Recent content by Xenosum

. Nvm I think I got it. The point is that you're supposed to contract the spinors along a single fermion line without ever crossing over the propagator. Then the contraction will be a direct contraction of spinor indices (the spinors will lie next to each other, not "the same number of terms...

Homework Statement This is more of a general question-- as the title suggests I'm not too sure how to place the terms given by the Feynman rules for fermions (since they involve operators and spinors, the order does of course matter). I've been reading Peskin & Schroeder and the rules are...

Okay but I'm also a little bit concerned about the transformations I have written down. In Peskin and Schroeder, it's claimed that "We can describe the infinitesimal translation x^{\mu} \rightarrow x^{\mu} - a^{\mu} alternatively as a transformation of the field configuration \phi(x)...

Well, I suppose it's possible that this transformation is only a symmetry for a massless scalar field field, ##m=0##... I don't know why I didn't think of this at the outset (derp) but in any case there was a number of things wrong with my procedure (most importantly, it's invariance in the...

Ah, I forgot to include the differentials in calculating the invariance of the kinetic term! Thanks! Though, it still seems like the same problem arises. When I expand out \mathrm{d}^4 x' (\partial_{\mu}' \phi(x')) (\partial'{}^{\mu} \phi'(x')) = \mathrm{d}^4 x (\partial_{\mu} \phi(x))...

I see, thanks, but I'm a bit thrown off because the condition of the mass term to remain invariant is m^2\phi(x) = m^2\phi^{'}(x^{'}) = m^2\lambda^{-D}\phi(x) So that ##D=0##. However the condition for the derivative term to remain invariant requires a different ##D## as shown above...

Homework Statement Take the action S = \int d^4x \frac{1}{2} \left( \partial_{\mu}\phi(x)\partial^{\mu}\phi(x) - m^2\phi^2(x) - g\phi(x)^p \right) , and consider the following transformations: x^{\mu} \rightarrow x^{'\mu} = \lambda x^{\mu} \phi(x) \rightarrow \phi^{'}(x) =...

Ah, I think I understand. Thanks!

I might be missing something extremely basic here, but what guarantees that <0|\phi(y) corresponds to the same particle that was created by \phi(x)|0> ? The way I understand it, \phi(x) is an operator whose states live in the so-called fock space-- the multi particle space-- and the...

Cool, thanks!

Homework Statement The Feynman Propagator is given by <0| T \phi(y)\phi(x) |0> , where T is the time-ordering operator. I understand that this turns out to be the solution to the inhomogeneous klein-gordon equation, etc., but is there any intuitive description of the propagator? Can this...

Homework Statement Consider the Lagrangian, L, given by L = \partial_{\mu}\phi^{*}(x)\partial^{\mu}\phi(x) - m^2\phi^{*}(x)\phi(x) . The conjugate momenta to \phi(x) and \phi^{*}(x) are denoted, respectively, by \pi(x) and \pi^{*}(x) . Thus, \pi(x) = \frac{\partial...

Homework Statement Silly question, but I can't seem to figure out why, in e.g. Peskin and Schroeder or Ryder's QFT, the Fourier transform of the (quantized) real scalar field \phi(x) is written as \phi (x) = \int \frac{d^3k}{(2\pi)^3 2k_0} \left( a(k)e^{-ik \cdot x} + a^{\dagger}(k)e^{ik...

Thanks a bunch. As a follow-up question I should like to ask whether this immediately implies the existence of a spin-1/2 particle. By this I mean to ask, since SU(2) acts on (spin-1/2) spinors, and the generators of SU(2) are the quantum operators corresponding to spin, what does this imply...

Er, nevermind I think I get it. But please correct me if I'm wrong. The point is that every algebra has a corresponding representation, which is given by a vector space equipped with a set of operators whose commutation relations (or whatever operations over which the algebra is defined) are...