Can superpositions of baryons with different electric charge and strangeness exist?

Question

I am trying to find out whether the following baryons can exist:

$$ |X\rangle = \frac{|u u u\rangle + |d d d\rangle + |s s s\rangle}{\sqrt{3}} $$ $$ |Y\rangle = \frac{|u u u\rangle + |d d d\rangle - 2|s s s\rangle}{\sqrt{6}} $$

I haven't found in any baryon-list such a quark-configuration, but I don't know of any reason why it shouldn't exist either.

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The question is motivated by the $\eta$-Mesons which have a quark representation in the following way: $$ |\eta'\rangle = \frac{|u \bar{u}\rangle + |d \bar{d}\rangle + |s \bar{s}\rangle}{\sqrt{3}} $$ $$ |\eta\rangle = \frac{|u \bar{u}\rangle + |d \bar{d}\rangle - 2|s \bar{s}\rangle}{\sqrt{6}} $$

However, of course each of their terms in the superposition has the same electric charge and strangeness content.

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Edit: Frobenius pointed out that my $|X\rangle$ and $|Y\rangle$ are superpositions of states with different electric charge and strangeness. This is a very good point. However it is not completly clear to me why such a superposition should not exist, given that there are many examples of states that exist in superpositions of different properties/quantum numbers.

For example, in atom physics, electrons can exist in superposision of different angular momentum quantum numbers; in particle physics states can exist in superposition of different masses (for example $|\eta\rangle$), in quantum optics photons can be in superpositions of different energies (frequencies). What makes electric charge and strangeness special?

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Edit2: Cosmas Zachos pointed out that there exist particles without a well-defined strangeness, namely Kaons (more precisely $K_0^S$ and $K_0^L$). Why shouln't baryons without well-defined charge exist?

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Edit3: Cosmas Zachos explains that electric charge conservation is general, in contrast to strangeness conservation. That makes me wonder, does such a state exist:

$$ |Z\rangle = \frac{|d d d\rangle + |s s s\rangle + |b b b\rangle}{\sqrt{3}} $$ (where $d$ is down, $s$ is strange and $b$ is beaty-quark) Which has an electric charge of C=-1e.

Answer 1

It's a good question, and the answer is surprisingly simple and physical. There is indeed no fundamental objection to having a superposition of particles of different charge. But it turns out this is not stable. This is basically due to wavefunction collapse, or more sophisticatedly, due to decoherence.

Imagine having a single particle in a superposition of being a proton and a neutron, $|\psi \rangle = |p^+\rangle + |n^0\rangle$. Now imagine a light wave (photon) $|\gamma \rangle$ passing by. So the initial state of the combined system is $\boxed{(|p^+\rangle + |n^0\rangle) \otimes |\gamma\rangle}$. This electromagnetic field will not notice the neutron, but it will interact with the proton and for example scatter a bit. So after some time we get the new state $\boxed{|p^+\rangle \otimes |\textrm{scattered } \gamma\rangle + |n^0\rangle \otimes |\gamma\rangle}$. But suppose we didn't know the initial state of the photon, then our effective description of the system is given by tracing out over the photon, but that effectively collapses our system into $|p^+\rangle$ or $|n^0\rangle$. In more physical terms, we can say the stray electromagnetic field measured the charge of the particle and hence collapsed it into a charge eigenstate (So you can reformulate the previous discussion purely in terms of measurements and wavefunction collapse if you prefer). You can imagine that in practice any particle will always be subjected to some background electromagnetic field and hence we should expect all particles to be in charge eigenstates.

EDIT: Well actually, there are also superselection rules, for example for charge, which comes down to the claim that you in fact cannot experimentally distinguish $|p^+\rangle + |n^0\rangle$ from the mixed state of $|p^+\rangle$ and $|n^0\rangle$. Depending on your viewpoint of quantum theory, that shows you shouldn't speak of superpositions of different charge. But I like the above reasoning I presented, cause it shows that even if you are willing to entertain superpositions $|p^+\rangle + |n^0\rangle$, then we expect it to rapidly decohere into charge eigenstates.

Answer 2

OK, since my name was taken in vain I suppose I am obliged to clarify my comment further. My invitation was to contrast charge oscillations to strangeness oscillations in the $K^0-\bar{K}^0$ system, not to use the latter to argue for the former. The finally mutated question I am addressing is “Why are there no charge oscillations and superpositions of differently charged states”? For simplicity take $e^+$ and $e^-$.

Let me first review why strangeness oscillations occur. The point is that strangeness (and CP) is not a completely conserved quantum number, i.e. “strangeness rotations” are a symmetry in the strong interactions, but not in the weak interactions. So the 2x2 hamiltonian of the $K^0-\bar{K^0}$ system is not diagonal: the celebrated doubly weak box diagram connects $K^0$ to $\bar{K^0}$, and provides small off-diagonal terms in that originally degenerate, diagonal hamiltonian. Diagonalizing it gives you the long and short propagation eigenstates, etc… and strangeness oscillations by 2. That is, $K^0$ and $\bar{K^0}$ interfere, because the hamiltonian connects them, and they can go to each other. All because S (and CP) is not a fully conserved charge. Neutrino flavor oscillation phenomena basically replicate this pattern in evident analogy.

By sharp contrast, which was the essence of my original comment, electric charge is fully conserved: it generates an exact symmetry of nature, and every hamiltonian. So now take $e^+$ and $e^-$. The Hamiltonian treats them identically, and, more importantly, because charge is absolutely conserved, there are no off diagonal terms: absolute charge conservation will never allow an $e^+$ to go to a $e^-$, and vice-versa. You may write a linear combination of these two, but since they will never interfere, you will never see any effects of quantum mechanics. They are really a mixture, not a state. You may write them as a state, abusively, but they are in two different superselection sectors, which any decent QM book covers, a direct sum $\oplus$. (If you had two particles together, an $e^+$ and an $e^-$, that would amount to a tensor product thereof, a 2-particle w.f., $|e^+\rangle\otimes|e^-\rangle$, not a superposition $|(e^+ + e^-)\rangle$.) There was nothing coherent to decohere, because these two guys will never confuse their identity with each other, propagate into each other, or otherwise know about each other. "Ships that pass in the night". A superposition of them, as I said, is a formal chimaera (a strictly mythological hybrid). When people write wave functions, they are usually interested in their interference.

Of course, as commented early on, any and all electric fields which couple to their opposite charges, will force them in opposite directions. Any interaction with a photon will radically change the identity of that chimerical state. So, if you could turn off any and all electromagnetic fields in the world, everywhere and forever, you might think you could make them interfere. No luck: there are virtual photons and $e^+$ - $e^-$ pairs filling up the QFT vacuum and which makes charge violating interference impossible and likewise single particle charge oscillations.

Answer 3

The electric charges of the states $\:|uuu⟩,|ddd⟩,|sss⟩\:$ are $\:+2,−1,−1\:$ respectively. More exactly these baryon states are the baryons $\:Δ++,Δ−,Ω−\:$. If $\:|X⟩\:$ or $\:|Y⟩\:$ would represent a baryon what would be the electric charge of this particle ? And electric charge is one of many quantum numbers. This problem is pointed out by @Cosmas Zachos in his comments.

The three baryon states $\:\Delta^{++}, \:\Delta^{-}, \: \Omega^{-}\:$ are members of the so-called Decuplet $\:\lbrace \Delta^{++},\Delta^{+},\Delta^{0},\Delta^{-},{\Sigma^{*}}^{^{\boldsymbol{+}}},{\Sigma^{*}}^{^{\boldsymbol{0}}},{\Sigma^{*}}^{^{\boldsymbol{-}}},{\Xi^{*}}^{^{\boldsymbol{0}}},{\Xi^{*}}^{^{\boldsymbol{-}}},\Omega^{-}\rbrace\:$. These ten baryon states are baryons and they consist a base for the 10-dimensional subspace $\:\boldsymbol{10}\:$ in the right hand side of equation $$ \boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}= \boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{10}\boldsymbol{\oplus} \boldsymbol{8}^{\boldsymbol{\prime}}\boldsymbol{\oplus}\boldsymbol{8} $$ Your $\:|X\rangle,\: |Y\rangle \:$ are of course baryons states in this 10-dimensional subspace. This subspace and every one of the other subspaces in the right hand side of above equation is invariant under $\:SU(3)\:$ applied to each factor $\:\boldsymbol{3}\:$ that is under $\:SU(3)\boldsymbol{\otimes}SU(3)\boldsymbol{\otimes}SU(3)\:$ applied to the product space $\:\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}$.