Hey guys a simple question about the postulate about equal a priori probabilites in thermal equilibrium in statistical mechanics.
I was thinking about the case of N non-interacting simple harmonic oscillators in thermal equilibrium. Shouldn't those phase space states that have single particle...
I'm not sure if I completely understand your confusion (or your notation) but the EM (no matter) action is S = \int d^4x\sqrt{|g|}\mathcal{L}=\int d^4x\sqrt{|g|}F_{\mu\nu}F^{\mu\nu}=\int d^4x\sqrt{|g|}g^{\alpha\mu}g^{\beta\nu}F_{\mu\nu}F_{\alpha\beta} This is the correct action because you can...
You've captured the idea perfectly. But what I want you to derive in particular is Feynman's statement that the uncertainty goes as the log of time. So in your case it would be to derive an uncertainty in the pool balls as a function of time. To do that you would probably want to assume the pool...
Hi Physics Forumers! I was listenining to the Feynman lectures and something Feynman said got me thinking. He was talking about the indeterminacy that exists in classical physics due to our uncertainty in the initial conditions:
I was wondering how he derived this. So I thought I would turn...
In and Out states only appear non-interecting to observers sitting at temporal infinity. To all other observers they appear to be "interacting" in the sense that they no longer have definite particle content. That is how they are able to be eigenstates of the full Hamiltonian. Equation (3.1.12)...
I think not but I'm not sure. In order to derive the spectral density function formula you are required to insert a basis of energy eigenstates which forces you to use the actual mass m and not the bare mass m_0. So that the free single particle propagator has the same pole location as the full...
I think actually that 3.1.1 doesn't apply to the in and out states although the notation seems to indicate this. If you read the paragraph under 3.1.7 Weinberg says:
"On the other hand, the transformation rule (3.1.1) does apply in scattering processes at t\rightarrow \pm \infty."
In...
I think we might be talking about different things. I'm not talking about the tree level Feynman diagrams derived from the standard lagrangian but the feynman diagrams derived from the diagrams the quantum action or quantum effective action.
So for example in \phi^3 the quantum action reads...
As far as I understand the tree level diagrams of the quantum action in QFT give the complete set of diagrams for a give process. Do the loop diagrams of the quantum action have any physical significance?
I thought maybe that by summing up all the diagrams in the quantum action might take us...
"I can't get the results such as
[H,a†]=ωa†
or
[H,a]=−ωa
, so how can I interpret them as creation and annihilation operators?"
I think part of the answer to this question is this: The creation and annhilation operators create multi-particle states in both the free and the interacting case. In...
"How do we interpret the in and out state? I can't understand that since they are the eigenstate of the full Hamiltonian, why they are used to describe the "free" particles?"
Saying they are not interacting was misleading. Rather in and out states appear to be eigenstates of the free...
There are somewhat analogous operators in the interacting theory
They are called in and out operators and they create "free" particles in the interacting theory that are eigenstates of the full hamiltonian. Look in Weinberg vol. 1
well if a particle has an antiparticle that is not itself then it is described by some kind of complex field (i think?) then writing down a real Lagrangian will force you to introduce a U(1) symmetry (i think) which means that it you should probably couple it to the EM field since that is what...