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Results tagged with quantum-field-theory
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user 133418
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.
2
votes
Accepted
Free Feynman propagator calculation
Combine the two fractions in the first equation to get:
\begin{equation}
\frac{2E_p - 2i\epsilon}{(p^0)^2 - (E_p - i \epsilon)^2}
\end{equation}
Yes, you then ignore the $\epsilon^2$ term in the deno …
- 3,912
0
votes
Why field redefinitions that leave Lagrangians unchanged are allowed?
The quantum field $\phi$ is not a measurable quantity. It's a crutch you use to calculate physical observables like cross sections. So there's nothing that prevents you from redefining your field, as …
- 3,912
3
votes
Accepted
Bhabha scattering Energy conservation
(2nd diagram) Since the photon is virtual, its exchange is the process that happens during interaction, neither before nor after.
Now, energy and 3-momentum is conserved at each vertex. Also keep in …
- 3,912
2
votes
What would a Lagrangian without a kinetic term represent?
There is no dynamical content without kinetic terms, and in fact it makes the theory sick.
Suppose you have
$$\mathcal L = \frac{X}{2}(\partial \phi)^2 -\frac{m^2}{2} \phi^2 -\frac{\lambda}{4!} \phi …
- 3,912
1
vote
Accepted
Electron and positron pair production and annihilation
A lone photon will never decay into an electron-positron pair. This process is kinematically not allowed because of 3-momentum conservation. You need a nucleus nearby for the process to materialize.
…
- 3,912
2
votes
Is the physics of a Quantum Field Theory dependent on the choice of Lagrangian?
As a corollary, two Lagrangians that look vastly different from each other, producing vastly different classical equations of motion, can actually describe the same physics, if the fields in the two L …
- 3,912
3
votes
What does it mean for the Hamiltonian to not be bounded from below?
Hamiltonians that are unbounded from below lead to instabilities of the physical system in question. Mathematically, it just simply means that the Hamiltonian (or crudely, energy) can take arbitrarily …
- 3,912
1
vote
Proof of overall momentum conservation
The number of 4-momentum conserving delta functions is one more than the number of (independent) internal momenta. After performing integrations over the internal momenta, you will be left with one re …
- 3,912
1
vote
Lagrangian with quadratic interaction
There is no absolute rule to be followed in order to determine if something is an "interaction" or not. You can interpret it in different ways, do field redefinitions, etc. At the end of the day, if y …
- 3,912
1
vote
How should I think about the word "coupling" in quantum field theory?
Is the fact that something is coupled to something else mathematically well-defined?
Sort of, because when you say that X and Y are coupled to each other, it means that X and Y interact with each ot …
- 3,912
1
vote
Accepted
QFT in A Nutshell Zee I.3.1
The 2003 edition of the book has a typo. In rightmost expression, the correct limits are $\int_{-\infty}^{\infty}$, not $\int_{0}^{\infty}$.
The second part of
\begin{equation}
\int_0^\infty \frac{dk …
- 3,912
1
vote
How do you derive a quantum field theory from a spacetime metric?
The standard (QFT) way of quantizing gravity is by applying what is known as the background field method. Here, you write the metric tensor $g_{\mu \nu}$ as a sum of some classical background spacetim …
- 3,912
0
votes
Why energy-positivity?
You are right: in a theory which describes a system not interacting with others, it does not matter if the energy is unbounded from below. However, that is a boring universe with no dynamics or intera …
- 3,912
10
votes
How does the absence of quadratic terms in the Lagrangian imply massless quanta?
I will expand a bit on one of the answers to be clearer why a quadratic term usually leads to the 'mass' term. Because the coefficient in front of the $A_\mu A^\mu$ term has dimension $2$ doesn't by i …
- 3,912
15
votes
Accepted
How can the graviton be both massless and self-interacting?
Gravitons couple to energy-momentum tensor $T_{\mu \nu}$, not mass (which captures only the $00$ component of $T_{\mu \nu}$). That's why photons (which are also massless, like gravitons) interact with …
- 3,912