Recent content by Diracobama2181

So I plan on defending my PhD in the summer. I want to continue my career as a physicist, however I have not applied for any postdocs yet, the reason being I technically have not gotten any published yet, however, I should be getting published within the next month. My question is what are my...

I am having difficulty writing out ##\bra{p',\lambda}\psi^{\dagger}(-\frac{z^-}{2})\gamma^0\gamma^+\psi\frac{z^-}{2})\ket{p,\lambda}## in momentum space. Here, I am working in light-cone coordinates, where I am defining ##z^-=z^0-z^3##, ##r'=r=(0,z^{-},z^1,z^2)##. My attempt at this would be...

By the logic of my argument, I simply mean that is showing the time of of travel less than the time of decay sufficient to show that the muon would have to reach the ground. I have recently contacted my professor however, and they said this form of argument dosen't work, and I would have to give...

i) The muon reaches the ground ii) To a ground observer, the decay time is dilated $$\Delta t_d=\frac{1}{\sqrt{1-\frac{0.999c^2}{c^2}}}\Delta\tau_d=22.4 \tau_d=4.5 *10^{-5}s>\Delta \tau_d$$ The time for the muon to reach the ground is $$\Delta t_g=\frac{10 km}{0.999c}=3.3*10^{-5} s< \Delta...

i) The muon reaches the ground ii) To a ground observer, the decay time is dilated $$\Delta t_d=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\Delta\tau_d>\Delta \tau_d$$ The time for the muon to reach the ground is $$\Delta t_g=\frac{10 km}{0.999c}< \Delta t_d$$ which is why it reaches the ground...

For example, after the Lagrangian is renormalized at 1-loop order, it is of the form $$\mathcal{L}=\frac{1}{2}\partial^{\mu}\Phi\partial_{\mu}\Phi-\frac{1}{2}m^2\Phi^2-\frac{\lambda\Phi^4}{4!}-\frac{1}{2}\delta_m^2\Phi^2-\frac{\delta_{\lambda}\Phi^4}{4!}$$. So if I were to attempt to find the...

Just wanted to check if I was on the right path. Thanks!

My guess would be to do an integral of the form $$\frac{\int d^4k}{(2\pi)^4}k^2(\frac{1}{(k^2-m^2+i\epsilon)}-\frac{1}{k^2-\Lambda_1^2+i\epsilon})(\frac{1}{(k^2-m^2+i\epsilon)}-\frac{1}{k^2-\Lambda_2^2+i\epsilon})$$ before Wick otating and integrating. Any help is appreciated. Thanks.

For the below integral in Euclidean space, $$\int d^4k_E k_E^2 \frac{1}{(k_E^2+M(\Delta))^2}= 2\pi^2\int dk_E k_E^5 \frac{1}{(k_E^2+M(\Delta))^2}$$ we find $$2\pi^2\int_0^{\infty} dk_E k_E^5 \frac{1}{(k_E^2+M(\Delta))^2}\xrightarrow{}2\pi^2\int_0^{\infty} dk_E...

In this case, the lagrangian density would be $$\mathcal{L}=\frac{1}{2}((\partial_{\mu}\Phi)^2-m^2\Phi^2)-\frac{\lambda}{4!}\Phi^4$$ whe $$\Phi$$ is the scalar field in the Heisenburg picture and $$\ket{\Omega}$$ is the interacting ground state. Was just curious if there were ways to do Feynman...

I know in the Heisenburg picture, $$\Phi(\vec{x},t)=U^{\dagger}(t,t_0)\Phi_{0}(\vec{x},t)U(t,t_0)$$ where $$\Phi_{0}$$ is the free field solution, and $$U(t,t_0)=T(e^{i\int d^4x \mathcal{L_{int}}})$$. Is there a way I could solve this using contractions or Feynman diagrams? Because otherwise, it...

Relevant Equations:: ##\ket{\vec{p}}=\hat{a}^{\dagger}(\vec{p})\ket{0}## for a free field with ##[\hat{a}({\vec{k})},\hat{a}^{\dagger}({\vec{k'})}]=2(2\pi)^3\omega_k\delta^3({\vec{k}-\vec{k'}})## $$ \bra{ \vec{ p'}} T_{\mu,\nu} \ket{ \vec...

$$ \bra{ \vec{ p'}} T_{\mu,\nu} \ket{ \vec {p}}=\bra{\Omega}\hat{a}(\vec{p'})(\partial^{\mu}\Phi\partial^{\nu} \Phi-g^{\mu \nu}\mathcal{L})\hat{a}^{\dagger}(\vec{p})\ket{\Omega}=\bra{\Omega}(\hat{a}(\vec{p'})\partial^{\mu}\Phi\partial^{\nu} \Phi\hat{a}^{\dagger}(\vec{p})\\...

Never mind. Figured it out. I would have to go to $$\lambda^2$$ in the expansion. Thanks.

I know $$ i\mathcal{M}(\vec {k_1}\vec{k_2}\rightarrow \vec{p_1}\vec{p_2})(2\pi)^4\delta^{(4)}(p_1 +p_2-k_1-k_2) $$ =sum of all (all connected and amputated Feynman diagrams), but what is meant by 1 loop order? In other words, when I take the scattering matix element...