Recent content by rgoerke

Yes, I believe so. In any case I don't really need to know explicitly how D^0 transforms since I can write everything in terms of D_0, D_{\mu}D^{\mu} = g^{\mu\nu}D_{\mu}D_{\nu} = D_0D_0 - D_1D_1

Hi, thanks for your responses. I'm getting from line 2 to 3 by taking the complex conjugate: \left(-iD_0\phi\right)^* =(-i)^*\left(D_0\phi\right)^*=(i)\left(D_0\phi\right)^* As for indicies, I have tried to be as careful as possible; if you see a mistake please point it out. This seems...

When solving for instanton solutions in a 1+1d abelian Higgs model, it's convenient to work in Euclidean space using the substitution x^0 \rightarrow -ix_4^E,\quad x^1 \rightarrow x_1^E The corresponding substitution for the covariant derivative is D^0 \rightarrow iD_4^E,\quad D^1 \rightarrow...

Thank you both, that will be a big help!

First year grad student here, I've taken two terms QFT. I'm studying some effective field theories, and one of the techniques I've seen used for writting down the effective lagrangian is identifying some fields or components of fields that are "small" and removing them from the lagrangian by...

Of course, that just comes from classical E&M. Thanks a lot, I understand now.

Thanks for your help, I will think about this. Again, thanks for your help. I'm finishing my undergrad and only recently have been introduced to perturbation theory. I have no doubt I'm making things overcomplicated, but it's still not clear to me how E_0^2 = -\alpha F^2/2 is equivalent to...

Right, but if I include the second order ( |F> = |0> + F|F'> + F^2|F''> ) term I get \frac{<F|D_z|F>}{F} = <F'|H_0|F'> + F^2<F''|H_0|F''> + \frac{1}{2}\alpha but the claim is only that \frac{<F|D_z|F>}{F} = \alpha to first order, so the F^2<F''|H_0|F''> should not contribute.

This is what I understand right now: The stark effect is when we perturb a system with hamiltonian H_0 by applying a constant electric field, so that H = H_0 - F D_z where F is the field, aligned in the z direction, and D_z is the z-component of the induced dipole. The first order...

Thanks to both of you! That's a big help.

I am led to believe (because it is in a paper I am reading) that \frac{1}{H - z} \left|\phi\rangle = \frac{1}{E - z}\left|\phi\rangle where H is the hamiltonian, \left|\phi\rangle is an energy eigenstate with energy E, and z is a complex variable. In attempting to understand this expression...