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Subatomic Particles and Lie Algebras

Background

Physics tries to bring simplicity and order to seemingly complex nature. Consider the case of matter. First the multitude of substances was reduced to a smaller but still large number of chemical compounds. Then compounds were explained as combinations of a relatively small number of elements. Elements were then interpreted as electrons and nuclei with the nuclei composed of subatomic particles. First the proton was identified and later the neutron. This seemed to be all that was needed to explain the composition of the hundreds of different nuclei. But the picture began to get complicated as mesons and antiparticles such as the positron were found. More particles were discovered, and as in the line from Lewis Carroll's poem The Walrus and the Carpenter,

Thick and fast
They came at last,
And more and more and more.
.

Soon there were several hundred distinct subatomic particles. The situation cried out for organization.

The Coding of Qualitative Information About Particles

Particles seemed to fall into groups and it was convenient to code that qualitative information about group membership as numbers, usually 0 and 1, 1 if the particle had a certain characteristic and 0 if not. Charge could be coded as +1, 0 and -1. Heavy particles associated with the nucleus of atoms such as proton, neutron and mesons were called baryons and given a baryon number B equal to 1. This distinguished them from the lighter particles such as the electron which were called leptons and had a baryon number equal to 0.

After the neutron was discovered and its properties, except for charge, were found to be so similar to the proton Werner Heisenberg conjectured the proton and neutron were simply different states of the same particle, Heisenberg characterized the property that distinguished a proton nucleon from a neutron nucleon as isospin and gave the proton an isospin of +1/2 and the neutron nucleon an isospin of -1/2.

When certain particle transitions were found not to occur even though they did not violate the conservation of energy and conservation of charge and the other conservation principles some physicists conjured up a strangeness property of particles and a corresponding conservation principle, the conservation of strangeness number to explain the nonoccurrence of transitions. The transitions did not occur because they did not conserve the strangeness number.

The code numbers for properties of particles could be arithmetically combined. For example, the sum of the baryon number B of a particle and its strangeness number S was defined as the hypercharge Y for a particle; i.e., Y = B + S. This process of coding qualitative information and carrying out arithmetic on the results seemed like a harmless but not very profound activity for physicists. But some interesting results started emerging. When the isospin and hypercharge of a family of particles were plotted the graph showed some simple geometric structures such as triangles and hexagons, as shown below.

Particle Characteristics
ParticleCharge
Q
Baryon
Number
B
Isospin
I3
Strangeness
Number
S
Hypercharge
Y = B + S
Baryons
Proton+1+1+1/20+1
Neutron0+1-1/20+1
Σ--1+1-1-10
Σ00+10-10
Σ++1+1+1-10
Λ00+10-10
Ξ--1+1-1/2-2-1
Ξ00+1+1/2-2-1
Mesons
Κ000-1/2+1+1
Κ++10+1/2+1+1
π--10-100
π000000
π++10+100
η000000
Κ-00-1/2-1-1
Κ000-1/2-1-1

In the table Κ0 represents the antiparticle of Κ0.

There is an interesting formula relating charge, hypercharge and isospin. It is:

Q = I3 + (1/2)Y

This is called the Gell-Mann-Nishijima formula.

When the data for baryon isospin and strangeness number is plotted the result is the hexagonal pattern below. The same plot for meson gives the same pattern.

Lie Algebras

Weight Diagrams

The Special Unitary Group of 2x2 matrices, SU(2), can be generated by the 2×2 identity matrix and the following three matrices;

1/20
0-1/2
             
0  1
0  0
             
0  0
1  0

Note that the trace, the sum of the elements on the principal diagonal, of each of these matrices is zero.

These matrices need to be given names and they will be referred to as I3, I+ and I-, respectively.

Consider now the vectors

p =
1
0

and

n =
0
1
.

The product of I3 with p is:

 
1/20
0-1/2
 
1
0
=  
1/2
0
      
= (1/2)
1
0
= (1/2)p.

Therefore p is an eigenvector of I3 and (1/2) is its eigenvalue. It is also true that n is an eigenvector of I3 but its eigenvalue is (-1/2) as shown below:

I3n =
1/20
0-1/2
 
0
1
=  
0
-1/2
      
= (-1/2)
0
1
= (-1/2)n.

In contrast,

I-p =
00
10
 
1
0
=   
0
1
 
= n.

Therefore p is not an eigenvector of I-.

Howerever,

I+p =
01
00
 
1
0
=
0
0
  = 0  
1
0
      
= 0p

so p is an eigenvector of I+ with an eigenvalue of 0.

The vectors p and n are eigenvectors of the identity matrix I, as are all vectors, because Ip = p and In = n and their eigenvalues are both +1.

There is a special matrix, called the Casimir operator, which for the above system is equal to:

I+I- + I32 - I3.

This matrix will be called IT2. Its value is:

3/40
03/4

Since this matrix is just (3/4)I it is easy to see that p and n are eigenvectors of it and the eigenvalues are both +1. It is also true that IT2 and I3.

When the pairs of eigenvalues, (+1/2,+1) for p and (-1/2,+1) are plotted the result is called a weight diagram. It is simple and symmetric as shown below.

The diagram is neat but not profound. SU(2) does not go very far in explaining subatomic particles.

The Special Unitary Group of 3x3 matrices, SU(3) is more interesting. It can be generated by the identity matrix and the following eight matrices;

I3 =
1/2   0   0
0   -1/2   0
0   0   0
             
Y =
1/3   0   0
0   1/3   0
0   0   -2/3

I+ =
0   1   0
0   0   0
0   0   0

I- =
0   0   0
1   0   0
0   0   0

U+ =
0   0   0
0   0   1
0   0   0

U- =
0   0   0
0   0   0
0   1   0

V+ =
0   0   1
0   0   0
0   0   0

V- =
0   0   0
0   0   0
1   0   0

In the matrix pairs I+ and I-, U+ and U- and V+ and V- one is the transposes of the other. Note again that the trace, the sum of the elements on the principal diagonal, of each of these matrices is zero.

Consider the products of I3 and Y; i.e., I3Y and YI3. The computation shows that I3Y and YI3 are equal to the same matrix. This is an important result. It means that I3 and Y have the same eigenvectors.

Consider now the vectors

u =
1
0
0

and

d =
0
1
0
.

and

s =
0
0
1
.

Some quick computations show that I3u = (1/2)u and Yu = (1/3)u. Therefore u is an eigenvector of both I3 and Y and the eigenvalues are 1/2 and 1/3. Another quick computation shows that I3d = (-1/2)d and Yd = (1/3)d. Likewise I3s = (0)s and Ys = (-2/3)s. So the pairs of eigenvalues of I3 and Y associated with their mutual eigenvectors u, d and s are (1/2, 1/3), (-1/2, 1/3) and (0, -2/3). These eigenvalue pairs may be plotted in a weight diagram as shown below.

The vectors u, d and s are the basis vectors for the column vector space. There is also a dual space of row vectors which has as basis vectors

u = (1, 0, 0), d = (0, 1, 0) and s = (0, 0, 1).

For this dual space the representations are the negative of the matrices in the primal space. Thus to find the result of the action of I3 on the vector u we take the product of u with -I3. Thus ρ(I3))u = u(-I3) = (-1/2)u. Therefore the eigenvalue of I3 associated with u is -1/2. Likewise the eigenvalue of Y associated with u is (-1/3). The eigenvalues of I3 and Y associated with d are (1/2,-1/3). For s the eigenvalues are (2/3,0). When these eigenvalue pairs are plotted in a weight diagram with those for (u,d) and s the results are as shown below.

While the above weight diagram is interesting the more interesting result comes from using combinations of vectors, such as (u,d. The effect of I3 acting on the pair (u,d is defined as follows:

I3(u,d) = (I3(u),d) + (u,I3(d)
= (I3u,d) + (u,dI3)
= (1/2)u,d) + (u,(1/2),d)
= (1/2)(u,d) + (1/2)(u,d)
= (u,d)
 

Thus (u,d) is considered an eigen(dyadic)vector of I3 with eigenvalue of +1. On the other hand Yu=(1/3)u and dY=(-1/3)d so

Y(u,d) = (Y(u),d) + (u,Y(d)
= (Yu,d) + (u,dY)
= (1/3)u,d) + (u,(-1/3),d)
= (1/3)(u,d) + (-1/3)(u,d)
= (0)(u,d)

so (u,d) is also an eigenvector of Y but with an eigenvalue of 0.

If the eigenvalue of I3 is identified with the component of isospin and the eigenvalue of Y with hypercharge then the particle in the table with isospin of 1 and hypercharge equal to 0 is the π+ meson. The convention is to represent the π+ meson as -ud.


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