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Results tagged with quantum-field-theory
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user 136463
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.
4
votes
Accepted
Off-shell QFT from on-shell amplitudes
Maybe a non-rigorous answer formulated in the language of perturbative QFT helps.
Examples can be produced in this context by performing non-trivial redefinitions of the fields. The off-shell Green f …
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7
votes
How many quantum fields are there?
The Standard Model of particle physics is the current well stablished theory for elementary particles. In it, fields are grouped together using symmetries (Lorentz and gauge) in what are called multip …
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3
votes
Can charged scalar have non-zero vev?
Charged scalars can't have a non-zero vev, because after spontaneous symmetry breaking the $U(1)_{em}$ subgroup of $SU(2)_L\times U(1)_Y$ survives, which means that the vacuum is invariant under it.
…
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5
votes
1
answer
741
views
Classical version of tree-level QFT correlation functions
Having read this previous question (and its answers) about the relation between tree-level quantum field theory and classical field theory I can see two facts that support the (perhaps too vague) stat …
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2
votes
Accepted
Uncertainty principle and measurement of the mass of Virtual particles
There are both virtual and real $W^{\pm}$, $Z$ bosons. Just as with photons (and for any other case), virtual $W^\pm$ and $Z$ bosons are some internal lines on Feynman diagrams and real $W^\pm$ and $ …
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0
votes
Accepted
Fermion anti-commutation relations
As you correctly state:
\begin{align}
\{\psi_{\alpha}({\bf x},t),{\bar \psi}_{\beta} ({\bf x'},t)\} &=
\{\psi_{\alpha}({\bf x},t),(\psi^{\dagger}({\bf x'},t) \,
\gamma^0)_{\beta}\} =
\{\psi_{\ …
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2
votes
How does Euclidean Quantum Field Theory describe tunneling?
The relation between tunneling and classical paths in imaginary time can already be seen in single particle quantum mechanics. In the path integral formulation, each path $x(t)$ contributes with ampl …
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1
vote
Fock Space, Propagator
It is unclear what is meant by "classical time-varying wave function". For example:
The quantum state of a particle without spin in three dimensions is a function $\mathbb{R}^3\to \mathbb{C}$. Its …
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3
votes
Can the Lagrangian of an effective field theory have higher derivative terms?
Yes, it can. In principle there isn't a reason not to include them. Of course, once we add them the theory will be different, just as it happens when you add any other type of higher dimensional opera …
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2
votes
Are non-renormalizable theories less predictive than renormalizable theories?
The effective lagrangian is usually written as an expansion in inverse powers of the energy cut-off $\Lambda$. That means that observables at some energy $E$ will be computed as the first terms of an …
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0
votes
Accepted
Proof of quantum correlation functions
First question
Using that $S=U_I(+\infty,-\infty)=U_I(+\infty, t_1)U_I(t_1,t_2)\cdots U_I(t_n,-\infty)$, as you state, you have that
\begin{align}
T\phi_{1I}\phi_{2I}\cdots\phi_{nI}S &=
T\phi_{1I}\p …
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5
votes
Why are derivatives in interaction terms treated differently from derivatives in the kinetic...
Informally speaking (I don't know if this can be stated more rigorously) we can treat the kinetic terms as having Feynman rule $\sim p^2$, in some sense.
Consider the massive case: $\mathcal{L}=|\par …
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14
votes
Accepted
What is wrong with a nonrenormalizable theory?
In the modern effective field theory point of view, there's nothing wrong with non-renormalizable theories. In fact, one may prefer a non-renormalizable theory inasmuch they tell you the point at whi …
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2
votes
Accepted
Why and how does the soft breaking of symmetry protect the mass of pseudo Nambu-Goldstone bo...
When the symmetry is broken explicitly by a relevant operator, its contribution is negligible at high energies. Thus, if $f$ is the scale of the breaking operator, at energies $\Lambda\gg f$ the symme …
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