Regarding the sign problem for the x_0 > y_0 case, I think the problem arise as you do not clearly specify the circulation direction of the contour. If you specify the direction of the lower-part contour (the case that we have a sign problem) to be clockwise, you will get the negative sign from the Residue theorem, and this should save everything. Thank you very much for these great videos!
Thank you for the nice video! around 24:58, when you "re" expands the factorization after considering p^0 = -E_p, I think we need to also multiply -i to the spatial -p(x-y) part in the exponential. So in this case, wouldnt the exponential be something like exp(iE_p(x-y) + ip(x-y))? If this is the case I dont see how we could get exp(i(E_p(x-y) - p(x-y)))
I think its not directly setting P^0=-E_P to P^0=E_P. For this we have to change the full four momentum P to -P and so both spatial and temporal both term have an extra minus. This change implies, P^0=-E_P to P^0=E_P. And obviously we can change three momentum P to -P because of the symmetric integral.
Hi Nick, I got stuck @37:45 - why is the time derivative only acting on "x" and not on "y" ? What makes the field's position "x" to be "better" than "y" ? Please help me !
[\del_\mu \phi(x),\phi(y)] = \del_\mu [\phi(x),\phi(y)] + [\del_/mu,\phi(y)] \phi(x) 2nd term in the RHS = 0 as derivative is a commutative operator. Commutator identity: [AB,C] = A [B,C] + [A,C] B
Regarding the sign problem for the x_0 > y_0 case, I think the problem arise as you do not clearly specify the circulation direction of the contour. If you specify the direction of the lower-part contour (the case that we have a sign problem) to be clockwise, you will get the negative sign from the Residue theorem, and this should save everything. Thank you very much for these great videos!
This is golden content for people like me who likes to study physics in amateur fashion.
Thank you for the nice video!
around 24:58, when you "re" expands the factorization after considering p^0 = -E_p, I think we need to also multiply -i to the spatial -p(x-y) part in the exponential. So in this case, wouldnt the exponential be something like exp(iE_p(x-y) + ip(x-y))? If this is the case I dont see how we could get exp(i(E_p(x-y) - p(x-y)))
true.
true..did u fine sol for this?
I think its not directly setting P^0=-E_P to P^0=E_P.
For this we have to change the full four momentum P to -P and so both spatial and temporal both term have an extra minus. This change implies, P^0=-E_P to P^0=E_P.
And obviously we can change three momentum P to -P because of the symmetric integral.
Amazing lectures - It's great to see the full mathematical detail. I can't wait to see more.
Hi Nick, I got stuck @37:45 - why is the time derivative only acting on "x" and not on "y" ? What makes the field's position "x" to be "better" than "y" ? Please help me !
[\del_\mu \phi(x),\phi(y)] = \del_\mu [\phi(x),\phi(y)] + [\del_/mu,\phi(y)] \phi(x)
2nd term in the RHS = 0 as derivative is a commutative operator.
Commutator identity: [AB,C] = A [B,C] + [A,C] B
@@rafidbuksh9554 OMG ! Thanks a lot. Yes, that makes sense. You made my day!
Thank you Nick. It is appreciated.
I can't believe they named a propagator after me
Love it! Just what I’m looking for!
27:00 How can you do p to -p here? You had not done it in previous video.
The solution is this: p goes to -p and the integral is the same! the 24:58 is resolved!
When p goes to -p doesn't the volume element d3p also pick up a minus sign?
@@photonsphere5920 yes , but then you also change the order in which you integrate therefore, the sign is absorbed by the integral.
anyone find a solution for the sign problem at 24:58 yet?
@NickHeumannUniversity please help
Gracias por los videos de verdad. ❤❤❤ Desde 🇵🇪
please upload more videos on qft
Hello! Where to start from in QFT?
Hey @caioishikawa8682, i concorde with you! in 24.58 there is an error in the exponential. This problem can be resolved?