# Trace of SU(3) generators

1. Jan 6, 2016

### Safinaz

Hi all,

The trace of two SU(3) generators can be calculated by:
$T_{ij} T_{ji} = \frac{1}{2}$, now how to calculate the trace of SU(3) generators:
$T_{il} T_{lk} T_{kj} T_{ji}$ ?

2. Jan 6, 2016

### Staff: Mentor

What's wrong about $∑ T_{ii}$?

3. Jan 6, 2016

### Safinaz

I didn't understand, can you be more clear please ?

4. Jan 6, 2016

### Staff: Mentor

The trace of any matrix is defined as the sum of its diagonal entries.

5. Jan 6, 2016

### ChrisVer

Tr(AB) = Tr(BA)
then use the comm. relations ?

6. Jan 6, 2016

### samalkhaiat

Are you asking about a generator or the generators? I don’t see any group index on $T$! Any way, for a single $3 \times 3$ traceless matrix, you have
$$\mbox{Tr}\left(T^{4}\right) = \frac{1}{2} \left( \mbox{Tr}\left( T^{2} \right) \right)^{2} .$$
For $SU(N)$, all the relevant traces can be obtained from the following relation
$$T^{a}T^{b} = \frac{1}{2} \Big \{ \frac{1}{N} \delta^{ab} \mathbb{I}_{N} + (d^{abe} + i f^{abe})T^{e} \Big \} , \ \ \ \ (1)$$
where $f^{abc}$ is the totally anti-symmetric structure constant:
$$[ T^{a} , T^{b} ] = if^{abc}T^{c}, \ \ a,b,c = 1,2, \cdots , N^{2}-1 ,$$
and $d^{abc}$ are totally symmetric constants satisfying
$$\big \{ T^{a} , T^{b} \big \} = \frac{1}{N}\delta^{ab} \mathbb{I}_{N} + d^{abc} T^{c} .$$
So, if you multiply (1) by $T^{c}$ and trace over, you get
$$\mbox{Tr}\left( T^{a}T^{b}T^{c} \right) = \frac{1}{4} (d^{abc} + i f^{abc}) . \ \ \ \ \ (2)$$
Then, you can keep going, multiplying (1) by $T^{c}T^{d}$, taking the trace and using (2) and the Jacobi Identities you obtain the following ugly relation
$$\begin{equation*} \begin{split} \mbox{Tr}\left( T^{a}T^{b}T^{c}T^{d} \right) =& \frac{1}{4N}\left( \delta^{ab}\delta^{cd} + \delta^{ad}\delta^{bc} - \delta^{ac}\delta^{bd} \right) \\ & + \frac{1}{8} \left( d^{abe}d^{cde} + d^{ade}d^{bce} - d^{ace}d^{bde} \right) \\ & + \frac{i}{8} \left( d^{abe}f^{cde} + d^{cde}f^{abe} \right) . \end{split} \end{equation*}$$
Another very useful identity for calculating traces in $SU(N)$ is
$$(T^{a})_{ij}(T^{a})_{kl} = \frac{1}{2} \left( \delta_{il}\delta_{jk} - \frac{1}{N} \delta_{ij}\delta_{kl} \right) , \ \ i,j = 1, 2, \cdots , N$$
Note that the left hand side is summed over the group index $a = 1, \cdots , N^{2}-1$. This identity can be used, for example, to calculate
$$\mbox{Tr}\left( T^{a}T^{b}T^{a}T^{c} \right) = - \frac{1}{4N} \delta^{bc} .$$

7. Jan 7, 2016

### vanhees71

Where is the relation in the second equation from below from? Is it really correct? Take the SU(2) adjoint representation, where
$$(T^a)_{ij} =-\mathrm{i} \epsilon_{aij}$$
and (Einstein summation relation implied)
$$(T^a)_{ij} (T^a)_{kl} =-\epsilon_{aij} \epsilon_{akl}=-(\delta_{ik} \delta_{jl}-\delta_{il} \delta_{jk}).$$

8. Jan 7, 2016

### Safinaz

Thanks guys for these inclusive answers.

9. Jan 7, 2016

### samalkhaiat

Okay, the question “where does it come from?” can be answered by deriving the identity. And since that identity is very important in QCD calculations, I will derive it in here.
Consider the algebra of $SU(n)$
$$[ \frac{\tau^{a}}{2} , \frac{\tau^{b}}{2} ]_{ij} = i f^{abc} \left(\frac{\tau^{c}}{2}\right)_{ij} , \ \ a = 1, 2, … , n^{2}-1, \ \ i = 1, … , n$$
Please observe the range of the of the group index $(a)$ and that of the fundamental matrix representation $(i)$.
We can always consider the normalization
$$\mbox{Tr}\left(\tau^{a}\tau^{b}\right) = 2 \delta^{ab} . \ \ \ \ \ \ \ (1)$$
Since the $n \times n$ hermitian traceless matrices $\tau^{a}$, together with the $n \times n$ identity matrix $\mathbb{I}_{n}$ form a complete set of $n \times n$ hermitian matrices, we can expand any arbitrary $n \times n$ hermitian matrix $M$ in terms of them (repeated indices are summed) :
$$M = C \ \mathbb{I}_{n} + C_{a} \tau^{a} . \ \ \ \ \ (2)$$
Using (1) and the tracelessness of the $\tau^{a}$’s, the coefficients $C$ and $C_{a}$ are calculated as follows
$$C = \frac{1}{n} \mbox{Tr}\left( M \right) = \frac{1}{n} \delta_{jk}M_{kj} , \ \ \ \ \ (3)$$
$$C_{a} = \frac{1}{2} \mbox{Tr}\left( M \tau^{a} \right) = \frac{1}{2} (\tau^{a})_{jk}M_{kj} . \ \ \ \ \ (4)$$
Substituting (3) and (4) back in (2), we find
$$M = \frac{1}{n} \delta_{jk} \ M_{kj} \ \mathbb{I}_{n} + \frac{1}{2} \tau^{a} \ (\tau^{a})_{jk} \ M_{kj} .$$
Okay, take the $il$ matrix element of the above
$$M_{il} = \frac{1}{n} \delta_{jk} \ M_{kj} \ \delta_{il} + \frac{1}{2} (\tau^{a})_{il} \ (\tau^{a})_{jk} \ M_{kj} .$$
Now, one left-hand-side substitute
$$M_{il} = \delta_{ik} \delta_{lj} M_{kj} .$$
Then, from the fact that $M_{kj}$ is arbitrary, we obtain
$$\delta_{ik} \ \delta_{lj} = \frac{1}{n} \delta_{jk} \ \delta_{il} + \frac{1}{2} \left(\tau^{a}\right)_{il} \ \left(\tau^{a}\right)_{jk} ,$$
which is just the identity in my previous post, with $T = \tau /2$. Warning: Please notice the range of the indices in the identity. $(i,j,k,l)$ run from $1$ to $n$, while $a$ is summed over from $1$ to $n^{2}-1$.
I can not believe you make such mistake! Read the warning regarding the indices. For $SU(2)$ the indices $(i,j,k,l)$ take values in $\{1,2\}$ while $a = 1,2,3$. The correct form of the generators in the adjoint representation is
$$\left( \frac{\tau^{a}}{2}\right)_{bc} = - i \epsilon_{abc} .$$

Last edited: Jan 7, 2016
10. Jan 8, 2016

### vanhees71

I don't get it. Of course, the adjoint representation is for spin 1 and thus 3-dimensional, but for spin 1/2 (the fundamental representation) 2D. So, how is the discrepancy between the two formulas resolved? As far as I see from your derivation, it's only valid in the fundamental representation(s) of SU(N).

11. Jan 9, 2016

### samalkhaiat

For $SU(2)$, the matrix $M$ in the previous post is $2 \times 2$, and the identity is just the following relation for Pauli matrices:
$$\frac{1}{2} \sigma^{a}_{ij}\sigma^{a}_{kl} = \delta_{il} \delta_{jk} - \frac{1}{2} \delta_{ij} \delta_{kl} , \ \ \forall (i,j,k,l) \in \{1,2\} .$$
From this, as you can check, we obtain the following correct relations
\begin{align*} \left( \sigma^{a}\sigma^{a} \right)_{il} &= \delta_{jk} \sigma^{a}_{ij}\sigma^{a}_{kl} \\ &= 2(2\delta_{il} - \frac{1}{2}\delta_{il}) \\ &= 3 \delta_{il}, \\ \mbox{Tr}(\sigma^{a}\sigma^{a}) &= 6 . \end{align*}
Yes, the adjoint representation of $SU(2)$ is 3-dimensional. This means that the generators matrices in the adjoint representation are $3 \times 3$, and there are 3 of them. So, your objects $(T^{a})_{ij} = - i \epsilon_{aij}$ with $a \in \{1,2,3 \}$ and $(i,j) \in \{1,2 \}$ are not correct because: (1) $\epsilon_{aij}$ is a $2 \times 2$ matrix, and (2) $T^{1} = T^{2} = 0$ and $T^{3} = \sigma^{2}$. So, $(T^{a})_{ij} = - i \epsilon_{aij}$ is not a representation, let alone adjoint representation.
You created the discrepancy by wanting the identity to apply to the single $2\times2$ matrix $-i \epsilon_{aij}$ which is not even a representation.
Yes, and I was stressing this fact all along, by specifying the range of different indices. However, this does not mean that the identity is good only for the fundamental representation (quarks). In fact, it is very useful in calculating the colour factors in gluon loop.
The sum over colour indices in gluon loop can be represented by either the trace factor $t_{2}(A)$:
$$\mbox{Tr}(A^{b}A^{c}) = t_{2}(A) \delta^{bc} ,$$
or by the eigenvalue of the quadratic Casimir operator in the adjoint representation $C_{2}(A)$
$$(A^{a}A^{a})_{bc} = C_{2}(A) \delta_{bc} ,$$
where $A^{a}$ is the generator in the adjoint representation, as required by the gluon field
$$(A^{a})_{bc} = - i f_{abc} = - \frac{1}{4} \mbox{Tr}\left( \tau_{c} [ \tau_{a} , \tau_{b} ] \right) .$$
Since the structure constant $f_{abc}$ of $SU(n)$ is totally antisymmetric, one can easily show that
$$\mbox{Tr}(A^{a}A^{b}) = \left( A^{c} A^{c} \right)_{ba} = f_{acd}f_{bcd} .$$
So, in order to calculate the value of the factor $t_{2}(A) = C_{2}(A)$, we need to compute
\begin{align*} f_{acd}f_{bcd} &= - \frac{1}{16} \mbox{Tr}\left( \tau_{d}[ \tau_{b} , \tau_{c} ] \right) \mbox{Tr}\left( \tau_{d}[ \tau_{a} , \tau_{c} ] \right) \\ &= - \frac{1}{16} \tau^{d}_{ij} \ [ \tau^{b} , \tau^{c} ]_{ji} \ \tau^{d}_{kl} \ [ \tau_{a} , \tau_{c} ]_{lk} . \end{align*}
Now, using our identity
$$\tau^{d}_{ij} \ \tau^{d}_{kl} = 2 \left( \delta_{il}\delta_{jk} - \frac{1}{n} \delta_{ij}\delta_{kl} \right) ,$$
we obtain
$$f_{acd}f_{bcd} = \frac{-1}{8} \mbox{Tr} \left( [ \tau^{b}, \tau^{c}] [ \tau^{a}, \tau^{c}] \right) - \frac{1}{n} \mbox{Tr}\left( [ \tau^{b} , \tau^{c}] \right) \mbox{Tr}\left( [ \tau^{a} , \tau^{c}] \right) .$$
The second term is zero because the trace of $[ A , B ]$ is zero for any two matrices $A$ and $B$. The first term, when expanded, consists of 4 terms each of which has a factor of the form $\tau^{e}_{ij}\tau^{e}_{kl}$. So, we can apply our identity to each term. I leave it for you as exercise to show that
\begin{align*} \mbox{1st term} &= \mbox{Tr}(\tau^{b}\tau^{c}\tau^{a}\tau^{c}) = - \frac{4}{n} \delta^{ab} \\ \mbox{2nd term} &= \mbox{Tr}(\tau^{b}\tau^{a}\tau^{c}\tau^{c}) = 4( n - \frac{1}{n}) \delta^{ab} \\ \mbox{3rd term} &= 4(n - \frac{1}{n}) \delta^{ab} \\ \mbox{4th term} &= - \frac{4}{n} \delta^{ab} . \end{align*}
This gives us
$$f_{acd}f_{bcd} = \frac{-2}{8} \Big \{ - \frac{4}{n} - 4 \left( n - \frac{1}{n} \right) \Big \} \delta^{ab} = n \delta^{ab} .$$
Which means that $t_{2}(A) = C_{2}(A) = n$.

12. Jan 9, 2016

### vanhees71

I think it's simply a misunderstanding. You talk about the fundamental representation and not about a general representation, and I thought it should be valid in any representation.

Of course the adjoint representation of su(2) (Lie algebra) is so(3) and you can use $(T^a)_{ij}=-\mathrm{i} \epsilon_{aij}$ as a basis of the Lie algebra in this representation.

13. Jan 9, 2016

### samalkhaiat

If and only if, the index $a$ and the pair $(i,j)$ take values in the same set $\{1,2,3\}$.

14. Jan 10, 2016

### vanhees71

Sure, that's the adjoint representation. You can "naturally" define it on the Lie algebra itself.