Tag Info

Hot answers tagged

4

Let us look at the instantons of an ordinary pure Yang-Mills theory for gauge group $G$ in four Euclidean dimensions: An instanton is a local minimum of the action $$ S_{YM}[A] = \int \mathrm{tr}(F \wedge \star F)$$ which is, on $\mathbb{R}^4$, precisely given by the (anti-)self-dual solutions $F = \pm \star F$. For (anti-)self-dual solutions, ...


3

Such operators are ill-defined in an interacting theory because whatever counterterms we try to subtract, their expectation value in any finite-energy state will diverge. The closest operators that are well-defined are densities of charge – number operators with signs labeling antiparticles – because the divergent contributions naturally cancel for them. ...


3

You can utilize the construction of the $Q$ space, as described in Reed and Simon vol.2, page 228-230. Oversimplifying, you can make the analogy $\lvert \phi\rangle \sim \lvert x\rangle$, but the associated momentum is not $\hat{p}$, but $\hat{\pi}$ (the canonical conjugate momentum of the field $\hat{\phi}$). With slightly more precision: the Fock space ...


3

I'd say that there is not a systematic summary of the status of symmetries on particle physics, but if any, it should be spread all over the PDG review. However, I'd like to comment on a few points. So far Lorentz symmetry is exact on all sectors.${}^\dagger$ Scaling (part of the conformal transformations) is broken once an energy scale is introduced in ...


3

See [1] M. Bos and V.P. Nair. Coherent State Quantization of Chern-Simons Theory. International Journal of Modern Physics A, A5:959, 1990. and chapter 20 of [2] V.P. Nair. Quantum Field Theory: A modern perspective. Springer, 2005. These references have geometric quantization of abelian Chern-Simons theory for $\Sigma=S^1 \times S^1$ (the first ref. ...


3

The Higgs mechanism has been understood in the framework of quantum field theory since the very beginning i.e. since the 1960s. Quantum field theory may be constructed as a quantization of its classical limit, i.e. of a classical field theory, so that's what's important for understanding some basic properties of the Higgs mechanism, too. But many other ...


3

Feynman diagrams are more than just the Lagrangian. They can be acquired by expanding the path integral of the theory into a perturbative series. There is a priori no reason to assume that all quantities needed in order to produce sensible results are consistent with gauge invariance. One possible issue is the problem of regularization: the way your ...


3

Hint: Recall that $$\tag{A} q\cdot (q+2p) ~=~(p^{\prime}-p)\cdot (p^{\prime}+p) ~=~p^{\prime 2}-p^2~=~m^2-m^2~=~0 .$$ So $$\tag{B} \Delta -(1-z)^2m^2~=~y(y-1)q^2 -2yz q\cdot p~\stackrel{(A)}{=}~y(y-1+z)q^2~=~-xyq^2. $$


2

If $\mathcal H$ is the Hilbert space of the QFT, then \begin{align} U:\mathrm{SO}(3,1)\to \mathscr U(\mathcal H) \end{align} where $\mathscr U(\mathcal H)$ is the set of unitary operators on $\mathcal H$. In other words, $U$ is a unitary representation on the Hilbert space of the theory. If $V$ is the target space of the fields begin considered, then ...


2

Let $f(x,y)\in L^2(\mathbb{R}^{2d})$ and $\Omega$ the vacuum of the symmetric Fock space $\Gamma_s(L^2(\mathbb{R}^d))$. Suppose there is no $f_1,f_2\in L^2(\mathbb{R}^d)$ such that $f(x,y)=f_1(x)f_2(y)$: then $f_s$ (the symmetrized of $f$) is an "entangled" two particle state of $\Gamma_s(L^2(\mathbb{R}^d))$. This is created by $$\frac{1}{\sqrt{2}}\int ...


2

The classical action (of particles or fields) has to be real, because this means a real classical Lagrangian. This is needed because (canonical) momenta are obtained (for instance for a particle) from the Lagrangian by $p_i = \frac{\partial \mathcal {L}}{\partial \dot x^i}$ , and momenta are real. In QM or QFT, the action has to be understood as a phase, ...


1

In your previous question, you say that these two images (a)  (b)  have been taken of the same area of your sample but with a different lens and lighting configuration. I think that I believe this: in this first image, there are two dark areas (the left shaped sort of like New York State, and the right shaped sort of like a heart), and in the second ...


1

The "is simply an excitation" might be a bit overstated, as it if it were that simple you probably wouldn't have needed to ask this question. It might be better stated as "can simply be modeled mathematically as an excitation". The "fields" are part of a mathematical model that attempts to explain our observations. Saying "where does a field come from" is ...


1

The quantum field has nothing to do with the wavefunction. This is a peculiar confusion that seems to arise quite often. The wavefunction $\psi(x)$ is a way of representing a quantum state $\lvert \psi \rangle$ in a Hilbert space $\mathcal{H}_{\mathrm{QM}}$ that is equipped with a position basis $\{\lvert x \rangle \rvert x \in \mathbb{R}\}$ by setting ...


1

You must first realize that a priori there doesn't need to be a relationship between the 2 things you're comparing here: A. The former, $|\phi(x)|^2= |\langle x|\phi \rangle|^2$ is the square of the position space amplitude of a state $|\phi\rangle$ living in Hilbert space B. The latter has two possible meanings depending on context: If we view $V(\phi)$ ...


1

Here is one tricky way to show that the Dirac representation is invariant under C,P,T-transformations. The free Dirac theory refers to the direct sum $\left( 0 , \frac{1}{2}\right) \oplus \left( \frac{1}{2}, 0 \right)$ of the Lorentz group irreducible spinor representations. As it can be shown, the representation $\left(\frac{n}{2}, \frac{m}{2}\right) \oplus ...


1

The question assumes that the nuclear force does have an attraction at long distances and a repulsive core at short distances. The reality is more complicated than that, and there is in fact no unambiguous way to decide whether this assumption is really correct. The strong force between two quarks is often modeled with a potential $V\propto r^n$, where ...


1

When I did the derivation I got this for delta: $\Delta = -y^2q^2 +yq^2 -2zpyq +(1-z)^2m^2$ In the text it says this: $\Delta = -xyq^2+(1-z)^2m^2$ Is there some information missing? I recommend looking up "how to use Feynman Parameters" on google to get more detail, but basically it looks like the derivation is given on this webpage, in example 2. It ...


1

A pretty exhaustive summary in the context of Standard Model already exists in the following source: ''Dynamics of the Standard Model'' - Donoghue, Golowich, Holstein, Chapter 3 - Symmetries and Anomalies A limited preview can be found here. (Embarrassingly though, the very first page of the chapter is excluded from Google's preview!) But here's the ...


1

They are very different. When you use a Higgs mechanism with a Yang-Mills action, symmetry breaking causes the gauge fields $A$ to gain mass. This is done in 4D. When you add a Chern-Simons term to a Yang-Mills action, you can see from the field equations that $\ast F$ becomes massive, not $A$. There is no symmetry breaking here. Also this is in 3D and ...


1

No $\hat\phi|0\rangle$ is not an eigenvector of $\hat\phi$. You can see this, for example, by writing out $\hat\phi$ in terms of creation and annihilation operators, then compare $\hat\phi|0\rangle$ against $\hat\phi^2|0\rangle$, and observe that one is not a scalar multiple of the other. So as you suspected, eq. 5 is not correct To obtain some analogy of ...


1

Below that regime, we have the strongly coupled regime where perturbative approaches fail, due to the large value of coupling constant $\alpha_S$. The same is related to the QCD $\beta$ function via this relation. The behavior as a function of the energy scale looks roughly like this. Any perturbation expansion in this regime would give a divergent series, ...



Only top voted, non community-wiki answers of a minimum length are eligible