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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...

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...

$$\hat{T}_{\mu v}(x)=e^{i\hat{P}x}\hat{T}_{\mu v}(0)e^{-i\hat{P}x}$$, so $$\bra{\overrightarrow{P'}}\hat{T}_{\mu v}(x)\ket{\overrightarrow{P}}=e^{iP'x}\bra{\overrightarrow{P'}}\hat{T}_{\mu v}(0)\ket{\overrightarrow{P}}e^{-i\hat{P}x}$$ Now, $$\partial^{\mu}\Phi=\int\frac{d^3 k_1}{2\omega_{k_1}...

For a single particle, $$Z=\frac{1}{h^2}\int_{-\infty}^{\infty} e^{-\beta \frac{P^2}{2m}}d^2p \int e^{-U(r)}drd\theta= \frac{1}{h^2}(\frac{2\pi m}{\beta}) 2\pi [\int_{0}^{r_0}e^{U_0}dr+\int_{r_0}^{R}dr]$$ $$ =\frac{1}{h^2}(\frac{2\pi m}{\beta}) 2\pi [e^{U_0}(r_0)+(R-r_0)]=\frac{\pi...

Let ##\phi_{+}=\frac{d^3k}{2\omega_k (2\pi)^3}\int(\hat{a}^{\dagger}(\overrightarrow{k})e^{ikx})## and ##\phi_{-}=\frac{d^3k}{2\omega_k (2\pi)^3}\int(\hat{a}(\overrightarrow{k})e^{-ikx})##. Then ##\phi^4=\phi_{1}\phi_{2}\phi_{3}\phi_{4}=(\phi_{1+}+\phi_{1-})(\phi_{2+}+\phi_{2-}...

I already know this quantity diverges, however I was wondering where to go from there. Any resource would be appreciated. Thank you. Useful Information: $$\phi=\int\frac{d^3k}{2\omega_k (2\pi)^3}(\hat{a}(\overrightarrow{k})e^{-ikx}+\hat{a}(\overrightarrow{k})^{\dagger}e^{ikx}))$$...

Thank you, this works. The math script just wasn't working when I tried previewing it in the homework forums, but worked when I actually posted it.

From this, I find $$\bra{P'} \phi^4 \ket{P} = \int \frac {d^3 k_1 d^3 k_2 d^3 k_3 d^3 k_4} {16 \omega_{k_1}\omega_{k_2}\omega_{k_3}\omega_{k_4} (2\pi)^{12} }\bra{0} a_{P'}(a_{k_1} a_{k_2} a_{k_3}a_{k_4}e^{i(k_1+k_2+k_3+k_4)x}+...)a^{\dagger}_{P}\ket{0}$$ (a total of 16 different terms) Right...

Is there a new code format for writing equations here? I usually use $$ or ## to delineated code, but that doesn't seem to work anymore. Thanks.

Part a was not much of a problem. I got that $$m=QR\omega \hat{z}$$. From that, I get $$A_{dip}=\frac{\mu_0}{4\pi}\frac{QR\omega}{r^2}\hat{\phi}$$ (using $$\theta=\frac{pi}{2}$$. My problem occurs in part b. I know there is a potential energy relation for two dipoles, but what would I use for a...

a) This would be true whenever |a_n> is an eigenvector of B_i. b) If this holds true for each eigenvector, then B_i and B_j must share the same basis. Therefore, they must commute. Is this reasoning correct? C) Despite commuting with the hamiltonian. the energy states can be degenerate, which I...

If two separate operators commute with the hamiltonian, then it implies they share a common set of eigenvectors. Using this reasoning, it can be shown that A and B commute. However, this proof fails for L_i. I just want to know why it fails. I believe it has to do with degeneracy.

Why is it the case that when some operators commute with the Hamiltonian (let's say A and ), it implies A and B commute, but even when each angular momentum component commutes with the Hamiltonian, it does not imply each the angular momentum components commute with each other?

Based on the conditions, I found that $$V(x)=\frac{a^2}{\pi^2} ρ_0sin(πx/a)$$ would be a solution to Laplace's equation for $$|x|\leq a$$ and $$V(x)=cx+d$$, where c and d are constants. From the boundary conditions, $$\frac{dV(a)}{dx}=\frac{a}{\pi} ρ_0cos(πa/a)=ac$$, $$c=\frac{a\rho}{\pi}$$ and...

Nevermind, figured it out. Thank you though.

$$<H(\omega)>=\sum_{j} χ_{HAj}h_j(\omega)$$ Where $$χ_{HA}=\frac{1}{2\hbar} Tr{{\rho}[{H(t)},{A(0)}]}$$. But $$[H(t),A(0)]=[H_o,A(0)]-[A(t)h,A(0)]=-h_0 cos(\omega t)[A(t),A(0)]$$. So $$χ_{HA}=-\frac{1}{2\hbar}Tr(\rho h_0 cos(\omega t)[A(t),A(0)])=-h_0cos(\omega t)χ_{AA}$$. Then...

Nevermind, figured it out. $$S_1^x$$ does not commute with $$S^2$$. However, $$S^2=J^2$$, so $$J^2$$ commutes with $$H$$,which I believe implies each component of $$J$$ also commutes.

$$H$$ can be rewritten as $$H=\frac{1}{2}(S^2-S_{1}^2-S_{2}^2-S_{3}^2-S_{4}^2)$$. Let's focus on the x component, $$J^x=\sum_{i}S_i^x$$. Now $$S_1^x$$ commutes with $$S^2_1, S^2_2, S^2_3, S^2_4$$, but does it commute with $$S^2$$? If not, what is the exact relation between $$S^2$$ and $$S_1^x$$?

Would T and P be different for $$\rho_{gas}$$ and $$\rho_{liquid}$$? Right now, after rearranging, I get $$T=\frac{P-\rho \beta P+\alpha \rho^2-\alpha \beta \rho^3}{R\rho}$$ which gives $$T=\frac{- \beta P+\alpha \rho-\alpha \beta \rho^2}{R}$$ when I let $$\rho$$ go to $$ \infty$$

Not sure where to actually start. Do I need to do a virial expansion? Any tips on on where to start would be greatly appreciated.

Using this new info, I get for a) that ## f(\theta)=\frac{\sqrt{4\pi}}{k} \sum_{l=0}^{\infty} \sqrt{2l+1}Y_{l0} (e^{i\delta _l})sin^2{\delta_l} ##, where $$Y_{l0}$$ is a spherical harmonic. For B, I can use $$\sigma=\int |f(\theta)|^2d\Omega=2\pi \int_{0}^{\pi}(\alpha+\beta cos(\theta)+\gamma...

I considered that, but I don't think that method would quite work since l is a summation to infinity.

So, I talked with my professor, and apparently, there was a typo. It should be that $$\frac{d\sigma}{d\Omega}=\alpha+\beta cos(\theta)+\gamma cos^2(\theta)$$.

a) I have $$d\sigma=-\beta sin(\theta)d(\theta)+2\gamma sin(\theta)cos(\theta) d\theta$$ and $$d \Omega=2\pi sin(\theta) d \theta$$ so $$\frac{d\sigma}{d \Omega}=-\frac{\beta}{2\pi}+2\gamma cos(\theta)=|f(\theta)|^2$$ b) $$\sigma(\theta)=\alpha+\beta cos(\theta)+\gamma...

I get $$B_2(T)=2\pi N\int_{0}^{\infty} (1-e^{-\beta E_0((\frac{r_0}{r})^{12}-2(\frac{r_0}{r})^6)})r^2dr$$ as the coefficient. I was just unsure how to evaluate it numerically from here. Any suggestions would be appreciated. Thank you.

Wait. Just figured it out. Thank you.

So $$(-i\hbar \nabla -\frac{q}{c}A+\frac{q}{c}\nabla \lambda)\psi'=(-i\hbar \nabla e^{-\frac{iq \lambda}{\hbar c}}\psi -\frac{q}{c}Ae^{-\frac{iq \lambda}{\hbar c}}\psi+\frac{q}{c}\nabla \lambda e^{-\frac{iq \lambda}{\hbar c}}\psi)=(-\frac{q}{c}\nabla\lambda e^{-\frac{iq \lambda}{\hbar c}}...

Perhaps it is a silly question. I was just wondering why it held. Thank you.

I know it is gauge invariant. I suppose it is not clear the issue I am having. I know it is the case that $$(-i\hbar \nabla -\frac{q}{c}A+\frac{q}{c}\nabla \lambda)\psi'=e^{-\frac{iq\lambda}{\hbar c}}(-i\hbar \nabla -\frac{q}{c}A)\psi$$. I want to know if it is trivially true that $$(-i\hbar...

Given the schrodinger equation of the form $$-i\hbar\frac{\partial \psi}{\partial t}=-\frac{1}{2m}(-i\hbar \nabla -\frac{q}{c}A)^2+q\phi$$ I can plug in the transformations $$A'=A-\nabla \lambda$$ , $$\phi'=\phi-\frac{\partial \lambda}{\partial t}$$, $$\psi'=e^{-\frac{iq\lambda}{\hbar c}}\psi$$...

The dubious assumption I am making is that the integral over the density of states is proportional to the volume in k space. Since $$\epsilon=\frac{(\hbar)^2k^2}{2m}$$ for part a, and $$\epsilon=(\hbar)\omega k$$ for part b, and $$V\propto k^d$$ for d dimensions in k space. So, $$\int...

Thank you. Just one more question. Would the limits of integration just be from 0 to $$\infty$$?

Taking this further, I substitute $$x=\sqrt{\frac{\hbar^2}{2mk_bT}}$$ and get that $$U=\int Z \epsilon D(\epsilon) e^{-\epsilon β}d\epsilon=\frac{gV4\pi}{(2\pi)^3}\int Z \frac{(\hbar)^2}{2m}\sqrt{\frac{\hbar^2}{2mk_bT}}^{\frac{5}{2}}x^4 e^{-x^2}dx=\frac{gV4\pi}{(2\pi)^3}Z...

I find that $$U=\int Z \epsilon D(\epsilon) e^{-\epsilon β}d\epsilon=\frac{gV}{(2\pi)^3}\int Z \frac{(\hbar)^2k^2}{2m}k^2 (4\pi)e^{-β\frac{(\hbar)^2k^2}{2m}}dk$$ where g=2s+1=2, $$Z=e^{βµ}$$ and $$D(\epsilon)=\frac{gV}{(2\pi)^3}k^2 4\pi$$ for the density of states From here, I can use $$c_v...