Deuteron is p-n, so naively should have zero electric quadrupole moment. However, experimentally it turns out quite large: $0.2859\ e\cdot fm^2$ 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? 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 - 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 (slides) 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)?

Could we distinguish experimentally dynamic (angular momentum) from static (e.g. shift of quarks) explanation of deuteron's quadrupole moment?

Update: The $l=2$, $m=0$ spherical harmonic used for the explanation: enter image description here

Update 2: To avoid bilocation claim for proton, quadrupoles are naturally obtained by shift of charges (like quarks). Here is suggestion from soliton particle model: structure of baryon requires some positive charge (e.g. +2/3), which needs to be compensated in neutron (explaining higher mass than proton), in deuteron this requirement is obtained by partial shift of charge from proton to neutron: enter image description here

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Jun 25, 2020 at 16:24

2 Answers 2

  1. First of all, regarding quarks and gluons: Of course, we should be able to understand the quadrupole moment of the deuteron starting from QCD, but this is complicated (or has the be done numerically, using lattice QCD). However, there is a large separation of scales, given by the fact that the neutron and proton are heavy, and the deuteron binding energy is small, $B\ll m_Nc^2$. This means that an effective theory of non-relativistic neutrons and protons interacting via short-range potentials is approriate.

  2. The deuteron has isospin zero, $I=0$, and total angular momentum $J=1$. Just based on symmetries, it can have a quadrupole moment. This follows from the Wigner-Eckardt theorem: A rank two spherical tensor $Q_{2m}$ can have an expectation value in a spin one state. The quadrupole moment can be defined as $Q=\langle J=M=1|Q_{20}|J=M=1\rangle$.

  3. An anti-symmetric two-nucleon wave function with the quantum numbers of deuteron can be constructed from a spin one, orbital angular momentum zero wave function. This wave function is spherically symmetric and has a vanishing quadrupole moment.

  4. However, there is nothing that prevents an $s$-wave ($L=0$) spin-one wave function with total angular momentum $J=1$ to mix with a $d$-wave ($L=2$) wave function with $S=1$ coupled to $J=1$. This wave function is proportional to the spherical harmonic $Y_{2m}$, and clearly has a non-vanishing quadrupole moment.

  5. In non-relativistic physics such a coupling naturally appears from the "tensor force" (or "spin-orbit coupling") potential $$ V = \left[3(\vec\sigma_1\cdot\hat{r})(\vec\sigma_2\cdot\hat{r}) -(\vec\sigma_1\cdot\vec\sigma_2)\right] (\vec\tau_1\cdot\vec\tau_2) V_T(r) $$ where $\sigma_i$ and $\tau_i$ are Pauli spin/isospin matrices acting on particle 1,2. In nuclear physics a tensor force is generated by one-pion exchange, because the basic pion-nucleon vertex is ${\cal L}\sim (N^\dagger \tau^a\sigma^i N) \nabla^i\pi^a$.

  6. Standard nuclear physics text books explain that the deuteron quadrupole moment can be quantitatively understood from the known pion-nucleon coupling.

  7. Note that this has nothing to do with three-body forces, the lightest system sensitive to three-body forces is the triton.

  8. Can we distinguish this explanation from an explanation of "quark wave functions" being deformed? Not really, if the nucleon-nucleon wave function has a $d$-wave component, then this obviously implies that the quark wave functions are modified, too.

PS.1.) Can we "prove" the explanation of $Q$ in terms of the $d$-state admixture? Yes, in the following sense: We can measure $sd$-wave mixing in elastic nucleon-nucleon scattering. This observable in unambiguously non-zero. I can use the measured $sd$-mixing parameter to fix the tensor interaction, and then compute the quadrupole moment of the deuteron. This works, see here and here.

P.S.2) One might hope that there is an even more direct check, and that one can measure the $d$-state probability directly. For example, study polarized deuteron photo-disintegration and measure the fraction of the deuteron spin that is carried by the $n,p$ in the final state. This is not quite right, in particular the $d$-state probability is not an observable, see here.


The deuteron is an isospin singlet (if it were not, one would expect $|nn\rangle$ and maybe $|pp\rangle$ to be bound states) with isospin wave function:

$$ |I=0, I_3=0\rangle = \frac 1 {\sqrt 2}[|pn\rangle-|np\rangle]$$

so the nucleons aren't in isospin eigenstates (that is: they're each already a mixture of neutron and proton, a la Schrodinger's cat). Moreover, they don't oscillate between $n$ and $p$ (with a radiating $\pi^+$ current) because they are in a stationary state.

With the isospin antisymmetric, the spin state is symmetric ($S=1$), which combines with an even spatial wave function to make an overall antisymmetric wave function per the Pauli exclusion principle.

The allowed spatial wave functions are the even order spherical harmonics satisfying $|l-1| \le 1 < l+1$ which is $l=0$ and $l=2$.

Though $S$-wave ($l=0$) is the lowest energy, there can be $D$-wave admixture ($l=2$). For a tensor polarized deuteron ($J_z=1$), the $S$-wave combination in the $|l, m, s_z\rangle$ basis is:

$$ |J=1, J_z=1\rangle = |0,0,1\rangle$$

while the $D$-wave is:

$$ |J=1, J_z=1\rangle = \sqrt{\frac 3{10}}|2,1,0\rangle + \sqrt{\frac 1{10}}|2,0,1\rangle + \sqrt{\frac 3{5}}|2,2,-1\rangle$$

which induces an electric quadruple moment via

$$ Y_2^0 \propto \frac{2z^2-x^2-y^2}{r^2}$$ $$ Y_2^1 \propto \frac{(x-iy)z}{r^2}$$ $$ Y_2^2 \propto \frac{(x-iy)^2}{r^2}$$

Polarized deuterium targets and/or deuteron beams can be used to understand both the nuclear and sub-nuclear (quark) physics of the system (https://iopscience.iop.org/article/10.1088/1742-6596/543/1/012008/pdf, https://www.jlab.org/conferences/qcd-frontier-2013/talks/tuesday/DeutTensSSF_QCDFrontier2013.pdf)

  • 1
    $\begingroup$ Thanks for bringing up DIS and the $b_1$ structure function of the deuteron. For non-insiders: I think this is the simplest object that one can measure in deep inelastic scattering (a microscopic probe, sensitive to the distribution of quarks), that is directly affected by the fact that the deuteron is not just two uncorrelated nucleons. $\endgroup$
    – Thomas
    Jun 23, 2020 at 15:54
  • $\begingroup$ Having said this, of course we already know that the deuteron is not just two non-interacting nucleons. So how does the $b_1$ structure function shed light on the origin of the deuteron quadrupole moment? $\endgroup$
    – Thomas
    Jun 23, 2020 at 15:56
  • $\begingroup$ In othet words: If some eventually measures a small but non-zero $b_1$, how would one use that information to decompose the quadrupole moment into a "nuclear" and a "quark" component? I don't think this is even a well defined question. $\endgroup$
    – Thomas
    Jun 23, 2020 at 15:57
  • $\begingroup$ I had never heard of $b_1$, back when I was in the game, it was $t_{20}$ that ppl wanted to measure with polarized targets. As far as PSE is concerned, I think the deuteron provides good irl pedagogy for understanding the Pauli exclusion principle, isospin, tensor forces, superposition of states, etc... $\endgroup$
    – JEB
    Jun 23, 2020 at 17:51
  • $\begingroup$ Thanks, so I have added above the l=2, m=2 spherical harmonic used for explanation - so we should imagine that proton is in both external parts at once to get the quadrupole? We could get any multipole with such magical trick - why should we limit to quadupole? $\endgroup$
    – Jarek Duda
    Jun 24, 2020 at 9:58

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