non-relativistic limit with our choice of longitudinal basis functions.

The first excited state (1P) is identified with $n=0, m=\pm1, l=0$ or $n=0, m=0, l=1$, noting that for heavy quarkonium, the term $\kappa^4/(m_q+m_{\bar q})^2 l(l+1)$ is small comparing to the remaining terms. There are four 1P states: $\chi_0$ ($0^{++}$), $\chi_1$ ($1^{++}$), $\chi_2$ ($2^{++}$) and $h$ ($1^{+-}$). Let us focus on $h$ and restrict the discussion to $m_j=0$. From Eq. (\[eqn:C\]), we conclude: $-1 = \mathsf C = (-1)^{m+l+1}(-1)^{s+1}$, where $s$ is the total spin, viz $s=0$ for singlet and $s=1$ for triplet. Apparently, for both sets of quantum numbers ($m=\pm1, l=0$ or $m=0, l=1$), $s=0$. From Eq. (\[eqn:LFP\]), $-1 = (-1)^j\mathsf P =
(-1)^{m}(-1)^{s+1}$, implying $m=0$. Therefore, the correct quantum numbers for $h$ meson ($1^{+-}$) are $n=0, m=0, l=1$ with a singlet spin configuration, which is consistent with the non-relativistic quantum number assignment $1{\,}^{1}\!P_0$. Note that the orbital motion is excited through the longitudinal direction but not the transverse direction. This cannot be obtained from IMA[^5].

Numerical Results {#sect:numerical_results}
=================

[ccc ccc ccc c]{}

& $N_f$ & $\alpha_s(0)$ & $\mu_\text{g}$ (GeV) & $\kappa$ (GeV) & $m_q$ (GeV) & rms (MeV) & $\overline{\delta_jM}$ (MeV) & $N_\text{exp}$ & $N_{\max}=L_{\max}$\

& & 0.6 & 