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Neutrinos are weakly interacting quantum mechanical point particles, with very small mass. Refraction is a classical mechanics phenomenon, happens to waves traveling in a medium and it is a collective synergy of many photons impinging on the field of the atoms and molecules of the medium. Individual photons are not refracted but are scattered. In synergy ...


4

$\Pi(x)$ and $\Phi(x)$ are not operators. They are operator valued distributions (distributions that when acting on the one-particle functions become operators). Therefore, it does not make sense to do their square since it is like doing the square of the $\delta$ function. The normal ordering is a prescription on how to cure this ill definition, and obtain ...


3

This is a particular example of a general theorem in effective field theory: if you have an operator that is proportional to the lowest order equations of motion, you can push that operator to higher order in perturbation theory by a field redefinition. This is especially useful if you are working to a fixed order in perturbation theory, in which case you ...


3

One way to define spacelike separation in special relativity is that any two events are spacelike separated if and only if there exists a reference frame in which the two events have the same time coordinate. So yes, if $x^0 = y^0$ the separation is spacelike. Alternatively you can work from the definition where two events are spacelike separated if (and ...


3

The difference is that in classical mechanics positions are exactly the fields you are looking at, whereas in general field theories they are the variables the actual fields depend on. In classical mechanics the solution of the dynamics is given by the knowledge of the position and the velocity $(q(t),\dot{q}(t))$ at any time $t$. Time plays the role of the ...


3

Notation: I will write a Poincaré transformation as ${x'}^\mu = {\Lambda^\mu}_\nu x^\nu + a^\mu$, the operator representing this transformation on the Hilbert space is $U(\Lambda, a)$. An infinitesimal transformation with ${\Lambda^\mu}_\nu = \delta^\mu_\nu + {\omega^\mu}_\nu$ and $a^\mu = \epsilon^\mu$ can be expanded as $$ U(\delta + \omega, \epsilon) = 1 ...


2

Locality is a physical requirement we impose (for good reasons). Locality is implemented in the theory by using fields, with local interactions in a Lagrangian density (ie, the Lagrangian only depends on products of fields and derivatives at a single point). I would definitely not say that locality occurs because Lagrangian densities show up in field theory, ...


2

According to these lecture notes (page 18 in the pdf) and a few others I have found, the integral can in fact be evaluated and yields the modified Bessel function of the second kind, of order 1 (see Wikipedia). The integral does not seem converge, but I am not sure if you can say that it diverges, because it oscillates (with increasing amplitude). The way ...


2

Energy and momentum are conserved at every vertex of a Feynman diagram in quantum field theory. No internal lines in a Feynman diagram associated with a virtual particles violate energy-momentum conservation. It is true, however, that virtual particles are off-shell, that is, they do not satisfy the ordinary equations of motion, such as $$E^2=p^2 + m^2.$$ ...


2

Yes, $\mu$ can be anything. Usually in renormalization, we measure (or define) the coupling constant $g$ at scale $\mu$, and then use this information to predict the coupling constant $g'$ at another scale $\mu'$. We require that $g'$ at $\mu'$ is independent of $\mu$. What I mean is that you should get the same $g'$ at $\mu'$ even if you use another $g$ ...


1

Since you start with a group $G$ which is a symmetry of the theory (ie of Lagrangian) there will be some generators for $G$. Call them $T_1,\ldots,T_N$. In the original Lagrangian, (before the Higgs gets a vev) you can do a transformation with any of the $T_i$'s and the Lagrangian will not change. After the field gets a vev, you can try the same and you will ...


1

I think one has to be very careful when talking about "particles popping in and out of existence". This interpretation is only sort of fine in flat-spacetime QFT, where the Minkowski metric is time-invariant, so has a global timeline Killing vector. The definition of a particle depends on the notion of there existing time invariance! Since black hole ...


1

You have found the reason that most 4D QFTs need to have renormalization: The propagator is (UV) divergent! To "cure" the theory, you need to regularize the theory (make the divergences expressible in some simple parameter like a momentum cutoff) - e.g. by dimensional regularization - and then proceed with some renormalization scheme. That the theory is ...


1

The use of the Lagrangian density is a convenience, and it is not directly related to causality or relativity, and neither strictly to quantum theories. What I mean is that it is possible to formulate non-relativistic (quantum or classical) field theories using exactly the same language. The difference between "mechanics" and "field theory" is that, instead ...


1

As to your question, yes, the QED Lagrangian is indeed invariant under charge conjugation. You may have found differently because your transformations under charge conjugation are faulty. The prefactors are correct, however, under charge conjugation $\psi$ goes to $\bar{\psi}$ and vice versa, i.e. $$ \hat{C} \, \psi \, \hat{C} = -i(\bar\psi \gamma^0 ...


1

So the problem here is that you are confusing quantum mechanics of a single harmonic oscillator with quantum field theory, which is quantum mechanics of a field and can also be considered as quantum mechanics of an infinite number of harmonic oscillators. In quantum mechanics of a single oscillator, the ground state $|0\rangle$ can be represented as: ...


1

The reason why the commutation relations between a field and its conjugate at equal times are of the form $$ \left[\phi(t,\textbf{x}),\pi(t,\textbf{y})\right]=i\hbar\,\delta^{(3)}(\textbf{x}-\textbf{y}) $$ is only to mirror and copy the canonical hamiltonian commutation relations $[q_i,p_j]=i\hbar\,\delta_{ij}$. No causality is involved, rather it is somehow ...


1

A representation of the Lorentz group implies a set of matrices $(S_{\mu\nu})_a{}^b$ (one for each $\mu,\nu$) that satisfy the Lorentz algebra, i.e. satisfies a relation of the form (I am not keeping track of signs) $$ [S_{\mu\nu} , S_{\rho\sigma} ] = i ( \eta_{\mu\rho} S_{\nu\sigma} + \cdots ) $$ What this means is that there is a vector space, with vectors ...



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