Can anyone explain this term is antisymmetric in SU(3)

[LAHLH]

Hi,

I'm reading the SU(N) chapter in Jones' Group theory book. In SU(3) we have these 3 component spinors which transform as $$\psi^{'}_{a}=U_{a}^{..b}\psi_{b}$$ and we have upper spinors defined by $$\psi^{a}=\epsilon^{abc}\phi_{[bc]}$$

Now if consider building up higher-dimensional reps, by taking Kronecker products. Starting with $$3 \otimes \bar{3}$$:

$$\psi_{a}\bar{\psi}^{b}=\left( \psi_{a}\bar{\psi}^{b}-\tfrac{1}{3}\delta^{b}_{a}\psi_{c}\bar{\psi}}^{c}\right)+\tfrac{1}{3}\delta^{b}_{a}\psi_{c}\bar{\psi}^{c}$$

Now I understand that the last term is invariant, which already suggests to me it's a singlet, but I can't see how this term is a completely antisymmetric combo of all three indices. Could anyone show me explicitley why this is?

 

[Avodyne]

Let $$\bar{\psi}^{c} =\epsilon^{cde}\phi_{[de]}$$

Then $$\psi_{c}\bar{\psi}^{c} =\epsilon^{cde}\psi_{c}\phi_{[de]}$$

So there are three indices which are completely antisymmetric due to the contraction with the epsilon symbol.

 

[LAHLH]

Thanks alot, I had convinced myself of it in a much more cumbersome way in the end.

Would you happen to know also why the first term $$\left( \psi_{a}\bar{\psi}^{b}-\tfrac{1}{3}\delta^{b}_{a}\psi_{c}\bar{\psi}}^{c} \right)$$ is constructed to be traceless. I understand why it is traceless, just not really why he emphasizes that we want it to be traceless. Do traceless tensors form their own invariant subspace too? just like antisymmetric/symmetric tensors? Since this first term is an object of mixed symmetry so on the basis of just antisymmetric/symmetric tensors forming invariant subspaces you wouldn't necessarily think this tensor would be an invariant subspace. Unless it's something to do with the tracelessness?

Finally is there anyway to see that this first term is of dimension 8? (other than just following the rules for Young Tableux quotient)

Thanks again.

 

[samalkhaiat]

Do traceless tensors form their own invariant subspace too? just like antisymmetric/symmetric tensors?

Yes. The prossess becomes more complicated when you deal with tensors that have both lower and upper indices like the mixed tensor (2,2).

Write

$$\phi_{a}\psi^{b} \equiv T_{a}{}^{b} = \left(T_{a}{}^{b} - \frac{1}{n}\delta_{a}^{b}T_{c}{}^{c} \right) + \frac{1}{n}\delta_{a}{}^{b}T_{c}{}^{c}$$

This separation ( traceless + scalar) is invariant under the action of the group SU(N). We simply divided the original $N^{2}$-dimensional representation space into two invariant subspaces of dimension $N^{2}-1$ and $1$, respectively.

Finally is there anyway to see that this first term is of dimension 8? (other than just following the rules for Young Tableux quotient)

Just count the number of components of each object. The prossess is as simple as writing,
9 = 8 + 1. For tensors of type (0,2) or (2,0) the divission is 9 = 6 + 3, the 6 represents the symmetric part and the 3 corresponds to the antisymmetric part;

$$T_{ab} = T_{(ab)} + T_{[ab]}$$

regards

sam

 

[LAHLH]

Thanks for the help samalkhaiat.

On a related note I'm trying to do q8.3 in Jones, not sure if you have the book, but basically it's looking at $$8 \otimes 8$$. So I know that the irrep of dimension 8 is carried by tensors of the form $$\hat{\phi}^{a}_{b}=\left( \psi_{a}\bar{\psi}^{b}-\tfrac{1}{3}\delta^{b}_{a}\psi_{c}\bar{\psi}}^{c} \right)$$ (where the hat means traceless, i.e. it is of the canonical form in the sense of syndey coleman). So from this I can deduce that $$8 \otimes 8$$ must be carried by tensors of the form $$\hat{\phi}^{i}_{j} \otimes \hat{\phi}^{k}_{l}$$ or in other words, $$\phi^{ik}_{jl}$$ (this time there is no hat because it is only traceless on ij or kl, not for e.g. il).

So first thing I need to do to get this into canonical form is make it completley traceless, but I don't quite know how. Judging from the answer in the back of the book Jones forms the completely traceless tensor as $$\hat{\phi}^{ik}_{jl}\rightarrow \phi^{ik}_{jl}-\delta^{i}_{l}\hat{\phi}^{k}_{j}-\delta^{k}_{j}\hat{\phi}^{i}_{l}-\delta^{k}_{j}\delta^{i}_{l}\phi$$ (where second and third terms are octets and last is singlet).

I can't see how such an object would now be traceless on all indices as it it supposed to be. e.g. tracing on il:
$$\hat{\phi}^{ik}_{ji}\rightarrow 0-3\hat{\phi}^{k}_{j}-\delta^{k}_{j}\hat{\phi}^{i}_{i}-3\delta^{k}_{j}\phi$$

 

[samalkhaiat]

Ok, so you want to decompose the tensor product $[8]\otimes [8]$. Let us consider an arbitrary SU(3) tensor of type (2,2). Let us call it $X^{ab}_{cd}$. Clearly, it has [81] components. Now we remove from it two octets and one singlet. This leads to a [64] components object ( = 81 – 8 – 8 – 1 ). If we can make it traceless in (ac) and (bd), then we can make the identification

$$[64]^{ab}_{cd} = ([8] \otimes [8])^{ab}_{cd}$$

The process is based on the simple identity

$$X = ( X - S ) + S$$

If we take S to be

$$S = [8] + [8] + [1],$$

[tex]X = [81][/itex]

then we can show that $T \equiv (X - S )$ is traceless in the pairs $(ac)$ and $(bd)$ and, therefore can be identified with $([8]\otimes [8])$. Indeed, it is easy to see that the following expression for S does the job we need;

$$S^{ab}_{cd} = (1/3) \delta^{a}_{c} X^{kb}_{kd} + (1/3) \delta^{b}_{d} X^{ak}_{ck} - (1/9) \delta^{a}_{c}\delta^{b}_{d}X^{ke}_{ke} \ \ (1)$$

To write this in terms of the octets $[8]^{b}_{d}$ and $[8]^{a}_{c}$, we only need to change the sign of the last term. Thus, starting from the arbitrary tensor $X^{ab}_{cd}$ we now have the following expression for the tensor $([8]\otimes [8])^{ab}_{cd}$;

$$T^{ab}_{cd} \equiv ([8]\otimes [8])^{ab}_{cd} = X^{ab}_{cd} - (1/3)\delta^{a}_{c}X^{kb}_{kd} - (1/3)\delta^{b}_{d}X^{ak}_{ck} + (1/9)\delta^{a}_{c}\delta^{b}_{d}X^{kj}_{kj} \ \ (2)$$

That is

$$([8]\otimes [8])^{ab}_{cd} = [81]^{ab}_{cd} - (1/3) \delta^{a}_{c}[8]^{b}_{d} - (1/3)\delta^{b}_{d}[8]^{a}_{c} - (1/9)\delta^{a}_{c}\delta^{b}_{d}[1]$$

Now you can repeat the process of subtracting two [8]’s and a [1], this time on the tensor $T^{ab}_{cd} \equiv ([8]\otimes [8])^{ab}_{cd}$ with respect to the cross indices (ad) and (bc). You will end up with a totally traceless, [47](= 64 – 8 – 8 – 1) components tensor. Again; by writing $T = (T - P) + P$, you find

$$(T - P )^{ab}_{cd} \equiv [47]^{ab}_{cd} = T^{ab}_{cd} - (1/3)\delta^{a}_{d}T^{kb}_{ck} - (1/3)\delta^{b}_{c}T^{ak}_{kd} + (1/9)\delta^{a}_{d}\delta^{b}_{c}T^{kj}_{jk}$$

Again in terms of the [8]’s and the [1], we have

$$[47]^{ab}_{cd} = ([8]\otimes [8])^{ab}_{cd} - (1/3)\delta^{a}_{d}[8]^{b}_{c} - (1/3)\delta^{b}_{c}[8]^{a}_{d} - (1/9)\delta^{a}_{d}\delta^{b}_{c}[1]$$

Next, divide $[47] \equiv \hat{G}$ into symmetric and anti-symmetric parts

$$\hat{G}^{ab}_{cd} = \hat{G}^{(ac)}_{bd} + \hat{G}^{[ac]}_{bd}$$

The second term can be written as $\epsilon^{ace}B_{(bde)}$ which is nothing but the totally symmetric decouplet [10] (try to prove it).

$$[47]^{ac}_{bd} = [37]^{(ac)}_{bd} + \epsilon^{ace}[10]_{(bde)}$$

Similar separation can be done on the lower indices resulting in

$$[37]^{(ac)}_{bd} = [27]^{(ac)}_{(bd)} + \epsilon_{bde}[\bar{10}]^{(ace)}$$

Finally, One might say; “give us a break! For god sake its only 1+8+8+10+10+27 = 8 x 8”.

regards

sam