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Good question. Some preliminary remarks.
The map "one particle" $\leftrightarrow$ "one field" holds, at best, in the weakly coupled regime, where fields are (by construction, cf. ref.1) interpolating fields for one-particle states. In a strongly-coupled theory, a single field may (and usually does) create many different particles, and ...
9
Even if there were some valid relaxed sense in which every field in every QFT has an associated particle, the important point is that fields are inputs (used to define the theory mathematically) and particles are outputs (phenomena that we derive from the theory). Particles are transient and not always sharply defined. As examples that challenge the ...
6
As usual in interacting field theory, there is no rigorous answer to your question directly in three dimensions. However, there is strong numerical evidence from the conformal bootstrap approach for the uniqueness of the Gaussian + Wilson-Fisher fixed points as the only two fixed points with conformal symmetry in your theory. In the paper https://arxiv.org/...
5
As an experimental particle physicist my knowledge of fied theory is on a working level, i,e, how it is used in order to calculate interaction crossections and decays for particle physics.
The course I took in field theory was back in 1964 and the professor used the book of Bogolyubov, and after struggling with creation and annihilation operators for some ...
5
It's worth remembering when reading Haag's book that it was published in 1965, well before the reality of quarks was accepted. There was no Standard Model at the time; instead they had piecemeal understanding of gauge fields, muddled together with a lot of S-matrix thinking. Bjorken's light cone scaling arguments didn't come along until 1968 and the ...
3
No. (Anti)commutators do not necessarily vanish at different times. In order to compute the (anti)commutator at different times you have to solve the dynamics of the system with $\psi(x,t)\to e^{iHt}\psi(x,0) e^{-iHt}$. What is true in a relativistic field theory, with Bose (Fermi) fields satisfying the spin-statistics relation however, is that the (...
3
There are definitely not-on-site symmetries which can be gauged, but it's not obvious how to do it on the lattice.
For instance, a free fermion in 1+1d has a U(1) x U(1) symmetry (vector and axial, if you will), of which only one of the two U(1)s is on-site in any formulation. Either one can be gauged however (at least in the field theory), but not both ...
2
As far as I can see, your formula is correct. Let's transform this formula slightly.
From $[\alpha_k,\alpha_{-k}] = 0$, it follows
$$
\varphi_k \equiv \frac{v_k}{u_k} = \frac{v_{-k}}{u_{-k}}.
$$
Last equality together with $|u_k|^2 - |v_k|^2 = 1$ and $|u_{-k}|^2-|v_{-k}|^2 = 1$ leads to equalities
$$
u_k = \frac{e^{i\gamma_k}}{\sqrt{1-|\varphi_k|^2}},
\ v_k =...
2
The explicit calculation of the full effective potential in terms of Feynman diagrams is first laid out in “Functional evaluation of the effective potential,” R. Jackiw, Phys. Rev. D 9, 1686 (1974). The results are nontrivial, in several different ways. For one thing, the structure of the one-loop contribution to the effective action is fundamentally ...
2
Well, the proof in Ref. 1 does strictly speaking not compute the quantum effective action $\Gamma[\Phi_{\rm cl}]$ directly, but rather the generating functional $W_c[J]$ of connected diagrams in 2 different ways:
As trees constructed from full propagators, 1PI vertices, and sources $J$, via a combinatoric argument.
As trees constructed from $\Gamma$-...
2
I think the Nielsen-Ninomiya theorem is more closely related to a gravitational anomaly than a $U(1)$ anomaly. For instance in 1+1d, two left-moving fermions of charge +3 and +4 and a single right-moving fermion of charge +5 has a vanishing $U(1)$ anomaly, since $3^2+4^2 =5^2$. However, it still cannot be put on a lattice, because it has a chiral central ...
1
I explained all this in I hope sufficient detail at
What is the Wilsonian definition of renormalizability?
and I urge the OP to read it. However,
let me give some quick remarks here.
There is not much to say about the first picture which just shows how the RG flow looks like near the UV fixed point that one is trying to perturb in order to construct a new ...
1
There is a collection of concepts here that do not necessarily have anything to do with each other. Quasi-particles in condense matter physics and particles in quantum field theory are different things. The different quasi-particles are caused by the diverse dynamics that one finds in condense matter scenarios. They usually have a finite size and are seldom ...
1
$U$ is an operator. It acts on the Hilbert space of your system. You can think of it as a generalization of a matrix: it is a linear transformation on a vector space (in this case, your Hilbert space). It is not just a matrix because your Hilbert space (the space of all your quantum states) may be infinite dimensional! Operators behave very similar to ...
1
The $M^{cd}$ are abstract operators that map abstract vectors in a Hilbert space to other abstract vectors.
If the abstract vectors in Hilbert space are represented as concrete functions on spacetime, then the abstract $M^{cd}$ are represented as concrete differential operators involving $\partial/\partial t$, $\partial/\partial x$, etc., and the identity ...
1
A field operator annihilates a particle, or creates an antiparticle. The adjoint does the opposite, it creates a particle or annihilates an antiparticle.
1
Step 1: Make things well-defined
Things work out better when we start with something well-defined. To make the problem well-defined, I'll treat the 2d space as a lattice with a finite but arbitrarily large number of sites. (Strong-coupling expansions are typically done using lattice QFT.) Then the integration variables $\phi(x)$ are ordinary real variables, ...
1
To quantize a classical field you promote the field and its conjugate momentum to operators and impose the canonical commutation relations, e.g. for a Klein-Gordon field $\phi (x)$
$[\phi (x), \pi (y)] = i \delta (x-y)$
where $\pi (x) = \frac{\partial \mathcal L}{\partial \dot \phi (x)} = \dot \phi (x)$ is the conjugate momentum.
As for a Dirac field $\psi$, ...
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