Physics 315, Problem Set 6
The generators of SU(3) may be obtained by writing down the most general 3x3 unitary
matrix with determinant equal to one and taking derivatives with respect to the eight independent parameters necessary to specify this most general group element. The typical form
of the SU(3) generators are defined from the so-called Gell-Mann matrices. They are:






0 1 0
0 −i 0
1 0 0






λ1 =  1 0 0  ; λ2 =  i 0 0  ; λ3 =  0 −1 0  ;
0 0 0
0 0 0
0 0 0






0 0 0
0 0 −i
0 0 1





λ4 =  0 0 0  ; λ5 =  0 0 0  ; λ6 =  0 0 1 
 ;
0 1 0
i 0 0
1 0 0




0 0 0
1 0 0
1 



λ7 =  0 0 −i  ; λ8 = √  0 1 0  ;
3 0 0 −2
0 i 0
The SU(3) generators are then given by Fi = λi /2. They satisfy the commutation relations [Fi , Fj ] = ifijk Fk , summation of repeated indices implied.
Unless specified, you will be able to use mathematica to perform many of the SU(3)related problems in this set. To utilize the SU(N) machinary:
• Run
“ln -s /var/phy/project/courses/phy315/HighEnergyPhysics
on your computer to link to the appropriate AddOn file.
∼/.Mathematica/Applications”
• Launch mathematica.
• Input “HighEnergyPhysics‘FeynCalc‘” into the mathematica window. The structure
constants are accessed by, for example, “SU3F[1, 4, 7]” for the i = 1, j = 4, k = 7
structure constant, f147 .
• Let me and problem@phy.duke.edu know if this doesn’t work.
1. (Not for mathematica) Show that the determinant of a general group element, eiαa Fa ,
where a = 1, ..., 8 is equal to one. You do not need to use an explicit matrix representation to do this.
2. (Not for mathematica) Show that the structure constants refered to above are antisymmetric.
3. (Not for mathematica) Prove that the structure constants satisfy, fijl flkm + fkil fljm +
fjkl flim = 0. This is the Jacobi identify. Summation is implied.
4. Examining the Gell-Mann matrices one notices the presence of Pauli-like matrices. In
fact, SU(2) is a subgroup of SU(3). There are three different (but non-orthogonal)
subgroups of SU(2) in SU(3) This can be seen by defining I-spin, U -spin, and V -spin:
I± = F1 ± iF2 ; I3 = F3
s
U± = F6 ± iF7
1
; U3 = − F3 +
2
3
F8
4
s
V± = F4 ∓ iF5
1
; V3 = − F3 −
2
3
F8
4
1
To see that one is redundant note that I3 + U3 + V3 = 0. Using the commutation
relations of the SU(3) generators above, find the commutators between and among
the nine I, U , and V operators. What patterns do you see?
5. In SU(2), only one of the three generators may be simultaneously diagonalized. We
use the eigenvalues of that diagonal generator
! to label!our states. For example, the
1
0
Pauli matrix σ3 has the eigenvectors
and
with associated eigenvalues
0
1
± 12 . In SU(3) on the other hand, we see that there are two matrices simultaneously
diagonalizable: F3 and F8 . The eigenvalues of both are needed to label the states.
The eigenvectors are obviously






0
0
1
 
 
 
0 , 1 , 0
1
0
0
(a) Find the three 2-component eigenvalues. These are called the “weights” m(i)
~
=
(m1 (i), m2 (i)); i = 1, 2, 3 of the representation. Plot them as though they were
three points in the (m1 , m2 ) plane. This is the “first fundamental” representation
of SU(3). Note the 120 degree symmetry.
(b) In the same way that ai , i = 1, 2 transforms as a 2 under SU(2), aj , j = 1, 2, 3
transforms as a 3 under SU(3). A vector bi , i = 1, 2, transforms as a 2 under
SU(2), which we have shown using the totally anti-symmetric Levi-Civita tensor
ij is equivalent to a 2. In SU(3), however, a 3, bj , j = 1, 2, 3 is distinct from a
3. The generators of 3 are the F i = −Fi∗ . Find the associated weights and plot
them as though they are three points in the (m1 , m2 ) plane. This is the “second
fundamental” representation of SU(3). Note the 120 degree symmetry.
(c) Noting that λ3 is just σ3 in the upper left 2x2 block, identify from your two
diagrams the two SU(2) doublets and two SU(2) singlets that are subsets of
these two SU(3) representations.
6. Find the roots ρ~(i) of SU(3). That is, solve the equation [Λ, X] = ρX for Λ = F3
and Λ = F8 . As we saw in class, this is equivalent to finding the eigenvalues and
P
P
eigenvectors of the i ai cijk , where Λ = j aj Xj and [Xi , Xj ] = cijk Xk . You will
find two pairs of (0, 0) eigenvalues; these correspond to F3 and F8 commuting with
themselves. The remaining six pairs correspond to the non-diagonal generators. Plot
them. Note the 120 degree symmetry.
7. From the corresponding eigenvectors to the roots above, find the six “raising and
lowering” operators.
8. The so-called “adjoint” representation may be found once the structure constants are
known via: (Ti )jk = −ifijk . Show that this representation indeed satisfies the commutation relations [Ti , Tj ] = ifijk Tk . Construct these for SU(2). Note their similarity
to the l = 1 SU(2) representation.
2