Recent content by iangttymn

That is cleaner. Thanks!

Alright … So the Lagrangian density \begin{eqnarray} \mathcal{L} \propto (\partial \phi)^2 \end{eqnarray} should be integrated over ##d^4x## to give units of action. The Lagrangian density has units \begin{eqnarray} [(\partial\phi)^2] = \frac{1}{l^2} [\phi]^2. \end{eqnarray} And...

I understand that. I'm looking for the non natural units version of \begin{eqnarray} \phi = \int \frac{d^3p}{(2\pi)^3 2E_{\boldsymbol{p}}} \big(a(\boldsymbol{p}) e^{ip_{\mu}x^{\mu}} + a^{\dagger}(\boldsymbol{p}) e^{-ip_{\mu}x^{\mu}}\big) \end{eqnarray}

Does anyone know how to write the classical solution to the Klein-Gordon equation in NON natural units? Where do all the c's and h's go?

I don't mean any changes in the fundamental physics. I just mean arbitrary conventional changes, like the ones I mentioned. The Klein Gordon equation is saying exactly the same thing in both cases - you just have to write it differently. I'm just trying to get a feel for the different things...

The two standard conventions for the Minkowski metric are diag(1,-1,-1,-1) and diag(-1,1,1,1). The physics comes out the same either way, but I'm trying to make a list of the things that change depending on the convention you use. The Klein Gordon equation is one - with the "mostly plus"...

Oh - one more thing. As I said, the states in this setup have one + index and - index, so they are v_{Ab} Srednicki says to convert these spinor indices to a spacetime index using \sigma^{\mu}_{Ab} where \sigma^{\mu}_{Ab} = (I, \sigma_1, \sigma_2, \sigma_3), so that v_{Ab} =...

No problem, and thanks. That helps. When I work things out according to what you said, it seems to make sense to give the vectors |m+, m-> two indices, a + index and a - index, each taking two values (up or down), rather than one spacetime index. Then the generators each have four indices...

Also, those definitions for J and K don't exponentiate to the standard Lorentz vector transformations.

Those definitions of J and K are not the canonical definitions for the vector representation. The canonical ones are the antisymmetric basis (times i). Is there a similarity transformation that gets you there from the definitions written here?

In (1/2,0), J_- = 0 and J_+ = (1/2)Pauli, which are 2x2 matrices. In (0,1/2), J_+ = 0 and J_- = (1/2)Pauli, which are again 2x2. So if you take both to have the 1/2, you would have J = J_- + J_+ = Pauli - again 2x2 matrices. I don't see how doing this gives you the canonical 4x4 vector...

With the Lorentz group (SO(1,3) = SU(2)xSU(2)), if you're working with the "true" SU(2) generators N^+_i and N^-_i (rather than the generators J_i and K_i), it is straightforward to find the form of the generators in either spinor rep (0,1/2) or (1/2,0) by setting either N^+ or N^- to zero and...