Recent content by diraq

You confused yourself too much. It is merely a DEFINITION of the general states \Psi_p{}. You can show that those states defined in this way are complete. Naturally, under this definition, the C matrix for U(L) is diagonalized. This is a nice way to simplify the analysis without loss of any...

On page 237, Weinberg checked eq. (5.7.23) with an example when \mathbf p is along the three direction. Below that equation the phase factor \exp([-a + b - \tilde{a} + \tilde{b}]\theta)=\exp([2\tilde b-2a]\theta). Under the transformation p^0\rightarrow -p^0;\mathbf p\rightarrow -\mathbf p, the...

Consider the Fourier transform of a complex function f(t): f(t)=\int_{-\infty}^\infty F(\omega)e^{-i\omega t} Here t and \omega are on real axis. Let's suppose f(t) is square integrable. Here are my questions: 1) Since f(t) is square integrable, so we have...

Thank you very much. I think I have got the idea from your help.

Thank you very much. I am still confused. In- and out-states are non-interacting but according to Weinberg they are still the eigenstates of the full Hamiltonian. It looks like that Weinberg took a different perspective. The asymptotic hypothesis that will turn off the interaction at...

Hi All, The S-matrix is defined as the inner product of the in- and out-states, as in Eq. (3.2.1) in Weinberg's QFT vol 1: S_{\beta\alpha}=(\Psi_\beta^-,\Psi_\alpha^+) When talking about the Lorentz invariance of S-matrix, the Lorentz transformation induced unitary operator U(\Lambda,a)...

You are right. It can be converted to an ODE using Lapace transformation. But the difficulty is transported to using the boundary condition to determine the coefficients which depends on the Laplace variable s. Plus, even if the coefficients can be determined, I need to do the inverse Laplace...

Hi guys, I have trouble when solving the following heat transport equation in half plane in frequency domain. (\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})\theta(x,y)=i\beta\theta(x,y),-\infty<x<+\infty,y\geq 0...

Thanks!

Well, let's do [A,P_1]. [A,P_1]=[J_2,P_1]+[K_1,P_1]. Using eq. (2.4.21) and (2.4.22), we have [J_2,P_1]=i\epsilon_{213}P_3 and [K_1,P_1]=-iH. Since \epsilon_{213}=-1, I got the result [A,P_1]=-i(P_3+H)=-i(P^3+P^0). If possible, please post your derivation. Thanks.

I checked and there is no such sign mistake. I read from the following notes: http://www.physics.buffalo.edu/professors/fuda/Chapter_3.pdf Probably it is better to consider A^2+B^2 rather than A and B individually. But I haven't checked the commutation relations.

Thank you very much. The problem is [A,P^1]=-iP^3-iH=[B,P^2]=-iP^3-iH, so [A,P^1]\Psi_{k,a,b}=-2i\Psi_{k,a,b}=[B,P^2]\Psi_{k,a,b}\neq 0.

I am reading Weinberg's Quantum theory of fields Vol I and have a question about the derivation on page 71. Right below eq. (2.5.37), it is written that A and B can be simultaneously diagonalized by \Psi_{k,a,b}. From the content, I inferred that \Psi_{k,a,b} is also the eigenstate of the...