Ah, yes that's it. I was just about there, but was having trouble commuting the partial derivatives correctly, essentially because of the issue you mentioned with using the wrong coordinate system. Thanks for your help!
If you want to define a covariant derivative which transforms correctly, you need to define it as ##\nabla_i f_j = \partial_i f_j - f_k \Gamma^k_{ij}##, where ##\Gamma^k_{ij}## has the transformation property
##\bar{\Gamma}^k_{ij} = \frac{\partial \bar{x}_k}{\partial x_c}\frac{\partial...
The path integral is a multi-dimensional integral over all of the variables in the configuration space. For that reason you need to add the contribution from every combination of the variables' values, which means you have to multiply the differentials together.
This is exactly the same as a...
In the case of small perturbations to a free Hamiltonian (even velocity-dependent ones), it makes sense to cast the perturbation term into this momentum basis, and compute the matrix elements of it between various free states. This is known as the Interaction Picture, and is a convenient way to...
However, it's important to remember that the sequence ##E^0_n + E^1_n + E^2_n + ...## will still have increasing powers of ##\lambda## in each term. This fact, plus the fact that ##\lambda## is small, is what makes each term of the sequence get progressively smaller, which is the thing that...
Pretty much. Think of it this way: it's clearly useful to be able to diagonalize an operator in general, because it gives you a useful basis to decompose your states into, where each basis state has a simple transformation rule under the operator. If the operator contains no degeneracy, then...
Hi, and welcome to PhysicsForums!
I guess I'm not sure what you mean when you say "remove degeneracies". The original subspace of states is still going to have the same singular eigenvalue under ##\hat{A}## that it did before, so the operator is still degenerate. The point is just that you've...
Very close, but not quite. In your bottom two, the arrows on the electron lines don't match up, so the diagram is not a valid one. The arrows at each vertex must continue out in the same direction they came in.
The remaining two diagrams are similar to the two-photon diagrams you have now...
You are correct that loop corrections to external lines are forbidden due to amputation, so that gets rid of 8-11. However, loop corrections to internal lines are allowed. You had a diagram like that in your first list, and it is still valid. That should get you two more with crossing. The...
The two right-hand diagrams have more than one loop in them, so they shouldn't be counted here. As for the other diagrams which do contribute, remember that in identical particle scattering, you have to put in the crossing diagrams as well. That should give you at least a few more of them.
I've never done PV regularization on phi-4 theory before, so I'm not exactly sure what the math is going to look like, but you're right that there should be no momentum dependence in the seagull diagram. Looking at your integral, though, my naive thought is that you've got a minus sign in your...
I should mention that I'm self-studying this material, not taking it as part of a course, but since this is still a homework-style problem I figured it'd be best to post here.
Homework Statement
In Peskin and Schroeder problem #11.2, they ask us to consider the Lagrangian:
$$\mathcal{L} =...
You know there must be a vertical reaction force somewhere in the diagram to counter the 10kN force--if there weren't, the whole thing would slide up the wall. So A and/or E must be providing a downward force.
As for A, you know the torque is zero on the member (with the fulcrum defined as...