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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.


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.


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?


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?


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.

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  • $\begingroup$ Nice Dean: Regarding your latest "Edit3", may I suggest that the state in question is rather $$\frac{\mid ddd \rangle + \mid sss \rangle + \mid bbb \rangle}{\sqrt{3}},$$ since down, strange, and beauty quarks all have equal charge (while the charge of charm quarks differs). $$ $$ Also, your choice of the symbol $\mid Z \rangle$ for denoting this state may not be optimal (I'd suggest $\mid M \rangle$ instead). $\endgroup$ – user12262 Jun 9 '16 at 17:46
  • $\begingroup$ @user12262 thanks a lot for pointing out my oversight in the quarks in the final state. I corrected it. $\endgroup$ – Mario Krenn Jun 9 '16 at 21:09
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    $\begingroup$ 1. Please do not let posts look like revision histories. 2. More importantly, do not continually change the question. The answerers addressed your specific concerns about some specific states. You editing in to ask about a different state is a different question, and should be asked as such. You can link back to this question for context, but after answers are given, you should not edit the question in a way that invalidates them. $\endgroup$ – ACuriousMind Jun 9 '16 at 21:15
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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.

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  • $\begingroup$ Thank you, this answer using decoherence is very nice. However I have a question about that: One can have electrons in superpositions at different locations (for example, the double-slit experiments with single electrons). Wouldn't your argument forbid superpositions (thus double-slit experiments) with charged particles? Maybe I'm just missing something. Thanks! $\endgroup$ – Mario Krenn May 6 '16 at 16:12
  • $\begingroup$ Good point! The difference is the following: if you have a superposition of two particles with different charge, then the weakest of background EM fields (think 'quantum foam') can collapse the superposition (by the above argument). However, if you have a superposition of two particles of the same charge, but separated by a distance $L$, then to collapse them using EM fields you need a photon that is localized up to order $L$ (otherwise the photon cannot 'feel' the difference), or by the uncertainty principle have momentum of the order $1/L$, and hence the necessary EM energy scales as $1/L$. $\endgroup$ – Ruben Verresen May 6 '16 at 16:36
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    $\begingroup$ Interesting try to solve this, but isn't your argument telling precisely the opposite? The further the electrons are apart, the longer the wavelength of the photon from vacuum fluctuation should be, right? But longer wavelength means less energy. So for destroying a superposition of two electrons far apart, one would need lower-energy photons than destroying superpositions of particles confined in a baryon Maybe again i'm missing something. Thanks, this is all very interesting. $\endgroup$ – Mario Krenn May 7 '16 at 7:15
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    $\begingroup$ Why doesn't this argument apply to quasiparticles with charge ±1/3? $\endgroup$ – Peter Shor Jun 9 '16 at 14:30
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    $\begingroup$ A quasiparticle is a superposition of different quantum states, each of which has some distribution of integer charges. So how can it act like it has $±1/3$ charge if stray electromagnetic fields are going make different spatial distributions of integer charges decohere? $\endgroup$ – Peter Shor Jun 9 '16 at 19:56
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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.

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    $\begingroup$ Does this argument hold water? There are anyons which behave like they have an electric charge of 1/3, and they're made out of superpositions of things with integer charge. See this talk or google them. So why doesn't the quantum vacuum decohere them? $\endgroup$ – Peter Shor Jun 9 '16 at 14:28
  • $\begingroup$ Yes, I tried to contrast this use of "superposition" in my parenthetical statement. The collective effects you are talking about involve Large numbers of particles, so hugely large tensor products of many-many-many electrons... the whole enchillada of QFT. My simple point is about just one electron and one positron that will never, by themselves, turn into each other. The Dirac sea, on the other hand is a hugely coordinated system. $\endgroup$ – Cosmas Zachos Jun 9 '16 at 15:08
  • $\begingroup$ Another way to say this is that Electric charge is not anomalous: the effect of the Dirac sea on it will never violate its conservation--EM is vectorlike anyway, so you cannot have a quantum violation of charge. Using the term "superposition" for the anyon collective excitations in Paul Fendley's talk takes the barking to a very different tree indeed! $\endgroup$ – Cosmas Zachos Jun 9 '16 at 16:46
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    $\begingroup$ In principle it could, if only you supplant the ccc with ddd, as ccc still violates charge... you want a charge -1 state. But the huge disparity of masses of the components would make its coherence problematic: Kaon components are basically degenerate, and neutrinos close in mass... I don't know what the oscillation length of this state would be. $\endgroup$ – Cosmas Zachos Jun 9 '16 at 18:05
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    $\begingroup$ OK, taking your conceit at face value and pretending your ddd component is close to a Δ, 1.2 GeV, the sss an Ω-, 1.7 GeV, and your bbb a freak fantasy state of 14GeV, and assuming Z is moving at PeV energies, so the neutrino oscillation formula "works" (!) the oscillation length still comes out a few thousand fermis! At distances beyond that you basically get a mixture. But, hey! this is a formal wisecrack exercise.... $\endgroup$ – Cosmas Zachos Jun 9 '16 at 19:15
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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 $\:\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}$.

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    $\begingroup$ Thank you, I see now the problem. The final thing I would be interested in: Is there a fundamental reason why one can not have a superposition of different states of charge or strangeness or bottomness? I'm asking, because in atomic physics, an e- can be in different eigenstates of the principal q.number or of angular momentum q.number; photons can be a superposition of different spin- and angular momentum eigenvalues. Particles can be in superpositions of different mass & energy. So it is not completly clear to me why this should not happen to charge aswell? $\endgroup$ – Mario Krenn May 6 '16 at 8:23
  • $\begingroup$ But already, as I pointed out in my answer, your $\:|X\rangle\:$ and $\:|Y\rangle\:$ are superpositions of different states of charge. Simply they don't correspond to particles. $\endgroup$ – Frobenius May 6 '16 at 8:30
  • $\begingroup$ Yes, I understand you argument. But now I want to know whether there is a good reason why different states of charge can not be in a superposition. It is not obvious for me, in particular because of my examples mentioned above. $\endgroup$ – Mario Krenn May 6 '16 at 8:51

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