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From a decoherence point of view, fields are more fundamental as they give rise to particle-like behavior from the wave behavior if interactions with the environment are strong. In the end though, quantum mechanics only describes correlations between macroscopic changes in detectors (or other materials), so whatever kind of ontology you want to take in the ...


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Simply because it is usually taught from historical, heuristic and pragmatic point of view, rarely from axiomatic point of view (e.g. Wightman axioms, as mentioned in a comment by ACuriousMind). This is because it is taught to be useful, as most QFT calculations boil down to scattering and decay amplitudes, and as Sean Carroll said: Heuristic QFT, on ...


2

Suppose I have two quantum modes $|\cdot \rangle_A$ and $|\cdot \rangle_B$. Suppose mode $A$ is in state $$\sum_n a_n|n\rangle_A$$ and mode $B$ is in state $$\sum_m b_m|m\rangle_B . $$ Then the state of the combined system is $$\sum_{n,m} a_n b_m \left( |n\rangle_A \otimes |m\rangle_B \right). $$ Note that the coefficients are multiplied. Now consider ...


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This looks to me like essentially a mean-field approximation. One is replacing $\psi$ with its expectation, so one is treating $\psi$ classically and $\chi$ quantumly. The back-action of $\chi$ on $\psi$ is ignored.


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Imagine a QFT with some particle content. Some of these fields will be massless and some massive. For simplicity, consider a massles scalar field and a massive scalar field with mass $M$. If we are working at some energy $E\ll M$, we won't see the massive field (as happened with the Higgs before LHC, for example). This is the IR CFT. Why IR? Because we ...


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I think that this nomenclature has nothing to do with the regulators. "UV CFT" and "IR CFT" actually refers to the end point of the RG flow which is triggered by a relevant operator that perturbs an UV fixed point, and it ends (barring certain non-unitary QFT) into a fixed point at lower energy scales, hence the name IR.


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You can identify states with fields, which are maps from Minkosky space-time to an operators space. By definition of fields you want this maps to satisfy some invariance proprieties under suitable transformation.


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Your idea 1) is the right idea: It's just the law of transformation of matrices generalized from the transformation of matrices: If we apply a general linear transformation $U : V \to V$ on a vector space, the matrices/operators on it transform as $M \mapsto U^\dagger MU$ For unitary operators $U^\dagger = U^{-1}$, so the transformation law becomes $M ...


2

$D_{\mu\nu} = A g_{\mu\nu}+B k_{\mu} k _{\nu}$ with A and B two unknown functions of the scalar k^2. The two tensor after A and B are the only possible Lorentz invariant tensors . Simply plugin and calculate the unknown functions.


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Some remarks: 1) Wavefunctions are not classical fields, and are not field operators. 2) Quantization transform classical fields in operators (so equations of movement, or equations of conservation of energy, are equations between operators). 3) Quantum Mechanics may be seen as Quantum Field Theory with zero spatial dimension. 4) Following the above ...


1

Consider an element $g$ of the symmetry group. Say $g$ is represented by a unitary operator on the Hilbertspace $$ T_g = \exp(tX) $$ with generator $X$ and some parameter $t$. It acts on an operator $\phi(y)$ by conjugation $$ (g\cdot\phi)(y) = T_g^{-1}\phi(y) T_g = e^{-tX}\phi(y) e^{tX} = \big[ 1 + t[X,\cdot]+\mathcal{O}(t^2)\big]\phi(y)$$ On the other ...


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Actually, all that is quite known from the foundational work by von Neumann and Birkhoff. In their formulation of QM they constructed the theory starting from the lattice of elementary properties (see my answer on quantum probabilities) for more details or "propositions" testable on a quantum system along the experimental common phenomenology of all quantum ...


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Ok, to expand my comment into an answer: There are only two finite-dimensional division rings (that admit division) containing the real numbers as a finite subring: the complex numbers and the quaternions (application of Frobenius theorem). Also, a vector space (and Hilbert spaces are vector spaces) is usually defined over a field, that is a non-zero ...


7

A fundamental postulate of QFT establishes that the theory admits a strongly continuous representation of (orthochronous proper) Poincaré group $\cal P$. A certain one-parameter subgroup of $\cal P$ describes time evolution (with respect to an inertial reference frame) which, as a consequence, turns out to be unitary since it is part of a larger unitary ...


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"Derivation" of Baryon Number Conservation - Consider the QCD Lagrangian (density) $$\mathcal{L} = \bar{\psi}(i\gamma^\mu D_\mu - m)\psi - \frac{1}{4}G^a_{\mu\nu}G_a^{\mu\nu}$$ where the symbols have their usual meaning. This is invariant under $U(1)$, which is nothing but a multiplication of $\psi$ by a global phase factor $e^{i \theta}$. This is ...


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Here is a proof following Ojima, "Lorentz Invariance vs. Temperature in QFT", Letters in Mathematical Physics (1986) Vol. 11, Issue 1 (1986) 73-80. The first two pages of the paper are available for free here, but the website wants money for more of the paper. (Click the orange "Look Inside" button if the paper doesn't open automatically.) Fortunately, the ...


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Comments to the question (v3): In contrast to QED with fermionic matter, in QED with bosonic matter, the full Noether current ${\cal J}^{\mu}$ (for global gauge transformations) tends to depend explicitly on the gauge potential $A^{\mu}$, see e.g. Refs. 1-2 and this Phys.SE post. The reason for this difference is because the QED Lagrangian for fermionic ...


2

1) Why don't we consider finite dimensional representations of this group? As you said, we ask (anti)unitarity, so it is impossible to find finite-dimensional representation. 2) Why associate the Lorentz group to fields? The essence of the answer is what Trimok already said in his comment: the "translational part" of the Poincarè group is already ...


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The assumption that $\mathcal{D}$ is invariant under $\phi(f)$ for each $f\in \mathcal{S}(\mathbb{R^4})$, the Schwartz space of functions of rapid decrease is one of the Wightman axioms. Its main use is for the vacuum expextation values $(\psi_0,\phi(f_1)...\phi(f_n)\psi_0)$ to make sense (where $\psi_0$ is the vacuum state), what would not happen in general ...


3

I think some of your confusion stems from the fact that there are two different kinds of vacuua in QFT. First there is the vacuum of the free theory, usually denoted $|0\rangle $, second there is the full (interacting) vacuum, usually denoted $|\Omega \rangle$. What we want to calculate are the different quantities in the full theory like: \begin{equation} ...


3

Lets start with the second question: Also, should I worry about taking $T\to\infty(1-i\epsilon)$ instead of $T\to \infty$ (which would the natural thing to do in (1))? And there lets start with asking ourselves "why" do we do the limit in the first place? The answer is -- we don't want the expectation for the $|\phi_a\rangle$ and $|\phi_b\rangle$ ...


2

Because muons don't carry a color charge, and hence don't participate in the strong interaction...


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For a scalar field $\phi$, the most widely used convention, based on my experience, is to write the Lagrangian with kinetic and potential terms, followed by interactions like so, $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2 - \frac{1}{2}m^2 \phi^2 - \sum_{n \geq 3} \frac{\lambda^n}{n!}\phi^n$$ where $\lambda_n$ are coupling constants. (We could not have a ...


1

We should distinguish between the different types of anomalies that can arise in quantum field theory. An anomaly is a symmetry of the classical action that is not preserved in the quantum theory. A gauge anomaly causes a gauge symmetry to be broken leading to a violation of a Ward identity which is needed to ensure that unphysical polarization states and ...


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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 ...


2

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. ...


3

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.


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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 ...


3

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 ...


5

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 $P$ ...


3

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 ...


4

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)\\ ...


4

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 ...



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