Polarized Dissociation and Spin Alignment of Moving Quarkonium in Quark-Gluon Plasma

We shall first consider contribution to dissociation rate from chromomagnetic coupling for a static quarkonium bound state in QGP and then generalize to a moving quarkonium bound state. We start with one-loop diagrams depicted in Fig. 2.

They give rise to the following contribution to color singlet self-energy

		$\displaystyle{\Sigma}_{ar}(P)=\frac{T_{F}}{N_{c}}(N_{c}^{2}-1)\int_{Q}D_{rr}^{mn}(Q)O_{ra}(P+Q)\langle S|{\mu}_{i}|O\rangle\langle O|{\mu}_{j}|S\rangle{\epsilon}^{ikm}q_{k}{\epsilon}^{jln}(-q_{l}),$		(11)
		$\displaystyle{\Sigma}_{ar}(P)=\frac{T_{F}}{N_{c}}(N_{c}^{2}-1)\int_{Q}D_{ar}^{mn}(Q)O_{rr}(P+Q)\langle S|{\mu}_{i}|O\rangle\langle O|{\mu}_{j}|S\rangle{\epsilon}^{ikm}q_{k}{\epsilon}^{jln}(-q_{l}),$

where $\int_{Q}\equiv\int\frac{d^{4}Q}{(2{\pi})^{4}}$. The color indices have been suppressed, with a factor of $(N_{c}^{2}-1)$ inserted from the color sum. $O$ and $D$ denote propagators for octet and gluons respectively. Note that since we take quarkonium as a probe, $O_{rr}$ is medium independent, see (2). It is clear then the second line corresponds to a medium independent contribution to dissociation rate, which we do not consider. The dissociation rate from the first line is given by

	$\displaystyle{\Gamma}$	$\displaystyle=-2\text{Re}[{\Sigma}_{ar}(P)]=2\frac{T_{F}}{N_{c}}(N_{c}^{2}-1)\int_{Q}D_{rr}^{mn}(Q)\text{Re}[O_{ra}(P+Q)]\langle S|{\mu}_{i}|O\rangle\langle O|{\mu}_{j}|S\rangle{\epsilon}^{ikm}q_{k}{\epsilon}^{jln}q_{l},$
		$\displaystyle=2\frac{T_{F}}{N_{c}}(N_{c}^{2}-1)\int_{Q}D_{rr}^{mn}(Q)\pi{\delta}(p_{0}+q_{0}-h_{o}^{(0)})\langle S|{\mu}_{i}|O\rangle\langle O|{\mu}_{j}|S\rangle{\epsilon}^{ikm}q_{k}{\epsilon}^{jln}q_{l}.$		(12)

We have assumed that all other factors except $O_{ra}$ are real, which we now justify. The explicit form of $D_{rr}$ is given by

$\displaystyle D_{rr}^{mn}(Q)=P_{T}^{mn}(Q)2\pi{\epsilon}(q_{0}){\delta}(Q^{2})\left(\frac{1}{2}+f(q_{0})\right),$

(13)

with $P_{T}^{mn}={\delta}_{mn}-\hat{q}^{m}\hat{q}^{n}$ being the transverse projector. It is clearly real. We will also drop the medium independent factor $1/2$. The transition amplitude between the singlet and octet states split into orbital and spin parts as

$\displaystyle\langle S|{\mu}_{i}|O\rangle\langle O|{\mu}_{j}|S\rangle=\lvert\langle S,L=0|O,L=0\rangle\rvert^{2}\langle S=1|{\mu}_{i}|S=0\rangle\langle S=0|{\mu}_{j}|S=1\rangle.$

(14)

Clearly, the orbital part contributes a real factor $\lvert\langle S,L=0|O,L=0\rangle\rvert^{2}$. The spin part and the remaining factors are calculated as

		$\displaystyle\langle S=1|{\mu}_{i}|S=0\rangle\langle S=0|{\mu}_{j}|S=1\rangle{\epsilon}^{ikm}q_{k}{\epsilon}^{jln}q_{l}P_{T}^{mn}$
	$\displaystyle=$	$\displaystyle q^{2}\langle S=1|{\mu}_{i}|S=0\rangle\langle S=0|{\mu}_{j}|S=1\rangle({\delta}_{ij}-\hat{q}^{i}\hat{q}^{j}).$		(15)

For $0$-state in the triplet, we have from (7)

$\displaystyle q^{2}\langle S=1,S_{l}=0|{\mu}_{i}|S=0\rangle\langle S=0|{\mu}_{j}|S=1,S_{l}=0\rangle({\delta}_{ij}-\hat{q}^{i}\hat{q}^{j})=\frac{g_{s}^{2}}{m_{Q}^{2}}\left(q^{2}-(\bm{q}\cdot\bm{l})^{2}\right).$

(16)

The spin summed counterpart is given by

$\displaystyle\sum_{s=0,\pm 1}q^{2}\langle S=1,S_{l}=s|{\mu}_{i}|S=0\rangle\langle S=0|{\mu}_{j}|S=1,S_{l}=s\rangle({\delta}_{ij}-\hat{q}^{i}\hat{q}^{j})=\frac{g_{s}^{2}}{m_{Q}^{2}}2q^{2}.$

(17)

We see that the spin part is also real, which justifies taking the real part on $D_{ra}$ before. To proceed, we note that for on-shell singlet state, we have $p_{0}=h_{s}^{(0)}=2m_{Q}-{\epsilon}_{B}$. For the color octet state, we may decompose the relative motion by plane wave basis $h_{o}^{(0)}=2m_{Q}+\frac{p_{\text{rel}}^{2}}{m_{Q}}+\frac{q^{2}}{4m_{Q}}$, with $p_{\text{rel}}$ being relative momentum of the pair⁵⁵5This amounts to ignoring the final state interaction between the quark pair in the octet.. The last term is the kinetic energy from motion of the center of mass. Ignoring the last term for now, we obtain the spin summed LO dissociation rate from the chromomagnetic dipole interaction as

	$\displaystyle\sum_{s=0,\pm 1}{\Gamma}_{s}$	$\displaystyle=2\frac{g_{s}^{2}}{m_{Q}^{2}}\frac{T_{F}}{N_{c}}(N_{c}^{2}-1)\int_{Q}2\pi{\delta}(Q^{2})f(q_{0})\pi{\delta}(p_{0}+q_{0}-h_{o}^{(0)})\absolutevalue{\langle S,L=0|O,L=0\rangle}^{2}2q^{2}$
		$\displaystyle=2\frac{g_{s}^{2}}{m_{Q}^{2}}\frac{T_{F}}{N_{c}}(N_{c}^{2}-1)\int\frac{d^{3}q}{(2{\pi})^{3}2q}f(q){\pi}\frac{m_{Q}p_{rel}}{(2{\pi})^{2}}\absolutevalue{\langle S,L=0|p_{\text{rel}}\rangle}^{2}|_{p_{\text{rel}}=\sqrt{m_{Q}(q-{\epsilon}_{B})}}2q^{2}.$		(18)

The Dirac delta function gives $p_{\text{rel}}=\sqrt{(q_{0}-{\epsilon}_{B})m_{Q}}\simeq\sqrt{q_{0}m_{Q}}$. Since $q_{0}\sim T\ll p_{\text{rel}}$ by our assumtpion, we can indeed ignore the kinetic energy from the center of mass motion. (3.1) is in agreement with [27]. The dissociation rate for spin $0$ state is easily obtained by noting that the integrand is isotropic in $\bm{q}$, so that we may substitute in (16) $\hat{q}^{i}\hat{q}^{j}\to 1/3\,{\delta}_{ij}$ to find that the dissociation rate for the spin $0$ state is $1/3$ of the spin summed counterpart. This is in line with our general discussion earlier that isotropic fluctuation of chromomagnetic fields lead to no spin alignment.

Next, we turn to the effective one-loop diagrams depicted in Fig. 3. In this case, we have similar representation as (11) for this contribution

$\displaystyle{\Gamma}$

$\displaystyle=2\frac{T_{F}}{N_{c}}(N_{c}^{2}-1)\int_{Q}D_{rr}^{mn}(Q)\pi{\delta}(p_{0}+q_{0}-h_{o}^{(0)})\langle S|{\mu}_{i}|O\rangle\langle O|{\mu}_{j}|S\rangle{\epsilon}^{ikm}q_{k}{\epsilon}^{jln}q_{l},$

(19)

with $D_{rr}^{mn}$ being resummed gluon propagator. Its explicit form in TAG is given by [42]

		$\displaystyle D_{rr}^{mn}={\rho}^{mn}(Q)\left(\frac{1}{2}+f(q_{0})\right),$
		$\displaystyle{\rho}^{mn}(Q)={\rho}_{T}(Q)P_{T}^{mn}+{\rho}_{L}(Q)\frac{q^{2}}{q_{0}^{2}}\hat{q}^{m}\hat{q}^{n},$		(20)

where ${\rho}_{T}=2\text{Im}[\frac{-1}{Q^{2}-G}]$ and ${\rho}_{L}=2\text{Im}[\frac{-1}{Q^{2}-F}\frac{Q^{2}}{q^{2}}]$ are transverse and longitudinal spectral functions. $F$ and $G$ are longitudinal and transverse components of self-energy given by

	$\displaystyle F$	$\displaystyle=-\frac{2m_{\text{g}}^{2}Q^{2}}{q^{2}}\left(1-\frac{q_{0}}{q}Q_{0}\left(\frac{q_{0}}{q}\right)\right),$
	$\displaystyle G$	$\displaystyle=\frac{1}{2}\left(2m_{\text{g}}^{2}-F\right),$		(21)

in the hard thermal loop (HTL) approximation⁶⁶6The self-energy is gauge invariant in the HTL regime.. $m_{\text{g}}^{2}=\frac{1}{6}g_{s}^{2}T^{2}(C_{A}+\frac{N_{f}}{2})$ is the gluon thermal mass. The spectral functions contain pole and cut contributions. In the limit $Q\gg m_{\text{g}}$, the spectral functions are dominated by pole contributions with the transverse pole reducing to the free theory spectral function (13) while the longitudinal pole decouples. The pole contribution is also suppressed by phase space, similar to the case of LO calculation. The cut contribution corresponds to the inelastic Coulomb scattering when expanded to first order in $F$ and $G$ [39]. When $Q\sim m_{\text{g}}$, the Coulomb scattering contains infrared (IR) divergence, which is partially screened by self-energy and the remaining divergence is cutoff by ${\epsilon}_{B}$, resulting in logarithmically enhanced contribution of the form $\ln\frac{T}{{\epsilon}_{B}}$ and $\ln\frac{T}{m_{\text{g}}}$ to the dissociation rate [40]. Below we shall focus on the following cut contribution (and loosely refer to it as NLO contribution)

$\displaystyle{\Gamma}$

$\displaystyle=2\frac{T_{F}}{N_{c}}(N_{c}^{2}-1)\int_{Q}{\rho}_{\text{cut}}^{mn}\left(\frac{1}{2}+f(q_{0})\right)\pi{\delta}(p_{0}+q_{0}-h_{o}^{(0)})\langle S|{\mu}_{i}|O\rangle\langle O|{\mu}_{j}|S\rangle{\epsilon}^{ikm}q_{k}{\epsilon}^{jln}q_{l},$

(22)

with ${\rho}_{\text{cut}}^{mn}=2{\pi}(P_{T}^{mn}{\beta}_{T}+\hat{q}^{m}\hat{q}^{n}{\beta}_{L})$. The explicit expressions of ${\beta}_{T/L}$ are given by

		$\displaystyle{\beta}_{L}=\frac{m_{\text{g}}^{2}x{\theta}(1-x^{2})}{\left[q^{2}+2m_{\text{g}}^{2}\left(1-\frac{x}{2}\ln\absolutevalue{\frac{x+1}{x-1}}\right)\right]^{2}+{\pi}^{2}m_{\text{g}}^{4}x^{2}},$
		$\displaystyle{\beta}_{T}=\frac{m_{\text{g}}^{2}x(1-x^{2}){\theta}(1-x^{2})/2}{\left(q^{2}(x^{2}-1)-m_{\text{g}}^{2}\left[x^{2}+\frac{x(1-x^{2})}{2}\ln\absolutevalue{\frac{x+1}{x-1}}\right]\right)^{2}+{\pi}^{2}m_{\text{g}}^{4}x^{2}(1-x^{2})^{2}/4},$		(23)

with $x=q_{0}/q$. Like in the LO case, the integrand is isotropic in $\bm{q}$, so the argument we have used for LO case still works, leading to an unpolarized dissociation rate with no spin alignment.

Fig. 3 contains only a subset of two-loop diagrams. The other diagrams are collected in Fig. 4. These diagrams can be cut in different ways to give rise to: interference correction to gluo-dissociation from diagrams (a)-(f); Compton like scattering from diagrams (a), (c)-(g)⁷⁷7Diagram in Fig. 3 with gluon self-energy from gluon loop also corresponds to Compton scattering in the t-channel in this terminology.. Although the diagrams contain the same powers of $g_{s}$ as those for Coulomb scattering, most of them are suppressed by extra powers of $rq_{0}\sim\frac{T}{m_{Q}v}\ll 1$ by our assumption. The exceptions are diagram (b) when the cross vertex is taken to be chromoelectric interaction and diagram (g). The former case vanishes for the following reason: the virtual color octet states sandwiching the chromomagnetic vertex inherit the spin states of the color singlet from chromoelectric vertices, thus are spin triplet states. Their matrix elements vanish on the chromomagnetic vertex. The latter case does give nonvanishing contribution to polarized dissociation rate. This contribution is, however, not IR enhanced like in the Coulomb scattering case. Below we shall restrict ourselves to the leading logarithmic contributions and leave more comprehensive studies for future work.

Now we consider dissociation of a moving quarkonium bound state. Since the spin states are defined in quarkonium rest frame, it is natural to think of the problem in the same frame, in which static quarkonium bound state dissociates in moving QGP. The chromoelectromagnetic fields seen by the quarkonium is a boosted version of the isotropic fields in the QGP frame, which are expected to be anisotropic. Through chromomagnetic coupling, we expect a polarized dissociation rate and nonvanishing spin alignment.

It is convenient to proceed in covariant form. We denote the quarkonium and QGP frames respectively by $n^{\mu}$ and $u^{\mu}$. An ambiguity arises in the definition of TAG, which can be either $A^{a}\cdot n=0$ or $A^{a}\cdot u=0$. We choose the former, which keeps the Feynman rules in Fig. 1 unchanged. The only change we need is to replace the gluon propagators in (3.1) and (19) by the following moving ones. Note that the gluons are in equilibrium so the fluctuation-dissipation theorem holds in general

$\displaystyle D_{rr}^{{\mu}{\nu}}=2\text{Re}[D_{ra}^{{\mu}{\nu}}]\left(\frac{1}{2}+f(Q\cdot u)\right).$

(24)

The replacement for the free gluon propagator (indicated by the superscript $(0)$) is straightforward

$\displaystyle D_{rr}^{{\mu}{\nu}(0)}$

$\displaystyle=P_{T}^{{\mu}{\nu}}(n)2{\pi}{\epsilon}(Q\cdot n){\delta}(Q^{2})\left(\frac{1}{2}+f(Q\cdot u)\right).$

(25)

Here the transverse projector defined as $P_{T}^{{\mu}{\nu}}(n)=n^{\mu}n^{\nu}-\eta^{{\mu}{\nu}}-\hat{q}_{n}^{\mu}\hat{q}_{n}^{\nu}$ with $\hat{q}_{n}^{\mu}=(-n^{\mu}n^{\nu}+\eta^{{\mu}{\nu}})Q_{\nu}/(Q\cdot n)^{2}-Q^{2})^{1/2}$ being covariant generalization of the unit vector $\hat{\bm{q}}$ in quarkonium frame. The factor $1/2$ will be dropped as medium independent contribution.

The replacement for the resummed gluon propagator is tricky. $D_{ra}^{{\mu}{\nu}}$ satisfies the following resummation equation

$\displaystyle D_{ra}^{{\mu}{\nu}}=D_{ra}^{{\mu}{\nu}(0)}+D_{ra}^{{\mu}{\alpha}(0)}{\Pi}_{ar,{\alpha}{\beta}}D_{ra}^{{\beta}{\nu}}.$

(26)

The free retarded gluon propagator in TAG is given by

$\displaystyle D^{(0)\,{\mu}{\nu}}_{ra}=\frac{i}{Q^{2}}P_{T}^{{\mu}{\nu}}(n)+\frac{i}{Q^{2}}\frac{Q^{2}}{(Q\cdot n)^{2}}\hat{q}_{n}^{\mu}\hat{q}_{n}^{\nu},$

(27)

On the other hand, the gluons are in equilibrium in QGP frame with the following self-energy

$\displaystyle{\Pi}_{ar}^{{\mu}{\nu}}(u)=P_{T}^{{\mu}{\nu}}(u)\,G(u)+P_{L}^{{\mu}{\nu}}(u)\,F(u),$

(28)

where the transverse projector $P_{T}^{{\mu}{\nu}}(u)$ are defined similarly as $P_{T}^{{\mu}{\nu}}(n)$ but with different frame vector. $P_{L}^{{\mu}{\nu}}(u)=\eta^{{\mu}{\nu}}-Q^{\mu}Q^{\nu}/Q^{2}-P_{T}^{{\mu}{\nu}}(u)$ is the longitudinal projector. The components of self-energy are covariant generalization of (3.1) as $G(u)=G(q_{0}=Q\cdot u,q=((Q\cdot u)^{2}-Q^{2})^{1/2})$ and $F(u)=F(q_{0}=Q\cdot u,q=((Q\cdot u)^{2}-Q^{2})^{1/2})$. For $n=u$, it is not difficult to arrive at (3.1). For $n\neq u$, the solution to the resummation equation is complicated.

To simplify the analysis, we assume the quarknonium is slow-moving in QGP and expand in the velocity. Working in the quarkonium frame, we parameterize $u^{\mu}=((1+{\delta}\bm{v}^{2})^{1/2},{\delta}\bm{v})$. We already know from the previous subsection that there is no spin alignment when ${\delta}\bm{v}=0$. It follows that the spin alignment arises at least from $\mathcal{O}({\delta}\bm{v}^{2})$⁸⁸8The naive contribution to spin alignment of the form $\mathcal{O}({\delta}\bm{v}\cdot\hat{\bm{l}})$ is excluded because the spin alignment is invariant under flipping of quantization axis.. We shall solve for the resummed gluon propagator up to $\mathcal{O}({\delta}\bm{v}^{2})$. (26) is solved as $D_{ra}=(I-D^{(0)}_{ra}{\Pi}_{ar})^{-1}D^{(0)}_{ra}$ with all quantities being matrices. To perform the expansion, we split the gluon self-energy as

$\displaystyle{\Pi}_{ar,{\mu}{\nu}}(u)={\Pi}_{ar,{\mu}{\nu}}(u)-{\Pi}_{ar,{\mu}{\nu}}(n)+{\Pi}_{ar,{\mu}{\nu}}(n)\equiv{\delta}{\Pi}_{ar,{\mu}{\nu}}+{\Pi}_{ar,{\mu}{\nu}}(n).$

(29)

The matrix inversion is performed using

$\displaystyle(M-{\delta}M)^{-1}=M^{-1}+M^{-1}{\delta}MM^{-1}+\cdots.$

(30)

to arrive at

$\displaystyle{\delta}D_{ra}^{{\mu}{\nu}}=D_{ra}^{{\mu}{\alpha}}(n){\delta}{\Pi}^{ar}_{{\alpha}{\beta}}D_{ra}^{{\beta}{\nu}}(n)+\cdots,$

(31)

with $D_{ra}(n)$ in the above being the resummed propagator given by

$\displaystyle D_{ra}^{{\mu}{\nu}}(n)=\frac{i}{Q^{2}-G(n)}P_{T}^{{\mu}{\nu}}(n)+\frac{i}{Q^{2}-F(n)}\frac{Q^{2}}{(Q\cdot n)^{2}}\hat{q}_{n}^{\mu}\hat{q}_{n}^{\nu}.$

(32)

The terms containing one more pair of ${\delta}{\Pi}^{ar}$ and $D_{ra}$ is suppressed by $m_{\text{g}}^{2}/Q^{2}$. As will be clear soon, the dominant contribution comes from the phase space with $q_{0},q\gg m_{\text{g}}$, allowing us to keep the leading term only. By the fluctuation-dissipation theorem (24), we find

$\displaystyle{\delta}D_{rr}^{{\mu}{\nu}}$

$\displaystyle=2\text{Re}[D_{ra}^{{\mu}{\nu}}(u)]\left(\frac{1}{2}+f(Q\cdot u)\right)-2\text{Re}[D_{ra}^{{\mu}{\nu}}(n)]\left(\frac{1}{2}+f(Q\cdot n)\right).$

(33)

Clearly, by (31) ${\delta}D_{rr}^{{\mu}{\nu}}$ is nonvanishing for spatial components only in quarkonium frame. $2\text{Re}[{\delta}D_{ra}^{{\mu}{\nu}}]$ can be interpreted as a correction to the spectral density, although strictly speaking it is only a consequence of our splitting of the self-energy.

We start with the LO dissociation for a moving quarkonium bound state. The LO dissociation rate written in quarkonium frame reads

$\displaystyle{\Gamma}$

$\displaystyle=2\frac{T_{F}}{N_{c}}(N_{c}^{2}-1)\int_{Q}D_{rr}^{mn(0)}(n)\pi{\delta}(p_{0}+q_{0}-h_{o}^{(0)})\langle S|{\mu}_{i}|O\rangle\langle O|{\mu}_{j}|S\rangle{\epsilon}^{ikm}q_{k}{\epsilon}^{jln}q_{l},$

(34)

with $D_{rr}^{mn(0)}$ being spatial components of $D_{rr}^{{\mu}{\nu}(0)}$. Plugging (3.1) and (25) into (34), we obtain

	$\displaystyle{\Gamma}$	$\displaystyle=2\frac{T_{F}}{N_{c}}(N_{c}^{2}-1)\int_{Q}2{\pi}^{2}\frac{1}{2q}{\delta}(q_{0}-q){\delta}(p_{0}+q_{0}-h_{o}^{(0)})f(Q\cdot u)\langle S|{\mu}_{i}|O\rangle\langle O|{\mu}_{j}|S\rangle q^{2}({\delta}^{ij}-\hat{q}^{i}\hat{q}^{j})$
		$\displaystyle\equiv{\Gamma}_{ij}\langle S=1|{\mu}_{i}|S=0\rangle\langle S=0|{\mu}_{j}|S=1\rangle\frac{m_{Q}^{2}}{g_{s}^{2}}.$		(35)

We used (14) and dropped the $1/2$ next to $f(Q\cdot u)$ because it does not contribute to the spin dependence of the dissociation rate. A normalization factor $m_{Q}^{2}/g_{s}^{2}$ is introduced for convenience. The spin part of the transition amplitude is obtained from (7) as

		$\displaystyle\text{spin 0-state}:\;\langle S=1|{\mu}_{i}|S=0\rangle\langle S=0|{\mu}_{j}|S=1\rangle=\frac{g_{s}^{2}l_{i}l_{j}}{m_{Q}^{2}},$
		$\displaystyle\text{spin summed}:\;\langle S=1|{\mu}_{i}|S=0\rangle\langle S=0|{\mu}_{j}|S=1\rangle=\frac{g_{s}^{2}{\delta}_{ij}}{m_{Q}^{2}}.$		(36)

We then express the splitting of dissociation rate as

$\displaystyle\overline{{\Gamma}}-{\Gamma}_{s=0}={\Gamma}_{ij}\left(\frac{1}{3}{\delta}_{ij}-{l}_{i}{l}_{j}\right),$

(37)

where $\overline{{\Gamma}}=1/3\,\sum_{s=0,\pm 1}{\Gamma}_{s}$. The explicit form of ${\Gamma}_{ij}$ is determined as follows

$\displaystyle{\Gamma}_{ij}=2\frac{T_{F}}{N_{c}}(N_{c}^{2}-1)\int_{Q}2{\pi}^{2}\frac{1}{2q}{\delta}(q_{0}-q){\delta}(p_{0}+q_{0}-h_{o}^{(0)})f(Q\cdot u)\langle S|{\mu}_{i}|O\rangle\langle O|{\mu}_{j}|S\rangle q^{2}({\delta}^{ij}-\hat{q}^{i}\hat{q}^{j}).$

(38)

The orbital part of the transition amplitudes is evaluated as

$\displaystyle|\langle S,L=0|O,L=0\rangle|^{2}=\int\frac{d^{3}p_{\text{rel}}}{(2{\pi})^{3}}\frac{64{\pi}a^{3}}{(1+a^{2}p_{\text{rel}}^{2})^{3}},$

(39)

with $a=(m_{Q}{\epsilon}_{B})^{-1/2}$ being the Bohr radius for the bound state. As in the static case, since $q\lesssim T\ll m_{Q}v$, the kinetic energy of the center of mass motion is suppressed with respect to the counterpart from relative motion of quark pair, so that ${\delta}(p_{0}+q_{0}-h_{o}^{(0)})\simeq{\delta}(-{\epsilon}_{B}+q_{0}-\frac{p_{\text{rel}}^{2}}{m_{Q}})$. It follows that

$\displaystyle\int\frac{d^{3}p_{\text{rel}}}{(2{\pi})^{3}}\frac{a^{3}}{(1+a^{2}p_{\text{rel}}^{2})^{3}}{\delta}(-{\epsilon}_{B}+q_{0}-\frac{p_{\text{rel}}^{2}}{m_{Q}})=\frac{1}{(2{\pi})^{2}}\frac{m_{Q}a^{3}p_{\text{rel}}}{(1+a^{2}p_{\text{rel}}^{2})^{3}}.$

(40)

evaluated at $p_{\text{rel}}=((q_{0}-{\epsilon}_{B})m_{Q})^{1/2}$.

To simplify the tensor integral of ${\Gamma}^{ij}$, we parameterize ${\Gamma}_{ij}=c_{1}{\delta}_{ij}+c_{2}{\delta}\hat{v}^{i}{\delta}\hat{v}^{j}$ by rotational invariance. Only $c_{2}$ is relevant for splitting of dissociation rate, which can be easily determined as

$\displaystyle c_{2}$

$\displaystyle=-\frac{1}{2}{\Gamma}^{ij}{\delta}_{ij}+\frac{3}{2}{\Gamma}^{ij}{\delta}\hat{v}_{i}{\delta}\hat{v}_{j},$

(41)

which is a scalar integral depending on magnitude of ${\delta}v$ only. It is calculated numerically. The dissociation rate in terms of $c_{2}$ reads

$\displaystyle\overline{{\Gamma}}-{\Gamma}_{s=0}$

$\displaystyle=c_{2}\left(\frac{1}{3}-\left({\delta}\hat{\bm{v}}\cdot\hat{\bm{l}}\right)^{2}\right).$

(42)

The coefficient $c_{2}$ is a function of $T$, magnitude of relative velocity ${\delta}v$, ${\epsilon}_{B}$ and $m_{Q}$. The spin dependence is entirely contained in the factor $1/3-({\delta}\hat{\bm{v}}\cdot\hat{\bm{l}})^{2}$.

Taking $J/\psi$ as an example, in the numerical integration, we take ${\alpha}_{s}=0.3$, $T_{c}=150$MeV and $m_{c}=1.3$GeV for charm quark. Using 1S wave function, $J/{\psi}$ has a definite mass $M=2m_{c}-\epsilon_{B}$ and binding energy $\epsilon_{B}=m_{c}C_{F}^{2}\alpha_{s}^{2}/4=0.052$GeV. The dependence of $c_{2}$ on ${\delta}v$ at different temperatures is shown in Fig. 5. We can see that $c_{2}$ is always negative, then the sign of the splitting of dissociation rate depends on the angle between ${\delta}\bm{v}$ and $\hat{\bm{l}}$: when $\absolutevalue{{\delta}\hat{\bm{v}}\cdot\hat{\bm{l}}}<1/\sqrt{3}$, $\overline{{\Gamma}}<{\Gamma}_{s=0}$, otherwise the result changes sign.