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So how to say on general that this is true for any weak force interaction? do you mean to ask: "why is this assumed for all weak interactions"? Weak interactions were classified by their "weakness". Lifetimes were long and interactions much more weak than electromagnetic or strong. The suggestion was made by Lee and Yang that as a class parity ...

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I would like to write a long answer, but almost all the information I think you require can be found here: https://en.wikipedia.org/wiki/Wu_experiment Whilst I'm reluctant to link Wikipedia, this page contains details about how parity is conserved in strong and electromagnetic reactions, but not in weak ones. Physically what is implied when parity is ...

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The transformation under time reversal of the forms electrondynamics is subtle because the gauge field 1-form $A = A_\mu \mathrm{d}x^\mu$ and the field strength $F = F_{\mu\nu}\mathrm{d}x^\mu\wedge\mathrm{d}x^\nu$ are not the correct physical objects to transform. This may be seen by observing that the Maxwell equations are $\mathrm{d}F = 0$ and ...

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The situation is impossible and therefore doesn't happen. When you claim to have two states $\psi_1$ and $\psi_2$ then I'll assume they are linearly independent, otherwise they aren't really two different states. Then you claim you have an Operator $O$ such that $POP^{-1} = \epsilon_3 P$ where $P$ is the parity operator and further that $P \psi_1 = ... 0 I think the first two things that is transformation of position and momentum operators are defined from definition...because by parity transformation sign of the position coordinates changes and time coordinate remain unchanged...so accordingly we get change in sign for position and momentum...once you do that then angular momentum should remain unchanged ... 2 Comments to the question (v1): In this answer let us focus on spatial point reflections and the spatial part of Minkowski spacetime. (There is, by the way, a similar issue with time reversal symmetry$T$.) A choice of point reflection$P_0$in affine 3-space inevitable has a fixed point${\bf r}_0$. (We can use the restricted Lorentz group to trade a ... 3 Is it true that if$u_n(x)$and$u_m(x)$are orthogonal (which is true), then$u_n(x)$and$u_m(x)f(x)$will be also orthogonal? No. The simplest example of this is the case$f(x) = u_n(x)/u_m(x)$for whatever$n$and$m$you're considering. More broadly, the result you're trying to prove is false. Consider the infinite square well between$\pm ...

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