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seen Mar 28 at 13:21

Doing a PhD in Theoretical Particle Physics (QCD)


Jul
26
awarded  Scholar
Jul
26
accepted Mixed symmetrization and antisymmetrization / Combinatorics
Jul
25
comment Mixed symmetrization and antisymmetrization / Combinatorics
That's not the point - I know that there isn't necessarily a full (block)-symmetry for the trace. Point is that when calculating the 5-generator trace, I get 123 terms, of which I can simplify 120 to 1 term using the existing symmetries and my short-hand notation. For example the 4-generators can also be written as $\frac{1}{4}N\left(d^{abx}d^{xcd}-f^{abx}f^{xcd}\right)+\frac{1}{2}\left( \delta^{ab} \delta^{cd}+3\delta^{(ab)}\delta^{(cd)}\right)$. It is the last term that in the case of the 5-generator trace correspondingly can be written as $10\delta^{(ab\,|}f^{cde]}$.
Jul
23
comment Mixed symmetrization and antisymmetrization / Combinatorics
Ok, I finally see what you mean; indeed $\delta^{(ab\,|}f^{cde]}$ is not related to $\delta^{(cb\,|}f^{ade]}$ in a straightforward way, but I don't see why it should. Simply interchanging antisymmetrised indices has an easy relation $T^{[ab]}=-T^{[ba]}$, but with mixed symmetries this is not necessarily the case. I got this sum from calculating the trace of 5 SU(N) generators in the adjoint representation $\text{tr}\,T^aT^bT^cT^dT^e$; using the other order you provides equates to calculating $\text{tr}\,T^cT^bT^aT^dT^e$, which is a different quantity.
Jul
22
comment Mixed symmetrization and antisymmetrization / Combinatorics
To clarify: in one of my calculations, I got a sum of 120 terms (exactly the one in my previous comment), which I was able to reduce to the 10 terms above. Now I try to reduce it to one term, using standard notation (instead of my own 'mixed notation' mentioned above). Thanks!
Jul
22
comment Mixed symmetrization and antisymmetrization / Combinatorics
How would you use it in tensor notation? For example when contracting indices (which is in the end what I will do), I can write $f^{[abc]}g_{bcd}$; how would I write this in your notation? It seems that $f^{a[3]}g_{abc}$ cannot be resolved unambiguously
Jul
22
comment Mixed symmetrization and antisymmetrization / Combinatorics
I don't see why you get the minus signs.. My original sum is $\delta^{ab}f^{cde}-\delta^{ab}f^{ced}-\delta^{ab}f^{dce}+\delta^{ab}f^{dec}+\de‌​lta^{ab}f^{ecd}-\delta^{ab}f^{edc}+\delta^{ac}f^{bde}-\delta^{ac}f^{bed}+\ldots$ which collapses to the 10 terms I mention in the start (ignoring constant factors in front) if you take the (anti) symmetry of $\delta$ and $f$ into account.
Jul
22
comment Mixed symmetrization and antisymmetrization / Combinatorics
Btw, the two terms you mention won't cancel, as in my definition $\delta^{cd}f^{abe}$ and $\delta^{dc}f^{abe}$ do have the same sign. This is because the sign only depends on the number of permutations of the second list of indices.
Jul
22
comment Mixed symmetrization and antisymmetrization / Combinatorics
Actually, I am not trying to decompose my tensor product in irreps, but rather I am looking for an straightforward notation, like one could write $\delta^{(ab)}$ as a shorthand for $\frac{1}{2}\left(\delta^{ab}+\delta^{ba}\right)$. But I'll try whether I can get something from your answer. Thanks anyway for your help!
Jul
22
comment Mixed symmetrization and antisymmetrization / Combinatorics
It doesn't, because $\delta$ and $f$ are fully symmetric resp. antisymmetric.
Jul
19
awarded  Editor
Jul
19
awarded  Promoter
Jul
19
revised Mixed symmetrization and antisymmetrization / Combinatorics
edited body
Jul
19
comment Mixed symmetrization and antisymmetrization / Combinatorics
Anybody an idea? I am a bit stuck here..
Jul
17
awarded  Student
Jul
17
asked Mixed symmetrization and antisymmetrization / Combinatorics