Recent content by Able

Thanks Vanadium 50

Can anyone explain to me what "twist" refers to in the context of QCD? More specifically, I have come across the terms "leading twist" and "higher twist".

Hi all. I'm looking at the decay of a neutral ρ meson to two neutral π's. I think it is disallowed but I can't figure out why. The J^{P} of the ρ is 1^{-} and 0^{-} for the neutral π's. The formula P(ρ^{0})=(-1)^{L}P(π^{0})P(π^{0}) then says that the decay is allowed provided the π's are...

Well I didn't off hand but I did some research. It seems to only be defined in conjunction with another function inside an integral. Then you use integration by parts to put derivatives on the other function. \int g(x)\delta^{\prime}(x-y)dx=-g^{\prime}(y) \int...

Ok, what about this H_{0}a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle=\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}\Bigr)a(x)a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle H_{0}a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle=\int...

Thanks for the reply strangerep. Acting the hamiltonian on the single particle state, we get a(x)a^{\dagger}(x) hitting the vacuum state. Using the bosonic commutation relation \left[a(x),a^{\dagger}(x^{\prime})\right]=\delta(x-x^{\prime}) we can write a^{\dagger}(x)a(x)+1 instead. a(x) kills...

Erm, how do I make that code into something legible?

Hi all. Here is an n-particle hamltonian. H=\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}+U(x)\Bigr)a(x)+\int d^{3}xd^{3}yV(x-y)a^{\dagger}(x)a^{\dagger}(y)a(x)a(y) Here is an n-particle state. \Bigr|\psi,t\Bigl\rangle=\int...