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3

It is a non-perturbative effect because it is 1-loop exact. The triangle diagram is actually the least insightful method to think about this, in my opinion. The core of the matter is the anomaly of the chiral symmetry, which you can also, for example, calculate by the Fujikawa method examining the change of the path integral measure under the chiral ...


0

In the case of symmetric double well potential (the Hamiltonian is even under parity) tunneling happens between two states localized at the two minima provided the barrier is finite. Those two states are superposition of the ground and 1st excited states of the Hamiltonian thus they are not eigenstates of the Hamiltonian. If we tune the barrier height as a ...


6

Spin-0 can be either massive or massless. Examples of known massive spin-0 particles are the pion $\pi^+$, kaon $K^+$, and also the recently discovered Higgs boson $H$. No known spin-0 particles are exactly massless, but the Goldstone boson arising from the spontaneous breakdown of a continuous internal symmetry is a good theoretical example. Spin-2 can ...


1

A particle interpretation of QFT answers most intuitively what happens in particle scattering experiments and why we seem to detect particle trajectories. Moreover, it would explain most naturally why particle talk appears almost unavoidable. [My italics: the answer to your question.] http://plato.stanford.edu/entries/quantum-field-theory/#TakSto The ...


0

Let us consider for a moment just a classical scalar field before quantization. For this field we have a Fourier decomposition: $$\phi = \int \frac{d^3 \!p}{...} (a_{\vec{p}}e^{ipx} + b^*_{\vec{p}} e^{-ipx})$$ Where $a_{\vec{p}}, b^*_{\vec{p}}$ are just numbers and the complex conjugation of $ b_{\vec{p}}$ is just a convention. In this description, the ...


0

While both the answers given in some sense are correct, the true reason has to do with energetic considerations. It is a matter of what is stronger and can be phrased as the following question: Will the wavefunction alter itself to accommodate the flux, or will the flux quantize itself because the wavefunction is trying to remain single valued? As an ...


2

As I showed in my article http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (published in J. Math. Phys.), in a general case you can use one real function instead of the four complex components of the Dirac spinor, as 3 out of four components can be algebraically eliminated from the Dirac equation, and the remaining component can be made ...


3

Here is some reasoning for representation both of particles and antiparticles by one (complex in general, not only in usual sense but also in a sense of the irreducible representation of the Lorentz group) field. All of QFT processes are described by S-matrix, which can be written in a form $$ S_{\alpha \beta} = \langle \beta | \hat{S} | \alpha \rangle , ...


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. ...


0

In fact, your question is not so clear. I try my best. The Yukawa potential is an exchange potential, so it is based on the particle which is exchanged between 2 interacting particles. So if that particle (a boson) is coupled to the Higgs boson, it will get mass and the potential between the 2 interacting particles should change from U(r)~1/r to U(r)~ ...


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 ...


0

I'm not sure what you're asking, but it's not really true to say that particles interact with the Higgs field. A quantum field like the electron field interacts with the Higgs field and the result is that the electron field is massive i.e. its excitations (electrons) have a mass. If you consider the Higgs boson, rather than the Higgs field, then the ...


2

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 ...


0

The key point in writing an action for spinors is the existence of a Clifford algebra (expanded by the gamma matrices) $$\{\gamma^a,\gamma^b\} = 2\eta^{ab}\mathbf{1},$$ where the index $a$ runs from $0$ to $3$ (or $0$ to $D-1$, where $D$ is the spacetime dimension). The whole basis for the algebra is given by the $\gamma$'s and all possible products... due ...


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 ...


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 ...


2

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 $V(\phi)$ ...


0

This Bardeen counterterm is an elusive beast, I must say. Yet I will share what I have found and understand: Define a $\mathrm{U}(1)$ gauge theory by writing its action in left- and right-handed chiral spinors as $$ S_{\mathrm{chiral}}[A] = \int \bar \psi_L (\mathrm{i} {\not \hspace{-4px} \partial} - \not \hspace{-5px} A)\psi_L + \bar \psi_R (\mathrm{i} ...


0

Yes, we can have entanglement between different degree of freedom of same particle or system. That is known as ''hybrid entanglement'' and that is experimentally demonstrated also. http://arxiv.org/pdf/1007.1322v1.pdf


0

A state like $ \frac{1}{\sqrt{2}}(a^\dagger_+(\vec p)a^\dagger_+(-\vec p) + a^\dagger_-(\vec p)a^\dagger_-(-\vec p))|0\rangle$ would be an example. It is both entangled in spin and entangled in momenta.


2

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 ...


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 ...


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 ...


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

In quantum theory, the S-matrix is unitary in order to preserve the concept of probabilities adding up to 100%. Unitarity is implemented via an appropriate reality constraint/condition. E.g. in the path integral formulation $$Z~=~ \int {\cal D}\phi~e^{\frac{i}{\hbar}S},$$ the action $S$ should be real-valued.


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, ...


0

Fields in QFT are promoted to operators. The Dirac field operator describes both a particle (electron) and and a anti-particle (positron), and there exist different creation and anihilation operators for the particle ($a^+_s(p), a_s(p))$ and the anti-particle ($b^+_s(p), b_s(p)$). The subscript $s$ indicates that you are creating or destroying a particle of ...


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 ...


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 ...


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. ...


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 ...


5

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, ...


0

I recommend reading this first: P. Cotta-Ramusino, E. Guadagnini, M. Martellini, and M. Mintchev. Quantum Field Theory and Link Invariants. Nuclear Physics B, B330:557, 1990. Also Guadagnini has a nice book("The Link Invariants of the Chern-Simons Field Theory") on the subject but it is hard to find.


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 ...


0

I recommend M. Bos and V.P. Nair. Coherent State Quantization of Chern-Simons Theory. International Journal of Modern Physics A, A5:959, 1990. Also, I wrote a short review of it: check the second section of my paper on Yang-Mills-Chern-Simons theory (http://arxiv.org/abs/1311.1853) .


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 ...


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 ...


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 ...


2

In order to show this, figure out how the transformations $C, P, T$ act on each element in the equation individually. Pay special attention to $C$. That's a tricky one! For a start on C you can check out this physics SE post.


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, ...


4

This formula follows the usual heuristic discretization rules (here written in 1D): $$\tag{1} \text{discrete var.}\qquad i\in\{1, \ldots,N\}, ~~x_i=i\Delta ~~\longrightarrow~~x~\in~[0,L] \qquad \text{cont. var.},$$ $$\tag{2} \text{sum}\qquad \sum_{i=1}^N ~~\longrightarrow~~ \int_0^L \! \frac{dx}{\Delta} \qquad\text{integral},$$ $$\tag{3} ...


0

When computing the effective potential $V(\phi)$ in the ordered phase ($\phi_b>0$), one has to use the classical propagator $G_c[\phi_b]$ given by the inverse of $$\frac{\delta^2 S[\phi]}{\delta\phi^2},$$ which is a functional of $\phi$. The vacuum energy is given by $V(\phi_b)$, where one should remember that $\phi_b$ is also computed consistently in ...


0

The simplest way to see that all products of your kind vanish is to notice that one of the objects (bilinear invariant) $T_{\mu\nu}$ is a self-dual 2-form while the other $T^{\mu\nu}$ is anti-self-dual, and their contraction without a complex conjugation has to vanish. A self-dual (anti-self-dual) antisymmetric tensor obeys $$T_{\mu\nu} = \pm \frac i2 ...


3

$X_i,Y_i,Z_i$ are three Pauli matrices acting on the $i$-th qubit where $i=1,2,3,4,5,6,7,8,9$ labels the qubit. In equation 4.1, the state is a superposition of tensor product of three states similar to $|000\rangle$. The latter is a state of three qubits, so if one takes the third power, it is a state of $3\times 3 = 9$ qubits. $X_1$ differs from $X_8$ by ...


1

Imagine a classical relativistic bosonic scalar field $\Phi(\vec x,t)$, its equation is the Klein-Gordon equation : $$\partial_0^2 \Phi(\vec x,t) - \sum\limits_i \partial_i^2 \Phi(\vec x,t) = 0 \tag{1}$$ Now, take the spatial Fourier transform of this equation, you get an equation for $\Phi(\vec k, t)$ : $$ \partial_0^2 \Phi(\vec k,t) + \vec k^2 ...


4

For a free theory, say for one scalar field for simplicity, which gives a a linear differential equation for the field $\phi$, one can cast the hamiltonian $$ H=\frac{1}{2}\int d^3x \dot\phi^2+(\partial_i \phi)^2+ m^2\phi^2 $$ in this form (basically by taking a Fourier Transform) $$ H=\mathrm{const}+\int \frac{d^3 k}{(2\pi)^3} ...


1

Let's look to the expression for field with mass $m$ and spin $s$ (for massless case following statements exist in similar form): $$ \tag 1 \hat {\psi}_{a}(x) = \sum_{\sigma = -s}^{s}\int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi )^{3} 2E_{\mathbf p}}}\left( u^{\sigma}_{a}(\mathbf p )e^{-ipx}\hat{a}_{\sigma}(\mathbf p ) + v^{\sigma}_{a}(\mathbf p ...


2

For simplicity, let us talk about a scalar field $\phi : \mathbb{R}^4 \rightarrow \mathbb{R}$. The action for a free scalar field is $$S[\phi] = \frac{1}{2}\int_{\mathbb{R}^4} \partial_\mu\phi\partial^\mu\phi - m^2\phi^2$$ and its classical equations of motion is the Klein-Gordon equation $$ (\partial_\mu\partial^\mu + m^2) \phi = 0 $$ Now that looks ...


0

I can't write comments right now So here is what i wanted to write Coupling constant's value is an indication of how strong or weak the intraction and also for ascribing correct dimensions to lagarangian( as mentioned by @winther) is i.e an interaction lagarangian of two particles with higher value of coupling constant will result in stronger interactions ...



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