1,702 reputation
528
bio website
location
age 29
visits member for 2 years, 5 months
seen 16 hours ago

As time goes on, I grow more disillusioned with quantum field theory.


Dec
15
comment Is the weak interaction Lagrangian invariant under parity transformations?
The question of whether weak interactions are invariant under parity is really asking whether the Lagrangian remains invariant once the all the fields have been written in a parity transformed manner. Mathematically, that's all there is to it.
Dec
15
comment Is the weak interaction Lagrangian invariant under parity transformations?
hmm.. you seem to be under the impression that texts claim that parity changes a left-chiral field into an all together different right-chiral field, when what parity really does is to re-express the same left-chiral field differently. While you are essentially correct about what parity does, no text claims that parity changes the identity of a field.
Dec
15
comment Is the weak interaction Lagrangian invariant under parity transformations?
Read the book (and other resources) carefully: none of the gamma matrices on their own change sign under parity. It is the whole bilinear $\bar\psi F \psi$ that transforms, and how it transforms is to be derived from the transformation law $\Psi\rightarrow\gamma^0\Psi$ alone.
Dec
15
answered Is the weak interaction Lagrangian invariant under parity transformations?
Nov
3
awarded  Revival
Oct
26
awarded  Popular Question
Sep
24
awarded  Autobiographer
Aug
29
awarded  Popular Question
Aug
12
answered Why can the Euler beta function be interpreted as a scattering amplitude?
Aug
9
answered Lagrangian and grassmann numbers
Aug
9
comment Lagrangian and grassmann numbers
Is $\Psi_R$ a two-component spinor like you say in the first paragraph, or is it a four-component one, as implied by how your $C$ is represented in terms of 4-dimensional gamma matrices?
Aug
9
comment A question about the asymptotic series in perturbative expansion in QFT
An asymptotic series happens when you expand a function around a singular point (that's why it has zero radius of convergence). The point of zero coupling is usually a branch point, which is singular. The reason for the appearance of the branch point is that for negative coupling the vacuum is unstable because the vacuum will create pairs of particles and antiparticles and "they will fly apart" forever. This leads to an imaginary part (i.e. a branch cut) in correlation functions. That's why (in your language) "things flying apart will lead to asymptotic series."
Aug
8
comment Distance between Newton's Rings fringes is does not seem linear
I wonder how you discovered this... do you have a picture?
Jul
30
answered Does spin-0 or spin-2 describe massive or massless particles?
Jul
15
comment How to replace $T$-product with retarded commutator in LSZ formula?
I believe there is a related mistake: in integrating by parts, you didn't include the surface terms. Especially important is from the time variable e.g. $\langle\varphi^\dagger(y)\varphi(x)\rangle\big|_{x^0=-\infty}^{x^0=+\infty} = \langle\varphi^\dagger(y)\big(\varphi_\text{out}(x)-\varphi_\text{in}(x)\big) \rangle$. Neglecting it, you can show all amplitudes vanish. See the remark on page 206, line 9-10 of Itzykson and Zuber.
Jul
15
comment How to replace $T$-product with retarded commutator in LSZ formula?
Thanks! I am working through the details, but can’t understand why $q_1^0$ and $\omega_{q,A}$ etc are distinguished. They must be equal since $A$ is an external particle on its mass shell.
Jul
14
revised How to replace $T$-product with retarded commutator in LSZ formula?
set equation number in tag.
Jul
12
revised Electrostatics with Yukawa mass
Corrected a typo in the last equation.
Jul
12
comment Electrostatics with Yukawa mass
Yes that's right, thanks! This final piece of information implies $C=-\big[\rho e^{-\mu R} (\mu R+1)\big]/\big[2 \mu^3\big]$.
Jul
12
asked Electrostatics with Yukawa mass