Timeline for Photon spin projection to arbitrary axis
Current License: CC BY-SA 3.0
6 events
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Jul 5, 2016 at 16:51 | comment | added | Luboš Motl | Massive vector bosons' states which are eigenstates of $\vec p$ with some eigenvalue along a different axis than $z$ cannot be simultaneously eigenstates of $J_z$, the total spin, $z$ component. But for massive vector particles, the Hilbert space basically tensor factorizes to the spin degrees of freedom and the center-of-mass (or momentum) ones, so even for $\vec p\neq 0$, it may be an eigenstate of $S_z$, the internal part of the spin, because $\vec S$ and $\vec p$ commute with each other. But that separation to 2 parts isn't possible for massless photons. | |
Jul 5, 2016 at 16:47 | comment | added | Luboš Motl | No, my argument is correct so it doesn't lead to any wrong conclusions. The first problem of photons - the absence of the rest frame - is avoided for massive spin-one bosons because they do have a rest frame. The second problem with the photons, the fact that they only have 2 transverse polarizations (to the momentum), is also avoided for massive vector bosons because those have all 3 polarizations, not just 2. But if you ask whether states of vector-one bosons with a well-defined $\vec p\neq 0$ in a non-$z$ direction may be eigenstates of $J_z$, my argument holds and the answer is No. | |
Jul 5, 2016 at 13:31 | comment | added | tparker | You say "if $\vec{p} \neq 0$ ... [the state] is therefore not an eigenstate of $J_x$". But massive spin-1 particles can be in an eigenstate of $J_x$ even if they have nonzero momentum. Doesn't your argument lead to the too-strong conclusion that even if there exists a rest frame, if you're not in it, then you can't have $S_z = 0$? | |
Jan 7, 2012 at 18:10 | vote | accept | kuzand | ||
Jan 7, 2012 at 12:56 | history | edited | Luboš Motl | CC BY-SA 3.0 |
added 679 characters in body
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Jan 7, 2012 at 12:46 | history | answered | Luboš Motl | CC BY-SA 3.0 |