Recent content by Thor Shen

the operator with hat likes the number operator as $$\hat{\pi}^+=\pi^{\dagger+}\pi^+$$ ,as well as $$\hat{\pi}^-=\pi^{\dagger -}\pi^-$$. And the operators without hat are satisfied with the boson commutation relation?

Well,thank you for your apply, but I still have some confusions. For example,the operator π+and π- act on the state |π+π->. π+π-|π+π->=|π+π->? (1) π+π-|0 0>=|π+π->? (2) <π+π-|π+π-=<π+π-|? (3) <0 0|π+π-=<π-π+|? (4) if we consider the crossing symmetry, we...

Note from mfb: I fixed the LaTeX formulas http://image.keyan.cc/data/bcs/2015/0428/w84h1446064_1430226899_628.jpg I met some trouble by using latex on this version, so I try to add the image from the pdf, the Latex code is as follow if any help. The lowest order lagrangian for ChPT is...

Thanks again firstly. In general, \delta^{ac}\delta^{bd}-\delta^{ab}\delta^{cd}=\varepsilon_{dae}\varepsilon_{bce}\neq 0, and a,b,c,d,e=1,2,3. Only when a=d or c=b are above deltas equal to 0.

Thank you for your recommendation. By the way , T^{I=2}(s,t)=<I=2,I_3=0|T^{abcd}|I=2,I_3=0> ? I and I3 are isospin and component of isospin for initial final state,respectively

Thank you for your reply. The traceless and symmetric for the tensor like the Young using in meson decomposed into octet and decuplet, right? Well, why the procedure that symmetrizing amplitude with subtracting its trace can stand for T2(s,t), <I=0,I3=0|Tabcd|I=0,I3=0>?(of course, I can get the...

Thank you so much! I am a newcomer in this field, so there are many definitions and notions I cannot catch up. Firstly I still cannot understand how to get the tensor 3\pi^a\pi^b-\vec{\pi}^2\delta^{ab} and find the way that symmetric amplitude subtracts its trace? Second problem is the...

How to use the standard techniques of projection operators to obtain the equation (1) by the first formula? Thanks

Actually, when I try to simplify the eq.2.3.153 for obtain the bracket in eq.2.3.154, I find we must use the two delta functions in eq.2.3.154, but I wonder whether we can use δ(1/4(q'+k')^2-m^2)=δ(k'^2+s-4m^2)δ(k'*q) directly, and the δ(k*q) is equivalent the δ(k'*q) because of the k' is...

http://d.kankan3d.com/file/data/bcs/2014/0508/w65h1446064_1399517186_873.jpg 1.How to deal with the delta functions in eq.2.3.153 to obtain the eq.2.3.154 by integrating over q'? 2.How to caculate the integral from eq.2.3.154 to eq.2.3.156, especially the theta function?

The diagram is from the proton-antiproton annihilation into two mesons.

Yes. Firstly, I write the same index for omitting the delta functions. But the two pairs will mislead using Einstein's reduction rule, the latter one should be \nu. Of course, the complete form should be written by Hepth.

Yes, the gauge bosons were gluons. I will take care next time,thanks!

the number stand for the index of particles (quarks and gluons) M=\bar{v}(p_2) ig_sT_{12}\gamma^\mu(12)u(p_1)\frac{-i}{p_7^2}\bar{u}(p_5) ig_sT_{56}\gamma_\mu(56)\bar{v}(p_6)\frac{-i}{m-\gamma^\mu p_{9\mu}}\bar{v}(p_3)...

if we define the \vec{p}'_1-\vec{p}_1=\vec{p}and \vec{r}_1-\vec{r}_2=\vec{r} then \int^{\infty}_{-\infty}d^3p\frac{1}{\vec{p}^2+\mu^2}e^{-i\vec{p}\cdot\vec{r}} =\int^{2\pi}_{0}d\phi\int^{1}_{-1}dcos\theta\int^{\infty}_{0}dp\frac{p^2}{p^2+\mu^2}e^{-ipcos\vartheta r} =4\pi^2\frac{e^{-\mu r}}{r}...