consider a group representation for the symmetric group in matrix form with element;

(

cos(θi) sin(θi)−sin(θi) cos(θi)

)

The group algebra is matrix multiplication. Thus the multiplication of two elements resultsin the sum of the two angles of the elements, θi + θj = θk.

(

cos(θi) sin(θi)−sin(θi) cos(θi)

) (

cos(θj) sin(θj)−sin(θj) cos(θj)

)

=(

cos(θk) sin(θk)−sin(θk) cos(θk)

)

The elements of the group, C3 are then;

(

1 00 1

) (

−1/2√

3/2

−√

3/2 −1/2

) (

−1/2 −√

3/2√3/2 −1/2

)

The first element is the identity, I. The next element represents a rotation by 2π/3, a1 andthe last element a rotation by 4π/3, a2. Thus a1a1 = a2, a rotation by 4/pi/3 and a1a2 = I,a rotation by 2π. Finally, a2a2 = a1, a rotation by 8π/3.

Table 1: The multiplication table of the simple group of 2 elements, C2

I aI I aa a I

Table 2: The multiplication table of the simple group of 3 elements, C3

I a ba b Ib I a

6

Figure 3: The geometry showing the symmetries involved in obtaining the group elementsof the Dihedral group D3. There are 3 reflections about the diagonals and 2 rotations aboutthe center of the figure. These symmetries plus the identity yield a total of 6 elements

An abelian group has communicative multiplication, ab = ba. All cyclic groups are abelian.The smallest non-abelian group has order 6 and consist of reflections about the diagonals ofan equilateral triangle and rotations about the center with angles of 2π/3 and 4π/3 3. Thisforms the Dihedral group, D3

Table 3: The multiplication table of the Dihedral group of 6 elements, D3

I a1 a2 a3 a4 a5

a1 I a4 a5 a2 a3

a2 a5 I a4 a3 a1

a3 a4 a5 I a1 a2

a4 a3 a1 a2 a5 Ia5 a2 a3 a1 I a4

The lowest order of a non-cyclic group is 4. An example of the Dihedral group, D3 is shownby the group Table 3. The D3 multiplication table shows 4 subgroups. A representation ofone of these subgroups is a set of rotations by π/2, π, and 3π/2. In a set of matricies therepresentation elements are;

(

1 00 1

) (

0 −11 0

) (

−1 00 −1

) (

0 1−1 0

)

7

5 Group representations

There are a number of ways a finite group may be represented. Supppose a group on a vectorspace V , and label a representation of that group by U . Let S be any non-singular operator(non-singular means the operator has an inverse) on V . Then another representation is;

U ′ = S−1US

U ′ has the same dimension and the same algebra. We determine if two representations areequivalent by determining their character. If the representation is placed in matrix form,the group character is the trace of a matrix of an irreducible representation.

χ = Tr [U ] =∑

i

gii

The above is true because;

Tr [S−1US] = Tr [U ]

A representation is reducible if it can be represented in a basis as a matrix in the form,

Mi =

(

Ai(mm) Bi(mn)0 Ci(nn)

)

The A, B, C, matricies have dimensions mxm, mxn, and nxn, respectively. For elementmultiplication;

Mi =

(

Ai(mm) Bi(mn)0 Cinn

) (

Aj(mm) Bj(mn)0 Cjnn

)

=

(

AiAj AiBj +BiCj

0 CiCj

)

Then operation on a vector;

(

Ai Bi

0 Ci

) (

UW

)

=

(

AiU +BiWCiW

)

The W subspace is invarient if B → 0. Thus the representation M = A⊕

C. Decom-position continues until irreducible representations are formed. Essentially one is findingthe subspaces of othorgonal vectors which represent the group. There is then a set of suchvectors for each irreducible representation of the group. There are nxn = n2 possibilitiesfor the matrix elements. Thus we expect;

∑

n2i = M

In the above M is the dimension of the representation of the group. Consider the groupD3(S3 Figure 3. There are 3 classes in this group.

8

1. The identity matrix I

2. The elements a1, a2, a3

3. The elements a4 a5

Suppose we have a matrix representation of the group as;

U →(

D1 00 D2

)

The group components D1,2 are irreducible representations of U . Again suppose an operationon U by the operator S. One might suppose this represents a rotation in the vector space.Then the operation on the vectors (group elements) transforms them into other vectors inthis space. If all the elements are reached by this transformation, then the group is irre-ducible, otherwise the group is reducible. In the above example Di represents subgroups ifthe identity is added to each subcomponent. In the above the Di are square matricies,and allessential properties of the group are contained in each representation. Suppose the followingrepresentation of the cyclic group, C3.

(

1 00 1

) (

1 00 a

) (

1 00 a2

)

A discete group has a finite number of classes. The group character of a class is unique to theclass. The character of the group elements is obtained by forming the trace of the elements;χ(1) = 2, χ(2) = 1 + a, and χ(3) = 1 + a2. The following is an equivalent representation.

(

1 00 1

) (

a 00 1

) (

a2 00 1

)

Now consider the permutation group, C2, with multiplication Table 1 above. There are twogroup elements, the identity, I, and element a. From the multiplication table, there are twopossible values for a, either ±1. These are the two inequivalent representations of the groupC2. The characters of the two representations are χ(1) = 2 and χ(2) = 0. The character isusually placed in tabular form - a character table, Table 4. The rows in the table are theireducible representations and the columns are the classses

The number of irreducible representations equals the number of classes. The permutationgroup has 3 classes, two being 1 dimensional and the third 2 dimensional.

As a further example, consider the regular representation of the rotation group. A regularrepresentation has the same dimensions as the order of the group. The rotation matricies are;

9

1 0 00 1 00 0 1

0 1 00 0 11 0 0

0 0 11 0 00 1 0

Identify the elements I with the first matrix, a with the second,and b with the third. Thisrepresentation is reducible into 2 subgroups. The irreducible representations are simply thenumbers;

1, e2πi/3 e4πi/3 or 1, e4πi/3, e8πi/3 = e2π/3

The second representation is interchange of elements 2 and 3.Multiplication by a complex number can be represented in matrix form. The first represen-tation above can be written;

(

1 00 1

)

1/2

(

−1√

3√3 −1

)

1/2

(

−1 −√

3√3 −1

)

This representation is irreducible. Note that the trace of all the matricies is the same.A crystallographic point group is composed of rotations, reflections, and inversions about apoint. Every molecule has a point group association. The number of classes represents thesubgroups or the number of symmetries of the system. We develop a group incorporating allthese operations for an amonia molecule. The ammonia molecule has the S3 (D3) symmetry,as can be seen in Figure 4. This is the permutation group of the 3 hydrogen atoms. Therepresentation of this group is irreducible and there are 3 group elements. The identity isa matrix with trace = 3. Then consider spatial reflections, for example in the (x, z) plane,This is obtained by the matrix;

1 0 00 −1 00 0 1

There are 2 other reflections each with trace = 1. There are also rotations with an element;

Table 4: The character table of the permutation group C3 after reducing the matrix to 1-D.The symbol w is the primitive third root of 1 (any complex number raised to the 3 power is1)

χ(1) 1 2 3χ(2) 1 a a2

χ(3) 1 a2 a

10

N

x

y

H

HH

z

Figure 4: The symmetry of the ammonia molecule representing 3 reflections and 3 rotations

−1/2 −√

3/2 0√3/2 −1/2 00 0 1

which has trace = 0. The character of the group is then (3, 1, 0). The number of classes arethe number of elements connected by group multiplication to the element. To find group el-ements conjugate to an element, a, find the elements bab−1 for each element, b, in the group.All elements conjugate to each other form a class of the group. Consider the multiplica-tion table for the group, D3(S3), Table 3. The elements conjugate to a1 are; a2 a3 a4, Thegroup character is the sum of the characters of the irreducible representations. Thus supposethere are gi elements in each class and g is the order of the group. Then the group character is

g =∑

|χi|2

The full group has the structure, M = I⊕

MR

⊕

MI . For the ammonia example using thegroup S4;

S = [(2)2 + (1)2] = 5

There are 2 irreducible representations excluding the identity. Then to apply the symmetryto the description of the ammonia molecule, consider a real 3-D vector. This vector trans-forms using the 3-D matricies described above which is not the irreducible matricies of thegroup. However, a vector in 3-D must have a 3-D operator. Thus the matricies are composedof additive irreducible representations. For the vector there is the rotation group and theidentity or MV = I

⊕

MR with character [3, 0]. However, an axial vector has a reflection

11

transformation represented by the matrix;

−1 0 00 1 00 0 −1

This has trace −1 be expressed by a 3-D. The group character contains the inversion androtation groups, MAV = MI

⊕

MR

6 Continuous groups

In the previous sections we discussed groups with a discrete set of elements. Now cosnidergroups with an infinite number of elements. Of these groups the Lie groups are of mostinterest. Suppose a transformation;

xi = fi(x1, x2, x3, θ)

In the above, θ is a parameter. In a Lie group, the function, f , must be analytic in θ.Consider the set of NxN unitary matricies (a matrix when operating on a vector preservesits length). This set of matricies forms a group defined by U(n). They can ge consideredrotation matricies. If we impose the condition that the determinant of the matricies is +1,we have have the special unitary group, SU(N). This is an NxN unitary matrix with N2−1independent parameters. For example, N = 2 means (2)2 − 1 = 3 independent parameters.For N = 3 there are 8 parameters, sometimes called generators. In SU(2) the generatorsare, sk. In SU(3) the generators are Mk.

7 Rotations

Previously we looked at rotations and determined that finite rotations are not additive.Consider a rotation about the x axis as shown in Figure 5. Note that we are using a righthanded system for coordinates and rotations. A rotation preserves the length of a vector (i.eIt is a unitary operation.). Suppose we rotate a vector in 3-D space about the x axis with arotation, ω. The coordinates (x, y, z) are rotated into (x′, y′, z′) by the matrix operation;

x′

y′

x′

=

1 0 00 cos(ω) sin(ω)0 −sin(ω) cos(ω)

xyz

= Rx

xyz

A rotation about z is;

12

ωx

ωx

ωx

xx’

y

y’

z

z’

Figure 5: Definition of a right handed rotation about the x axis

x′′

y′′

x′′

=

cos(ω′) sin(ω′) 0−sin(ω′) cos(ω′) 0

0 0 1

x′

y′

z′

Then note that RxRz 6= RzRx.

The rotation groups in 2 (SO(2)) and 3 SO(3)) dimensions are isomorphic, having the samecommutation relations and structure constants. The rotation about the z axis in 3-D iswritten;

Rz =

cos(ω) sin(ω) 0−sin(ω) cos(ω) 0

0 0 1

This contains the subgroup SO(2), as is obvious from the earlier discussion concerning irre-ducible subgroups obtained by diagonalizing the matrix representation.

Continuous group algebra is equivalent to Lie algebra, which involves commutators as willbe seen below. Thus as an example, consider the rotation group SO(2). As above a groupelement can be written;

R(2) =

(

cos(ω) sin(ω)−sin(ω) cos(ω)

)

= I cos(θ) + iσz sin(θ) = eiσzθ

In the above, I is the 2-D identity matrix, and σz is the Pauli spin matrix. Note thatR(θ1)R(θ2) = R(θ1 + θ2) as long as the rotations are about the same axis. Then;

R(sj) = eiǫjsj → 1 + iǫjsj − (1/2)(ǫjsj)2 + · · ·

Here si are the group generators. Then R−1(si) = R(−si) since RR−1 = 1

R−1(sj) = 1 − iǫjsj − (1/2)(ǫjsj)2 + · · ·

13

However, suppose an infinitesmal transformation ω → 0 so that sin(ω) → ω, cos(ω) → 1,and ωω′ → 0. The rotation matrix is then;

Rx =

1 0 00 1 ω0 −ω 1

We find that in this case;

RxRz =

1 ω′ 0−ω′ 1 ω0 −ω 1

= RzRx

The rotation operator is unitary. Therefore, when operating on a wave function

U = ei~ω·~L = I + i~ω · ~L − (~ω · ~L)(~ω · ~L)/2 · · ·

In the above I is the unit matrix, and the meaning of the exponential operator is shown byits series expansion. Remember one must maintain the order of the operators. To obtainthe infinitesmal operator matricies written above, take the 1st two terms in the expansionassuming ω is small and use the L operators defined below.

Lx =

0 0 00 0 10 −1 0

Ly =

0 0 10 0 0−1 0

Lz =

0 1 0−1 0 00 0 0

Now we note that LxLy − LyLx = Lz and in general one obtains the commutation rules[Li, Lj] = Lk These form the generators of a continuous group, SO(3), and define the Liealgebra. We have already noted that this group is homomorphic to the group SU(2). Theabove operations are not the lowest representation (irreducible representation) of this group.The operators of the irreducible representation are the Pauli spin matricies. When one addsangular momentum operators, for example, one obtains a higher order representations of theSU(2) group.

Return to the unitary operation; U = eiωiLi, where the Li are the generators as describedabove. This transformation is to be interpreted as;

14

U = I + iωiLi − ω2iL

2i /2 + · · ·

Up through 2nd order in the expansion;

I + ω2/2

0 0 00 −1 00 0 −1

=

1 0 00 1 + ω2/2 00 0 1 + ω2/2

Although this only shows the series to 2nd order, one can at least see the beginning con-vergence of the series to the harmonic forms sin(ω) and cos(ω). Thus an infinite number ofinfinitesmal rotations reproduce a finite angle rotation.

8 Structure constants of the Lie group

Look at a series of small rotations which can be written, R−1i R−1

j RiRj. As infinitesmal op-erations this is;

R−1i R−1

j RiRj ≈ (1 − iǫisi)(1 − iǫj)(1 + ǫi)(1 + ǫj) ≈(1 + ǫiǫj [sj , si]) = 1 +

∑

k

ckij sk

This operation remains near but not equal to the identity because the operators do notcommute. The operation does result in an element in the group, however.

([sj , si]) =∑

k

ckij sk

The ckij are the structure constants of the group and form the Lie algebra and sk the gener-ators of the group. Lie groups are formed in higher dimensions, for example SU(3). Thereare more group generators in higher dimensions, in the case of SU(3) there are 8 Mk, butthe commutator relations remain. Suppose generators of the rotation group SO(3) are Mi.The commutator relations are;

[Mi,Mj] = icijkMk

As above, the structure constants are cijk of the Lie group algebra. An irreducible repre-sentation in SU(2) of the generators of this group are the Pauli spin matricies. Every 2-Dmatrix with zero trace and be represented by a linear combination of Pauli sipn matricies.Then using the Jacobi identity for the commutators;

[[si, sj ], sk] + [[si, sk], si] + [[sk, si], sj] = 0

15

The structure constants are constrained to be;

∑

m

[cmij cnmk + cmjkc

nml + cmklc

nmj ] = 0

For example, look at a representation for SU(2).

U =

(

a b−b∗ a∗

)

U−1 =

(

a∗ −bb∗ a

)

U−1U = |U | = a∗a + b∗b = 1

In SO(3) this forms the 3 Cayley-Klein parameters which are connected to rotations aboutthe 3-D set of axes in coordinate space. Observe that the transformation U transforms a2-D spinor as;

(

α′

β ′

)

=

(

a b−b∗ a∗

) (

αβ

)

α′ = aα + bβ

β ′ = −b∗α + a∗β

Thus SU(2) is homomorphic to the rotation group in 3-D. A one-to-one correspondance be-tween the elements is an isomorphism. If the elements obey the same multiplication table thegroups are homomorphic. Now look for irreducible representations of SU(2). This requiresextensive algebra. Begin by describing the above transformation on the spinor in polynomialform. Without following the development one finds a unitary representation for the rotationgroup and identifies;

a = e−iα/2cos(β/2)e−iγ/2

b = −e−iα/2sin(β/2)eiγ/2

In matrix form this is;

(

e−iα/2 00 eiα/2

) (

cos(β/2) −sin(β/2)sin(β/2) cos(β/2)

) (

e−iγ/2 00 eiγ/2

)

=(

e−iα/2cos(β/2)e−iγ/2 −e−iα/2sin(β/2)eiγ/2

eiα/2sin(β/2)e−iγ/20 eiα/2cos(β/2)eiγ/2eiα/2

)

The first matrix on the left is a rotation about the z axis, the second a rotation about they axis and the third a rotation about the x axis. The correspondance of the elements repre-senting the rotation about the z axis is;

16

Uz =

(

e−iα/2 00 eiα/2

)

→

cos(α sin(α) 0−sin(α) cos(α) 0

0 0 1

However, the correspondance is not one-to-one. As α goes from 0 to 2π, α/2 goes from 0 toπ. Thus SU(2) and SO(3) are homomorphic but not isomorphic. Every 2-D unitary matrixof determinant 1 corresponds to a 3-D rotation. The a, b amy be muliplied by a phase toobtain the standard representation of the rotation elements;

Dj(α, β, γ)m,m′ =∑

λ

(−1)λ

√

(j +m)!(j −m)!(j +m′)!(j −m′)!(j −m′ − λ)!(j +m− λ)!λ!(λ+m′ +m)!

eim′α cos2j+m−m′−2λ(β/2) sin2λ+m′−m(β/2) emγ

This representation is (2j + 1) dimensional. It may be shown that;

Dl00 = Pl(cos(β))

Dlm0 = (−1)m(

√

4π2l + 1

)Y ml (β, α)

Dl0m = (

√

4π2l + 1

)Y ml (β, α)

The components transform among themselves as;

Y m′

l (θ′, φ′) =∑

m

Y ml (θ, φ)Dl

mm′(R)

This makes connections to integrals involving the rotation operators and their addition(Clebsh-Gordon coefficients).

9 Classical interpretation of Clebsch-Gordon coefficients

Recall that in many cases of a composite system, one must add angular momentum compo-nents to obtain the system angular momentum. This for 2 angular momentum terms one has;

Y ml (θ, φ)Y m′

l′ (θ, φ) =∑

LM

√

(2l + 1)(2l′ + 1)(2L+ 1)4π

(

l l′ Lm m′ M

) (

l l′ L0 0 0

)

Y ML (θ, φ)

The terms;

17

L

l

l’

z

mm−1

m+1

P

Length to beprojected on z

Figure 6: The classical geometry for vector addition of angular momentum. The Clebsch-Gordon coefficient is the probability of finding the projection of the addition of the vectorson the z axis equal to M

(

l l′ Lm m′ M

)

are the Wigner 3-J symbols and are related to the Clebsch-Gordon coefficients by;

(

l l′ Lm m′ M

)

= (−1)l−l′−M 1√2L+ 1

(l,m, l′, m′|L,−M)

Classically the Clebsch-Gordon coefficients are the probability that the z components of l,and l′ are m and m′ when the angular momentum vectors add to L and the z component ism+m′ = M . This is illustrated in Figure 6. The probability is found to be;

P = (2L+ 1) [

(

l l′ L0 0 0

)

]2

10 Projecting angular momentum symmetry when com-

bining states

As an example of combining angular momentum consider 2 particles in a central potentialwell. This is the way initial wave functions are obtained for multi-particle states in quantummechanics. These wave functions are then used in a diagonalization process to include theresidual inter-particle interaction. To simplify, assume 2 particles in a cnetral well. Neglectthe residual potential, we only need here the wave fumctions before diagonalization. Thereare 2 wave functions, ψi(~ri). In these wave functions the angular momentum eigenfunctionis Y m

l and the spin eigenfunction is jµ. The magnitude of the nuclear (or electron spin

18

depending on the problem of interest) is J = 1/2.

ψi = AiRl, i(ri) Ymi

lijµi

i

The product wave function is

Ψ =∑

ARl1(r1)Rl2(r2) Ym1

l1Y m2

l2jµ1

1 jµ2

2

Then combine the angular momentum terms.

Y mi

li jµii

=∑

(li, mi, ji, µi|Ji,Mi) JMi

i

Rewrite the product wave function in an expansion of angular momentum terms as;

Ψ =∑

ARl1(r1)Rl2(r2) (−1)l1−j1+M1+l2−j2+M2

√

(2J1 + 1)(2J2 + 1)(

l1 j1 J1

m1 µ1 M1

) (

l2 j2 J2

m2 µ2 M2

)

JM1

1 JM2

2

Ψ =∑

ARl1(r1)Rl2(r2) (−1)l1−j1+M1+l2−j2+M2+J1−J2+M√

(2J1 + 1)(2J2 + 1)(2J + 1)(

l1 j1 J1

m1 µ1 M1

) (

l2 j2 J2

m2 µ2 M2

)

JM1

1 JM2

2

(

J1 J2 JM1 M2 M

)

The 3-j symbols can be contracted into 6-j symbols. As illustrated in the handout.

11 Representations of continuous groups

We developed the definition of a continuous group in terms of a unitary rotation in 3-Dcoordinate space. The infinitesmal operators of the rotational transformation were foundto have an algebra defined by the commutation rule [Si, Sj ] = i ǫijk Sk. Because the groupis continuous, an infinite number of such infinitesmal transformations must be applied toreproduce a finite transformation.The lowest representation of this algebra, SU(2), is obtained by the 2-D Pauli matricies.There are 2 eigenvalues ±1, and these matricies are unitary and traceless. Now there areN-dimensional representations of this group, i.e. we can find N dimensional matricies thatsatisfy the algebra, where N is an arbitrary integer > 0. If N = 3 we found that the regularrepresentation of this group is a set of 3 × 3 matricies. Different representations can bespecified by the eigenvalue, S(S+ 1), of S2. here S is the maximum eigenvalue of S3. In thecase of the irreducible representation S3 = 1/2 so that the eigenvalue of S2 is 3/4. The valueof S3 for the regular representation is 1 so the eigenvalue of S2 is 3. Therefore, rotationalsymmetry for various angular momentum values ~L is described by a representation of SU(2).

19

Remember that rotational symmetry requires that the system is independent of the spinprojection on the z axis, m value. However, the energy will depend on the total value of theangular momentum. Thus, while the interaction is independent of the projection, S3, it candepend on S2. Indeed, the higher the eigenvalue of S2 the higher the self energy (mass) ofthe particle.

12 Isospin

Symmetries other than rotations can behave in a way similar to rotations in coordinate space.One of these symmetries is Isospin, which is an unfortunate misnomer. The only correspon-dance of Isospin to angular momentum is that it has the same algebraic structure, SU(2). Theunderlying reason that a particle system is invarient under a rotation in Isospin space is thatthe u and d quarks have almost the same mass and interact similarly. We will discuss this inmore detail later. For the moment, we observe an experimental fact that the strong interac-tion between neutrons and protons is almost identical. That is, without an electromagneticforce, protons and neutrons would be indistinguishable. The small mass difference between aneutron and a proton would be assigned to the self energy of the electromagnetic field. Thisbegs the question about summetry and symmetry breaking - something to be discussed later.

Thus we have 2 particle states for this species of hadrons which are indistinguishable, justas 2 spin states are indistinguishable for rotational invariance. The algebra is the same asrotations, and expressed by the group, SU(2). The lowest representation has 2 states, but wefind a hiearchy of representations. While the interaction is independent of the projection, Tz

it can depend on T 2 just as for the case of angular momentum. Here we define the Isospin op-erator T corresponding to S and T3 coresponding to S3. We write a nuclear wave function as;

ψN =∑

C(l,m)Rl,τ Yml (θ, φ) σ1/2 τ 1/2

Here τ is the “spin” vector similar to the angular momentum spin vector σ. For a multi-nucleon wave function, the angular momentum is combined to provide an overall symmetryby using the Clebesh-Gordon coefficients which force the wave function to be a represen-tation of SU(2). In a similar way the isospin vectors are combined using Clebesh-Gordancoefficients. As an example consider the wave function for the deuteron. It is composed ofa neutron and proton with spin of 1 and 0 angular momentum. Combining two spin 1/2systems there are two possibilities, spin 1 or spin 0. Spin 1 gives a symmetric wave functionupon interchange of the nucleons and spin 0 gives an anti-symmetric wave function. In asimilar way we combine the isospin vectors to produce an isospin of 1 or 0. To obtain anoverall antisymmetric wave function as is needed for a 2-baryon system a spin 0 must becoupled with an isospin 1 wave function and a spin 1 must be coupled with an isospin 0 wavefunction. Because the deuteron spin is known to be 1 the later wave function represents thedeuteron ground state, composed of a neutron and proton (isospin 0). There are no bound

20

states for the isospin 1 system consisting of a spin 0 system composed of pp, np, nn (i.e. anisospin 1 system)

13 Gellmann-Nishijima relation

Gell-mann and Nishijima found a connection between Baryon number and charge, T3.

Q = [(1/2)B + T3]

In this equation B is the baryon number, Q the charge and T3 the 3rd component of isospin.As an example consider the ∆++(1232), a resonant state as defined above. This state has acharge Q = 2, baryon number = 1, and T3 = 3/2. This equation is valid for nucleon statesbut not for hyperons. Note the commutation relations [Q, T3] = 0 and [Q, T ] 6= 0 showingthat isospin is not invarient when applying the electromagnetic interaction.

14 Combining SU(N)symmetries

Suppose we have a system that is symmetric under 2 or more SU(N) symmetries. Theseare combined, as previously indicated, so that the overall system can be represented by aSU(N) representation. The dimensions of this representation can be obtained by a graphicaltechnique called Young’s tableaus.

Suppose we have a system that has a symmetry SU(N) × SU(N ′). Let a box symboldenote the fundamental representation of SU(N). A column of (N - 1) boxes is the conjugaterepresentation. One builds a tableau by combining the boxes using the rules;

1. Form the tableaus of the two fundamental representations

2. Choose one of the representations as the base

3. Label rows of boxes as shown in the figure below

4. Construct correct tableaus (length of a row ≤ length of the row above)

5. Keep adding boxes on this row until exhausted, then move to the row below

6. Two boxes with same label cannot be in the same column

7. Keep only N +N ′ rows

8. Keep only tableaus with row 1 > row2 > row 3 · · ·

21

N

N−1

N−2

N−3

N−1

N

N+1

N+1

N+2

N+2

N+3

N+3

N+4Hook

Figure 7: An example of a Young’s Tableau

X X X

X

X X

2 2 2 2

2

2

2

1

1

3

3

3

1

4 Hook

SU(2) SU(2) SU(2)

Figure 8: An example of a Young’s Tableau

Fill the boxes with the numbers shown in Figure 7. The dimension of a tableau is obtainedby evaluating the product of the boxes divided by the number of hooks. Hooks are countedas shown in the figure.

The dimension of a tableau is N !no.hooks

where the number of hooks is determined by drawingall lines through a row of boxes and then down through a column and counting the numberof boxes that are crossed and multipling these numbers together.

Consider the example of finding the symmetry and dimension of SU(2) × SU(2) × SU(2)These are combined in the figure in steps. The lowest tableau is not possible so it is dis-carded. There are three possible tableaus having dimensions;

22

(2)(3)(4)(1)(2)(3)

= 4

(2)(3)(1)(3)(1)

= 2

15 Example

Suppose the example of the deutron which we solve in relative coordinates. This examplehas 2 interacting nucleons each with spin, ji, and isospin, τi. To a good approximation, thedeuteron has no amgular momentum term, Y 0

0 . Thus combine the spin and isospin to obtain;

jµ1

1 jµ2

2 =∑

(j1, µ1, j2, µ2|J,M) JM

τ ν1

1 τ ν2

2 =∑

(τ1, ν1, τ2, ν2|τ, ν) τ ν

Ψ =∑

AR(r) (j1, µ1, j2, µ2|J,M) (τ1, ν1, τ2, ν2|τ, ν) JM τ ν

The spin and isospin can also be combined but we do that in a different way. The spinand isospin wave functions each have SU(2) symmetry. Thus we expect the deuteron wavefunction to have the form;

SU(2)⊗

SU(2)⊗

SU(2)⊗

SU(2)

Consider the first two SU(2) terms represent the spin of the 2 nucleons, and the second theirisospins. Thus we expect 6 possibilities.

• Two neutron-neutron states, one with spin 0 and one with spin 1

• Two proton-proton states, one with spin 0 and one with spin 1

• Two neutron-proton states, one with spin 0 and one with spin 1

This symmetry is combined using Young’s tableaus, Figure 9.

Because of the Pauli excliusion principle the nn and pp state with spin 1 cannot exist. Thereare 4 states with spin 0. The interaction in this spin state is found to be weaker that thatfor the spin 1 case so these states are unbound. There is one remaining state with spin 1.This np state forms the deutron. It has dimension of 3 representing the 3 projections of thespin on the z axis.

23

X

X X

X

X

X 2 X

XX 22

X 2 X

X

2 2

SU(2) SU(2) SU(2)

2

2

nn and pp spin 1 Dimension 5

nn spin 0 Dimension 3

pp spin 0 Dimension 3

pn Spin 1 Dimension 3

pn spin 0 Dimension 1

pn spin 0 Dimension 1

J Τ

Figure 9: Young’s Tableau for two nucleons showing the dimensional representations

24

16 More on the Gellmann-Nishijima relation

In the last lecture we found that the Gell-mann and Nishijima relation related Baryon num-ber, charge, and the third component of isospin.

Q = [(1/2)B + T3]

But this relation was not valid for hyperons. Now we also know that hyperons are differentfrom nucleons and nuclear resonances. They have long lifetimes, which are more like life-times associated with the weak interaction. Thus Pais suggested that these paticles possessedanother quantum number which was called strangeness, S. The Gell-mann and Nishijimarelation was then extended to include strangeness through the expression;

Q = [(1/2)(B + S) + T3]

An assignment of strangeness or anti-strangeness was assigned to each of the hyperons andhyperon resonances. The quanty B + S is called hypercharge, Y, something we used earlierin the lecture.

17 Extension of SU(2) to SU(3)

The fact that neutrons and protons can’t be distinguished under the strong interaction isnow extended to include hyperons. The fundamental doublet;

(

ud

)

This is extended to include the lowest mass hyperon Λ, to form a triplet, forming the lowestfundamental representation of the special unitary group, SU(3).

udΛ

This symmetry is violated more strongly than that of isospin. One certainly observes thatthe mass of the Λ is much larger than that of the nucleon, 1115 to 939 MeV. Still this tripletof states is an approximate symmetry. In this case we have an operator, U , that transformsa triplet state, ψ into an equivalent state;

ψ′ = Uψ

The operators U must be 3 × 3 unitary, traceless matricies as are the Pauli matricies in 2dimensions. There are 8 independent Hermitian, traceless 3×3 matricies λi, and they satisfy

25

SU(3) Octet for J = 0 MesonsP

K −1 K_

+1−1

K +1Y

K

I

− 0

+z

+0

− 0 ππη

π

Figure 10: An SUf (3) weight diagram for mesons

the commutation rules;

[λi, λj] = i ǫijk fifk λk

where the fifk are the structure constants of the SU(3) group. A graphical representationof the group structure is a weight diagram of the group elements. This is illustrated by theweight diagram for mesons and baryons in Figures 10 and 11, respectively.

It is important to note here that the application of the SU(3) group applies to flavor. Thisis a reference to a class of quark families which will be discussed later. The more commonlyknown application of SU(3) applies to color which is a different quantum number of quarksthat will also be discussed later. The Strong interaction, which includes nucleons and hy-perons, is approximately symmetric under SUf(3)

18 Parity

The normal modes of a string have either even or odd symmetry.This also occurs for stationary states in Quantum Mechanics. The transformation is calledpartiy. We previously found for the harmonic oscillator that there were 2 distinct types ofwave function solutions characterized by the selection of the starting integer in their seriesrepresentation. This selection produced a series in odd or even powers of the coordianteso that the wave function was either odd or even upon reflections about the origin, x = 0.Since the potential energy function depends on the square of the position, x2, the energyeignevalue was always positive and independent of whether the eigenfunctions were odd oreven under reflection. In 1-D parity is the symmetry operation, x → −x. In 3-D the strong

26

+1−1

Y

I+

z

+0

− 0

dss

udd

uusΣΣ

Ξ Ξ

N Puud

uss

udsdds ΛΣ

−1

+1

Baryon Octet

Figure 11: An SUf (3) weight diagram for baryons

Even Odd

interaction is invarient under the symmetry of parity. Parity is a ciscrete symmetry, U(1),which is homomorphic to charge symmetry.

~r → −~r

Parity is a mirror reflection plus a rotation of 180, and transforms a right-handed coordinatesystem into a left-handed one. Our Macroscopic world is clearly “handed”, but “handedness”in fundamental interactions is more involved.Vectors (tensors of rank 1), as illustrated in the definition above, change sign under Parity.Scalars (tensors of rank 0) do not.One can then construct, using tensor algebra, new tensors which reduce the tensor rankand/or change the symmetry of the tensor. Thus a dual of a symmetric tensor of rank 2 isa pseudovector (cross product of two vectors), and a scalar product of a pseudovector and avector creates a pseudoscalar.We will construct bilinear forms below which have these rotational and reflection character-istics.

19 Time reversal

Time reversal is the mathematical operation;

27

t → −t;

with exchange of initial and final states. Macroscopically T is not a good symmetry. How-ever, For Quantum Mechanics;

T †HT = H;

Hψ = i~∂

∂tψ;

THψ = [Tψ];

H[Tψ] = −i~ [Tψ].

Thus Ψ and [T †ψ] are not equivalent, and T † requires t → −t and i → −i. However,one constructs observables in Quantum Mechanics by bilinear forms, ( i.e. by products twooperators and wave functions, as discussed later) so that microscopic time reversibility holds.

20 Charge conjugation

Charge Conjugation changes a particle to its anti-particle, but without change to its dynam-ical variables. The symmetry is based on the asumption that for every particle there is anantiparticle which has Q→ −Q, B → −B, L→ −L, etc.

An eigenstate of C must have:

Q = B = L = S · · · 0.

Thus a π0 is an eigenstate of C but K0 is not since it contains S, a quark of the 2nd gener-ation. The strong and electromagnetic interactions are invarient under C. Under the weakinteraction the operation C is not a good symmetry.

21 The operations of P and T

The operations of reflection and time reversal in classical systems is shown in Table below.

28

µ+ µ−s

s

s

s

s s s

s

s s

s

s

µ−

νe νµ_

νe

_νµ

νµ

e−

e−e+

µ+ e+ νe

_νµC Violation in

νe

_

C CP

Name P TTime + -Position - +Energy + +Momentum - -Spin + -Helicity - +Electric Field - +Magnetic Field + -

Obviously some parameters are invarient but some change sign under this combined opera-tion.

22 The operations of P and C

The operation of CP is composed of the simultaneous operations of C and P . Suppose onewishes to distinguish a galaxy from an anti-galaxy. It is not sufficient to find C violationbut one needs CP violation as well.

The weak interaction violates C and P but CP is experimentally conserved except for flavorchanging decays. In flavor changing weak decay CP is not preserved.

The K0 and K0 are eigenstates of strangeness but not of CP. However states of the weakinteraction ( as presently defined) are invarient under CP. Thus the two possible CP eigen-states of the K0 (K0) have different masses and decay widths. However it was found thatthe decay of CP eigenstates does not preserve CP.

29

νµ νµ

νµ νµ

νµC and P Transformations for π µ+ +

π+ π+

π−π−__

µ µ

µµ

+ +

−−

s s

ss

CPC C

P

P

23 The operations of P and T

The operations of reflection and time reversal in classical systems is shown in Table below.

Name P TTime + -Position - +Energy + +Momentum - -Spin + -Helicity - +Electric Field - +Magnetic Field + -

Obviously some parameters are invarient but some change sign under this combined opera-tion.

24 Bilinear forms of Dirac wave functions

We recall that a Dirac wave function has 4-components, and that ~γ, ~α and β are 4 × 4matricies used in the dirac equation. As an aside note that the current ~j = cψ~αψ leadsto an expectation value of the velocity. We write the following bilinear forms that have thevarious listed transformation properites;

In the above,

~γ =

(

0 −~σ~σ 0

)

30

Bilinear Form Transformation Property

ψ ψ Scalar

ψ γn ψ Vector

ψ γ5 ψ Pseudoscalar

ψ γ5γn ψ Pseudovector

State Energy Helicity Chirality

1 > 0 + 1 +12 > 0 - 1 -13 < 0 + 1 -14 < 0 - 1 +1

γ0 =

(

−I 00 I

)

γ5 = −i(

0 II 0

)

where ~σ are the Pauli spin matricies, I is the 2× 2 idenity matrix, and γ5 = γ1γ2γ3γ5. Notethat the γi are the components of a relativistic 4-vector, and ψ is the adjoint of the Diracwave function ψ.

The helicity of the wave function is defined as the direction of the particle spin vector relativeto the momentum vector. It measures the handedness of a particle and is a pseudoscalarinvarient under T .

Σ = ~σ · ~p/|~p|

The chirality operator, γ5, operates on the helicity states to produce the chirality of thestate. We then find the helicity and chirality of the eigenvectors for the various Dirac states

For E > 0 states, a spatial reflection inverts the momentum vector and changes the sign ofthe helicity. The Chirality of a state is optained by the projection operator;

31

P± = 1/2(1 ± iγ5)

The projection operator has the properties;

P+ + P− = 0P 2± = 1P+P− = P−P+ = 0

25 The operations of P, C, and T

Conservation of the simultaneous application of C, P, and T is expected under very generalconditions. In all Lorentz invarient quantum field theories, CPT is a good symmetry. Thismeans that if CP is violated then T must be violated as well. Direct searches for T violationare difficult as null experiments are not easily designed.

26 Zero mass equation

In the case of zero mass the Dirac equation has the form;

[i ∂∂t

+ iγ0~γ · ~∇]ψ = 0

Which looks like ;

[E −(

0 ~σσ 0

)

]ψ = 0

If we divide the 4-component Dirac wave function into 2 two component wave functions de-scribed by upper and lower components ψu and ψl respectively, the zero mass dirac equationforms 2 equations;

Eψu − ~σ · ~p ψl = 0

Eψl − ~σ · ~p ψu = 0

Then a specific chirality state is not a state of specific parity, and can be described by a2-component wave function;

32

P

T

+

+

−

−

+ +

− −

+

+−

−

C

Symmetry Operations

33

ψ− = P− ψ =

(

0φ

)

ψ+ = P+ ψ =

(

φ0

)

and ;

γ5 ψ− = ψ−

γ5 ψ+ = −ψ+

Thus chirality is a good symmetry for massless particles. It represents the direction of thespin relative to the momentum vector, and divides masselss Dirac states into left and righthanded doublets.

27 Lagrangian

The lagrangian is a Lorentz scalar, and is composed of bilinear forms in a way to make thescalar. Thus for example a vector form must be contracted (dot product) with a vector.Two pseudo-scalars can be multipled together, etc.

28 Quarks

Historically, spectroscopy of the baryons and mesons led to a classification scheme of theseparticles in terms of their spin, parities, and a quantum number called flavor. We havealready discussed nucleon and strangeness quantum numbers which we now identify by dif-ferent flavors. In order to match the spectroscopy of the states we propose to construct abaryon of 3 substructures which are called quarks. Obviously quarks will have some mul-tiple of 1/3 of the electronic charge and have baryon number of 1/3. The introduction of3 possible states (proton like, neutron like, and strangeness like) gives an SU(3) symmetrywhich for 3 substructures has the symmetry SU(3)f (flavor in tis case);

SU(3) × SU(3) × SU(3)

Apply Young’s Tableau to optain the dimensions 10s + 8m,S + 8m,A + 1 The sub-labelsrepresent the symmetry of the states. Note that the final state must be anti-symmetric uponexchange of quarks because they are Fermions. The spin wave function could presumablyprovide the remaining component to obtain the overall anti-symmetry. The symmetric 10dimensional representation presents a problem. The weight diagram is shown in Figure 12

34

J = 3/2p +

∆∆∆∆ + +−− 0

ΣΣ +**

Σ * −0

Ξ *Ξ *− 0

Ω −

Figure 12: The ten dimensional representation of SU(3)

One of the components of this representation is the ∆++. The substructure wave function issymmetric and the spin wave function must also be symmetric because the spin of the ∆ is3/2. The only way to preserve ani-symmetry is to introduce another quantum number, color,which can be anti-symmetrize the wave function. Thus initially we used SUf(3). Now theSU(3) symmetry is the color symmetry of the quarks. There are 3 colors which are composedto provide and anti-symmetrize a color component of the wave function having dimension 1.

29 Classical Electromagnetism

Maxwell’s equations define electromagnetism;

~∇ · ~E = ρ/ǫ ~∇ · ~B = 0

~∇ × ~E = −∂ ~B∂t

~∇ × ~B = µ ~J + (1/c2) ∂~E∂t

Charge is conserved as is expressed by the equation of continuity.∂ρ∂t

+ ~∇ · ~J = 0 with

~J = ρ~V . This can be written in a Lorentz invarient form by reconizing that a 4-vector canbe formed. Thus;

xµ = (~x, t); and jµ = (~j, ρ)

35

This allows a 4-divergence of the form ∂∂xµ

= (~∇, ∂∂t

)

The electromagnetic fields , ~E, ~B, can be expressed in terms of potentials. Thus ~B = ~∇× ~A

and ~E = −~∇φ − ∂ ~A∂t

~A = (µ/4π)∫

d3x (~Jr ) ~E = (1/4π)

∫

d3x (ρr )

The potentials form a Lorentz 4-vector ( ~A, φ). Not only are Maxwell’s equations determinedby the 4-vector, but we take these potential forms as more fundamental than the fields them-selves, since potentials are used in the Lagrangian formulation. (also see the Bohm-Aharonoveffect described below) The fields are components of a Lorentz 4-component tensor of rank 2.

Fµ ν = ∂µAν − ∂µAν =

0 −Ex −Ey −Ez

Ex 0 −Bz By

Ey Bz 0 −Bx

Ez −By Bx 0

Thus in fully covarient form maxwell’s equations are written,

∂µ Fµν = jν for the inhomogeneous eqations

∂λ F µν + ∂ν F λµ + ∂µ F νλ ≡ 0 for the inhomogeneous eqns.

However, the 4-vector potential is not unique as the same field tensor is obtained under thepotential transformation;

Aµ → Aµ + ∂µ χ = ( ~A − ~∇χ, φ +∂χ∂t

)

Here χ is an arbitrary scalar function of position and time. This is called a gauge transfor-mation. Now for the transformation to keep the 4-vector potential relativisticly invarient,the transformation must obey a restricted class of gauge transformations (Lorentz gauge)such that;

∂µ Aµ = 0

∇2χ− (1/c2)∂2χ∂t2

= 0

36

BSource

Screen

path A

path B

Figure 13: A path integral around a Magnetic field

30 Bohm-Aharanov effect

Classically the force on a charge is independent of the gauge. However, in QM the phaseof the wave function depends on the gauge properties. In global symmetry the phase mustbe the same at all points and at all times and this connects the spatial values of the vectorpotential. Consider the plane wave function of a non-interacting particle wave.

ψ0 = Aei(~p·~x)/~

In the presence of a static vector potential ~B = ~∇ × ~A = 0. But the kinetic energy termhas the form;

−(~2/2m)[~∇ +iqA~c

]2 ψ = Eψ

The solution is ψ = eiφψ0

Now choose a path around the magnetic field as shown in the figure. Here φ = (q/~c)∫

path

~A·

~dl As the phase depends on the path length it will not necessarily return to the same value.This can be related to a geometric effect of moving a vector around a surface in 3-D. Thusthe potentials have an effect that is not contained in the fields.

31 Example of the non-relativistic Hamiltonian

Remember that any invariance (conservation) must be associated with a symmetry byNoether’s theorem. Consider a non-relativistic particle in the presence of an E-M field.The Hamiltonian for a charge, q, moving in an electromagetic field is;

H =(~p− q ~A)2

2m + V (x) + qφ

37

Here, ~p is the momentum, and the 4-vector potential of the electromagnetic field is ( ~A, φ).

We identify H = E (~p − q ~A) as the kinetic energy, so that the momentum ~p contains the

momentum of the charge as it interacts with the field. Thus (~p−q ~A) represents the conjugatemomentum to the coordinate, ~x. Apply the relations for one of the 3 coordinates;

ux = dxdt

= ∂H∂px

px = −∂H∂x

Operating on the Hamiltonian gives;

duxdt

= (1/m) [px − qAx]

so that ;

m duxdt

= [−∂H∂x

− qdAxdt

]

After working through the algebra and using a vector identity, the non-relativistic force ona charge particle moving with velocity ~u is obtained.

m duxdt

= −∂V (x)∂x

+ qEx + q/c (~u× ~B)x

32 Gauge transformation and symmetry

A particle in the presence of an E-M field has a conjugate momentum of, ~p → ~p− q ~A, andan effective energy, E → E − qφ. In Lorentz invarient form;

pµ → pµ − qAµ

In QM the momentum operator is pµ → i ∂µ so that the Dirac equation becomes;

[γµ(i∂µ − qAµ) − m]ψ = 0

Note here the summation convention, i.e. if an index is repeated it means summation overthe index. This γµ∂µ is a contraction producing a scalar (4-D dot product). We wouldcombine this into a Lagrange density (Lagrangian per unit volume) in the form;

L = ψ[γµ(i ∂µ − qAµ) − m]ψ − (1/4)FµνFµν

38

The first term represents the particle Lagrangian plus the interaction of the particle withthe field, and the second term is due to the energy contained in the electromagnetic field itself.

First note that the Lagrangian is invarient under the transformation ψ → eiα ψ for α aconstant. This is a global symmetry as the phase, eiα, represents a rotation of the systemwhich is everywhere the same (independent of spatial position). This symmetry provides aconservation principle. To see this suppose a small spatial variation, α → α + δα(x). Putthis into the Lagrangian density and vary the action.

δA =∫

d4x [ψγµψ] ∂µ(δα) = 0 to 1st order

Integrate by parts to get

δA =∫

d4x [∂µ(ψγµψ)] (δα) = 0

Which means for an arbitrary function, δα

∂µ(ψγµψ) = 0

This represents a conserved 4-vector current, jµ which may be explicitedly written;

∂P∂t

+ ~∇ · ~J = 0

Here P = ψψ is the probability density and ~J the probability current. This conservation lawstates that probability is conserved. Now

∫

P dx3 counts the number of particles minus thenumber of anti-particles so the symmetry leads to Lepton (Baryon) number conservation. Ifwe vary the fields, Aµ, we obtain a charge-current density which is conserved (conservationof charge) as previously shown.

However, a gauge transformation can depend on the spatial location, i.e. the scalar functionχ can depend on (~x, t). Thus suppose we consider the LOCAL transformation ψ → e−iqχ(x).Substitution into the Dirac equation yields;

e−iqχγµ[i∂µ + q∂µχ − qAµ]ψ = 0

Here we see the form Aµ + ∂µχ which is a gauge invarient. The dirac equation with the EMfield has a Global symmetry, U(1) group (charge conservation), and a local symmetry dueto gauge invarience. Suppose we choose to use the symmetry as the fundamental propertyof the EM field. Then the symmetry requires that the field be a gauge field. This invarienceis directly related to the zero mass of the photon. Quantum electrodynamics is an abeliangauge theory with the symmetry group U(1) and one gauge field SU(2)(photon)

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33 Weak decay

The electron has probability of 12[1 + v

c] to have left-handed helicity, and a probability of

12[1− v

c] to have right handed helicity. If there are no factors favoring helicity in weak decay,

the longitudinal polizarition of an electron will be;

[

−12(1 + v

c ) + 12(1 − v

c )]

= −vc .

For positrons the probabilities are reversed. The neutrino (zero mass) is left-handed, andthe anti-neutrino right-handed.

For π decay at rest;

π ss

eν__ν

eµ µ(( ))

Pion Decay at Restto Either µ or e

The probability of the decay is then;

[

−12(1 − v

c )]

= m2

m2π + m2 .

Which correctly predicts the rate for both e and µ decay. We then connect the left-handedelectron with the left-handed neutrino and put them in a doublet. The right-handed elec-tron component is placed in a singlet. This provides the symmetry of weak isospin, and inanalogy to nuclear isospin we define a weak hypercharge;

Y = 2(Q − T Y3 ).

34 Weak current

A weak hypercharge current can be introduced;

jYµ = ψ†γµY ψ.

Here Y is the weak hypercharge. For the electromagnetic interaction;

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Q = T Y3 + 1/2 Y

emµ = j3µ + 1/2jY

µ

An Electromagnetic interaction current can be composed of a linear combination of the 3rdcomponent of weak isospin, SU(2), coupled to the left-handed weak doublets and the weakhypercharge, U(1), coupled to both left- and right-handed singlets. This couples the electro-magnetic and weak interactions.

jYµ = −2(eRγµeR) − (χLγµχL).

One then couples the 3 three weak isospins to 3 vector gauge bosons and the weak hyper-charge to one vector gauge boson, B.

[gjµ ·W µ + g′jYµ B

µ].

The neutral components mix to produce a linear combination, one component being themassless photon and the other the massive, Z0.

Aµ = Bµ cos(θW ) + W 3µ sin(θW )

Zµ = −Bµ sin(θW ) + W 3µ cos(θW )

This is the standard model of the Electromagnetic and Weak interactions.

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