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.

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...

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...

Nevermind, I think I figured it our. I mistakenly assumed the $$e^{-iE_n T}\to 0$$ as $$T\to \infty$$, but that is not the case, which is why the substitution is needed.

In P&S, it is shown that $$e^{-iHT}\ket{0}=e^{-iH_{0}T}\ket{\Omega}\bra{\Omega}\ket{0}+\sum_{n\neq 0}e^{-iE_nT}\ket{n}\bra{n}\ket{0}$$. It is then claimed that by letting $$T\to (\infty(1-i\epsilon)) $$ that the other terms die off much quicker than $$e^{-iE_0T}$$, but my question is why is this...

Apologies, that should be $$\Pi^{\mu}\partial^{v}\phi$$ where $$\phi$$ is a field and $$\Pi^{\mu}=\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}$$. Originally had this posted to the High Energy subforum since this was originally a quantum field theory question, but someone moved it here.

I know the tensor can be written as $$T^{\mu v}=\Pi^{\mu}\partial^v-g^{\mu v}\mathcal{L}$$ where $$g^{\mu v}$$ is the metric and $$\mathcal{L}$$ is the Lagrangian density, but how would I write $$T_{\mu v}$$? Would it simply be $$T_{\mu v}=g_{\mu \rho}g_{v p}T^{\rho p}$$? And if so, is there a...