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Scattering, Perturbation and asymptotic states in LSZ reduction formula

The first question we have to ask is: what is a one particle state in an interacting theory? It is reasonable to require that they are states that are both momentum eigenstates and energy eigenstates. ...
user1379857's user avatar
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32 votes
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Quantum Field Theory in position space instead of momentum space?

The most important reasons we use momentum space Feynman rules are: In position space, the Feynman rules generate convolutions of propagators. Because of the convolution theorem, the momentum space ...
AccidentalFourierTransform's user avatar
14 votes
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What actually means to compute things at tree level?

Suppose you want to compute a correlation say in Euclidean signature $$ \frac{1}{Z}\int D\phi\ \prod_i \phi(x_i)\ \exp\left(-\frac{1}{\hbar}S(\phi)\right) $$ with $$ S(\phi)=\frac{1}{2}(\phi,A\phi)+g\...
Abdelmalek Abdesselam's user avatar
14 votes
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Is an interacting QFT Hilbert space a physical particles Fock space?

Haag's theorem says that the Hilbert space on which interacting relativistic quantum fields can be defined cannot be a standard Fock space. The finite time dynamics happens in this space, hence not in ...
Arnold Neumaier's user avatar
12 votes

Reflection positivity for general fields

This is one of the first hits for "reflection positivity" on google, so someone better answer it! Reflection positivity is just positivity of the Hilbert space norm along with the fact that complex ...
Ryan Thorngren's user avatar
11 votes
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What does it mean for QFT to be unitary?

Maybe this is a boring answer but unitarity of a QFT means states at different times can be related to each other by \begin{equation} \left | \Psi(t_2) \right > = U \left | \Psi(t_1) \right > \...
Connor Behan's user avatar
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11 votes

What physical processes other than scattering are accounted for by QFT? How do they fit into the general formalism?

We have a model (the standard model) that is formulated as a particular QFT, that seems to describe every experiment we've ever done on Earth (excluding anything gravitational). My point being that ...
QCD_IS_GOOD's user avatar
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10 votes

Allowed Field Re-definitions in QFT

After the field redefinition $$ \phi \rightarrow \phi + \frac{\partial_{\nu}\partial^{\nu}}{v^2}\phi\tag{1} $$ there is an apparent violation of unitarity because the propagator decays too fast in ...
AccidentalFourierTransform's user avatar
10 votes

Scattering, Perturbation and asymptotic states in LSZ reduction formula

Doubt 1: You cannot simply put $t=\pm\infty$ since all formulas become meaningless unless the limit is carefully done. The proof of the LSZ theorem by Haag and Ruelle shows that one needs the ...
Arnold Neumaier's user avatar
10 votes
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The use of $a^\dagger(\mathbf{k}) = -i \int d^3x e^{ikx}\stackrel{\leftrightarrow}{\partial}_0 \phi(x)$ in the derivation of the LSZ-formula

OP asks good questions. Let us try to sketch the logic of the LSZ reduction formula. In the Heisenberg picture, a free real field $\hat{\varphi}(x)$ has a Fourier expansion $$ \hat{\varphi}(x)~=~\...
Qmechanic's user avatar
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8 votes
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Why this loop carries an integral if there is no undetermined momenta?

The momentum conservation at the vertex is NOT $p_1-p_2-k=0$ but rather $p_1-p_2-k+k=0$ which leaves $k$ completely free. The vertex has two half-edges carrying the momentum $k$ but in opposite ...
Abdelmalek Abdesselam's user avatar
8 votes
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LSZ reduction formula (Schwartz)

It's implicit because the operators $a_p$ are being treated as time-dependent. If instead we were in a picture where the operators did not have explicit time dependence, then we would need to ...
Technically Natural's user avatar
8 votes
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The reasoning of the definition of $S$-matrix

As lurscher already said correctly, the idea is to choose an initial and a final time that is (in time) away from the actual time-dependent interaction in order to make sure that the interaction does ...
Frederic Thomas's user avatar
7 votes

Relation between scattering matrix and an effective Hamiltonian

It's been some time since this question has been asked, but allow me to post a new answer. Its piece of my thesis, where I've put the derivation (since its so hard to find). I hope it'll help anyone, ...
Michał Jan's user avatar
7 votes

What actually means to compute things at tree level?

In principle in QFT you want to calculate the interaction picture unitary time evolution operator. Then you can use the operator to evolve quantum states just like you would in normal quantum ...
gautampk's user avatar
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7 votes

How does a perturbation theory make sense in quantum field theory?

A sketch of the philosophy: Step one: Introduce a regulator. Everything is finite. Step two: Set up the perturbative expansion. This is a purely algebraic step, where there is no notion of "this is ...
AccidentalFourierTransform's user avatar
7 votes

Status of particles in interacting QFT

If you have a free particle theory then the number of particles is well behaved because they are just the Fock states. The trouble is that when you turn on interactions between the particles then the ...
John Rennie's user avatar
7 votes

In Feynman functional integrals why do we integrate the action over all time?

Short answer (assuming proper time-ordering of $x$ and $y$): $$ \langle \Omega| \phi(x) \phi(y) | \Omega\rangle = \frac{ \langle 0 | U(-\infty, x^0) \phi_I(x) U(x^0, y^0) \phi_I(y)...
Darkseid's user avatar
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7 votes
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Optical theorem in QFT

You might be assuming the matrix element $T_{ii}$ to be real. If so, then $$ \lvert S_{ii} \rvert^2 = 1 + \lvert T_{ii} \rvert^2 > 1 $$ Without such an assumption, $$ \begin{align*} \lvert S_{ii} ...
Styg's user avatar
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7 votes
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Why can we not define asymptotic states in CFTs?

The problem with defining asymptotic states in a CFT is the same as in any quantum field theory with massless particles: operators don't decouple fast enough at large distances. Let me explain: In a ...
M.Jo's user avatar
  • 841
7 votes

Spin-$J$ Amplitude $A_J(s,t) = - \frac{g^2(-s)^J}{t-M^2}$?

The intuitive proof is straightforward. The group $SU(2)$ is simple, and so any representation is contained in the tensor product of sufficiently many copies of the fundamental (à la Young). More ...
AccidentalFourierTransform's user avatar
7 votes
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The interpretation of the quantum field

(This answer is written from the perspective of lattice QFT, where everything is mathematically well-defined. Lattice QFT is not fundamental, of course, but most of the QFTs we use in physics are not ...
Chiral Anomaly's user avatar
7 votes

Weinberg, Effective Field Theories

Unitarity says that the sum of the probabilities for all possible out states (for a given in state) must be 1. Since the probability is the square of an amplitude, if any individual amplitude has a ...
Andrew's user avatar
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7 votes
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Peskin & Schroeder LSZ formula missing in- and out states

The question is not really about LSZ, but rather about the relation between in/out states and free states and how the ${\cal S}$-matrix is defined. A good reference for this is Weinberg's The Quantum ...
Gold's user avatar
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6 votes

Why should there be one-particle states in an interacting quantum field theory?

An interacting quantum field theory on Minkowski space should have a representation of the Poincare group as unitary operators, and the irreducible representations of the Poincare group are exactly ...
octonion's user avatar
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6 votes
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Significance of LSZ reduction formula

The premise of the question is that you either use LSZ formula or "directly" compute $\langle f | Te^{-i\int d^4xH_I} |i\rangle$. Implying that these are two unrelated approaches. While the LSZ ...
Kostya's user avatar
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6 votes
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Crossing Symmetry for Particles with Spin

General concepts Crossing symmetry is basically the CPT theorem applied in the context of the LSZ formula, using microcausality to re-order the field operators. The role of the CPT theorem is to ...
6 votes
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Why amplitudes are rational functions?

After the OP explained in the comments what exactly they're looking for, I will attempt an answer. There are a few separate facts that need explanation: Tree amplitudes are rational functions of ...
Anonjohn's user avatar
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6 votes

Scalar Electrodynamics or QED?

In scalar electrodynamics you still have the four-vector potential $A_{\mu}$ but no spinors, that is no Dirac fermions. Spinors are solutions to the Dirac equation (QED) that take their name from the ...
SuperCiocia's user avatar
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6 votes
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Trouble deriving expression for differential scattering cross section from $S$-matrix

I did this calculation some time ago myself. Since I am not sure what is your problem precisely I just give you my old notes. As far as I remember they should be fairly detailed and approximately ...
jkb1603's user avatar
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