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8

Maybe the simplest way to think about this is that the Sun is in approximate thermal equilibrium and would absorb any photon, of any frequency, that is incident upon it. This is essentially the definition of a BB. There are many radiative processes that can absorb (and hence emit) radiation at all frequencies, not just those corresponding to atomic ...


8

You seem to be confusing regularization with renormalization. Regularization is the process of removing (or, more properly, parameterizing) infinities in loop integrals. Often in elementary texts a "cutoff" representing an energy scale above which the theory is assumed to be invalid is discussed, and counterterms are added to the Lagrangian in order to make ...


4

Black body radiation is given by Planck's formula (see link for variables) Here is the measured irradiance of the sun and the attempt to fit it with the black body formula: The effective temperature, or black body temperature, of the Sun (5,777 K) is the temperature a black body of the same size must have to yield the same total emissive power. ...


4

The mistake is that $W^a_\mu [\tau_a, \tau_b] W^{\mu b} = 0$ does not imply $[\tau_a, \tau_b] = 0$, this is true for any antisymmetric object $A_{ab}$. In your example, you can see this in the following way: Let $A_{ab}$ be antisymmetric: $A_{ab} = -A_{ba}$. Then: $$ \begin{align} W^a_\mu A_{ab} W^{\mu b} &= W^b_\mu A_{ba} W^{\mu a} \qquad ...


3

Firstly, renormalisation isn't really inherently to do with the UV divergences you get: it's more to do with the idea that interactions with other fields change a particle's energy from its inherent 'bare' mass, so the measured mass isn't the same as the value that appears in the quadratic part of the action. You play a renormalisation game in perturbation ...


2

My first question is related perhaps to notation choice or a typo. Should we use the same function used for the filed to write the Fourier transform? Or we should put Φ(p,t) instead of ϕ(p,t)? This is a standard awful physicist convention. Clearly the functional forms of the two "phis" are different, but the symbol being used is the same. This ...


2

The word "fluctuation" shows up in a lot of different contexts, most of the time without a formal definition and sometimes meaning many different things. For example, in the context of statistical mechanics, S.F. Gull writes: Suppose we use the Gibbs algorithm to set up an equilibrium ensemble, and calculate the ensemble average of a quantity of ...


2

$\newcommand{\ket}[1]{\lvert #1 \rangle}$The charge and baryon number operators are not bounded because you can create states of arbitrary charge and baryon number: Let $a^\dagger$ be any creation operator that creates a bosonic charged particle state (let's say with unit charge), and let $\ket{\Omega}$ be the vacuum. Then, $$\ket{n_e} = ...


2

What I think one needs to internalize conceptually is that the program of renormalization is always favourable (and almost always required) in physical theories, be they fundamental or effective phenomenological ones (including condensed matter field theories), be there infinities or not. I think the last point is by far the most important. Yes, ...


2

No, a magnetic monopole a la the Dirac string does not "violate" gauge symmetry. Rather, the statement "we have a magnetic monopole" means only that we are forced to consider the gauge theory not on the whole spacetime, but on the spacetime with the location of the magnetic monopole removed. Why? Because, at the location of the magnetic monopole, the curl of ...


2

Well you want to go from QFT to Classical mechanics. Let's do this in three steps 1. QED to Dirac Equation QED lagrangian with electric dipole is $\mathcal{L} = \bar{\psi}\left(\gamma\cdot\Pi - \frac{\mathrm{i}d}{2}\sigma^{\mu\nu}\gamma^5 F_{\mu\nu}\right)\psi\\$ Where $\Pi\equiv \partial - \mathrm{i}eA$. This implies the hamiltonian $$\mathcal{H} = ...


1

Maybe this would be better as a comment, since it is not a full answer, but I don't have enough reputation for that. The most important ambiguity is that there is an infinite number of functions that have the same asymptotic expansion. As an example, if $f(g)$ has some asymptotic expansion in $g$ as $g \to 0$ than $f(g) + e^{-1/g^2}$ has exactly the same ...


1

No, one can only conclude that the symmetrization $$[ \tau_a,\tau_b] + (a\leftrightarrow b)~=~0$$ is zero, which is indeed true because the commutator is antisymmetric.


1

For the unboundedness see ACuriousMind's answer. About the associated projections, for unbounded operators there is the notion of affiliation. An unbounded, closed and densely defined operator $A$ is affiliated with a von Neumann algebra $M$ if all its spectral projections are in $M$, that is $$A = \int\lambda\text dE(\lambda),$$ with the spectral measure ...


1

There is no well established theory predicting the neutrino mass (even if of Majorana type). A particle is its own anti-particle if the field describing the particle is a real field. That means obviously that the electric charge is 0 but not necessarily that other charges are zero. For instance, if the neutrinos are Majorana neutrinos, they would still have ...


1

A more colloquial way of understanding this is to write down the expressions for loop corrections in a SUSY gauge theory. The important ingredient is that in addition to the gauge bosons you will also have gauginos, i.e. fermions in the adjoint representation. To understand how a vacuum diagram vanishes thing about how a gauge boson in 4D has two ...


1

I guess we can assume that for any "physical" theory, the S-matrix must be invertible. We like to think of the S-matrix as being unitary i.e. $S^\dagger \cong S^{-1}$. For any matrix, so long as the eigenvalues are non-zero, there is a way to normalize the states such the eigenvalues become unimodular. But if there were zero-eigenvalues, then the matrix ...



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