A message from our CEO about the future of Stack Overflow and Stack Exchange. Read now.

New answers tagged

1

I suspect the integral is in Minkowski signature which requires more care, but let me take the simpler Euclidean case. If the points $x_1,x_2,x_3,x_4$ in $\mathbb{R}^4$ are distinct, then the integral $$ \int_{\mathbb{R}^4\backslash\{x_1,x_2,x_3,x_4\}}\frac{d^4 x_5}{(2\pi)^8}\ \frac{1}{(x_1-x_5)^2(x_2-x_5)^2(x_3-x_5)^2(x_4-x_5)^2} $$ converges rigorously in ...


2

Recall that, by definition, $$ Z[J]=e^{-iW[J]}=\langle\Omega|e^{-iHT}|\Omega\rangle $$ where $H$ is the Hamiltonian of your system, and $\Omega$ is the vacuum state. Therefore, $W[J]$ can be interpreted as the energy of the vacuum $E_\Omega$ in the presence of a source $J$, where the origin is chosen at $E=0$ for $J=0$. In other words, $W[J]$ is how much ...


2

Is there a neutral current interaction between an antineutrino and an anti-strange quark? The easier formulation of the question is to ask "is there a neutrino-strange quark interaction?" If the answer is "yes" for particles it will be "yes" for the antiparticles. Since the neutrino has zero charge and is only weakly interacting it has to be a neutral ...


6

The whole point of QFT is that it is a framework that allows you to define Lorentz (co)variant scattering amplitudes. In fact, under some general hypothesis it is the only framework with that property. The expression in the OP is not manifestly Lorentz covariant, although it turns out to be, after a very cumbersome analysis. See ref.1 for a detailed ...


-1

The expected probability density for Compton scattering is \begin{equation*} \langle|\mathcal{M}|^2\rangle= 2e^4\left( \frac{\omega}{\omega'}+\frac{\omega'}{\omega} +\left(\frac{m}{\omega}-\frac{m}{\omega'}+1\right)^2-1 \right) \end{equation*} where $\omega$ is the angular frequency of the incident photon, $\omega'$ is the angular frequency of the ...


4

1) A normalization condition for one-particle states $| \vec p \rangle$, which is Lorentz invariant is $\langle \vec p | \vec q \rangle = 2 E_p (2 \pi)^3 \delta^{(3)} (\vec p - \vec q)$ (The factor $2$ is conventional) However, if we write $\delta^{(3)} (\vec k) = \int \frac{d^3x}{(2 \pi)^3} exp (i \vec k \cdot \vec x)$ we can state $\delta^{(3)} (\vec 0) ...


0

From your diagram: You can interchange the 3 lines connecting the 2 vertices in $3!=6$ different ways. This argument is the same as counting how many different ways 3 people can sit in 3 seats. The complete rules for calculating the symmetry factors of Feynman diagrams can be found in Chapter 4 of Peskin & Schroeder.


1

From the Boltzmann factor perspective $e^{\frac{i}{\hbar}S}$, if we exponentiate the path integral measure $\rho= e^{\frac{i}{\hbar}\frac{\hbar}{i}\ln\rho}$ in the path integral $Z=\int\!{\cal D}\phi~\rho~e^{\frac{i}{\hbar}S}$, the anomaly in the measure is formally a 1-loop effect, cf. the $\hbar$/loop-expansion. This seems to be essentially Bilal's ...


1

Just to fill in a few of the blanks of TwoBs answer: Using \begin{align*} \sum_{s} \epsilon^s(k)^{\mu}\epsilon^s(k)^{\nu *} = -\eta^{\mu\nu} + \frac{1}{2E^2}(k^{\mu}\bar{k}^{\nu}+k^{\nu}\bar{k}^{\mu}), \end{align*} from TwoBs notes, we would like to calculate \begin{align*} \sum_{s,s'} |\mathcal{M}|^2 = 4e^4 \left( \frac{p_{\mu}p'_{\nu}}{p\cdot k}-\frac{p'...


0

Another reasoning leading to the conclusion that one cannot just simplify the expression for $\mathcal{M}$ (first equation) using the Lorentz gauge condition $\epsilon . k=0$ and subsequently replace in the simplified expression (second equation) $\sum \epsilon_\mu \epsilon_\nu^*$ by $ -\eta_{\mu \nu}$ is the following. The matrix element of the first ...


1

You can use a program I wrote, Package-X for Mathematica, which allows you to directly calculate diagrams like this. By the way, I don't think you wrote down your amplitude correctly. Use LoopIntegrate in conjunction with FermionLine to initiate the computation of the integral Then replace the dot products with the relevant on shell conditions (p.p->mμ^...


0

The triangle you mention is not a divergent integral, which is why it does not appear in the paper you mention. I wrote a Mathematica program Package-X to help solve integrals like this, and you can get a result for your integral by simply running this: << X` (*Load Package-X*) ScalarC0[p1^2, p2^2, p3^2, m, m, 0] // C0Expand The integral in which ...


0

Answer: Casimir trick In a typical collider experiment the momentum vectors are $$ p_1=\begin{pmatrix}E\\0\\0\\p\end{pmatrix}\qquad p_2=\begin{pmatrix}E\\0\\0\\-p\end{pmatrix}\qquad p_3=\begin{pmatrix} E\\ \rho\sin\theta\cos\phi\\ \rho\sin\theta\sin\phi\\ \rho\cos\theta \end{pmatrix} \qquad p_4=\begin{pmatrix} E\\ -\rho\sin\theta\cos\phi\\ -\rho\sin\theta\...


1

One node of the triangle is given by the current $j^{\mu 5}$ with momentum $q$, the other two nodes represent photons with momenta $p$ and $q$. The current $j^{\mu 5}$ has two fermions and one photon. The fermions give two of the internal lines of the diagram.


2

In the first expression that you write for $i\mathcal{M}$, you have simplified to the second line using a gauge condition (Lorentz gauge) $\epsilon_\mu k^\mu=0$. This is a legal choice but it is incompatible with replacing both polarizations sums with $-\eta_{\mu\nu}$, since you have partially fixed the gauge in this way. It's simple to check that squaring ...


3

Consider a generic Feynman diagram with, $L$ loops, $N_f$ number of internal fermion lines (or fermionic propagators) and $N_b$ number of internal boson lines (or bosonic propagators), different kinds of vertices and the $i^{th}$ kind appears $N_{v_i}$ times, and number of derivatives in each vertex be $h_i$. Now suppose that we have written down the ...


1

It seems that OP is not questioning the standard convention to divide each term in the Lagrangian with its symmetry factor, e.g., $${\cal L}~=~-\frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi -\frac{1}{2}m^2\phi^2 - \frac{\lambda}{3!}\phi^3.$$ Rather OP is assuming the above standard convention, and asks if the vertex factor$^1$ is $-\frac{i}{\hbar}\lambda$ ...


1

It is conventional to write interactions normalized by the number of permutations of identical fields. So, there will be a $\frac{1}{n!}$ factor for each interaction with $n$ identical fields. This factor is then canceled by the $n!$ ways of permuting the $n$ identical lines coming out of the same internal vertex. The diagram is therefore associated with ...


1

The power counting of momentum $p$ has nothing to do with the divergence power like $1/\epsilon^a$, actually what we are doing here for calculating loop diagrams is translating(mapping) divergence into the language of Gamma function. I think we can call it Gamma function regularization. I will show you examples as the following, as we know the $\beta$-...


1

There is no trivial way to start with a list of initial-state and final-state particles and determine which channels are involved. You just have to start enumerating diagrams, and working out if they are allowed or not. To do that you use a small set of tools Each fundamental interaction has an allowed set of diagrams. You have to know or look these up. ...


0

The $\hbar$-weights are a direct consequence of eq. (B9) in my Phys.SE answer here: A vertex has $\hbar$-weight$=-1$. An internal propagator comes with $\hbar$-weight$=+1$ (rather than $-1$) because an internal propagator is accompanied by 2 source differentiations (which each carry $\hbar$-weight$=+1$). By the same token, an external leg (=source+...


1

To start with, one has to keep in mind that detectors detect, by recording interactions in a medium. As neutrinos leave no track in the detectors, the final state neutrinos are useless for detecting. In principle, if the neutrino beam were mono-energetic, using conservation of energy and momentum one could fit for a missing neutrino, by just detecting the ...


1

In my view, one of the best tools to calculalte complicated loop diagrams is dispersion relation method. You should to the following steps: Write down the expression for diagram with you Feynman rules Carefully simplify numerator: do trace, make contractions, etc (please check numerator, there is a trace, isn't it?). You will obtain an expression $N^{\mu\nu}...


Top 50 recent answers are included