# Tag Info

2

I think you're confusion is with notation since unfortunately, two notation often used to denote projected spinors. One notation is to write: $$\psi \equiv \left( \begin{array}{c} \psi _L \\ \psi _R \end{array} \right)$$ In this notation $\psi _L$ and $\psi _R$ are two component Weyl spinors. However, a second notation ...

3

It's not a sufficient explanation. There are asymptotically free theories which are not strongly coupled in the IR. The rate at which the coupling gets strong is important. In QCD, it seems to get strong very quickly near the confinement scale, so that beyond a certain scale, you only see hadrons. It is not really understood how this works. The ...

4

To be honest, I think that the route you describe (and which is also used in many textbooks) is not physically well motivated at all. You have begun with a theory of a fermion with a global symmetry which maps physical states to different physical states. This theory has the property that specifying initial conditions on a spacelike surface completely ...

2

In principle one has to calculate the pole of correlation functions involving gauge invariant operators like $\text{Tr}F_{\mu\nu}F^{\mu\nu}$. The problem is that due to asymptotic freedom, QCD is not solvable perturbatively at low energies. This is why nonperturbative techniques like lattice QCD are used to calculate such spectra. A key achievement in this ...

3

Not all irreducible representations (irrep's for short) of the Poincaré group lead to a Lagrangian. One example (see my comment to Julio Parra's answer) are the zero-mass, "continuous-helicity" (sometimes called "infinite-helicity") representations. There is, however, a way to begin from a positive energy irrep of the Poincaré group (i.e. a 1-particle ...

1

Alex Nelson's answer is much better that mine, but it doesn't address your question at all. (a) Does it form a group? No, it doesn't. See bellow, to find out what does 'look like' (but isn't) a group. (b) What are the elements of the group? The group-like structure is the following. Being sloppy, effective action satisfies the folowing semigroup ...

2

I'm not sure such a thing exists. Usually reps only helps you classify the kind of particles you have (i.e the quantum numbers that identify them) and how they transform under the corresponding group. I believe how to represent this particles mathematically and what is their dynamics is a different matter. The only thing similar I know about is that some ...

6

You are right, it is wrong to think that in gauge theory "gauge transformations are just a redundancy". This becomes true only if one abandons locality, ignores all boundary effects, all instanton effects, hence most of what is interesting about gauge theory. Of course forming gauge equivalence classes (say of observables) is something one wants to do every ...

1

This is how I understand this issue. First, I believe you may agree that imposing gauge invariance is a sensible thing to do. If we want our fields to be invariant under some kind of transformation it better be local, since two separate space-time points shouldn't be related in any unnecessary way, otherwise we may violate causality. A different issue is ...

3

Suppose $a$ and $a^{+}$ operators satisfy $$\left\{ a,a\right\} =0\mbox{ and }\left[a,a^{+}\right]=1$$ We have basically $a^{2}=0$ and $aa^{+}=a^{+}a+1$. Now consider $aaa^{+}$. $$0=aaa^{+}=a\left(a^{+}a+1\right)=aa^{+}a+a=a^{+}aa+2a=2a.$$ So we get $a=0$.

0

In theory, it should be (lumpy), however: Even if you could observe the peaks, or nodes and antinodes of the EM vectors of a single photon, its energy would be absorbed by the instrument you employed to observe it. Fast enough doesn't cut it. If the instrument absorbs no energy from the photon, it will also not be detected. Intensity for light is not ...

7

There are really several questions here: (a) What is the renormalization group? Specifically the law of composition, etc. (b) How does the equation the OP gave relate to this? Short Answer It's a semigroup (see references below). The equation you wrote, $$\tag{1} \left[\mu \frac{\partial}{\partial \mu} + \beta \frac{\partial}{\partial g} + m \gamma_{m^2} ... 2 I have to admit that I have no idea about the model you are working on, but the standard way to determine whether a gauge theory is confining or not is to calculate the vacuum expectation value expectation value of Wilson loops. The latter are gauge invariant operators that describe parallel transport around a closed loop in spacetime. If the vacuum ... 0 It is small wonder there is no definitive answer to this question here. But I can pass along the best answer I got. I asked the same question, in a slightly different format, to a blog "Of Particular Significance" that is popular with Physics Stack Exchange. My question was whether the famous Goldstone Mexican Hat potential had a value of 245 GeV at the ... 0 What is really being said is that it is not an uncaused event in the vacuum, but in simplistic terms, it could be said to be more a property of the vacuum. No event is taking place as such. Describing it as an event is trying to put it in simplistic terms, which while it may help in understanding to some degree, is not an exact description. You could equally ... 0 So, let me try to rephrase your question a little. The "electrostatic" case in, for example, plasma physics refers to the case when |v| \ll c, so that the coupling to the vector potential \vec{A} is negligible and we can consider the pure situation of the scalar potential \phi. We can TOTALLY write a Lagrangian quantum density for this as ... 1 Let us quickly run through the standard KK compactification. We start with a d+1 dimensional theory$$ S = \frac{1}{16\pi G_{d+1}} \int d^{d+1}x \sqrt{G} R_{d+1} $$More general actions on the d+1 dimensional space can be considered, but this will suffice for our purposes. The metric G_{MN} can be decomposed as$$ ds^2 = G_{MN} dx^M dx^N = e^{2\Phi} ...

0

In the standard quantization of the free electromagnetic field, the field operators satisfy the (equal time) commutation relations $$[E_i(\mathbf{x}, t), B_j(\mathbf{y}, t)] = -i \hbar \epsilon_{ijk} \partial_k \delta^3(\mathbf{x}-\mathbf{y})$$. Please, see for example the following article by Stewart. This implies the existence of an uncertainty ...

1

As previous answers have correctly noted gamma matrices do not forma a basis of $M(4,\mathbb{C})$. Nevertheless you can construct one from them in the following way 1 the identity matrix $\mathbb{1}$ 4 matrices $\gamma^\mu$ 6 matrices $\sigma^{\mu\nu}=\gamma^{[\mu}\gamma^{\nu]}$ 4 matrices $\sigma^{\mu\nu\rho}=\gamma^{[\mu}\gamma^{\nu}\gamma^{\rho]}$ 1 ...

2

I believe in the last line, the plane-wave functions $u_k(x)$ should carry different coordinates and momenta, e.g $$[a(k)^\dagger,a(k')]u_k(x)u_{k'}(x')$$ You may note that the commutator $[\phi(x),\pi(x')]=i\hbar\delta(x-x')$ holds if one choses $[a_k,a_{k'}^\dagger]=\delta_{kk'}$. However, this indirect reasoning is no proof that this choice is unique. ...

0

Spin one fields are in the adjoint representation because of the specific transformation behaviour of such modes under gauge symmetries. This is basically a symmetry requirement on the lagrangian. Regarding your second question: the two multiplets are different. In one case we have a central charge and in the other we do not. However, the issue is ...

2

There are two different kinds of symmetry breaking involved in your question. The first would be spontaneous symmetry breaking. In this case we are dealing with a theory that is invariant under a certain symmetry group, but its vacuum is not. The breaking of the symmetry corresponds to a specific choice of the vacuum, the freedom of choosing a vacuum results ...

2

Leibniz rule holds for covariant derivatives, both in gauge theories and gravity. Mathematically, a derivation is one for which the Leibniz rule holds. How does it work for non-abelian covariant derivatives. I will give you an example. Let $\Phi^\dagger \Phi$ be invariant under local non-abelian gauge transformations. Then \partial_\mu (\Phi^\dagger ... 0 Your question is riddled with ^'s in equations making it hard for me to understand the body of your question. If I understand your question "why is there no weak isospin vacuum angle in analogy with the one in QCD?," then I can answer it easily: Suppose we write that CP-odd term in the Lagrangian. Then, to remove it, all you need to do is to look for a ... 0 As mentioned in the comments,  C  and  M  are matrices in different spaces. Explicitly showing the flavor space matrices, in the lepton sector perhaps, we have, \begin{align} \overline{\psi} _L M ^\dagger \left( \psi _L \right) ^c & = \overline{\psi} _L M ^\dagger C\bar{ \psi _L} ^T \\ & = \left( \begin{array}{ccc}\overline{ \psi _{e, L} } ... 3 To complement V. Moretti's excellent answer, it's worth emphasizing that the dimension of the four-by-four complex matrices \mathbb C^{4\times 4}, when seen as a vector space over \mathbb C, is 4\!\times\!4=16. As such, a set of four matrices (i.e. vectors in \mathbb C^{4\times 4}) can never be a basis for it. It's also worth saying that the general ... 5 No they do not, due to dimensional reasons, but they are generators of the algebra. That is, I and the products of \gamma^a (products of one, two, three and four matrices) form such a basis. NOTE ADDED. As Emilio Pisanty correctly remarked (also making some further interesting comments) GL(4, \mathbb C) is not a linear space so questions about bases ... 4 The form of the propagator is correct. The expressions from your Wikipedia link are complicated because they show the propagator for the massive theory, where Susskind's argument fails because the propagator can involve any function of the dimension zero combination m^2 |x-y|^2. The "simple" massless result is recovered in the m\to0 limit; for example ... 1 An experimentalist's view: I do not see the need to search further for why the three quarks add up to the electron charge than that given by the group structure of the Standard Model. The SM is very successful in organizing into beautiful symmetries the particle and resonances data gathered the last sixty years or so. There is no experimental reason to ... 5 This is because the path integral {\cal Z} is an infinite-dimensional version of a Grassmann-odd Gaussian integral\int \!\mathrm{d}^n \bar{\theta} ~\mathrm{d}^n\theta ~e^{\sum_{i,j=1}^n\bar{\theta}_i ~M^i{}_j ~\theta^j}~\propto~\det(M), $$where the indices i,j can be interpreted as DeWitt's condensed notation. 4 In the first case, the vertex is a vertex in the common sense (used to construct diagrams). In the second case, the gauge field is not dynamic (in a path integral formulation, you do not integrate over), it is a background field that is fixed. In that case, we are interested on the effect of this non-dynamical field on the electron field. This is useful to ... 1 This means that you assume the energy of the initial state is approximately conserved and the initial state goes over smoothly into the state from which it originates when first adding the perturbation to your theory. Adiabatic here means "without changing the energy". 1 In general it means varying (or turning off an interaction in your particular question) a parameter on a time scale that is much larger than the smallest energy separation of your Hamiltonian. More explicitly : Suppose you have a Hamiltonian H with energy levels E_n and suppose that \left|E_a-E_b\right| (a\neq b) is your smallest level splitting ... 0 If gravity is a fundamental force, then the Higgs mechanism is also. This is is true whether they are related or not. The Higgs mechanism is certainly the source of the inertial mass that inspired Newton to quantify what a force is, and how it behaves. Gravity, like the Higgs mechanism, can add mass/energy to matter in bulk (like constant acceleration), ... 2 The conformal group is defined for any spacetime you want. The conformal group of d-dimensional Euclidean space, which has isometry group SO(d), is SO(d+1,1). The conformal group of d+1 dimensional Minkowski space, whose isometry group is SO(d,1), is SO(d+1,2). The defining property of the conformal group is that its the set of transformations that leave the ... 1 The \sqrt{\frac{\hbar c^2}{V}} is there because of the fact that Mandl & Shaw the quantise the field in a box and not in free space. So when you take the continuum limit you have that$$ \int d^3k \rightarrow \sum_k \sqrt{\frac{\hbar c^2}{V}} to get the right dimensions. The \frac{1}{\sqrt{2\omega_k}} comes from Lorentz invariance, particularly ... 2 You are always allowed to introduce a new integration variable as long as its not its already being summed over. This might be more clear in discrete form: \begin{align} \int d x \, f (x) & \rightarrow \Delta x\sum _i \,f ( x _i ) \\ & = \big( N \Delta y\sum _j g ( y _j ) \big) \Delta x \sum _i f ( x _i ) \\ \end{align} where  N \Delta ... 0 I think I got answer myself. The vanish commutator and EPR paradox are not correlated. The vanishing commutator simply says, once a measurement at x was done, the obtained state will not bring uncertainty for measurement of y, by the simutaneous eigenstate property. Not like one measures momentum, the state becomes |p \rangle, then measure position ... 2 Non-conservation of charge in Majorana terms The Dirac mass term is m\bar\psi \psi where one field-factor \bar\psi is complex conjugated (aside from other transpositions included in the Dirac conjugation) and the other is not. So one may assign a fermion number 1 to \psi which means that \bar\psi automatically carries -1 and in the product, the ... 2 The field is not interpreted as a wave function but as an operator \hat{\psi} which creates/annihilates particles. This quantisation procedure is done by expanding the field in its Fourier components which depend on the momentum, spin, etc, a_s(p), a_s^\dagger(p), b_s(p), b_s^\dagger(p) (the fact that these \hat{\psi} operators are not self ... 1 What confused me was the explanation from the tangentbundle homepage (second yellow box in OP). The generalization is straightforward, for simple zeros we have:\bigg\vert\frac{\mathrm{d}f(x)}{\mathrm{d}x}\bigg\vert_{x_0}\delta[f(x)] = \delta(x-x_0)$$integrate$$\int \mathrm{d}x\,\bigg\vert\frac{\mathrm{d}f(x)}{\mathrm{d}x}\bigg\vert_{x_0}\delta[f(x)] ...

2

The notation $$\frac{ \partial f_i}{ \partial x ^i }$$ means the diagonal elements of the matrix: $$J _{ ij} = \frac{ \partial f _i }{ \partial x ^j }$$ where $f_i$ is the component of the vector $\vec{f} (x)$. I found this very confusing a few weeks ago so. Here is the proof I wrote up for the ...

3

Again assuming it only has a zero $x^i=x_0^i$ what you have is $$\delta(f(x^i)) = \frac{\delta(x^1-x_0^1)}{\left|\frac{\partial f}{\partial x^1}\right|_{x^i=x_0^i}} \frac{\delta(x^2-x_0^2)}{\left|\frac{\partial f}{\partial x^2}\right|_{x^i=x_0^i}}\cdots \frac{\delta(x^n-x_0^n)}{\left|\frac{\partial f}{\partial x^n}\right|_{x^i=x_0^i}} = \prod_{j=1}^n ... 1 This is a Gaussian integral:$$\int_{-\infty}^\infty e^{-x^2} dx =\sqrt{\pi}$$(which is not very hard to prove). Now you only need to do a substitution to bring it into this form. 2 I) The un-gauge-fixed QED Lagrangian density reads$$\tag{1} {\cal L}_0~:=~-\frac{1}{4}F_{\mu\nu}^2 + \bar{\psi}(iD\!\!\!\!/ \ \ -m)\psi.$$The gauge-fixed QED Lagrangian density in the R_{\xi}-gauge reads$$\tag{2} {\cal L}~=~ {\cal L}_0 +{\cal L}_{FP}-\frac{1}{2\xi}\chi^2 , $$where the Faddeev-Popov term is$$\tag{3} {\cal L}_{FP}~=~ ...

3

Figured it out. If I write \begin{align}\bar{u} \gamma^{\mu} u &= \frac{1}{m} \bar{u} \gamma^{\mu} \gamma^{\nu} p_{\nu} u = \frac{1}{m} \bar{u} \left( \{\gamma^{\mu},\gamma^{\nu}\}-\gamma^{\nu} \gamma^{\mu} \right) p_{\nu} u \\ &= \frac{1}{m} \bar{u} \left( 2 g^{\mu \nu} - \gamma^{\nu} \gamma^{\mu} \right) p_{\nu} u = \frac{2}{m} \bar{u} u p^{\mu} ...

1

In Landau's theory, the order parameter $M$ should make $G(M)$ minimal. $$\frac{\partial G}{\partial M} = 2 B(T) M + 4 C(T) M^3=0$$ $$\frac{\partial^2 G}{\partial M^2} = 2 B(T) + 12 C(T) M^2 > 0$$ Hence, $M=0$ or $M = \pm M_0 = \pm \sqrt{ - \frac{B(T)}{2 C(T)}}$ When $T$ is higher than the critical point $T_c$, the groud state of system satisfies $M=0$. ...

3

An approach alternative to that discussed by David Bar Moshe is to start from a different coordinate system in the Rindler wedge $W_R$: $$ds^2 = e^{2y}(−g^2dt^2+dy^2)$$ here $t, y \in \mathbb R$. The relation with the standard spatial coordinate in $W_R$ is $x=e^y$, where $x>0$ is related with the alternate form of the (same) metric: $$ds^2 = -g^2 x^2 ... 7 Essentially, what the Wick theorem tells you is that the moments of a multivariate gaussian distribution are determinate by the second moments; for instance, for a 3D gaussian in (x,y,z) space, the quantity$$ \langle xyzx \rangle  can be calculated in terms of $\langle xy\rangle$, $\langle xz \rangle$, $\langle xx\rangle$ and $\langle yz\rangle ... 1 Starting from $$\tag1 \mathcal{P}\sim\frac{ \mathrm{i} }{ p ^2 - m _0 ^2 + M ^2 ( p ^2 )}$$ and as Jeff points out, by the Optical theorem*,$M^2$(for a particle that decays) can have a nonzero imaginary part. Hence one defines the physical mass$m\$ of the particle, not through m ^2 - m _0 ^2 - M ^2 ( m ^2 ) ...

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