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Doing a PhD in Theoretical Particle Physics (QCD)


1d
comment Gauge Field Tensor from Wilson Loop
@crackjack If you want, I could call it the natural emergence of 'a set of gauge-equivalent functions' instead :-).
1d
comment Gauge Field Tensor from Wilson Loop
@crackjack Why do I imply it should be unique? Of course it is not, I can gauge it into another function. But this doesn't change anything to the natural emergence. All possible parallel transporters are related by a gauge transformation, so whatever function I have in there, from the moment it enters the Lagrangian it is equivalent, as the Lagrangian is gauge invariant.
2d
comment Gauge Field Tensor from Wilson Loop
It's cool, I never realised lattice QCD could be relevant to me :-). Is the paper yours?
2d
comment Gauge Field Tensor from Wilson Loop
Ok, thank you, your edit clarifies it for me. To get the field tensor to emerge naturally, one first has to discretise spacetime on a grid with spacing $\epsilon$. We consider a loop around an elementary 4Dcube, leading to something of the form $1+\epsilon^4 F_{\mu\nu}$. Taking the continuum limit, we divide by the volume and have the expression $\lim_{A\rightarrow\infty} A + F_{\mu\nu}$ (we don't care about the infinite term as we put it in the normalisation of the path integral and gets divided out). Is that it?
2d
comment Gauge Field Tensor from Wilson Loop
It's not about the term being allowed or not - I know how to gauge a QFT - but about the natural emergence of the correct kinetic terms based on geometric arguments. See the edit. This is not the 'standard' way to construct a Lagrangian.
Aug
16
comment Gauge covariant derivative in different books
If you want to know in which formulas this sign convention possibly makes a difference, just substitute $g\rightarrow -g$.
Aug
16
revised Gauge covariant derivative in different books
added 1312 characters in body
Aug
16
answered Gauge covariant derivative in different books
Aug
16
comment Gauge Field Tensor from Wilson Loop
Any idea on the $\epsilon^4$ factor?
Aug
16
revised Gauge Field Tensor from Wilson Loop
Added (a lot of) clarification at the end.
Aug
15
comment Gauge Field Tensor from Wilson Loop
Yes, but that doesn't really solve the problem of the $\epsilon^4$. See eg. Peskin & Schroeder eq. 15.66. How to interpret this $\epsilon$? To make the identification with the Lagrangian it should be 1, but this kind of fine-tuning doesn't make sense to me, and plus $\epsilon$ was taken to be infinitesimal...
Aug
15
awarded  Commentator
Aug
15
comment Gauge Field Tensor from Wilson Loop
Where I am having difficulties with is that using this method, a factor $\epsilon^4$ comes in front of $F_{\mu\nu}^2$, with $\epsilon$ an infinitesimal parameter being the length of one side of the small rectangular loop. How do I interpret this $\epsilon^4$ physically? I would need to fine-tune the loop to unit length in order to get the correct factor in front of the tensor in the Lagrangian..
Aug
15
awarded  Supporter
Aug
15
revised Gauge Field Tensor from Wilson Loop
Corrected the typo in the power of F_{\mu\nu} in the last equation: was 4, should be 2.
Aug
15
asked Gauge Field Tensor from Wilson Loop
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.