# Tag Info

## Hot answers tagged quantum-field-theory

6

Because a pair of particle and anti-particle can be created from the vacuum, it means that infinite number of pairs of particle and anti-particle can be created from the vacuum. So when you consider relativistic quantum theory it's impossible to only consider a finite number of particles. When you calculate Feynman diagram, you are actually only doing ...

6

The short answer: Peskin & Schroeder assume that the vacuum is invariant under translations and Lorentz transformations (which implies $P^\mu|\Omega\rangle=0$). Alternatively, translation invariance implies $\vec P|\Omega\rangle=0$ in all reference frames, and combining this with invariance under boosts gives $P^0|\Omega\rangle=0$. The less short ...

2

Things have intrinsic spin. There's no "class of objects" that has spin and another class of objects that doesn't. Everything is a quantum object with a quantum state, and spin is a number that tells you how the state of the object transforms under rotations. It is different from "classical" angular momentum in that spin is not the operator associated to ...

2

Be aware that the term "virtual particle," although very common, is ambiguous and potentially misleading. What's relevant here are the electromagnetic field modes (ordinary functions, which are coefficients of the quantum creation and annihilation operators), which are one thing people mean by "virtual particles." These modes are indeed affected by the ...

2

I talked to my friend about this, and he said something along the lines of "the pair production particles aren't 'real' particles, and they can't really interact with anything", though I'm not too sure I understand what that means. Your friend is correct. A virtual particle is a mathematical construct, it is a creation of the mathematical model used ...

2

Comments to the question (v2): The idea to consider the planar large $N_c\to \infty$ limit in $SU(N_c)$ QCD goes back to Ref. 1. In light-cone membrane theory, pioneered in Ref. 2, the group $SU(\infty)$ is naturally identified with area-preserving diffeomorphisms ${\rm SDiff}_0(T^2)$ on the torus $T^2$ connected to the identity. Concretely, OP's proposal ...

2

First note that $$\frac{\partial}{\partial m} \frac{i}{\not p - m} = \frac{i}{(\not p-m)^2}$$ Now the exact answer for finite $q$ for that diagram is $$\mathcal{A} = \left( \frac{im}{v} \right) \frac{i}{\not p-m} \frac{i}{\not p + \not q - m} G(q)$$ where $G(q)$ is the propagator for whatever ...

1

In the path integral, $\phi$ is not an operator, but rather a dummy integration variable which runs over all possible classical field configurations. You can pass between the two formalisms using the following relation:  \left< 0 \right| T \left\{ \hat{a} \hat{b} ... \hat{z} \right\} \left| 0 \right> = \frac{\int D\phi e^{i S[\phi] / \hbar} \cdot ...

1

It all depends on what application you are interested in. When discussing chiral perturbation theory it is implied that you are interested in low energy processes. If you are considering processes such as low energy (~100 MeV) pion scattering, pion decay, etc. then the typical energies involved are O(100 MeV) which is much smaller then $\Lambda_{QCD}$. This ...

1

I am not sure this is the answer you are looking for but you may have heard, in QM especially, that an excited system eventually goes back to its ground state. This is true of subatomic particles like neutrons decaying into proton + extra stuff and of atoms too which, once in an excited electronic state, eventually release this extra energy as a photon. The ...

1

Entropy is from statistical physics, a single particle or a single degree of freedom does not have an entropy. Edit: There are a lot of different things that are called entropy. So I'm not sure I feel comfortable with the above as a blanket statement. And there is a way to see time from the dynamics of quantum mechanical systems, so if the above statement ...

1

A (generalized) 't Hooft-Polyakov monopole and a Dirac monopole with a Dirac string attached are two types of magnetic monopoles, which differ in several ways, as OP and user ACuriousMind correctly states. On one hand, (generalized) 't Hooft-Polyakov monopoles are regular, soliton-like, finite-energy solutions to the classical Euler-Lagrange field ...

1

With respect to the Casimir effect, why can't the wavelengths of the virtual particles between two plates just “pass through” the plates themselves? Did you mean vacuum fluctuations? They aren't the same thing as virtual particles. Virtual particles are "field quanta". It's like you divide up the electron's electromagnetic field into abstract chunks and ...

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