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.

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.

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). — user12262, Jun 09 '16 at 17:46
@user12262 thanks a lot for pointing out my oversight in the quarks in the final state. I corrected it. — Mario Krenn, Jun 09 '16 at 21:09

Ruben Verresen · Answer 1 · 2016-05-06T12:59:10.237

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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.

edited May 06 '16 at 12:59

answered May 06 '16 at 12:07

Ruben Verresen

8,811

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! – Mario Krenn May 06 '16 at 16:12
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 $1/L$. – Ruben Verresen May 06 '16 at 16:36
(So in other words you can allow for the superposition of an electron between different positions if you can shield the area from photons with more energy than $1/L$. This can indeed be done. In the other case, with for example the superposition of the proton and the neutron, even the weakest photon will collapse the state, so even vacuum fluctuations should do the job.) – Ruben Verresen May 06 '16 at 19:08
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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. – Mario Krenn May 07 '16 at 07:15
It is true that the argument implies that the farther the electrons are apart, the easier the superposition decoheres. Still for any fixed distance, there is a minimum energy that a photon needs to have to collapse it. This is different for the superposition of two particles of different charge: distance doesn't enter it at all, and a superposition of a proton and neutron decoheres no matter how low the energy of the photon is, even if the photon and neutron are on the same spot! – Ruben Verresen May 07 '16 at 14:23
Remember what I wrote above in my main post: $\boxed{(|p^+\rangle + |n^0\rangle) \otimes |\gamma\rangle} \to \boxed{|p^+\rangle \otimes |\textrm{scattered } \gamma\rangle + |n^0\rangle \otimes |\gamma\rangle}$. You see this equation is independent of the distance between the proton and the neutron! This is different for a superposition of two electrons, e.g. if we start with $(|e^-,x=0 \rangle + |e^-,x=L \rangle) \otimes |\gamma, k=0\rangle$ where the photon has approximately zero momentum, then there will be no decoherence because it can't distinguish $ x=0$ from $x=L$. – Ruben Verresen May 07 '16 at 14:26
As you see, i have added a bounty to this question. I somehow doubt that your argument is complete at the moment (because of the counterintuitive conclusion, see my comment above); and in particular because decoherence (actually dephasing, as there is no fundamental decoherence involved) behaves differently usually. I would like to see such an argument in a published/trustworthy source. Do you know of any? – Mario Krenn Jun 05 '16 at 07:55
And your answer does not explain why Kaons exist, but the |X> and |Y> not. – Mario Krenn Jun 05 '16 at 07:56
I think the point about Kaons is interesting. I don't really see the analogy of strangeness with electric charge though: electric charge quantifies the coupling to the electromagnetic field, which was crucial to my argument. Strangeness doesn't play a similar role. Nevertheless, the question remains: can you have particles that do not have a well-defined 'weak' or 'strong' charge? My argument still goes through to show they are not stable, but there is a subtlety: if you have a superposition of particles of different electric charge (or mass) then they will collapse immediately [cont ...] – Ruben Verresen Jun 05 '16 at 13:59
This is because you can have an arbitrarily weak electromagnetic (or gravitational) field. It is different for the weak and strong interaction, since the carriers have a finite mass. Hence I would think that if you were to create a state that is in a superposition of particles with different weak charge, then it could be stable as long as you keep away all background radiation above a certain energy threshold. (This is similar to my answer to your question about how you can have a stable superposition of an electron on two different positions: to collapse it requires a minimum energy.) – Ruben Verresen Jun 05 '16 at 14:06
Anyway the main point of my argument was to show why you cannot have a superposition of different electric charge or mass. My last two comments were about the similarities and differences when trying to extend it to the weak and strong charges, but I am less certain about those cases. However it seems you still disagree with me concerning the electric charge (or mass) superposition? I don't see the problem, I feel I have addressed your objection regarding the superposition of an electron at different positions. If there is a remaining confusion, can you be more explicit about what it is? – Ruben Verresen Jun 05 '16 at 14:08
Thanks again. I still think that there is a problem, namely that superpositions that are further away need longer wavelength - that does not make sense to me. And the whole idea about your "decoherence" from vacuum fluctuations - i have never heared anything like that; unfortuantly I'm no expert therefore: For the bounty I have therefore explicitly asked for a source. Do you know one? – Mario Krenn Jun 05 '16 at 18:09
Just to be clear, for superpositions of different charge, distance is irrelevant. But indeed for superpositions of the same charge but at different positions, then for a given distance you need at least a certain energy $\sim 1/L$ to collapse the state. It's because lower energy photons (i.e. longer wavelength) can't tell the two particles apart and hence cannot cause collapse. But yeah I have no resources for these statements, I just wrote down what made sense to me. It would be interesting to see what references there are to explains these things, so I hope something nice comes from it. – Ruben Verresen Jun 05 '16 at 20:22
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Why doesn't this argument apply to quasiparticles with charge ±1/3? – Peter Shor Jun 09 '16 at 14:30
Do you mean the quasiparticles themselves are on a microscopic level superpositions of different charges? I'm not very familiar with the FQHE, but at least in the SSH model, with fractional e/2 quasi-particles, I would say that such a quasi-particle is not a superposition of different electric charges (which can be traced back to the fact that we cannot create a quasi-particle in isolation). But I guess I am misunderstanding your point? – Ruben Verresen Jun 09 '16 at 17:19
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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? – Peter Shor Jun 09 '16 at 19:56
I fear there is a language trap here: in this question "superposition" is a synonym for "linear combination of w.f.s". The quasiparticle "superpositions" you are bringing in are collective multiparticle states, definitely not linear combinations--if you represented in Fock space they'd tensor an infinity of elementary excitations. I fear nobody else on this page has broached collective multiparticle excitations. – Cosmas Zachos Jun 09 '16 at 21:38
@RubenVerresen I asked a new question on what we were discussing here, in case you are interested in: http://physics.stackexchange.com/questions/262049/what-is-the-influence-of-qed-vacuum-in-electron-double-slit-experiments?lq=1 – Mario Krenn Jun 12 '16 at 09:22

Cosmas Zachos · Answer 2 · 2016-06-10T14:10:57.210

4

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.

edited Jun 10 '16 at 14:10

answered Jun 09 '16 at 00:25

Cosmas Zachos

62,595

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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? – Peter Shor Jun 09 '16 at 14:28
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. – Cosmas Zachos Jun 09 '16 at 15:08
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! – Cosmas Zachos Jun 09 '16 at 16:46
Thanks Cosmas this is interesting. I have to think more about it. in the meantime, in order to see whether I understood it correctly, I edited my post (edit3) to add the state $|Z>$. According to your argument, such a state could exist, right? – Mario Krenn Jun 09 '16 at 17:01
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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. – Cosmas Zachos Jun 09 '16 at 18:05
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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.... – Cosmas Zachos Jun 09 '16 at 19:15
Thank you Cosmas, I understand the difference better now. The conservation of electric charge in any case, in contrast to strangeness quasi-conservation is the key. Is the reason that photons are massless (thus vacuum can create virtual photons basically all the time), and strangeness couples to weak-interaction (and Z and W bosons are massive thus unlikely to be produced)? About your last paragraph, the decoherence reasoning - same as for Rubens answer: Wouldn't that say that electron double slit experiments should not exist, because the background field measures the electrons all the time? – Mario Krenn Jun 11 '16 at 19:26
Didn’t get the point on massless photons… Both electric and weak charges are conserved, but Strangeness is not… it is not as though it gets an exemption because the weak bosons are massive--the vacuum is full of them, as well. But S is only part of the weak Y, and the d quark has the same Y as the s, so strangeness is still violated. As for decoherence in the double slit experiment, it is minuscule, and they take special pains to avoid it. But if an EM interaction records the welcher Weg values of the electron, yes, it has then decohered, as it has been "observed". – Cosmas Zachos Jun 11 '16 at 22:00
Weak hypercharge. – Cosmas Zachos Jun 11 '16 at 22:11
@CosmasZachos your answer motivated me to ask a separate question, but similar to this one; i would be interested in your thoughts if you are interested: http://physics.stackexchange.com/questions/262052/can-coherent-superpositions-of-a-neutron-and-antineutron-exist?lq=1 – Mario Krenn Jun 12 '16 at 09:24
@CosmasZachos actually, it resulted also in a second question motivated by your answer. This one is I find very exciting, understanding the decoherence mechanism you describe: http://physics.stackexchange.com/questions/262049/what-is-the-influence-of-qed-vacuum-in-electron-double-slit-experiments?lq=1 – Mario Krenn Jun 12 '16 at 09:26

score 1 · Answer 3 · answered May 06 '16 at 05:53

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 $\:\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}$.

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? — Mario Krenn, May 06 '16 at 08:23
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. — Frobenius, May 06 '16 at 08:30
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. — Mario Krenn, May 06 '16 at 08:51

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

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