Timeline for Lorentz Algebra Representation and QFT
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 31 at 13:08 | comment | added | Quillo | Closely related physics.stackexchange.com/q/799610/226902 | |
| Apr 8, 2015 at 17:29 | history | tweeted | twitter.com/#!/StackPhysics/status/585857118047227904 | ||
| Apr 8, 2015 at 15:14 | vote | accept | Quantization | ||
| Apr 8, 2015 at 15:00 | answer | added | ACuriousMind♦ | timeline score: 17 | |
| Apr 8, 2015 at 14:39 | comment | added | Quantization | @ACuriousMind: Yes, that is what I am confused about. Are you saying the right hand side (finite dimensional representation) is not a finite dimensional "Hilbert" space? So $\psi_a(x)$ has a infinite dimensional representation in the Hilbert space and finite dimensional representation in classical vector space? Also, you are saying Wightman axiom tells us, if I find an explicit infinite dimensional representation of $\psi_a(x)$ and $U(\Lambda)$, left hand side will give exactly same result? Thank you ACuriousMind! | |
| Apr 8, 2015 at 14:22 | comment | added | ACuriousMind♦ | Your definition of a "vector" is also off, already in the QM case - is $V^i$ is a vector under rotation, it transforms as $V\mapsto D(R) V$, or, in components, $V^i \mapsto D(R)^i_j V^j$. That in QFT the "operator transformation" $U(\lambda)\phi U(\lambda)^\dagger$ and the "vector transformation" $\phi\mapsto L(\Lambda)\phi$ coincide is one of the Wightman axioms. | |
| Apr 8, 2015 at 14:19 | comment | added | ACuriousMind♦ | If you know that there is no finite-dimensional unitary rep of the Lorentz group, what missing piece are you searching for? What is your specific question? (You seem confused by the fact that there are two "simultaneous" representations of the Lorentz group in QFT. Those that act directly on the (classical) fields as finite-dimensional representations (your $L(\lambda)$), and the unitary representations upon the Hilbert spaces of state under which the fields transform as operators (by your $U\Lambda$). You don't get a complete analogy to QM because this doesn't happen in QM. | |
| Apr 8, 2015 at 14:08 | history | edited | Quantization | CC BY-SA 3.0 |
added 4 characters in body
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| Apr 8, 2015 at 14:03 | history | asked | Quantization | CC BY-SA 3.0 |