7

The poles of the spectral density give you the particles of the system, whether they are fundamental or bound states. There is no fundamental difference between these two types: they behave exactly the same, phenomenologically, and they are indistinguishable theoretically. This becomes even clearer in light of the concept of duality: the same theory often ...


4

What happens in between is everything and nothing. There is no privileged clearcut answer what happened that would be physically meaningful. It's really the very basic point of quantum mechanics that only results of measurements are physically meaningful facts or observables; all other data are fictitious or uncertain. By the very definition of your problem, ...


4

Yes. In something like: $$ e^-+\mu^-\rightarrow e^-+\mu^- $$ you start with two free particle states, and end with 2 free particle states. In between, the electron and electromagnetic fields (and perhaps the weak field) go through all possible configurations that conserve energy an momentum. That is something that cannot be solved exactly, so it is ...


4

Consider the Dyson series for the S-matrix: $$ S = \lim_{t\to\infty} U(-t, +t) = \mathcal{T}\exp\left(-i\int_{-\infty}^{\infty}\mathrm{d}^4 x \ \mathcal{H}(x)\right) $$ Expanding out the first few terms, we see that this is $$ S = \mathcal{T}\left[\color{red}1-i\int \mathrm{d}^4x \ \mathcal{H}(x) + \frac{-i^2}{2!}\iint \mathrm{d}^4x \ \mathrm{d}^4y \ \...


3

Personally, I've always found the concept of symmetry factors more confusing than they're worth. I find it much easier and faster to simply count the number of contractions. The diagram you are looking at includes the fields $$\phi_1 \phi_2 \phi_x\phi_x\phi_x\phi_x \phi_y\phi_y\phi_y\phi_y$$ $\phi_1$ can contract with 8 possible fields (any of the $\phi_x$ ...


3

OP's main issue seems to be that the symmetry factor $S$ depends on whether the external legs are distinguishable ($S=4$) or indistinguishable ($S=8$), respectively. On one hand, P&S on p. 93 consider the distinguishable situation$^1$ where there are not integrated over the external positions/momenta. On the other hand, OP apparently considers the ...


3

Bound vs. extended states In condensed matter physics bound states are the states with the wave functions decaying towards infinity, as opposed to the extended states, as, e.g., plane waves. In this sense the definitive answer to the question can be given only by the exact diagonalization of the Hamiltonian and studying the behavior of the wave functions. ...


2

Quantum mechanics does not predict interactions the same way that classical mechanics does. There are no trajectories for particles to interact. In QM the integration is necessary because the prediction is a probability curve for the interaction to occur at (x,y,z,t). Probabilities have to integrate to 1 over all the available phase space , by definition.


2

The fact that the particle is described by a superposition of states, does not mean that one has to measure both states at once. That's how superposition works, one one makes a measurements and then the wavefunction collapses on one of the possible superimposing states. This means that the neutral pion, when measured, can be found in the $|d\bar{d}\rangle$ ...


1

They are probability amplitudes, not probability ranges. They are also called “matrix elements”. They are complex numbers, and their squared magnitudes are related to the probability that the initial state $i$ in the Feynman diagram becomes the final state $f$.


1

If you denote the external points by $x,y$ and the internal points by $1,2,3$ than the diagrams of the last line in your picture are $ x1, 11,12,23,23,23,3y$ and $x1,12,12,12,23,33,3y$ with integration over the internal points and what is between brackets is contracted. Yes, they are the same.


1

The question is a bit vague, thus the answer will also be quite vague. First of all when you say that you want to "calculate the process" you have to be more specific. What do you want do to? For example: You may want to calculate the probability that the system will go from state $1$ to state $2$(in the case you mentioned, these states are two-...


1

Heuristically, the counterterms are also infinite, and are engineered so as to cancel out the infinities from the bare amplitude. How these values are chosen depends on your choice of regulator and subtraction scheme, but the same general principle holds - rather than the unphysical bare parameters that we used when we made our first stab at writing the ...


1

If I assume that your diagram is to be read from left to right, then it seems that two quarks respectively coming from the two colliding protons, annihilate. However, for quarks to annihilate, one needs to be an anti-quark. So what one would expect instead is a gluon exchange between the two quarks with enough energy to try and kick the quarks out of the ...


1

My understanding is that this is due to momentum conservation. A gamma ray cannot create an e-/e+ pair without interacting with a nucleus in order to conserve momentum. This is why the virtual gamma ray here can only create a virtual (off-shell) e-/e+ pair. The change that occurs in the electron after the photon emission is just a change in energy/momentum. ...


1

Your result is correct. As you know, each component of a Feynman diagram combines multiplicatively. Meanwhile, the kind of loop that you have here does not depend on the momentum of the propagator that it attaches to. Loosely speaking, this is because the loop both "absorbs and emits" its own momentum into the vertex. It doesn't affect and is not ...


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