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From the hyperfine calculation we find that the mass of the vector mesons are heavier than pseudoscalar meson, as the following proves.

The magnetic moment is proportional to the spin and inversely proportional to the mass. The hyperfine interaction correction is proportional to the dot product of the two moments, $$ \Delta E_{i j}=k \frac{\vec{S_i} \cdot \vec{S_j}}{m_i m_j}, $$ so the mass of the meson is $$ \begin{gathered} m=m_i+m_j+\Delta E_{i j} \\ S^2=\left(\vec{S_i}+\vec{S_j}\right)^2=S_i^2+S_j^2+2 \vec{S_i} \cdot \vec{S_j}. \end{gathered} $$

So we predict that vector mesons are heavier than pseudoscalar mesons.

However, strong forces favour the aligned spins, and it’s said here that the binding energy is larger for aligned spins.

Does the strong force increase or decrease with aligned spins?

Since vector mesons have aligned spins, they should have higher binding energy, hence smaller mass. How is that consistent with the hyperfine correction?

Cosmas Zachos
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L L
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    Vector mesons, like the ρ, are lighter than pseudo vector ones, like the a1. Vectors are spin triplets, and pseudovectors spin singlets. – Cosmas Zachos Jun 04 '23 at 01:32
  • " How is that consistent with the hyperfine correction?" googling, I can find no recent papers with hyperfine corrections to QCD calculations. Do you have a recent paper? – anna v Jun 04 '23 at 04:15
  • @CosmasZachos what is a1? $\rho$ is heavier than $\pi$. In the light quark model, the vector mesons are heavier than their corresponding pseudovector mesons. – L L Jun 04 '23 at 10:26
  • @annav The hyperfine structure consideration is shown above. It’s from our lecture notes and is used to explain that pseudovector mesons are lighter than the vector mesons – L L Jun 04 '23 at 10:28
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    $\pi$ is not a pseudovector but rather is a pseudoscalar. – Martino Jun 04 '23 at 10:36
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    Please check your PDG. Do you mean pseudovectors like the a1, or pseudoscalars, like the π? The latter is light as a pseudoGoldstone boson… – Cosmas Zachos Jun 04 '23 at 11:07
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    Unlike the ρ, the a1 has L=1. It gets its J=1 from it, and not S, so it’s heavier. – Cosmas Zachos Jun 04 '23 at 11:13
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    @CosmasZachos Sorry, i wanted to compare vector meson and pseudoscalor mesons. I’ve corrected my question – L L Jun 04 '23 at 16:26

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Let me review for you the relevant neutral mesons consisting of the two lightest quarks you seem to be confused about the name/labels of; I'm sure your instructor has listed them for you. Writing them in spectroscopic notation might help you.

In increasing mass (MeV) order, and $J^P$, L, S, they are:

state symmetry $J^P$ $L$ $S$
$\pi$ (135) pseudoscalar $0^-$ $0$ $0$
$\rho$ (770) vector $1^-$ $0$ $1$
$a_0$ (980) scalar $0^+$ $1$ $1$
$a_1$ (1260) pseudo vector $1^+$ $1$ $0$

They might help you compare and contrast keeping L or S invariant, and increasing the other.

  • The pseudoscalars are meant to defy all quark model systematics, as they are the effective pseudogoldstons of the dynamically (spontaneous) breaking of chiral symmetry of QCD. Their masses are described by Dashen/GOR formulas instead!

Trying to fit them by any and all deprecated constituent quark model mechanisms, including obsolete failed ones such as this, is bound to unleash agita and umbrage among mainstream particle physicists today...

PS As an aside, your Deuteron analogy is quite poor. What actually lowers the energy of the spin triplet ortho-Deuteron is the statistics which allows L=0 , instead of L=1 for the spin singlet para-Deuteron! This is what lowers the energy of the stable state, and not the hyperfine interaction. For deprecated quark models, you could use such vague arguments to see why vector mesons are lighter than pseudovectors.

rob
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Cosmas Zachos
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  • I’ve corrected my question. Would you mind comparing pseudoscalor and vector mesons? In my lecture notes, they compare them by using the hyperfine structure but it seems to violate the fact the nuclear force favours aligned spins – L L Jun 04 '23 at 16:30
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    Your lecture notes are not sound. Pseudoscalars are light by virtue of Goldstone's theorem and not suspect nuclear models! – Cosmas Zachos Jun 04 '23 at 17:34