# How to understand observed electric quadrupole moment of deuteron?

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Posted (edited)

Deuteron is p-n, so naively should have zero electric quadrupole moment. However, experimentally it turns out quite large: 0.2859 e⋅fm2 from https://en.wikipedia.org/wiki/Deuterium#Magnetic_and_electric_multipoles

This Wikipedia article explains it by adding l=2 angular momentum states - should we imagine it as a hidden dynamics?

Maybe as oscillations between 'pn' and 'np' by some pi+ exchange? (but shouldn't it make it a linear antenna producing EM waves?)

To describe e.g. deuteron-proton scatterings they neglect quark structure, but require three-body force ( https://en.wikipedia.org/wiki/Three-body_force) - would including quarks into considerations allow to focus only on two-body forces?

But what happens with quarks when biding proton and neutron into deuteron? I am working on soliton particle model suggesting that there is a shift of charge from proton to neutron for binding of deuteron, like uud-udd slightly shifting quark u toward right, d toward left - is such explanation of quadrupole moment allowed (e.g. by QCD)?

Edited by Duda Jarek

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5 hours ago, Duda Jarek said:

I am working on soliton particle model suggesting that there is a shift of charge from proton to neutron for binding of deuteron, like uud-udd slightly shifting quark u toward right, d toward left - is such explanation of quadrupole moment allowed (e.g. by QCD)?

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So what happens with quarks when proton and neutron bind into deuteron?

How exactly deuteron gets the quadrupole moment? Saying that it's due to angular momentum means some dynamics - what kind of dynamics?

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A zero electric quadrupole moment implies a spherical nuclear charge distribution. Why is that the naive expectation for deuterium?

That number is not large compared to some other values on the list

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Posted (edited)

Naively we have proton-neutron with only proton charged - shouldn't it have electric multipoles as proton: zero electric dipole and quadruple moment?

Quadrupole moment grows with square of distance, to get 0.2859 e fm^2 for deuteron, e.g. spitting its charge into 1/2 - 1/2 e they would need to be in ~0.76 fm distance.

How could we get such charge distance for 'pn'? Thinking about it as 6 quarks it seems more doable (?)

ps. However, I have seen some papers claiming nontrivial structure of charge in neutron (?): positive core, negative shell, e.g.

Edited by Duda Jarek

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2 hours ago, Duda Jarek said:

Naively we have proton-neutron with only proton charged - shouldn't it have electric multipoles as proton: zero electric dipole and quadruple moment?

If it's not a spherical nucleus, no.  Filled shells represent spherical nuclei.

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Quadrupole moment grows with square of distance, to get 0.2859 e fm^2 for deuteron, e.g. spitting its charge into 1/2 - 1/2 e they would need to be in ~0.76 fm distance.

How could we get such charge distance for 'pn'?

The radius of a nucleon is about 0.8 fm, so I don't see a problem with that.

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Thinking about it as 6 quarks it seems more doable (?)

I would imagine you can get a better result, but I don't see a problem so far.

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ps. However, I have seen some papers claiming nontrivial structure of charge in neutron (?): positive core, negative shell, e.g.

Again, more complex models will likely give a more precise answer, but the naive model gives something fairly close to the correct value.

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The current explanation of this quadrupole moment using angular momentum is that single proton is simultaneously on both sides of neutron:

This probability cloud means that if being able to measure position of proton, we should get 1/2 - 1/2 statistics.

Doesn't it mean that there is some 'pn' - 'np' oscillation?

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3 hours ago, Duda Jarek said:

This probability cloud means that if being able to measure position of proton, we should get 1/2 - 1/2 statistics.

Doesn't it mean that there is some 'pn' - 'np' oscillation?

What would be the oscillation frequency? Why don’t you get radiation emission at that frequency?

There is no trajectory here. Why should this be treated as an oscillation? When we look at electron states, we say the electron is at all points in its orbital until measured. Not oscillating between them. Why should the proton be different?

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Exactly - this dynamical approach to explain qudrupole moment in 'pn' with angular momentum is quite controversial if thinking about it.

So the question is: why not consider static instead?

Seeing deuteron as (among others) 6 quarks, e.g. minimization of Coulomb energy says positive (fractional) charges should have tendency to shift far apart - as in quadrupole.

Can we distinct experimentally these two possibilities: angular momentum explanation from e.g. quark-level shift of charge from proton to neutron?

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On 6/26/2020 at 6:55 AM, Duda Jarek said:

Exactly - this dynamical approach to explain qudrupole moment in 'pn' with angular momentum is quite controversial if thinking about it.

The controversy would be assigning classical behavior to a quantum system, which is not confined to this problem.

The solution that puts it in the l=2 state ~5% of the time is a conclusion driven by experiment.

On 6/26/2020 at 6:55 AM, Duda Jarek said:

So the question is: why not consider static instead?

Seeing deuteron as (among others) 6 quarks, e.g. minimization of Coulomb energy says positive (fractional) charges should have tendency to shift far apart - as in quadrupole.

Can we distinct experimentally these two possibilities: angular momentum explanation from e.g. quark-level shift of charge from proton to neutron?

“Tendency” is not an absolute. The l=0 ground state of hydrogen has a probability distribution that has the electron overlapping the proton some fraction of the time.

Static, if applied to position, is not allowable in QM. There is no such thing.

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Ok, let us see deuteron as "|np> - |pn>" to get quadrupole moment.

Measuring proton's position, we would get pn or np ... but having electron in orbital, wouldn't it "measure" proton's position through EM interactions as there is slight difference between pn and np?

Ok, we can say we have superposition:" |pn with electron seeing pn> - |np with electron seeing np> ", but would such electron see quadrupole this way?

Here is some paper with deuteron based atoms, and they consider quadrupole: https://www.researchgate.net/profile/Nir_Nevo_Dinur/publication/263316174_Improved_estimates_of_the_nuclear_structure_corrections_in_mD/links/543ce7860cf2c432f7422a6b/Improved-estimates-of-the-nuclear-structure-corrections-in-mD.pdf

Regarding charge shift, imagining deuteron as 6 quarks/charges - calculating e.g. Schrodinger equation for 6 charges and averaging, you would also see repulsion in statistics - shifting away charges as in quadrupole.

So this is in fact question if quarks take part in nuclear binding, e.g. by some (statistical?) shifts to reduce energy.

Currently for e.g. deuteron-proton or neutron scattering, there are fitted ~40 parameters models including 3-body forces ( https://en.wikipedia.org/wiki/Three-body_force ) and neglecting quark structure - we could fit anything for modelling 3 nucleons with 3-body forces.

Why not try to go toward quarks to understand nuclear binding? Would we still need 3-body forces in this case?

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44 minutes ago, Duda Jarek said:

Ok, let us see deuteron as "|np> - |pn>" to get quadrupole moment.

Measuring proton's position, we would get pn or np ... but having electron in orbital, wouldn't it "measure" proton's position through EM interactions as there is slight difference between pn and np?

Why are they different? What splits the degeneracy?

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"|np> - |pn>"  means superposition of these two possibilities ... they are destroyed by measurement, interaction ... like electromagnetic with electron in orbital - not true?

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41 minutes ago, Duda Jarek said:

"|np> - |pn>"  means superposition of these two possibilities ... they are destroyed by measurement, interaction ... like electromagnetic with electron in orbital - not true?

You are claiming they are not degenerate and are distinguishable. I am asking for justification for your claims.

This treatment makes no such distinction

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Posted (edited)

To get quadrupole, proton in these two cases ('pn' and 'np') needs to be in a bit different places (shifted by ~fm).

Electron in orbital interacts with proton e.g. through Coulomb, for which there is a tiny difference between these two cases - they are distinguishable for electron in orbital. Not true?

You can says that we have superposition " |pn with electron seeing pn> - |np with electron seeing np> ", but in this case such electron doesn't see quadrupole - which seems required in deuteron-based atoms.

Edited by Duda Jarek

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1 hour ago, Duda Jarek said:

To get quadrupole, proton in these two cases ('pn' and 'np') needs to be in a bit different places (shifted by ~fm).

Or you flip the deuteron around. Why does it matter? The solution to the electron orbitals is 3D, so it already accounts for this.

1 hour ago, Duda Jarek said:

Electron in orbital interacts with proton e.g. through Coulomb, for which there is a tiny difference between these two cases - they are distinguishable for electron in orbital. Not true?

The electron does not follow a trajectory. You seem to be thinking about this in terms of classical physics, which won’t work. Are you allergic to applying quantum mechanics to the problem?

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The problem with getting quadrupole is that it requires at least two charges in a distance, while (ignoring quarks) we have only one - here it is explained by "proton bilocation", while it seems nobody considers (believes?) in quarks in nuclear physics level (? why?).

I didn't say anything about electron's trajectory, |pn with electron seeing pn> can be treated e.g. with Schrodinger.

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1 hour ago, Duda Jarek said:

The problem with getting quadrupole is that it requires at least two charges in a distance, while (ignoring quarks) we have only one - here it is explained by "proton bilocation", while it seems nobody considers (believes?) in quarks in nuclear physics level (? why?).

The proton is not a point particle

The original treatment of the deuteron’s quadrupole moment happened a few decades before quarks were proposed. Nobody believed in quarks at that time.

1 hour ago, Duda Jarek said:

I didn't say anything about electron's trajectory, |pn with electron seeing pn> can be treated e.g. with Schrodinger.

Go ahead and do it then. Show a different result for the two cases.

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So what happens with quarks in nuclei?

Schrodinger equation for shifted proton gives shifted orbitals ... without need of quadrupole moment, which is needed for deuterium.

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3 hours ago, Duda Jarek said:

So what happens with quarks in nuclei?

You’d have to find papers that analyze this.

3 hours ago, Duda Jarek said:

Schrodinger equation for shifted proton gives shifted orbitals ...

Please stop doing this. You haven’t shown this to be true. Stop stating it as if you have.

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