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19

The paths of the Feynman path integral are not actually taken. The phrase "takes every possible path" is a mangled statement of the mathematical instruction to take the integral of $\exp(-\mathrm{i}S)$ over all possible paths for the action $S$ to get the probability amplitude of something happening. It is a fact of quantum mechanics that this integral ...


18

Even though there is a single photon in a volume of your choice the light is still a wave. An experiment was performed which proved this. In this experiment a Michelson interferometer was set up and the incident light is so weak that only one photon was in the whole setup at a time. A photographic plate was used to detect the interference pattern. Now just ...


6

Correction, a single photon does not have a circular polarization. It has spin +1 or -1 to the direction of its motion. Qualitatively Left and right handed circular polarization, and their associate angular momenta. The way the classical wave emerges from the quantum mechanical level of photons is given in this blog entry, and it needs quantum ...


6

Let me explain @ACuriousMind 's answer with some verbiage. The short, regrettably oracular, answer is that the Fabri-Picasso theorem does not hold in a finite superconductor, since translational invariance fails at its boundaries. Really, I do appreciate this is aggressively obscure: will strive to explain. First of all, if you have a chunk of warm ...


5

I would like to add a few things to ACuriousMind's answer. What Greene most certainly intends to say is every path(even faster than light ones i.e. those which are not time-like everywhere) contributes to the propagation amplitude. In fact,since every path in space-time contributes with equal weight, there are also paths which go "back in time" and "come ...


4

Method One: \begin{eqnarray*} & & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}+i\epsilon}e^{-ik\cdot x}\\ & = & \frac{i}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\cdot\mathbf{x}}\int dk_{0}e^{-ik_{0}x_{0}}\frac{1}{[k_{0}+(|\mathbf{k}|-i\epsilon)][k_{0}-(|\mathbf{k}|-i\epsilon)]}\\ & = & \frac{i}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\...


4

Field interaction does not "give rise to particles". In fact, field interaction makes it particularly difficult to speak of particles. To understand how particle states and fields are interrelated, we must employ quantum field theory. This answer of mine roughly sketches how a particle state is defined in QFT - and the fact is that such particle states are ...


3

Let us look at coherent states $$ |\alpha\rangle~=~e^{-|\alpha|^2/2}e^{\alpha a^\dagger}|0\rangle $$ $$ e^{-|\alpha|^2/2}\sum_{n=0}^\infty \frac{(\alpha)^n (a^\dagger)^n}{n!}|0\rangle $$ If you have a classical system it means overlap between states is small. We then look at over lap $\langle\alpha'|\alpha\rangle$ $$ \langle\alpha'|\alpha\rangle~=~e^{-(|\...


2

The conventional way to handle bound states in relativistic quantum field theory is the Bethe-Salpeter equation. The hydrogen atom is in QFT usually treated in an approximation where the proton is treated as an external Coulomb field (and some recoil effects are handled perturbatively). The basics are given in Weinbergs QFT book Vol. 1 (p.560 for the Bethe-...


2

The way to see to this a la Peskin and Schroeder is by staring at the renormalized propagator in equation 7.75, \begin{equation} P _{ \mu \nu } \equiv \frac{ - i g _{ \mu \nu } }{ q ^2 ( 1 - \Pi ( q ^2 ) ) } \end{equation} In $ 4 $ dimensions, $ \Pi ( q ^2 ) $ doesn't contain a pole (as expected) and so, \begin{equation} P _{ \mu \nu } = \frac{ - i g _{...


2

Ok you say in an act of faith, and then QTF experts say that there are no virtual particles, they are just a calculation trick. The "trick" part is the assignment of the name "photon", "gluon" , "Z", "graviton", to the mathematical model that allows to predict the interactions of the four forces seen and classified as electromagnetic, strong, weak, ...


2

The diagrams you are drawing are not allowed. In the last one you have an electron going into an muon, by the emission of a photon. Try to isolate that part. If it works one way, it should also work the other way - a muon should be able to decay into an electron by the emission of a photon. This cannot happen. Muons decay to electrons in the weak interaction,...


2

Light never completely behaves as a particle. Light never completely behaves as a wave. As pointed out by hsinghal, the Michelson interferometer showed that, even at the "single photon" level, we still see wave behaviors. These behaviors are well modeled by quantum mechanics, which treats light as neither a pure wave nor a pure particle. As you "add ...


1

I'm afraid that linear polarization is not as interesting an example as you may have hoped. First, the answer: in quantum optics, whether or not a quantum state exhibits linear polarization is independent of the photon count for that state. A single-photon state can be linearly polarized. Now, the explanation: in quantum electrodynamics (QED) it is ...


1

If you're only considering QED, there is only one vertex for this interaction -- electron and positron annhilate to photon, photon pair-produces muon and antimuon. Any electron, photon, muon vertex will have flavor conservation issues. If you're not JUST considering QED, there is a second diagram, identical to the first, with the photon replaced with a $...


1

As you said waves have particle nature. The correct statement will be to say that the quantized fluctuations of the field can be visualized as particles. An intuitive example would be to imagine yourself in a pond with no ripples. In such a situation you will not feel any thing but if the pond has ripples then you would feel as if something is hitting you i....


1

Someone please correct me if I'm wrong, but I suspect it might just be a bit ambiguous, like "beam splitter". The $\pi/2$ part of the name refers to the amount of rotation in the Bloch sphere, but single-qubit quantum operations have an extra degree of freedom on top of the amount of rotation and the axis of rotation: the global phase factor. For example, ...


1

If you just take the empty bandstructure, you will see that any periodic arrangement of atoms (conductors, semiconductors, insulators) features a set of allowed bands and forbidden regions, so called bandgaps. Fully occupied bands can not contribute to electrical current. There are no free places, where carriers could move. Only partially occupied levels ...


1

In a simple word, particles are thought to be excitations in fields. Particles are not interactions but we notice particles when they interact because we are capable of noticing the change in interaction. (How I see is it's not the EM field that creates photon but photon itself is an excitation of the quantum electromagnetic field.)


1

As alluded to in the other answer here, the integral can basically be evaluated in several ways. One of them is the one that OP has himself followed. (PS - OP, congratulations on completing that feat!) Let me present here a way of computing this using Schwinger parameterization. We will use $$ \frac{1}{a} = \int_0^\infty d\tau e^{- \tau a} ~, a > 0~. $$...



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