Recent content by redtree

The Hilbert space for the derivation is: ##\mathcal{H}=L^2(X_m^+,\lambda)## where λ denotes the invariant measure over ##X_m^+##. This space does not include photons because they are not represented by the orbit ##X_m^{+}##. Thus, it would seem that the resulting derivation would not apply to...

What would be the explicit formulation of ##\textbf{S}## in the spacetime algebra?

Question on the Pauli-Lubanski vector formulated in geometric algebra ##\textbf{W}##. From \url{https://en.wikipedia.org/wiki/Relativistic_quantum_mechanics#Relativistic_quantum_angular_momentum}, ##\textbf{W} = \star (\textbf{M} \wedge \textbf{P})##, where ##\textbf{M} = \textbf{X} \wedge...

שלום בני

For anyone coming across this thread in the future, it seems the above hyperlink has been taken down. The best mathematical proof I've found for the decomposition of total angular momentum into orbital and spin angular momentum can be found at...

Can you prove that mathematically in a simple and intuitive way? The only proof I've found is https://www-user.rhrk.uni-kl.de/~apelster/Vorlesungen/WS2021/v6.pdf. [link broken, see below]

Why is that?

Clearly $$\begin{equation} \begin{split} \epsilon^{\mu\nu\rho\sigma} P_{\nu} J_{\rho\sigma} &= \epsilon^{\mu\nu\rho\sigma} P_{\nu} (M_{\rho \sigma} + S_{\rho \sigma})\\ &= \epsilon^{\mu\nu\rho\sigma} P_{\nu} M_{\rho \sigma} + \epsilon^{\mu\nu\rho\sigma} P_{\nu} S_{\rho \sigma}\\...

I understand why symmetric and anti-symmetric tensors contract to zero. I apologize, but then I don't see how spin come out as non-zero.

Is this correct? I apologize if the Latex syntax is wrong for Physics Forums. If it is, maybe you can cut and paste into another Latex editor....? (or you can tell me what I did wrong and I can repost) $$\begin{equation} \begin{split} W^{\mu} &= \frac{1}{2} \epsilon^{\mu \nu \rho \sigma}...

My understanding is that ##\mathbb{Z}_2## is abelian and therefore a normal subgroup.

I apologize, but I don't see it.

Given $$M_{\rho \sigma} = i (x^{\rho} \partial_{\sigma} - x^{\sigma} \partial_{\rho})$$ and $$W^{\mu} = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} P_{\nu} M_{\rho \sigma}$$ Why does ##W^{\mu}## pick up only the spin part of the total angular momentum?

I just want to make sure I understand this correctly. For an infinite-dimensional representation, the generators of translation can be written as ##i \frac{\partial}{\partial_{\mu}}= i \partial_{\mu}##, where the generators of the Lorentz group can be written as ##i (x^{\mu}\partial_{\nu} -...

All of the formulations of the Lie algebra of ##\frak{so}(3)## (or ##\frak{su}(2)##) utilizing raising/lowering operators that I have seen in the literature involve complexification to ##\frak{su}(2) + i \frak{su}(2) \cong \frak{sl}(2,\mathbb{C})##. I have found explicit derivations in a...