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Is quantum wave superposition of electrons and quarks possible?

If not, can different types of elementary particles be mixed in wave superposition?

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"Superposition" refers to linear combinations of eigenstates of a single wave function, not combinations of different wave functions. – Dmitry Brant Aug 3 '12 at 13:08
@DmitryBrant I think that is wrong. See my answer bellow. – Diego Mazón Aug 3 '12 at 23:07

This is a good question. No experiment has shown mixing between leptons (such as the electron) and quarks. I'm using the word "mixing" here in the same you used superposition (though more often the word superposition is used to refer to energy states). There is certainly experimental evidence for the superposition of other particles (e.g. neutrino oscillations).

Mixing between leptons and quarks is also not allowed by the standard model since it would violate a good number of symmetries (the symmetries themselves being assumptions of the standard model).

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Since electrons and quarks have different electromagnetic charge and the electromagnetic charge is conserved in time ($[Q,H]=0$), it is not possible create a state with no defined electromagnetic charge from a state with well defined charge. In equations, if one has an initial state with charge $q$ $Q|\Psi(t=0)\rangle =q|\Psi(t=0)\rangle$:

$$Q|\Psi(t)\rangle=Qe^{-itH}|\Psi(t=0)\rangle=e^{-itH}Q|\Psi(t=0)\rangle=q|\Psi(t)\rangle$$ Thus the only the option would be to have such state with no defined charge from the beginning of the universe. I think there is none theoretical reason explaining why these states were not present in the very beginning. But we have never observed them.

Nonetheless, these vectors belong to the Hilbert space because they are linear combinations of vectors that are part of the Hilbert space. Hence there are vectors in the Hilbert which do not correspond to physical states.

You will find more information under the name of superselection rules.

On the other hand, we have actually "observed" superpositions of different elementary particles such as neutrinos of different flavours through neutrino oscillations. Since the Sun, for instance, produces electron neutrinos and we can detect muon neutrinos in the Earth, this indicates that family leptonic number is not a symmetry of nature and therefore this number is not conserved, in contrast with the electromagnetic charge. Because in the Standard Model with massless neutrinos family (flavour) leptonic number is conserved, we have to modified the Standard Model to allow neutrino oscillations. The easiest way to do that (and probably the only consistent one) is to add neutrino masses. Arguably, neutrino oscillations is the first indication that the Standard Model (with massless neutrinos) is not the complete theory of non-gravitational interactions, but nowadays most people denominate Standard Model to the Standard Model with neutrino masses, just because the introduction of neutrino masses is not difficult to achieve. But this is in fact the only evidence we have that the Standard Model is an effective theory, in contraposition to fundamental theory, of the electroweak and strong interactions.

P.S. (after Ron Maimon clarification, not for very beginners in quantum mechanics): A necessary condition to say that two states (such as two states with different electromagnetic charge) are separated by a superselection rule is that for all physically observables $O$: $$\langle\Psi _1|O|\Psi _2\rangle=0$$ This is in contrast with selection rules which only demand the previous condition for the Hamiltonian $O=H$. Therefore, two states with different momentum or angular momentum (as long as both have integer ang. momentum or both have semi-integer ang. momentum) have not to be separated by a superselection rule because they can be connected by operators such as angular momenta or boosts. On the other hand, the electric charge commutes with the Hamiltonian and with the rest of observables (momentum, angular momentum, etc.) and one cannot find any observable that connects states with different charge. Therefore the charge is a good candidate to superselector operator (perhaps under the condition that no Higgs mechanism be present that Ron Maimon mentions).

In addition, as far as I know, superselection rules do not explain why some linear combinations of physical states are not physical states. The only they explain is that if they did not exist in the beginning, them they cannot be created by interaction/evolution.

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The charge superselection rule is not a simple consequence of charge conservation--- it fails if you have a Higgs mechanism. Momentum is also conserved, and yet momentum can be superposed. The superselection rule is weird, it says that no local operator can produce a superposition of charges, that is that you can only measure uncharged operators locally. – Ron Maimon Aug 4 '12 at 0:57
I know 'superselection' is tricky (by the way, I'd like to read something from you about it). Because of that, I have only mentioned it as a key term to look up. I think there is no contradiction between what I wrote and what you say: if at $t=t_0$ a state is an eigenstate of momentum operator, then at $t=t_1$ it is still an eigenstate. So one needs to start with a superposition. If you know why the initial state cannot be a superposition of states with different electrical charge, then you should post it. I'd like to read it. Your comment is too short to understand it. At least, for me. – Diego Mazón Aug 4 '12 at 1:43
It's not hard, but it's tricky--- you need the right framework (I think I wrote this as an answer to a question along the lines "what is a superselection rule?"). Consider a path integral for a membrane which is along the x-y plane at a certain z, elastically bombarded by Brownian atoms. The center of mass in z of the membrane will never ever shift, so it is a superselection sector. If you allow a nonlocal operator, you can just move the whole membrane with no energy cost, but you can't do it locally if the membrane is infinite. Similarly, every QED local operator you have doesn't alter Q. – Ron Maimon Aug 4 '12 at 3:37

You have very good question and you should think about it more.

Up to now its not observed experimentally. Quantum mechanics is a wave mechanics.

Now imagine waves - only those types of waves exhibit interference which are the same in nature - for example transverse waves will mix with transverse and longitudinal with longitudinal. You can look at elementary particles as DIFFERENT TYPES of WAVES in the same medium (vacuum) so particles who belong to the same wave type (electrons for example) will show mixing-superposition patterns.

To sum up superposition happens only between waves of similar type. If they are not similar then they either pass through each other or interact.

Thank you for thinking about this message.

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Yes, you can. But decoherence is rapid. As for not being able to have superpositions across charge superselection sectors, how else can you explain Higgs condensates?

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Is quantum wave superposition of electrons and quarks possible?

It depends on what you define as "quantum wave", and "quarks". In s-wave scattering of electrons off protons, since protons are made up of quarks, the s-wave electron is superimposed over the quarks in order to interact with the proton.

If not, can different types of elementary particles be mixed in wave superposition?

The wave we are talking of here is a probability wave, i.e the probability of finding a particle at (x,y,z,t) with fourmomentum (p_x,p_y,p_z, E). If one creates a mixed beam one has also the probability of being particle A or particle B ( which is true for many beams since they cannot be pure of one particle type). So their probabilities are superposed.

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"the s-wave electron is superimposed over the quarks". Is this a quantum a superposition or simply a classical statistical mix? – Diego Mazón Aug 6 '12 at 17:25
@drake S waves are quantum states in the solution of two body scattering, the l=0 state, which means 0 angular momentum quantum number. They have the characteristic that the probability to go through the center of the nucleon is non zero. – anna v Aug 6 '12 at 17:29
Thanks but I was asking about the system electron plus quarks. In other words, about the density matrix of the system, are there quantum correlations between electrons and quarks? – Diego Mazón Aug 6 '12 at 17:35
Quarks are not free, they are in groups of two (mesons) or three( nucleons). The scattering matrix formulation is equivalent to talking about density matrix elements of a quantum system, and demonstrates superposition as you are talking of an electromagnetic interaction. When QCD is considered in the density matrix formulation for electron proton scattering life is more complicated: example – anna v Aug 6 '12 at 17:45

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