# Tag Info

1

Entropy is from statistical physics, a single particle or a single degree of freedom does not have an entropy. Edit: There are a lot of different things that are called entropy. So I'm not sure I feel comfortable with the above as a blanket statement. And there is a way to see time from the dynamics of quantum mechanical systems, so if the above statement ...

1

I am not sure this is the answer you are looking for but you may have heard, in QM especially, that an excited system eventually goes back to its ground state. This is true of subatomic particles like neutrons decaying into proton + extra stuff and of atoms too which, once in an excited electronic state, eventually release this extra energy as a photon. The ...

2

I talked to my friend about this, and he said something along the lines of "the pair production particles aren't 'real' particles, and they can't really interact with anything", though I'm not too sure I understand what that means. Your friend is correct. A virtual particle is a mathematical construct, it is a creation of the mathematical model used ...

1

It all depends on what application you are interested in. When discussing chiral perturbation theory it is implied that you are interested in low energy processes. If you are considering processes such as low energy (~100 MeV) pion scattering, pion decay, etc. then the typical energies involved are O(100 MeV) which is much smaller then $\Lambda_{QCD}$. This ...

3

There are apparently several thousand references to "SU(\infty)" on arxiv.org, and some of them are definitely talking about gauge fields or Yang-Mills. I suspect that some of the time, this will just be a way of talking about the large N limit of SU(N), i.e., not referring to a literal SU(∞) field theory, but rather the N→∞ limit of some quantity in SU(N) ...

2

First, terminology: You are not "determining the gauge group", what you are doing in gauge fixing is determining a smooth choice of (hopefully only) one representant of an equivalence class of field configurations called the gauge orbit. Geometrically, you are seeking a section which intersects each gauge orbit exactly once. The problem of finding a gauge ...

1

1 : When $A$ is restricted to $[-\pi,\pi)$, the group of gauge transformations is the compact $\mathrm{U}(1)$ group. (Roughly speaking, to be compact, a group needs to be bounded.) In the nonperiodic theory, the gauge group is the group of real numbers under addition, which is noncompact. 2 : The most important distinction between the compact and noncompact ...

1

Periodicity is compactness. If $A_{x,\alpha}$ only varies between $-\pi$ and $+\pi$, the range of $A_{x,\alpha}$ is obviously compact. The difference only exists on the lattice. The continuum limits of the two theories are identical. Compact QED simply shows one way confinement can arise. I don't understand your question.

1

No, instead of the metric, the Vierbein enters, pulling back the gamma matrix, defined in the usual way in the tangent space, to the spacetime manifold. $$\bar \psi \, \gamma^\mu D_\mu \psi = \bar \psi \, \gamma^\alpha {e^\mu}_\alpha D_\mu \psi$$ This can be considered as the insertion of "half a metric", if one wishes. The Vierbein captures the ...

4

Heuristically, one approach to justifying Weinberg's application of Cauchy's formula is to treat the nonanalytic integrand as the boundary behavior of a meromorphic function (kind of like Fourier series): perform all internal integrations until you are left with an integral over energy, solve the Cauchy-Riemann equations within the upper half plane ...

1

To answer the first part of your question, in the transformation rule $U(\Lambda)\Psi_{p,\sigma}=\sum_{\sigma'}C_{\sigma\sigma'}(\Lambda,p)\Psi_{\Lambda p,\sigma'}$ the matrices $C_{\sigma\sigma'}(\Lambda, p)$ are invertible, and lie within some continuous group of operators (looking ahead, it's OK to think of this group as a finite dimensional Lie group). ...

2

In the first paragraph you quoted he is saying that in general it may be the case that the $\Psi_{p,\sigma}$ fields are reducible. Say $\sigma$ is an index going from 1 to 3, maybe after you do a suitable transformation, the first component is the state of a spinless particle and the next two are the states of a spinor. Or maybe all three components refer to ...

3

The Dirac equation is more restrictive than the Klein-Gordon equation. For every solution to the Dirac equation, its components will be a solution of the Klein-Gordon equation, but the converse isn't true: if you form a spinor whose components are solutions of the Klein-Gordon equation, it might not solve the Dirac equation. If we start with the ...

0

At a time when my hammer was the Dirac's delta distribution, I conjectured that the answer was Feynman Integral is a generalization of a Dirac's delta, the use of this delta being to find the extreme of the action. Given a function $f(x)$, find a Dirac measure $\delta_f$ concentrated in the critical points of $f$. The answer is obviously $\delta(f'(x))$, ...

4

For the simplest case of interaction, the quadratic one, the interacting representation is roughly speaking the Fock one, but with a different mass (representations with different masses are proved to be inequivalent in the second volume of the Reed-Simon book). For $(\varphi^4)_3$, with spatial cut off $g(x)$, the construction is a bit more involved (and ...

0

The original Unruh & Wald (1983) paper addresses precisely this question. (You can find it here: http://www2.kau.se/tp/marcus/physics/lectures/unruhwald.pdf.) They discuss measurement of the thermal radiation by the accelerated observer via a two-state system, where the system jumps to a higher energy level on absorbing a quantum of the thermal ...

1

US patent 7411772: Casimir effect conversion Abstract: Techniques in which a 1.sup.st force, field, or effect caused by a Casimir effect is converted into a 2.sup.nd force, field, or effect. The 1.sup.st force, field, or effect might be distinct from the 2.sup.nd force, field, or effect only in the net vector or might be distinct in other ways. For ...

1

Experimental Applications Take the LHC. We need to calculate what the various probabilities are regarding which particles are going to emerge from the collisions, for how long will they exist and in what way will some decay into stable particles such as electrons and the lowest mass quarks. Without QFT we could not calculate the probabilites of these ...

2

What makes the case of the Higgs field different from that of other particles is that the Higgs field in the vacuum has a nonzero expectation value. So, if the electromagnetic field is in its lowest energy state then that means that the field strength will be zero on average (there are still quantum fluctuations, but on average it is zero). But for the Higgs ...

1

Non zero quark masses break chiral symmetry. This breaking is assumed to be small enough (certainly for the up and down but also for the strange quark) to do a meaningful Taylor like expansion in it. The relevant scale to which the masses need to be compared is about 1 GeV. It is no guarantee the series will converge but there are predictions that follow ...

2

In quantum field theory it is actually not obvious how many fields there are since fields can have components. If we have two fields $A$ and $B$, we can consider them to be merely components of the same field. Or reversely, if $A_1$ and $A_2$ are components of a field we can relabel them $A$ and $B$. However, it rubs physicists the wrong way to split fields ...

1

Every particle has a corresponding field that permeates all of space in the same way the Higgs has a field that does so. The spin up electron. The spin down electron. The spin up positron. The spin down positron. The up quarks (all three colors and both spins). The down quark (all three colors and both spins). Same for the charm, strange, top and bottom. ...

0

The basis change for the Clifford algebra $$\{\Gamma_A, \Gamma_B\}_+~=~2\eta_{AB} \tag{1.12}$$ can be viewed as a linear coordinate transformation in the (complexified) Minkowski space$^1$ $V$ of dimension $2n$. The longitudinal and temporal coordinates go into light-cone coordinates. The transversal coordinates pair up two and two and are transformed in ...

1

Ref.1 is already in eq. (3.1) considering a functional integral over the scalar field $\phi:M\to \mathbb{R}$. Here $M$ is spacetime. For a rigorous treatment of functional integrals, Ref. 1 points in the beginning of Section 3 to its Ref. [3.2]. In this answer we will just take an intuitive heuristic approach, and try to construct the functional integral as ...

1

We have to change the renormalization scale to a typical scale of the process we are considering because otherwise perturbation theory becomes rather useless, since process far away from the renormalization scale get large higher order corrections due to the "large log" $\ln(-p^2/\mu^2)$, where $p$ is the process scale and $\mu$ the renormalization scale. ...

0

Another way to find an equivalence is to express the Green's function via the equation solutions - write down its spectral representation: $$G(x_1,x_2,t_1,t_2)=\sum_n \psi^*_n(x_1)\psi_n(x_2)e^{iE_n(t_1-t_2)}$$ or something like that.

0

There are Feynman integrals that don't come from diagrams. These integrals occur naturally when trying to use tensor reduction to remove the numerator in some Feynman integrals. The generalization of a graph/diagram is called a Matroid and in Feynman integrals (as you described them) where the momentum flow in the denominators can not come from a graph, it ...

9

First, the term "propagator" is usually defined as the Green's function of the first type, not the second type, i.e. as a solution to the diffential equation $\hat L G = \delta$. At any rate, those definitions are ultimately equivalent – when the details are correctly written down – because the Green's function defined as the correlator in the second ...

1

If you plug the propagator into the equation of motion you'll get a $\delta$ function. The second "definition" is just a prescription to calculate the Green's function in a free field theory.

1

If you want to think of a free on-shell particle as a mode that propagates like a soliton far from interactions then you can think of it as a thing that interacts with the vacuum so as to make more of itself. But really you should stick to asking questions the formalism is designed to answer. The vacuum is not a thing like a house that provides space for ...

0

Eh nevermind, I figured it out. $(E^2-|\mathbf{p}|^2) = (E+|\mathbf{p}|)(E-|\mathbf{p}|)\approx2E(E-|\mathbf{p}|)$ when E is very close to $|\mathbf{p}|$ (considering that the term is in the denominator), which is the limit we are making the approximation in. And it seems like the last equality in line two was probably a typo in P&S, and should have been ...

2

Here we assume that OP's question asks about $\phi^4$-theory in 1+1D, where the lagrangian density reads $$\tag{1} {\cal L}~=~\frac{1}{2}\dot\phi^2 -{\cal U}, \qquad {\cal U}~:=~ \frac{1}{2} \phi^{\prime 2} + {\cal V},\qquad \phi \in C^1(\mathbb{R}^2),$$ where the $\phi^4$-potential density $$\tag{2} {\cal V}(\phi)~\propto~(\phi^2-v^2)^2~ \geq~ 0$$ ...

3

Let's start with qft. Imagine taking the action $S$ and adding a constant. The new action is $$S_{new} = S + M^4 \int d^4 x$$ where M is a mass scale needed for dimensions. This last term in the action does not depend on any fields so it does not contribute to any dynamics. One way to express this is that the only effect of the ...

1

Use the following expansion of effective action $$\Gamma[\phi_c]=\int d^4x [-V_{eff}(\phi_c)+(\partial_\mu\phi_c)^2A(\phi_c)+...]$$ where $...$ represents higher order derivatives of $\phi_c$. When $J\rightarrow 0$, $\phi_c(x)=constant$. This constant is the VEV $\langle\phi_c(x)\rangle=\phi_0$ and we obtain from above $$\Gamma[\phi_0]=-VT. ... 0 The energy density of the state \pm v is going to be something like \propto μ^4, if you are using the basic \varphi^4 theory. While the energy of the domain wall is finite, the energy of the vacuum state is not, and so the transition to the vacuum state iver all space will be infinite. 3 You should work out the minimum energy state of your system (classically) to find the vacuum expectation value. I assume you're working with the standard \phi^4-Lagrangian$$\mathcal L=\frac{1}{2}(\partial \phi)^2-\frac{1}{2}m^2\phi^2-\frac{\lambda}{4}\phi^4 $$which corresponds to the Hamiltonian$$\mathcal ...

4

The Lorentz group is the group of matrices that conserve the quadratic form: $$\mathscr{Q}(X,\,Y) = X^T\,\eta\,Y\tag{1}$$ where here $X$ and $Y$ are $1\times 4$ column vectors, the $4\times 4$ group member matrices act on these from the left and $\eta$ is the Minkowski (pseuso) metric. Therefore, $\Lambda\in O(1,\,3)$ if and only if: ...

1

I highly recommend Richard D. Mattuck A Guide to Feynman Diagrams in the Many-Body Problem. You can read some pages here. It's a very surface level introduction, but the first 3 or so chapters are presented at what he calls a "kindergarten" level so you shouldn't have any problems understanding it. However, the last part is most definitely not ...

1

There are many resources on many-body Green's functions (propagators) both on-line and in print. You may want to search "quantum field methods in many-particle systems" or "quantum field methods for condensed matter systems" or variations thereof. In any case, I personally recommend the oldie-but-goodie book by Fetter and Walecka, Quantum Theory of ...

3

Massless particles can carry and do carry confined charges (gluon in QCD) and they may carry charges under spontaneously broken generators (photon is transforming e.g. under the generators of $SU(2)$ associated with the W-bosons) but they cannot carry charges under unconfined $U(1)$ force like electromagnetism. The reason may be explained in different ways. ...

3


2

The claim is that $|0(\theta)>$ lies outside the Hilbert space built on the original vacuum |0>. To check that this true, consider the overlap of the new vacuum $|0(\theta)>$ and the (unnormalized) basis states $(a_k^\dagger)^n|0>$ generated from |0>, taking into account that $a_k(\theta) = a_k + \theta_k$, $a_k = a_k(\theta) - \theta_k$, and ...

1

The Feynman propagator is the green's function of the klein gordon equation therefore $G(x,z) = \int G(x-y) (\Box_y +m^2) G(y-z) dy^4$ is equivalent to $G(x,z) = \int G(x-y) \delta(y-z) dy^4$ if you use the delta function to perform the integral you get $G(x,z)=G(x-z)$. Now as far as intuition goes I would say that the reason that you cannot simply multiply ...

1

The defining property of the fundamental representation of the Lorentz group $\mathrm{SO}(1,3)$ $$M^T\eta M = \eta \quad \forall M\in\mathrm{SO}(1,3)$$ and hence the defining property of the Lorentz group itself does not make sense in representations other than the fundamental, because those are not naturally equipped with a metric "$\eta$" from a physics ...

0

It is so, because Dyson formula was designed for the interaction picture in Quantum Mechanics. Interaction picture may seem strange, especially in comparison with the 'natural' Heisenberg picture, but has been proven useful, especially in perturbative computations. The basic idea is to split the total Hamiltonian into two parts: $$H = H _0 + H _{int}$$ ...

0

Comments to the questions (v2): The Hamiltonian $H(t)$ in QFT often depends on time $t$, e.g. if there are interactions in the interaction picture, or if there are external sources. In particular, Hamiltonians at different times need not commute. For an explicit example see e.g. my Phys.SE answer here. Well, if the Hamiltonian $H(t_1)=H(t_2)$ is the same ...

2

Indeed, the QFT notion of locality is that observables at space-like separation commute, i.e. $$(x - y)^2 < 0 \implies [\mathcal{O}_1(x),\mathcal{O}_2(y)] = 0$$ for all local observables $\mathcal{O}_1,\mathcal{O}_2$, which are generically polynomials in the fields and their derivatives. This is our notion of locality because, classically, we know that ...

-1

The essence of the Higgs mechanism is that it allows the breaking of the (gauge) symmetry to grow a mass for the gauge (vector) bosons, which are necessarily massless in the unbroken symmetry. The Higgs scalar and the two degrees of freedom of the massless vector boson combine to form the three degrees of freedom of a massive vector boson. Goldstone's ...

0

Your questions are answered in detail in the 1st chapter of this book: W. Greiner, Relativistic Quantum Mechanics, http://iate.oac.uncor.edu/~manuel/libros/Modern%20Physics/Quantum%20Mechanics/Relativistic%20Quantum%20Mechanics.%20Wave%20Equations,%203rd%20ed.%20-%20W.%20Greiner.pdf It treats the Klein-Gordon equation (KGE), its current interpretation, and ...

3

I think the answer is neither one sadly. If you do a taylor expansion you find $$\partial_\mu e^{i t^a \phi_a} = \sum_n \left(\frac{i^n}{n!}\right) \partial_\mu (\phi^a t_a)^n = i \partial_\mu (\phi^a t_a) - \frac{1}{2}\left( \partial_\mu (\phi^a t_a) \phi^b t_b + \phi^a t_a \partial_\mu (\phi^b t_b) \right) + \cdots$$ The second ...

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