touqra
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What does superselection rule mean?
zbyszek said:Physically meaningful solutions of an equation E must be invariant under transformation T. Solution s1 and solution s2 are invariant under T. Superposition s1+s2 is also a solution of E, but it is not invariant under T. Thus s1+s2 is not a miningful solution.
vanesch said:Decoherence tries to explain superselection as a dynamical phenomenon.
In the example of charge, decoherence theory tells you that states with different charge couple differently to the EM field, so very quickly a (forbidden by the superselection rule) superposition of states with different charges will entangle with the EM field, so that all interference effects on the particle alone will be suppressed, and it will look *as if* the superposition was forbidden and one had to apply projection.
Careful said:In my mind a state only ``exists'' when it is dynamically stable over some respectable period of time.
vanesch said:If s1 and s2 are quantum states, and T is a linear operator representing the invariant transformation, I fail to see how s1 and s2 could possibly be invariant, but not their sum ?
zbyszek said:Example: T stands for Galilean transformations, s1-represent a state of N massive
particles, s2- a state of N+1 massive particles. All particles have the same mass.
Under the transformation s1 acquires an overall phase that depends on the total mass,
same with s2.
s1+s2 acquire overall phase and the relative phase so it transforms differently under T
than s1 and s2.
vanesch said:Well, that's then a "dynamically stable state", no![]()
LanceV said:Hello Forum!
I am confused about this topic, too.
The post of @vanesch seems to suggest, that the superselection rules are additional postulates to the quantum theory, while in @zbyszek's post there seems to be a mathematical way to deduct them?
Or rephrased into a more direct question:
Why does the weak isospin "generate" (?) a superselection rule, while the usual spin does not? The transformation law is the same for a spin 1/2 particle?
What is the correct terminology anyways? The superselection rule selects the quantum nubers of a state, that a given state can interfere with?
vanesch said:As you point out, in as much as isospin is mathematically of an identical form as normal spin, we can have superpositions of |z+> and |z-> in normal spin, which make up, say, |x+>. But we can't have that in isospin ; we can't have states which are the superposition of a proton and neutron state.
vanesch said:This is what this particular superselection rule says: no superpositions of states with different charges. There are other superselection rules (I'm no expert).
vanesch said:But in decoherence theory, one simply explains the *apparent* validity of superselection rules by the extremely rapid decoherence of the "forbidden" superpositions. In other words, the superpositions are not "forbidden" by any law of nature, but the different components interact so quickly in different ways with the environment that they get almost instantaneously irreversibly entangled, which makes them *appear* almost always as mixtures, and which makes it impossible to observe any interference effects which would testify of their superposition.
LanceV said:Why can't we have it? Interestingly I never thought about this, but now I am unsure. Am I missing something obvious? Or is it just an observation?
touqra said:What does superselection rule mean?
Read my selection rule: You can choose to do theoretical physics without knowing anything about superselection rules. Just ignor it! What you need to know is the following
Let G be a symmetry group and M(0), M(1/2), M(1), ... are its multiplets (representations). Then the combination
a_{1}M(0) + a_{2}M(1/2) + a_{3}M(1) + ...
with atleast two nonvanishing a's, is not allowed by the symmetry group G.
Examples
1) Lorentz group G = SO(1,3):
M(0) is the scalar (spin = 0) representation, M(1/2) is the spinor (spin = 1/2) rep. and M(1) is the vector (spin = 1) rep.
We know that Lorentz transformations do not mix, say, scalars with spinors (supersymmetry does).
2) Nuclear I-spin G = SU(2)
M(I=0) = trivial rep. is the isoscalar (singlet) \Lambda.
M(I=1/2) = fundamental rep. is the isospinor (doublet) (P,N).
M(I=1) = adjoint rep. is the isovector (triplet) (\Sigma^{+}, \Sigma^{-}, \Sigma^{0}).
SU(2) transformations mix particles occupying the same multiplet, for example, neutron with proton (in real life, we don't see such mixing because isospin is not an exact symmetry, it is broken by mass-difference and em-interaction), But particles belonging to different multiplets don't get mixed by SU(2) transformations.
regards
sam