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Results tagged with quantum-field-theory
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user 84967
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.
0
votes
Accepted
Nucleon scattering and the meson 4-momentum in scalar Yukawa
In the COM you have $k^2=-(\vec p-\vec p')^2<0$. But $k^2$ is a scalar, and thus independent of the frame of reference. Therefore, we conclude that $k^2<0$ in any frame of reference. In particular, it …
3
votes
Accepted
Why 1PI graphs are enough to discuss renormalizability?
Because any correlation function $G_i$ can be decomposed into (simpler) irreducible functions $\Gamma_i$. For example, the two point function is
$$
G_2(p)=\frac{1}{\Gamma_2(p)}
$$
, the three point fu …
4
votes
Accepted
Interacting Hamiltonian commutes with momentum operator?
According to Heisenberg's equations of motion,
$$
[H,P]\propto \partial_t P
$$
which equals zero because $P$ is time-independent. This assumes nothing about the nature of $P,H$, (except for the fact t …
2
votes
Accepted
Is the sign of an amplitude in QFT meaningful?
The phase of a quantum-mechanical probability amplitude is completely arbitrary – this is in fact one of the fundamental postulates of quantum mechanics.
A quantum-field-theoretic scattering amplitude …
2
votes
Real scalar field and its quantum state: why are the diagonal components static here?
It is not true that $\dot\rho(\phi_1,\phi_1)=0$: the object $\rho(\phi_1,\phi_1)$ does not exist.
It doesn't make sense to talk about $\rho$ for $\phi_1=\phi_2$. The object $\rho$ is a distribution (a …
4
votes
1
answer
141
views
Modifying a renormalisable theory$.$
Apparently, if we take a certain renormalisable theory, then any modification consistent with the symmetries must render the theory non-renormalisable. Is this claim true? Has it been discussed rigoro …
4
votes
Green's function in Hamiltonian vs. Path Integral QFT
In general, $|\psi\rangle$ is the ground state of your theory, i.e., the state with the lowest energy.
In free theories, $|\psi\rangle$ is written $|0\rangle$, and is defined through $H_0|0\rangle=0$ …
4
votes
Accepted
Is Lorentz invariant differential measure arbitrary?
Yes: any scalar $A$ makes $\delta(k^2+A)$ covariant. A different choice for $A$ changes the form of $\widetilde{\mathrm d k}$, which in turns changes the form of the creation/annihilation operators $a …
1
vote
Accepted
Energy-momentum relation for dressed particles and how interactions change mass
But from the mathematics, I can't see why, in general, $m_\lambda$ cannot be equal to $m$.
The physical mass $m_\lambda$ may be equal to $m$, the bare mass. This is precisely what happens in some …
3
votes
Accepted
What ideas in Special Relativity are preserved in Quantum Field Theory, and what ideas aren't?
Q1: the bullets are correct. To add some details,
Lorentz covariance, or more properly, Poincaré covariance, is one of the basic pillars of the theory, and one of the main motivations to use fields …
2
votes
Accepted
LSZ and current conservation
Contact terms vanish in the on-shell limit $k^2\to m^2$. See M. Srednicki's book, chapters 67, 68 for a more or less detailed discussion.
In a nutshell, a Dirac delta $\delta(x_1-x_2)$, in Fourier sp …
6
votes
Virtual particles creation operator
Virtual particles do not exist for a limited time, they are not off-shell states, and you cannot create those. They are Wick contractions of fields in the interaction picture, in the Dyson expansion o …
1
vote
Accepted
Energy-momentum relation and quantum field theory
I assume that by "energy-momentum relation" you mean
$$
p^2=m^2
$$
where $ab\equiv a^0 b^0-\boldsymbol a\cdot\boldsymbol b$. Here, $p=(E,\boldsymbol p)$.
Note that this relation is kinematical, not d …
12
votes
Do fields describing different particles always commute?
No.
In full generality, the super-commutator of two fundamental fields is identical to the Dirac bracket of the corresponding classical variables (modulo the standard obstructions). If the system is …
2
votes
What is the physical meaning of the canonical momentum field in quantum field theory?
A bit of personal rant first: the operator $\phi(x)$ does not create a particle at $x$, and insisting that it does is of no help to anyone. It may make it easier to digest the new formalism for someon …