Tag Info

New answers tagged

0

Like OP says, using dimensional regularization to wave away the naturalness problem would be quite reckless. I will elaborate below: Case 1: If you honestly believe that there is no new degree of freedom beyond the standard model (at least till the Planck scale, where all hell might break loose, for all you care) AND you believe in the 't Hooft-Veltman ...


1

It's fullfilled for any local quantum system that the entanglement entropy of a region A in a mixed state is extensvie, Could you give a reference, please? The entanglement entropy measures not just the entanglement but also the classical entropy. It is a measure of entanglement, only if you start with a pure state and then trace over some subsystem. ...


1

The problem with such an integral is that it is both UV and IR divergent. We therefore need to introduce two regulators (to regulate both divergences). To regulate the UV divergence, we use dimreg. To regulate the IR divergence, we give the particle a mass and then take the massless limit. Doing both of them, the integral becomes $$ I = \lim_{m^2 \to 0} ...


3

To put it simply: Renormalizability is the feature that the theory you know at low energy scales can be extrapolated to "arbitrarily high" energy scales, without losing consistency. Now some observations: When you say that a theory comes with a cutoff and seemingly doesn't work beyond that cutoff, then you're seeing something (that is scale dependent) ...


6

Unitarity says that the probabilities of any event is less then $1$. This is obviously an essential requirement for a given quantum theory and if a theory is not unitary then for it to describe Nature, it is necessarily missing some information that will fix this issue (such as new states and/or interactions). Renormalizability just says that the theory ...


0

Take a look at Collision theory by Goldberger&Watson (1964). Its an old classic book covering variety of topics in scattering theory within relativistic QM and QFT.


0

disclaimer: I am not sure about this answer but nevertheless I'm quite confident about it (and would like to be corrected otherwise, learning something). The one-particle states refer to asymptotic states at infinity where interactions are supposed to switch-off, that is the gauge coupling is sent to zero. In this limit a $U(1)^N$ gauge symmetry survives ...


2

I would check if the corresponding Hamiltonian is self-adjoint. The time evolution operator is $$ U(t,t') = \text{e}^{i(t-t')H} \, .$$ Unitarity is equivalent to requiring that probability is conserved along the time evolution, $$ \frac{\text{d}}{\text{d}t} \langle \psi |\psi\rangle = i \langle \psi | H |\psi \rangle -i \langle \psi | H^\dagger |\psi ...


2

Correct me if I'm wrong, but your line of thinking goes like this... Since quantum fields do not commute in general one can have finite variances for, e.g., particle number. Since the vacuum states defines a probability distribution we can find the corresponding entropy. However, here we are dealing with quantum physics. The entropy is in general $S ...


3

The field theory is fully analogous for Hermitian and non-Hermitian fields The Hermitian operator $\varphi$ still creates and/or annihilates particles and the number of these particles $N$ is still well-defined (at least if we ignore interactions and problems with loops and divergences). The only difference from the non-Hermitian field is that the ...


0

One 'standard' approach to quantization which always works is the path integral: \begin{equation} \langle \mathcal O \rangle = \frac{\int [dQ dP] \mathcal O\exp \left( \int dx dt (\mathcal H + i P \dot Q)\right)}{\int [dQ dP] \exp \left( \int dx dt (\mathcal H + i P \dot Q)\right)} \end{equation} where $\mathcal O$ is whatever operator you are interested in. ...


0

In the following all integrals are done over the whole real line. The only one I can give an in depth explanation for, in terms of calculation, is the free particle. $$U(x,t;x_0) = <x|U(t)|x_0> = \int dp <x|e^{-ip^2/2m\hbar}|P><P|x_0>$$ where the $P$ is a momentum eigenstate and the Dirac notation indicates a jump from $x_0$ into $P$ ...


2

Typically, when you calculate quantum effects, you will put some cut-off $\Lambda$, and typical integrals, say for a $ \lambda \phi^4$ theory, are, at first order, going in $\log \Lambda$ (ex : $\int \frac{d^4k}{k^2 (p-k)^2}$) While defining renormalized quantities at some energy scale $p_0$, you may remove the cut-off, and get typically equations like (at ...


3

The second-quantised description of the electromagnetic field in terms of oscillators holds in QED as well. The part that is modified is the single particle description of charged particles. In other words, (virtual and real) pair-creation is permitted in QED. So for energy scales less than $2mc^2$ as well as low intensities (see Schwinger limit), where ...


1

There is so called optical theorem, which states that for the unitary theory must be $$ Im (M_{k_{1}, k_{2} \to k_{1}, k_{2}}) = 2E_{cm}|\mathbf p_{cm}|\sigma_{total}(k_{1}, k_{2} \to all), $$ where $cm$ denotes center of mass frame, $\mathbf {p}_{cm}$ - momentum of one particle at CM frame, $M$ is amplitude of scattering and $\sigma_{total}$ is total cross ...


5

The QFT is strongly based on the group theory formalism. Often when people say about some QFT theory they primarily say about the symmetries of the theory - invariance of lagrangian of theory (or about covariance of equations of motion) under sets of transformations. The group theory formalize these statements and help to construct theories which corresponds ...


7

Yes, your confusion is wholly caused by you thinking classically ;) In a hand-wavy way, particles are certain localized excitations of the quantized fields. The QFT picture contains the particle picture in the perturbative approach known as Feynman diagrams (and, relatedly, the LSZ formalism). There, we are given the action of our theory dependent on some ...


3

The type of field that you have depends on the way that your field transforms. The fields that you encounter in quantum field theory usually are: Scalar fields, these describe spin-0 particles such as the Higgs boson. Spinor fields, these describe spin-1/2 particles, these describe for example the elementary fermions, like the leptons and quarks. Vector ...


4

Fundamental fermions like quarks and leptons are described by the spinor field, while gauge bosons like photons are described by the vector field. They together with the Higgs bosons are currently what we have in the Standard Model for elementary particles.


1

What you need is Kirchhoff's Matrix-Tree Theorem which expresses ${\rm det}\ A$ as a sum of trees. You can find an easy "Fermionic" proof of this theorem and a list of original references in my article "The Grassmann-Berezin calculus and theorems of the matrix-tree type" (arXiv version here if you do not have access to the journal).


3

In canonical quantization, a quantum field is a linear combination of so-called "creation and annihilation operators". That means that the field $\phi$ creates and destroys particles of type $\phi$. The state $|0\rangle$ is the vacuum: the state with no particles. If $\phi$ is a quantum field that creates and destroys particles, it must be that $\langle 0 ...


1

Classical fields emerge when there is a large (but not definite) number of particles in a coherent state. For a simple example, for a scalar field $\phi(x)$ we can write a state that describes a classical configuration as something like $$ \exp\left(\int d^D p\; \tilde\phi(p) a^\dagger_p\right)|0\rangle. $$ Note that this isn't an eigenstate of particle ...


5

Here is the formal answer on your question based on particular result of Pauli theorem. Calculations are rather cumbersome, but they are general. Arbitrary fermionic field (with invariance under discrete transformations of the Lorentz group) It can be shown that each Poincare-covariant fermion field with spin $s = n + \frac{1}{2}$ and mass $m$ can be ...


3

In curved space, and even more so in expanding curved space, the concept of a vacuum wherein there are no particles has significant ambiguity. This is all covered very thoroughly in chapter 3 of "Quantum Fields in Curved Space" by Birrell and Davies. In normal flat space, we can define a vacuum state with no particles that every inertial observer would agree ...


0

As @TwoBs and @Trimok mentioned, in the case of the breaking $U(1)^n\to U(1)^{n-k}$, the charges don't change. This is however true only in a basis the broken fields are diagonals (only charge under one U(1). As an example, consider $U(1)^3$ and the following three fields with their charges: \begin{aligned} \Phi_1:& (1,1,0)\\ \Phi_2:& (1,-1,0)\\ ...


3

You can't violate the conservation of energy (at least not on long timescales) so you can't just create matter from nothing. If you're going to create matter from energy then that energy has to come from somewhere. One example of this is Hawking radiation. Although we often talk about a black hole event horizon as though it were an object, it is not. A ...


3

This is a situation where knowing the history of the terminology can be helpful. The QFT/string theory terminology comes from algebraic geometry, where the term moduli space is used for any space whose points correspond to some kind of geometric object. The projective space $\mathbb{P}(V)$, for example, is the moduli space of lines in the vector space $V$. ...


1

I think this calculation is the answer for your question: $$\sum_{\alpha,\mu}\frac 1 2 \omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)=\frac 1 2 \sum_{\alpha<\mu}\omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)+\frac 1 2 \sum_{\mu<\alpha}\omega_{\alpha \mu} \left(x^\mu ...


3

Well, there is the Kawai-Lewellen-Tye (KLT) relations, which says that a closed string amplitude is roughly speaking a product of two open string amplitudes. See e.g. Ref. 1. References: Z. Bern, Perturbative Quantum Gravity and its Relation to Gauge Theory, Living Rev. Relativity 5 (2002) 5; Section 3.1.


3

The question is not what is possible, but what is useful to obtain analytic results. In multi-loop integrals one is often interested in analytic results, because it can be very hard to confirm numerical results to a level where you trust them completely. Now the two-loop integrals usually involve propagators of the type $$ F(q_1, q_2) = \frac{i}{(p - q_1 - ...


2

Both ways are possible. Since you seem to be a mathematician, let me try an analogy from mathematics. Say that you are accomplished in commutative algebra. Now, you want to study algebraic geometry. Sure, you can start with sheaves of local rings and cohomology of schemes, instead of "at the bottom" with classical algebraic varieties defined by polynomial ...


2

This very much depends on what you want to do in the area of quantum theory. If you want to solve specific mathematical problems and to have only a very rough conception of why you are doing what you are doing, then you can in principle omit classical mechanics. But if you want to have a well-rounded knowledge of the subject, you should know some basics of ...


1

This is one of the few times when using the Einstein convention isn't helpful: recall that in Noether's theorem you need to sum over the independent parameters of your transformation, which is usually trivial but, for Lorentz transformations, you need to be careful because of the antisymmetry of $\omega_{\mu\nu}$, which implies that you have only 6 ...


1

Yes, you could use LHC data for that purpose, and that is exactly what they are doing at LHC. Technically, however, this is a lot harder than you can imagine. An experiment like ATLAS or CMS (the large LHC detectors) does not give you useful event counts by looking at the raw data. There are far too many error sources in both the physics at the interaction ...


0

Gauge quantum field theories are a robustly confirmed paradigm of contemporary particle physics. The Standard model is a gauge quantum field theory and LHC basically does not do anything else than testing it. There is so much data confirming at least certain ranges of validity from various experiments that the theory received the epithet "standard". If the ...


0

Following this ref, one sees that, in some basis where the current is diagonal ($3.2.16$), then a term like $\chi \eta$ is just a part of the mass term ($3.2.17$).


0

$$S_{\alpha \beta} = \langle\beta^+|\alpha^- \rangle$$ The normalization of states don't have any relation to the S Matrix. The interpretation of S-matrix in terms of Rates and Cross-sections may be have some implicit relation to the normalization of the In- and Out- States. If we set a box with very big volume, than we can normalize the In- and Out- ...


2

Just realize that you can form ordinary Dirac spinors from 2-spinors by using charge conjugation, $i\sigma_2\eta^*$, that gives a right- handed field that can fit in the right-handed slot (forming a 4 component Majorana field) $$ \Psi_1=\left(\begin{array}{c}\eta \\ i\sigma_2\eta^*\end{array}\right) $$ And analogous for $\Psi_2$ in terms of $\chi$. Then you ...


6

The fundamental representation of a Lie group $G$, as commonly used in this context, is the smallest faithful (i.e. injective) representation of the group. We do not require fermions to belong to the fundamental rep, it is just the case that, in the Standard model, they always either belong to the fundamental or the trivial representation (as that is ...


4

The Standard Model Yukawa interactions must be $SU(3)\times SU(2) \times U(1)_Y$ gauge invariant. The down-type Yukawa interaction is $$ \mathcal{L} \supset -y_d \bar Q \phi d_R + \text{h.c.}. $$ This is indeed gauge invariant. The $\bar Q d_R$ form a colour singlet ($3^* \times 3$), the $\bar Q \phi$ form an $SU(2)$ singlet ($2^*\times2)$, and the whole ...


5

The reason is that the $SU(2)$ invariant in $\mathbf{2}\otimes\mathbf{2}$ (or in their complex conjugate $\mathbf{2}^*\otimes \mathbf{2}^*$) is given by contracting the two $\mathbf{2}$ with the anti-symmetric $2\times 2$ matrix $\epsilon_{ab}$, as $i\tau_2$ is. In the case at hand the two $\mathbf{2}^*$ are $\bar{Q}$ and the $\Phi^*$. You could form another ...


2

First of all, it is not always true. Most of the time, people work with positive definite Hilbert spaces. Second, the physical subspace of the Hilbert space has to be positive definite because the (squared) norms appear as probabilities and those can't be negative. Third, one often works with an extended Hilbert space that also contains unphysical states ...


5

In addition to JeffDror answer : In the case of scalar fields, this is definitely the case. In order to get a massless field, you need to fine-tune one parameter of the Lagrangian. In the case of the a $\varphi^4$ theory defined by the parameters ($m_\Lambda$, $g_\Lambda$, $\Lambda$), i.e. the bare mass, interaction and the UV cut-off, one need to fine-tune ...


9

It is possible for particles to get masses at loop level while they are absent at tree level as long as there is no (non-anomalous) symmetry that forbids it. However, in most models if there is a particle doesn't have a bare mass then its due to a symmetry which then protects it from getting masses at loop level. This is often a subtle topic due to chiral ...


0

I suggest you to read http://arxiv.org/pdf/1408.0287.pdf for a nice discussion on the RG dependence of the effective potentials and the RG invariant statements that one can make. About your point, I think you are simply misunderstanding what people mean: they are calculating at what point in $\Phi$, not in $\mu$, the potential cross zero.


2

If you are considering massless neutrinos there is no such a diagram since all interactions would preserve flavor. If you take instead massive neutrinos, you are probing lepton flavor violation within the SM since the new interaction with $\varphi$ respects flavor. It is thus very very small, being controlled by the neutrino masses. In turn, it is therefore ...


4

The $SU(2) \times U(1)$ electroweak gauge theory has 4 symmetry generators, of which the vacuum breaks only three -- corresponding to $W^+$, $W^-$ and $Z$. The vacuum is symmetric under the generator of electromagnetism, hence the photon cannot interact with the vacuum. Think of the vacuum as a medium -- if it was not symmetric under EM, then it could ...


2

My apologies if this answer is more simplistic that you were looking for, but you did ask why phases are related to CP-violations. Most QFTs are CPT invariant, so a CP-violation is also a T-violation. If you consider an S-matrix element to be the amplitude for a transition from $a$ to $b$, i.e. $\langle a|b\rangle$, then the T-version should be $\langle ...


0

You are making your life harder than it needs to be. I will sketch the solution for you. You already know that $$ U = U_0(k) e^{-\imath\sqrt{1+k^2}t} $$ and likewise for $V$. Then plug this into the expression for $u(x,t)$, and set $t=0$. You obtain $$ u(x,0) = e^{-x^2} = \int U_0(k) e^{-\imath k x } d\!x $$ which gives you immediately $U_0(k) = \alpha ...


0

I wonder if there can be an error in your derivation. OK, you uncoupled the equations. Then you could consider a 2-dimensional Fourier expansion of $v$ into exponents, say, $\exp(i(\omega t +k x))$. Then the dispersion relation (what you get when you substitute the exponents into your linear differential equation for $v$) would be $-\omega^2+k^2+1=0$, so ...



Top 50 recent answers are included