New answers tagged

1

The general problem that the Higgs mechanism solves is giving mass to spin-one particles. It turns out that finding relativistic, unitary theories of spin-one massive particles is non-trivial. There are a few known ways of doing it (this paper has a pretty good list of sources), but the oldest and easiest is probably the Higgs mechanism. In contrast, there ...


0

The coherent state path integral is basically a recipe for converting a Hamiltonian into a Lagrangian. In condensed matter, we often start with a "microscopic" Hamiltonian description of a material at the level of individual atoms/electrons, and want to convert that into a Lagrangian so that we can more easily do QFT. In high energy, it's usually easier to ...


-1

If we did have decay of massless particles then there'd be no reason a photon wouldn't decay from an energy of $h\nu$ into two photons of energy $h\nu/2$. Eventually all our photons would be red-shifted into oblivion and we'd have no light left in the universe.


2

1) Universal covering groups are groups with the property of being simply connected. Each algebra has a unique covering group. The other groups, $\{G\}$, associated to the same algebra can be obtained from the covering group in the following way $$G=\frac{\tilde G}{Ker(\rho)},$$ where $Ker(\rho)$ is the kernel of the group homomorphism $\rho:\tilde ...


-2

Well, in the NJL model you get rid of the gluon. They are considered to be frozen in the low energy limit where you are working because the mass is higher than the energy. Thus you are only working with quarks, and you consider interaction between quarks via effective coupling constants. But in standard model, gluons acquire a mass because they carry a ...


0

The electromagnetic radiation of gamma rays and X-rays is generated through accelerative processes involving atomic nuclei. The involved photons are the so called hard radiation and could destroy human tissue. The acceleration of electrons generates photons in the range of UV radiation, visible radiation and IR radiation. It is possible to send informations ...


1

Of course the SM Higgs gives mass to both fermions as well as the gauge bosons. However, the latter is much more fundamental and predictive than the former. Point is, it is enough for a scalar to transform non-trivially under a gauge symmetry to contribute to its associated gauge bosons masses (after taking vev). This contribution is constrained by the ...


0

Each of those spectral regions involve electromagnetic waves (photons) at a range of frequencies, and it is that which determines how they are generated, how they interact with matter (i.e. Particles with charge, one by one or in various groupings and configurations like atoms, molecules, bulk material composed of different molecules, different temperatures, ...


1

Particles are described by quantum fields, and the quantum field determines the mass, spin and charge. So for example all electrons (and positrons) have the same mass, spin and (magnitude of) charge because they are all excitations of the electron quantum field. Individual electrons can have different energy and momenta, but I'm guessing you wouldn't ...


1

Most people prefer to do research in the Path Integral Formalism (PIF), instead of the Operator Formalism (OF), because it is "easier." Easier means that whatever property you have in the OF, if you have a PIF for that theory, you can have more properties in the PIF. For example, Noether's theorem is only valid, as far as I know, in a Lagrangian formulation. ...


0

If the vacuum of the theory is supersymmetric - i.e. SUSY is not broken - then it is annihilated by the SUSY generators. On the other hand, using the SUSY algebra one can show that the hamiltonian can be written in terms of the SUSY generators. This implies that the vacuum $|0\rangle$ is supersymmetric if and only if $\langle 0|H|0\rangle=0$, i.e. the vev ...


0

Quantum entanglement is a physical phenomenon that occurs when pairs or groups of particles are generated or interact in ways such that the quantum state of each particle cannot be described independently — instead, a quantum state must be described for the system as a whole. Measurements of physical properties such as position, momentum, spin, ...


10

Quantum entanglement is the property of two objects $A,B$ – more precisely two subsystems – or a relationship between these two objects whose quantities or observables aren't independent of each other. It means that there exist some quantities $a_j$ and $b_k$ describing $A,B$, respectively, such that the probability distribution for these observations ...


4

Essentially you are asking why only scalars are allowed to develop a vacuum expectation value (VEV). A scalar (as the name suggests) does not point to any direction -it has spin 0- therefore it can have a VEV without breaking the Lorentz symmetry. On the other hand, a boson with higher spin, e.g. a vector (spin 1) would spontaneously break Lorentz by ...


0

It helps to remember that invariant quantities are seen as scalars to the transformation (they have no indices in the target space). In the other hand, covariant quantities are objects that transform in a certain way. Example: Vectors in $R^{2}$, under rotation $R_{ij}$, transform covariantly since $v'_{i}=R_{ij}v_{j}$, but it's length is invariant since ...


0

First I would recommend you to check this video: https://www.youtube.com/watch?v=hHTWBc14-mk The video mentions Renormalization in the context of one-loop Feynman Diagrams. In slide 14 (check slide 12 for a picture) of http://www.physics.indiana.edu/~dermisek/QFT_08/qft-II-11-1p.pdf you can check an example of an infinity (a divergent integral), and also ...


2

Short answer to your question: yes. There's no reason for the path integral and the operator formalism to be equivalent. A simple example are all the non-Lagrangian theories. We know (some of) them through their operator algebra, but there's no corresponding Lagrangian. This might sound strange to most people, like myself not long ago, but it is not hard ...


1

There are two mistakes. As AccidentalFourierTransform pointed out, the coefficient $7.181\times 10^{-16}$, when converted from MeV to eV, should give $7.181\times 10^{-46}$. Mega means a million, and it to the fifth power gives $10^{30}$, not just $10^{15}$. In this way, the OP has to add a $10^{-15}$ factor to his result. That makes his result $10^{-3}$ ...


1

Hints: The starting point is the 2-point relation $$T(\phi(x)\phi(y)) ~-~:\phi(x)\phi(y): ~=~ C(x,y)~{\bf 1}, \qquad C(x,y)~\equiv~\langle 0 | T(\phi(x)\phi(y))|0\rangle,\tag{1} $$ cf. this Phys.SE post. The relevant Wick's theorem is a nested Wick's theorem $$ T(:\phi(x)^n::\phi(y)^m:)~=~\exp\left( ...


3

Frankly I find this so-called pedagogical article quite unintelligible and fail to see what the author wanted to say about these two operations. I also can't make any sense of the "derivation" of 7.3 based on the chiral projection being a "numerical matrix" and therefore commuting with charge conjugation operator. Moreso the remark: Elaborate statements ...


0

While the answer provided above is correct, there's maybe simple / less technical way to see this. I put it here hoping it would help other curious minds. In any gauge theory based on a simple non-abelian group, gauge bosons mediate interactions between (i.e., connect) particles belonging to the same multiplet (example: in the SM, the fact that charged ...


2

You are basically asking a circular question of nomenclature. The Dirac quantum field is is a bispinor compactly packaging several degrees of freedom, such as the left- and right-handed Weyl spinors you wrote down the Lorentz transformation properties of. We call both left- and right-handed electrons "the electron", collectively, but of course they are ...


1

It does't really make much sense to talk about a tree-level truncation (it helps for calculations, but that's it) or to take the first Feynman diagram as a true representation of reality. By the way, in your $e^- e^- \to e^- e^-$ example, the whole notion of spatial separation is ill-defined since this is a t-channel process. If going from virtual to real ...


2

General comments to the question (v1): Any textbook derivation of the correspondence between $$\tag{1} \text{Operator formalism}\qquad \longleftrightarrow \qquad \text{Path integral formalism}$$ is just a formal derivation, which discards contributions in the process, cf. e.g. this Phys.SE post. Rather than claiming complete understanding and existence ...


1

I consider the scattering process $A+B \to 1 + 2$. The differential cross-section is always given by \begin{equation} \begin{split}\label{eq1} d\sigma &= \frac{1}{(2E_A)(2E_B)|v_A - v_B|} \frac{d^3p_1}{(2\pi)^3} \frac{1}{2E_1} \frac{d^3p_2}{(2\pi)^3} \frac{1}{2E_2} \left| {\cal M} \right|^2(2\pi)^4 \delta^4( p_A + p_B - p_1 - p_2) \end{split} ...


0

Particles are simply high momentum wave states interacting weakly with matter, i.e. their "existence" is observer dependent. Trying to derive them from some form of free field equation is therefor useless and so is the assumption that they are a general phenomenon. They are a highly likely phenomenon for energies that are much higher than the typical em ...


4

You are asking four questions, whose answers are routinely provided in textbooks. We consider it in QCD since there is no fundamental reason to exclude it, and topological configurations such as instantons, etc.. might well generate it in an effective low energy theory: the rule of thumb is that anything that is not prohibited has to emerge out of the ...


2

$|q\phi|\ll mc^2$ means that the rest mass of the particle is way larger than the electrostatic potential energy. This is the usual assumption of non-relativistic mechanics, which is equivalent to $v\ll c$. You can see this with the virial theorem: $$ \frac{1}{2}mv^2\sim q\phi $$ which means that $q\phi\ll mc^2$ iff $v\ll c$. ...


1

Take for example $q$ a vector field: $$ q^a=A^\mu $$ where $a=\mu$ is a vector index. The conjugate momentum is $$ \frac{\partial\mathcal L}{\partial A^\mu} $$ and, as it is an upper index in the denominator, it makes sense to write it as $\pi_\mu$. Also, you can use the definition of vectors and covectors to prove that $\pi_\mu$ transforms as a covector, ...


0

Subalgebras whose root system is not a subset of the root system of the original algebra are called special subalgebras. Therefore, the generators are not a subset of the original's group generators. That is not quite right. A special subalgebra is one such that their step operators do not form a subset of the algebra step operators.That is what is ...


0

It results from the combination of two facts: $i$) the embedding of $U(1)_Y$ into $SU(2)_R \times U(1)_{B-L}$, and $ii$) the normalization of the $U(1)$ charges. Let me take the simpler chain: $SU(2)_L \times SU(2)_R \times U(1)_{B-L} \to SU(2)_L \times U(1)_Y$. $i$) At the scale of the left-right symmetry, the hypercharge gets merged into $SU(2)_R \times ...


3

Let me first describe the basic idea without even mentioning supergravity. Consider some classical field theory of two fields $\phi$ and $\sigma$, with an action $S[\phi,\sigma]$. Suppose this theory has a continuous symmetry (with parameter $\epsilon$): $$S[\phi,\sigma]=S[\phi+\delta_\epsilon\phi,\sigma+\delta_\epsilon\sigma].$$ Say I find some particular ...


0

Crossing symmetry tells you that you should not only exchange $$p_2\leftrightarrow -p_4$$ in the amputated matrix elements, but also replace the wavefunction polarizations $$ u^{\pm}(p_2)\rightarrow v^{\mp}(p_4) $$ (where the spin polarizations have been reversed), and finally multiply the amplitude for a factor $-1$ (Since you are crossing a fermionic ...


1

The current limits on proton lifetime are $\tau_\mathrm p^\text {exp}\gtrsim 10^{34}$ years. In unified models, we can naively estimate the proton lifetime with this relation: \begin{equation} \tau_\mathrm p = \frac{M_\text{GUT}^4}{\alpha_U^2 M_\mathrm p^5}\,, \end{equation} where $M_\mathrm{GUT}$ is the mass scale of your proton-decay mediating gauge ...


0

Q1: The mass splitting he is referring to is between the components of a vector-like quark (VLQ) doublet, $Q=\begin{pmatrix} U \\D \end{pmatrix}$, i.e. between U and D. There is such a splitting because they have different charges, so they get different loop corrections to the tree- level mass term $M_Q \bar{Q}{Q}$. What Sher is saying is simply that we pass ...


9

The fact that the theory is not gauge invariant implies that all degrees of freedom of $A_\mu$ must have physical meaning: This is not the theory of photons where only transverse degrees of freedom make sense. This way you must tackle some non-trivial issue like the negative norm associated with temporal modes. This could be avoided by adding a mass to ...


1

I think the question was already answered here, look at either my answer of the one by Adam. The punch line is that that the expectation value of a field (such as the field $\phi$ at the bottom of the mexican hat potential) is fixed by the way the source $j$ that couples to $\phi$ in the path-integral for the functional generator is sent to zero. As there ...


2

As it was noted in the comments, the derivatives $\partial_\mu$ is outside the correlator in the Ward identity so that the Stokes theorem trivially applies. As for the second question, one can perform the spectral decomposition of $\langle j^\mu(x)\mathcal{O}(0)\rangle$ getting something along the lines of $$ \langle j^\mu(x)\mathcal{O}(0)\rangle\sim \int ...


0

Without the factor $1/2$ for a complex field all observables constructed out of the Lagrangian in the standard way vie Noether theorem, like the energy $H:= \int T_{00} dx$ or the momentum $P_i = \int T_{i0} dx$, turn out automatically to be the ones of a system of identical particles of two types, {\em proper particles} and {\em anti particles}. E.g., $$H ...


2

Comments to the question (v2): OP is considering the higher-derivative Lagrangian density $$ {\cal L}_1~=~ \frac{1}{2}(\partial\phi)^2 +\frac{g\phi}{2} (\partial^2\phi)^2,\tag{1} $$ where $g$ is a coupling constant. We use Minkowski sign convention $(+,-,-,-)$. Quantum mechanically, the model is not unitary and therefore ill-defined, cf. the Ostrogradsky ...


3

After thinking about it some more, I think I have a resolution. Consider a single Weyl fermion, which has equation of motion $$\sigma^\mu \partial_\mu \psi = 0.$$ Just like the Dirac equation, the Weyl equation is linear in momenta and hence has negative energy solutions. We then perform the usual procedure to regard these as positive energy solutions ...


0

The rule of thumb is to note that $$M(k_1,k_2)=-M(k_2,k_1)$$ so that $M(k,k)=0$, i.e., you just need the Pauli Exclusion principle! The requirement of antisymmetry fixes the relative sign, so that's all you need here. Equivalently, there is a rule that the sign of a certain diagram is $(-1)^n$ where $n$ is the number of crossing fermion lines. This fixes, ...


0

First of all, what's the motivation for the Yang-Mills action and how should I understand the coupling constants $\theta$ and $g$? I would say motivation comes from experiments. For instance it is an experimental fact that the electric charge is conserved. The associated current is also conserved, in the sense of $$\partial_\mu J^\mu=0.$$ Therefore we ...


1

Many answers discuss the transition from the unstable state to the stable one. Let me thus discuss the issue of choosing the ground state itself. I will suppose a two dimensional Mexican hat potential. As Numrock realised, it has degeneracy. There is nothing which lift this degeneracy in principle. Then you can change the ground state without energy. This ...


4

I think your problem is mostly a problem of notation. If you write two Weyl spinors inside a Dirac spinor, you should use different symbols to avoud confusion, i.e. $$\psi = \begin{pmatrix} \xi_L \\ i \sigma_2 \xi_L^* \end{pmatrix}.$$ Now, your object $\Psi$ has a left-chiral component $\xi_L$ and a right-chiral component $i \sigma_2 \xi_L^*$. (A Dirac ...


0

Okay, this wasn't as hard to find an answer to as I expected. However, any clarifications/ critisisms are welcome. Weinberg basically gives the answer in section 5.6 of his QFT book: A general tensor of rank N transforms as the direct product of N (1/2, 1/2) four-vector representations. It can therefore be decomposed (by suitable symmetrizations ...


3

It does involve bound states. For an electron in the Coulomb field $A(r)=-\frac{Ze}{r}$ that arises from a nucleus (of mass $m$), the lowest energy is $E_0=m\sqrt{1-(2Z\alpha)^2}$ where $\alpha$ is the fine structure constant. This is positive for $Z\le\frac{1}{2\alpha}$, which is what Weinberg means by "moderate" fields. The assumption so far was that the ...


2

Yes, it is possible to match the QM with a QFT in 1+0 dimensions. However, the Fock vacuum $|0\rangle$ (which is annihilated an annihilation operator $a|0\rangle=0$) is naturally related to the coherent states $$\hat{a} |z\rangle~=~ z|z\rangle \tag{1}$$ rather than position eigenstates $$\hat{q} |q\rangle~=~q|q\rangle.\tag{2}$$ [Of course, it is possible ...


2

Echoing knzhou's comment, the expression $$ \langle 0|\mathcal{T}x(t_i) x(t_f) | 0 \rangle = \frac{1}{2\omega} e^{-i\omega|t_i-t_f|} \tag{1} $$ only depends on $t_i,t_f$, while $$ \langle x_f,t_f|x_i,t_i\rangle \tag{2} $$ is a function of $t_i,t_f$, but also of $x_i,x_f$, that is, it depends on four variables, and therefore is more general than $(1)$. This ...


3

We do observe spontaneous symmetry restorations in nature. This is called an emergent symmetry. See e.g. this post. A system posses an emergent symmetry if it appears symmetric at large (coarse-grained) scales although the apparent symmetry is explicitly broken by the microscopic description (typically the Hamiltonian or Lagrangian). I can give two examples ...



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