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Everything in nature is some field configuration.Urs Schreiber
“There are no particles, there are only fields”
In its mature form, the idea of quantum field theory is that quantum fields are the basic ingredients of the universe, and particles are just bundles of energy and momentum of the fields. In a relativistic theory the wave function is a functional of these fields, not a function of particle coordinates. Quantum field theory hence led to a more unified view of nature than the old dualistic interpretation in terms of both fields and particles.
What is Quantum Field Theory, and What Did We Think It Is? by S. Weinberg
We have no better way of describing elementary particles than quantum field theory. A quantum field in general is an assembly of an infinite number of interacting harmonic oscillators. Excitations of such oscillators are associated with particles. The special importance of the harmonic oscillator follows from the fact that its excitation spectrum is additive, i.e. if $E_1$ and $E_2$ are energy levels above the ground state then $E_1 + E_2$ will be an energy level as well. It is precisely this property that we expect to be true for a system of elementary particles.
A. M. Polyakov, “Gauge Fields and Strings”, 1987
Undoubtedly the single most profound fact about Nature that quantum field theory uniquely explains is the existence of different, yet indistinguishable, copies of elementary particles. Two electrons anywhere in the Universe, whatever their origin or history, are observed to have exactly the same properties. We understand this as a consequence of the fact that both are excitations of the same underlying ur-stuff, the electron field.
Quantum Field Theory by Frank Wilczek
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Before reading anything else, you should read the "Bird’s Eye View" of quantum field theory by Robert Klauber.
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Life Cycle of a Theoretical Physicist
Write down a Lagrangian density $L$. This is a polynomial in fields $\psi$ and their derivatives. For example $$L[\psi] = \partial_\mu\psi\partial^\mu\psi - m^2\psi^2 + \lambda\psi^4$$ Write down the Feynman path integral. Roughly speaking this is $$\int e^{i\int L[\psi]} D\psi$$ The value of this integral can be used to compute “cross sections” for various processes. Calculate the Feynman path integral by expanding as a formal power series in the “coupling constant” $\lambda$. $$a_0 + a_1\lambda + a_2\lambda + \cdots$$ The $a_i$ are finite sums over Feynman diagrams. Feynman diagrams are a graphical shorthand for finite dimensional integrals. Work out the integrals and add everything up. Realise that the finite dimensional integrals do not converge. Regularise the integrals by introducing a ``cutoff'' $\epsilon$ (there is usually an infinite dimensional space of possible regularisations). For example $$\int_\mathcal{R} {1\over x^2} dx \longrightarrow \int_{|x|>\epsilon} {1\over x^2} dx$$ Now we have the series $$a_0(\epsilon) + a_1(\epsilon)\lambda + \cdots$$ {\bf Amazing Idea:} Make $\lambda$, $m$ and other parameters of the Lagrangian depend on $\epsilon$ in such a way that terms of the series are independent of $\epsilon$. Realise that the new sum still diverges even though we have made all the individual $a_i$'s finite. No good way of fixing this is known. It appears that the resulting series is in some sense an asymptotic expansion. Ignore step 8, take only the first few terms and compare with experiment. Depending on the results to step 9: Collect a Nobel prize or return to step 1.There are many problems that arise in the above steps
[Problem 1] The Feynman integral is an integral over an infinite dimensional space and there is {\bf no} analogue of Lebesgue measure. [Solution] Take what the physicists do to evaluate the integral as its definition. [Problem 2] There are many possible cutoffs. This means the value of the integral depends not only on the Lagrangian but also on the choice of cutoff. [Solution] There is a group $G$ called the group of finite renormalisations which acts on both Lagrangians and cutoffs.QFT is unchanged by the action of $G$ and $G$ acts transitively on the space of cutoffs. So, we only have to worry about the space of Lagrangians.
[Problem 3] The resulting formal power series (even after renormalisation) does not converge. [Solution] Work in a formal power series ring.Lectures on Quantum Field Theory by R. E. Borcherds, A. Barnard
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In the quantum case, axiomatic formulations of field theory assert that fields $\varphi$ are operator-valued distributions [8]. Distributions are continuous linear functionals which map a space of test functions $\mathcal{T}$ onto the complex numbers: $\varphi: \mathcal{T} \rightarrow \mathbb{C}$. In quantum field theory (QFT), $\mathcal{T}$ is chosen to be some set of space-time functions; usually either the space of continuous functions with compact support $\mathcal{D}(\mathbb{R}^{1,3})$, or the space of Schwartz functions $\mathcal{S}(\mathbb{R}^{1,3})$. (The distributions with which these test functions are smeared are called tempered distributions.) In either of these cases one can represent the image of the map $\varphi$ on a space-time function $f$ as: \begin{align} \varphi(f) = (\varphi,f) : = \int d^{4}x \ \varphi(x)f(x) \label{int_rep} \end{align} which gives meaning to the $x$-dependent field expression $\varphi(x)$. Since $\varphi$ is an operator-valued distribution in QFT, only the smeared expression $\varphi(f)$ is guaranteed to correspond to a well-defined operator. The derivative of a distribution $\varphi'$ is defined by: \begin{align} (\varphi',f) := -(\varphi,f') \end{align} and is itself also a distribution [8]. By applying the integral representation in equation~\ref{int_rep}, one can interpret this definition as an integration by parts where the boundary terms have been `dropped': \begin{align} \int d^{4}x \ \varphi'(x)f(x) = -\int d^{4}x \ \varphi(x)f'(x) \end{align} Although this shorthand notation is useful, and will be used for the calculations in this paper, it can also be slightly misleading. Sometimes it is incorrectly stated that integration by parts of quantum fields can be performed, and the boundary terms neglected. However, distributions are generally not point-wise defined, so boundary expressions like: $\int_{\partial \mathbb{R}^{3} } \varphi(x)f(x)$ are often ill-defined. Therefore, when manipulations like this are performed one is really just applying the definition of the derivative of a distribution, there are no boundary contributions. This makes the question of whether spatial boundary term operators vanish a more subtle issue in QFT than in the classical case.
The physical rationale behind using operator-valued distributions as opposed to operator-valued functions in QFT is because operators inherently imply a measurement, and this is not well-defined at a single (space-time) point since this would require an infinite amount of energy [9]. Instead, one can perform a measurement over a space-time region $\mathcal{U}$, and model the corresponding operator $\mathcal{A}(f)$ as a distribution $\mathcal{A}$ smeared with some test function $f$ which has support in $\mathcal{U}$. If one were to smear $\mathcal{A}$ with another test function $g$, which has different support to $f$, then in general the operators $\mathcal{A}(f)$ and $\mathcal{A}(g)$ would be different. But the interpretation is that these operators measure the same quantity, just within the different space-time regions: $\text{supp}(f)$ and $\text{supp}(g)$.
As well as differentiation it is also possible to extend the notion of multiplication by a function to distributions. Given a distribution $\varphi$, a test function $f$, and some function $g$, this is defined as: \begin{align} (g\varphi,f) := (\varphi,gf) \label{dist_prop} \end{align} In order that $g\varphi$ defines a distribution in the case where $f\in \mathcal{D}$, it suffices that $g$ be an infinitely differentiable function. For tempered distributions, in which $f\in \mathcal{S}$, it is also necessary that $g$ and all of its derivatives are bounded by polynomials [8].
Besides the assumption that fields are operator-valued distributions, axiomatic approaches to QFT usually postulate several additional conditions that the theory must satisfy. Although different axiomatic schemes have been proposed, these schemes generally contain a common core set of axioms\footnote{See[8] ,[9] and [10] for a more in-depth discussion of these axioms and their physical motivation.}. For the purpose of the calculations in this paper, the core axioms which play a direct role are: ….
Generally the product of distributions is not well-defined, and so one must first introduce a regularisation procedure in order to make sense of such products [9].
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a property which is well-established in Quantum Electrodynamics (QED) [14], as well as other gauge theories [15] – charged states are non-local. This means that it is not possible to create a charged state by applying a local operator to the vacuum. However, by virtue of the Reeh–Schlieder Theorem, a charged state can always be approximated by local states as closely as one likes in the sense of convergence in some allowed topology on H. Often this topology is chosen to be the weak topology and so convergence means weak convergence.
Boundary terms in quantum field theory and the spin structure of QCD by Peter Lowdon
The comment regarding the impossibility to create a charged state using local operators is true because gauge interactions have infinite reach. E.g. the photon is massless and thus an electrically charged particle, like an electron, is always surrounded by a cloud of photons, even infinitely far away. In this context a charged particle is called an infraparticle.
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For many more questions and answers see: http://www.mat.univie.ac.at/~neum/physfaq/physics-faq.html
The physical rationale behind using operator-valued distributions as opposed to operator valued functions in QFT is because operators inherently imply a measurement, and this is not well-defined at a single (space–time) point since this would require an infinite amount of energy [9]. Instead, one can perform a measurement over a space–time region U, and model the corresponding operator A(f) as a distribution A smeared with some test function f which has support in U. If one were to smear A with another test function g, which has different support to f , then in general the operators A(f ) and A(g) would be different. But the interpretation is that these operators measure the same quantity, just within the different space–time regions: supp(f ) and supp(g).
Boundary terms in quantum field theory and the spin structure of QCD by Peter Lowdon
Strictly speaking, quantum field theory (at least in most of the fully relevant non-trivial instances of this theory that we know is mathematically inconsistent, and various 'tricks' are needed to provide meaningful calculational operations. It is a very delicate matter to know whether these tricks are merely stop-gap procedures that enable us to edge forward with an mathematical framework that may perhaps be fundamentally flawed at a deep level, or whether these tricks reflect profound truth that actually have a genuine significance to Nature herself. Most of the recent attempts to move forward in fundamental physics indeed take many of these 'tricks' to be fundamental.
page 610 in Road to Reality by R. Penrose
Common signs of mathematical problems in QFT are that the perturbation series does not converge and that we need renormalization procedures to get rid of infinites. However, these problems are actually not as severe as they may seem.
The non-convergence of the perturbation series can be traced back to the fact that the perturbation theory does not include contributions from non-perturbative effects, like instantons.
The appearance of infrared divergences can be traced back to our simplified assumption that spacetime is infinite.
Lastly and most importantly, one can show that the appearance of ultraviolet divergences has its origin in the fact that we physicists handle Distributions improperly.
For more on this, have a look at: How I Learned to Stop Worrying and Love QFT by Robert C. Helling.
This property of QFTs is emphasized by Weinberg in his QFT books.
Weinberg's view of QFT is nicely summarized in chapter 2 of http://sites.krieger.jhu.edu/jared-kaplan/files/2016/05/QFTNotes.pdf
Basically, all the textbooks on quantum field theories out there use an old framework that is simply too narrow, in that it assumes the existence of a Lagrangian.
This is a serious issue, because when you try to come up e.g. with a theory beyond the Standard Model, people habitually start by writing a Lagrangian … but that might be putting too strong an assumption.
We need to do somethingWhat is Quantum Field Theory by Yuji Tachikawa
See also https://youtu.be/XM4rsPnlZyg?t=9m32s around 9:30
It is good to keep in mind, as much as physicists love Lagrangians, there are many many very interesting theories that you can almost prove that there is no Lagrangian description for them. That's kind of a warning sign that if you really want to understand quantum field theory, a Lagrangian cannot be the starting point. Duality and emergent gauge symmetry - Nathan Seiberg
The Lagrangian approach is good for approximations and to get a geometric point of view. Whereas a description without a Lagrangian, i.e. in terms of operators and correlation functions is useful for exact solutions and yields an algebraic point of view.
There is a lot of evidence that probably neither of them will end up to be the fundamental one. Duality and emergent gauge symmetry - Nathan Seiberg
There are many theories, which have more than one Lagrangian. So that's the opposite. Either we have no Lagrangian at all or we have more than one Lagrangian.
Duality and emergent gauge symmetry - Nathan Seiberg
See: The Evolution of Quantum Field Theory by Gerard ’t Hooft