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Results tagged with quantum-field-theory
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user 266156
Quantum Field Theory (QFT) is the theoretical framework describing the quantisation of classical fields which allows a Lorentz-invariant formulation of quantum mechanics. QFT is used both in high energy physics as well as condensed matter physics and closely related to statistical field theory. Use this tag for many-body quantum-mechanical problems and the theory of particle physics. Don’t combine with the [quantum-mechanics] tag.
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Peskin and Schroeder spinor high-energy limit (5.26 and A.20)
Yet another way to go about it... In the relativistic limit,
$$\sqrt{p\cdot \sigma}\equiv\sqrt{E+\mathbf{p}\cdot\boldsymbol\sigma}\approx\sqrt{E+E\hat{p}\cdot\boldsymbol{\sigma}}=\sqrt{2E}\sqrt{\frac{ …
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3
votes
Accepted
Spinor indices in $\not p = m_P$ (mass renormalization)
I'm not familiar with Schwartz's book, but I know that in Peskin and Schroesder, they first show that the fourier-space two-point correlation function is given by
$$\int d^4x\langle\Omega|T\psi(x)\bar …
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2
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Looking for a self-contained book on QFT
If you're looking for a philosophy of science approach, Weinberg's The Quantum Theory of Fields is a classic textbook that examines why QTF is the way it is. It builds up the theory from fundamental i …
1
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Which configurations are important in lattice QCD?
The issue with perturbation theory in QCD is that it only works when the interaction strength is small compared to the energy scale you're considering. In terms of field configurations in a monte carl …
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7
votes
Accepted
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 explicit …
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4
votes
$S$-matrix from LSZ
You're not quite using the LSZ reduction formula properly. As part of LSZ, you're supposed to take the residue of the poles as all external momenta go on-shell. If you start by including propagators f …
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3
votes
1
answer
74
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Can we eliminate gauge degrees of freedom in QFT by quantizing the field strength directly?
In Matthew Schwartz's Quantum Field Theory and the Standard Model, he says (section 8.6, page 132) that it is possible to avoid introducing the redundant gauge degrees of freedom in QED by quantizing …
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4
votes
Three integrals in Peskin's Textbook
Just a note on the third integral.
$$\frac{1}{4\pi^2}\int_m^\infty dE\sqrt{E^2-m^2}e^{-iEt}.$$
If you don't want to explicitly do the calculation, as in Alex's answer, there is a plausibility argument …
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5
votes
An integral in Peskin & Schroeder's Quantum Field Theory p. 27
I'll just add a little more information about pushing the contour upward, since this was very confusing to me at first. We may push the contour of integration up, as long as we keep the endpoints fixe …
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$\phi^4$-theory, S-matrix Feynman diagram to first order from Peskin and Schroeder
Suppose we have a general term of the form
$$\langle 0 | a_{p_1} a_{p_2} \cdots a_{p_{n-1}}a_{p_n} a_{p_1}^\dagger a_{p_2}^\dagger \cdots a^\dagger_{p_{m-1}}a_{p_m}^\dagger | 0 \rangle$$
We can elimin …
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4
votes
Accepted
Infinitesimal generator of change of basis (Fock Space)
The step where you went wrong is in writing
$$\hat{U}(\theta(\vec{r}))=1-i\theta(\vec{r})G$$
for infinitesimal transformations. This would imply $U(\theta(\vec{r}))$ is an operator-valued function of …
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1
vote
1
answer
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What's wrong with this "proof" that QFT violates causality?
In An Introduction to Quantum Field Theory, by Peskin and Schroeder, when discussing the quantized real Klein-Gordon field ($\phi=\phi^\dagger$), they show the commutator $[\phi(x),\phi(y)]$ vanishes …
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Spin-1 polarization vectors (massive particle)
A vector representing the spin state of a spin-1/2 particle is two-dimensional. Spin is associated with the group $SU(2)$. The Pauli sigma matrices form a basis for a two-dimensional representation of …
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1
vote
Chiral symmetry breaking and appearance of hadrons (from Schwartz's QFT book)
At lower temperatures, quarks are always bound with other quarks because of color confinement. Only color-neutral particles can exist by themselves. A free particle with a color charge has too much en …
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2
votes
Accepted
Why commutator of positive and negative parts of scalar field is equal to the Feynman propag...
Just evaluate the $p^0$ part of the integral. For $x^0>y^0$, we can close the contour below and enclose the pole at $p^0=E_\mathbf{p}$ (there's a pole at this point because $p$ is on-shell, so $p^2-m^ …
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