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The polarization of the Λ (and
$\bar{\Lambda }$ ) hyperons can be measured by its parity-breaking weak decay Λ→p+π−. The daughter proton is emitted preferentially along the Λ polarization in its rest frame. The angular distribution of the daughter proton reads$ \begin{eqnarray}\frac{{\rm{d}}N}{{\rm{d}}{\Omega }^{* }}=\frac{1}{4\pi }\left(1+{\alpha }_{H}{{\mathscr{P}}}_{\Lambda }\frac{{\mathit{\boldsymbol{n}}}^{* }\cdot {\mathit{\boldsymbol{p}}}^{* }}{|{\mathit{\boldsymbol{p}}}^{* }|}\right), \end{eqnarray} $
(1) where αH is the hyperon decay parameter,
$\begin{eqnarray}{{\mathscr{P}}}_{\Lambda}\end{eqnarray}$ is the Λ global polarization; n*, p* and Ω* are the Λ polarization, the proton momentum and its solid angle in the rest frame of the hyperon, respectively, labeled by the superscript ′*′. We note that Eq. (1) is Lorentz invariant by observing$ \begin{eqnarray}\begin{array}{lll}{\mathit{\boldsymbol{n}}}^{* }\cdot {\mathit{\boldsymbol{p}}}^{* } & = & -{n}_{\mu }{p}^{\mu }=-n\cdot p, \\ {E}_{{\rm{p}}}^{* } & = & \frac{1}{2{m}_{\Lambda }}({m}_{\Lambda }^{2}+{m}_{{\rm{p}}}^{2}-{m}_{\pi }^{2}), \\ |{\mathit{\boldsymbol{p}}}^{* }| & = & \frac{1}{2{m}_{\Lambda }}\sqrt{[{m}_{\Lambda }^{2}-{({m}_{{\rm{p}}}-{m}_{\pi })}^{2}][{m}_{\Lambda }^{2}-{({m}_{{\rm{p}}}+{m}_{\pi })}^{2}]}, \end{array}\end{eqnarray} $
(2) where pμ and
${p}_{\Lambda }^{\mu }$ are the four-momenta of the proton and the hyperon in any frame, respectively, and nμ is the space-like four-vector of the hyperon polarization in a general frame. We now focus on the lab frame and the hyperon rest frame. We use pμ,${p}_{\Lambda }^{\mu }$ and nμ to label quantities in the lab frame; all quantities with the superscript ′*′ are in the hyperon rest frame. The Lorentz transformation of the Λ polarization is,$ \begin{eqnarray}{n}^{\mu }={\Lambda }_{\, \nu }^{\mu }(-{\mathit{\boldsymbol{v}}}_{\Lambda }){n}^{* \nu }, \end{eqnarray} $
(3) where
${\Lambda }_{\, \nu }^{\mu }(-{\mathit{\boldsymbol{v}}}_{\Lambda })$ is the Lorentz transformation with${\mathit{\boldsymbol{v}}}_{\Lambda }={\mathit{\boldsymbol{p}}}_{\Lambda }/{E}_{\Lambda }$ . The Λ polarization in the rest frame n*ν has the form n*μ = (0, n*) where n* is the three-vector of the polarization with |n*|2 <1. From Eq. (3) we have$ \begin{eqnarray}{n}^{\mu }=({n}_{0}, \mathit{\boldsymbol{n}})=\left(\frac{{\mathit{\boldsymbol{n}}}^{* }\cdot {\mathit{\boldsymbol{p}}}_{\Lambda }}{{m}_{\Lambda }}, {\mathit{\boldsymbol{n}}}^{* }+\frac{({\mathit{\boldsymbol{n}}}^{* }\cdot {\mathit{\boldsymbol{p}}}_{\Lambda }){\mathit{\boldsymbol{p}}}_{\Lambda }}{{m}_{\Lambda }({m}_{\Lambda }+{E}_{\Lambda })}\right).\end{eqnarray} $
(4) We can also express n*μ in terms of nμ,
$ \begin{eqnarray}{n}^{* \mu }={\Lambda }_{\nu }^{\mu }({\mathit{\boldsymbol{v}}}_{\Lambda }){n}^{\nu }, \end{eqnarray} $
(5) or explicitly,
$ \begin{eqnarray}{n}^{* \mu }=(0, {\mathit{\boldsymbol{n}}}^{* })=(0, \mathit{\boldsymbol{n}}-\frac{{\mathit{\boldsymbol{p}}}_{\Lambda }(\mathit{\boldsymbol{n}}\cdot {\mathit{\boldsymbol{p}}}_{\Lambda })}{{E}_{\Lambda }({E}_{\Lambda }+{m}_{\Lambda })}).\end{eqnarray} $
(6) The polarization four-vector of a particle is always orthogonal to its four-momentum, n·pΛ = n0EΛ-n·pΛ=0, so we can express n0 in term of n, n0 = n·vΛ. One can verify that nμ in Eq. (4) satisfies n0=n·vΛ. From (n0)2-|n|2 = -|n*|2 and n0=n·vΛ, we can solve for |n|2 giving
$ \begin{eqnarray}|\mathit{\boldsymbol{n}}{|}^{2}=\frac{|{\mathit{\boldsymbol{n}}}^{* }{|}^{2}}{1-|{\mathit{\boldsymbol{v}}}_{\Lambda }{|}^{2}{(\hat{\mathit{\boldsymbol{n}}}\cdot {\hat{\mathit{\boldsymbol{v}}}}_{\Lambda })}^{2}}.\end{eqnarray} $
(7) We see that when
$|{\mathit{\boldsymbol{v}}}_{\Lambda }{|}^{2}{(\hat{\mathit{\boldsymbol{n}}}\cdot {\hat{\mathit{\boldsymbol{v}}}}_{\Lambda })}^{2}\to 1$ , |n|2→∞, i.e. |n|2 is not bounded. In case of transverse polarization, i.e.$\hat{\mathit{\boldsymbol{n}}}\cdot {\hat{\mathit{\boldsymbol{v}}}}_{\Lambda }=0$ , we have |n|2 = |n*|2 < 1.In the lab frame, a 3-dimensional vector (e.g. impact parameter, global angular momentum, beam direction) can be written as a = axex+ayey+azez, where (ex, ey, ez) are the three basis directions.
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In this section we describe briefly the method used in the STAR experiment for measurement of the Λ hyperon polarization [21]. From Eq. (13), we can determine the Λ polarization in its rest frame by taking the event average of the direction of the proton momentum
${\hat{\mathit{\boldsymbol{p}}}}^{* }$ . We then make a projection onto the direction of the global angular momentum eL,$ \begin{eqnarray}{{\mathscr{P}}}_{\Lambda }=\frac{3}{{\alpha }_{H}}{\langle {\hat{\mathit{\boldsymbol{p}}}}^{* }\cdot {e}_{L}\rangle }_{{\rm{ev}}}=\frac{3}{{\alpha }_{H}}{\langle \cos {\theta }^{* }\rangle }_{{\rm{ev}}}\end{eqnarray} $
(8) where θ* is the angle, in the Λ rest frame, between the proton momentum and the global angular momentum corresponding to the reaction plane. We have the following relation
$ \begin{eqnarray}\cos {\theta }^{* }=\sin {\theta }_{{\rm{p}}}^{* }\sin ({\phi }_{{\rm{p}}}^{* }-{\psi }_{{\rm{RP}}}), \end{eqnarray} $
(9) where
${\theta }_{{\rm{p}}}^{* }$ and${\phi }_{{\rm{p}}}^{* }$ are the polar and azimuthal angles of${\hat{\mathit{\boldsymbol{p}}}}^{* }$ , respectively, and ψRP is the azimuthal angle of the reaction plane. We integrate over${\theta }_{{\rm{p}}}^{* }$ in Eq. (1) to obtain$ \begin{eqnarray}\frac{{\rm{d}}N}{{\rm{d}}{\phi }_{{\rm{p}}}^{* }}=\displaystyle {\int }_{0}^{\pi }{\rm{d}}{\theta }_{{\rm{p}}}^{* }\sin {\theta }_{{\rm{p}}}^{* }\frac{{\rm{d}}N}{{\rm{d}}{\Omega }^{* }}=\frac{1}{2\pi }+\frac{1}{8}{\alpha }_{H}{{\mathscr{P}}}_{\Lambda }\sin ({\phi }_{{\rm{p}}}^{* }-{\psi }_{{\rm{RP}}}), \end{eqnarray} $
(10) which gives the polarization in terms of the azimuthal angle of the daughter proton,
$ \begin{eqnarray}{{\mathscr{P}}}_{\Lambda }=\frac{8}{\pi {\alpha }_{H}}{\langle \sin ({\phi }_{{\rm{p}}}^{* }-{\psi }_{{\rm{RP}}})\rangle }_{{\rm{ev}}}, \end{eqnarray} $
(11) with
$ \begin{eqnarray}{\langle \sin ({\phi }_{{\rm{p}}}^{* }-{\psi }_{{\rm{RP}}})\rangle }_{{\rm{ev}}}=\displaystyle {\int }_{0}^{2\pi }{\rm{d}}{\phi }_{{\rm{p}}}^{* }\frac{{\rm{d}}N}{{\rm{d}}{\phi }_{{\rm{p}}}^{* }}\sin ({\phi }_{{\rm{p}}}^{* }-{\psi }_{{\rm{RP}}}).\end{eqnarray} $
(12) In the STAR experiment, the azimuthal angle of the reaction plane cannot be directly measured. It is determined from the measurement of the event plane given by the direct flow. This introduces a reaction plane resolution factor in the denominator of Eq. (11),
${R}_{{\rm{EP}}}^{(1)}={\langle \cos ({\psi }_{{\rm{RP}}}-{\psi }_{{\rm{EP}}}^{(1)})\rangle }_{{\rm{ev}}}$ , where${\psi }_{{\rm{EP}}}^{(1)}$ is the azimuthal angle of the event plane determined by the direct flow.
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In this section, we introduce alternative methods for measurement of the Λ hyperon polarization. The advantage of these methods is that the polarization can be measured using the proton momentum in the lab frame.
We start with the formula for the Λ polarization vector in its rest frame,
$ \begin{eqnarray}{\overrightarrow{{\mathscr{P}}}}_{\Lambda }=\frac{3}{{\alpha }_{H}}{\langle {\hat{\mathit{\boldsymbol{P}}}}^{* }\rangle }_{{\rm{ev}}}.\end{eqnarray} $
(13) We can project the above onto the direction of the global polarization, which we assume to be along the y-axis, see Fig. (1).
We now try to evaluate
${\langle {\hat{\mathit{\boldsymbol{P}}}}^{* }\rangle }_{{\rm{ev}}}$ . To this end, we use the following Lorentz transformation of the proton momentum,$ \begin{eqnarray}p=\mathit{\boldsymbol{p}}^{* }+\frac{\mathit{\boldsymbol{p}}_{\Lambda }(\mathit{\boldsymbol{p}}^{* }\cdot \mathit{\boldsymbol{p}}_{\Lambda })}{{m}_{\Lambda }({E}_{\Lambda }+{m}_{\Lambda })}+\frac{{E}_{{\rm{p}}}^{* }}{{m}_{\Lambda }}\mathit{\boldsymbol{p}}_{\Lambda }, \end{eqnarray} $
(14) where
${E}_{{\rm{p}}}^{* }$ is determined by the masses of the proton, pion and Λ, as in Eq. (2). We take the event average of 〈p〉ev,$ \begin{eqnarray}{\langle \mathit{\boldsymbol{p}}\rangle }_{{\rm{ev}}}={\langle \mathit{\boldsymbol{p}}^{* }\rangle }_{{\rm{ev}}}+{\left\langle \frac{\mathit{\boldsymbol{p}}_{\Lambda }(\mathit{\boldsymbol{p}}^{* }\cdot \mathit{\boldsymbol{p}}_{\Lambda })}{{m}_{\Lambda }({E}_{\Lambda }+{m}_{\Lambda })}\right\rangle }_{{\rm{ev}}}, \end{eqnarray} $
(15) where we have used 〈pΛ〉ev = 0.
In order to evaluate the second event average in the right-hand-side of Eq. (15), we make two assumptions: (1) pΛ and p* are statistically independent, so we have
${\langle \mathit{\boldsymbol{p}}_{\Lambda }(\mathit{\boldsymbol{p}}_{\Lambda }\cdot \mathit{\boldsymbol{p}}^{* })\rangle }_{{\rm{ev}}}\approx \mathit{\boldsymbol{e}}_{i}{\langle \mathit{\boldsymbol{p}}_{\Lambda }^{i}\mathit{\boldsymbol{p}}_{\Lambda }^{j}\rangle }_{{\rm{ev}}}{\langle \mathit{\boldsymbol{p}}_{j}^{* }\rangle }_{{\rm{ev}}}$ , where$\mathit{\boldsymbol{p}}_{\Lambda }=\mathit{\boldsymbol{e}}_{i}\mathit{\boldsymbol{p}}_{\Lambda }^{i}$ with i = x, y, z; and (2)${\langle \mathit{\boldsymbol{p}}_{\Lambda }^{i}\mathit{\boldsymbol{p}}_{\Lambda }^{j}\rangle }_{{\rm{ev}}}={\langle |\mathit{\boldsymbol{p}}_{\Lambda }^{i}{|}^{2}\rangle }_{{\rm{ev}}}{\delta }_{ij}$ . Eq. (15) then becomes$ \begin{eqnarray}\begin{array}{c}{\langle {\mathit{\boldsymbol{\hat p}}}_{x}^{* }\rangle }_{{\rm{ev}}}\approx \frac{1}{|\mathit{\boldsymbol{p}}^{* }|}{\left(1+{\left\langle \frac{|\mathit{\boldsymbol{p}}_{\Lambda }^{x}{|}^{2}}{({E}_{\Lambda }+{m}_{\Lambda }){m}_{\Lambda }}\right\rangle }_{{\rm{ev}}}\right)}^{-1}{\langle \mathit{\boldsymbol{p}}_{x}\rangle }_{{\rm{ev}}}, \\ {\langle {\mathit{\boldsymbol{\hat p}}}_{y}^{* }\rangle }_{{\rm{ev}}}\approx \frac{1}{|\mathit{\boldsymbol{p}}^{* }|}{\left(1+{\left\langle \frac{|\mathit{\boldsymbol{p}}_{\Lambda }^{y}{|}^{2}}{({E}_{\Lambda }+{m}_{\Lambda }){m}_{\Lambda }}\right\rangle }_{{\rm{ev}}}\right)}^{-1}{\langle \mathit{\boldsymbol{p}}_{y}\rangle }_{{\rm{ev}}}, \\ {\langle {\mathit{\boldsymbol{\hat p}}}_{z}^{* }\rangle }_{{\rm{ev}}}\approx \frac{1}{|\mathit{\boldsymbol{p}}^{* }|}{\left(1+{\left\langle \frac{|\mathit{\boldsymbol{p}}_{\Lambda }^{z}{|}^{2}}{({E}_{\Lambda }+{m}_{\Lambda }){m}_{\Lambda }}\right\rangle }_{{\rm{ev}}}\right)}^{-1}{\langle \mathit{\boldsymbol{p}}_{z}\rangle }_{{\rm{ev}}}.\end{array}\end{eqnarray} $
(16) We choose a coordinate system as in Fig. 1: the impact parameter vector is along the x-axis, the global orbital momentum is along the y-axis, and the beam direction is along the negative z-axis. In the coordinate system used in the experiment, the beam direction is along the negative z-axis, and the impact parameter vector (reaction plane) is at an azimuthal angle ψRP relative to the x-axis. In the new coordinate system, we have
$\mathit{\boldsymbol{p}}_{\Lambda, {\rm{p}}}^{x}=|\mathit{\boldsymbol{p}}_{\Lambda, {\rm{p}}}^{T}|\cos ({\phi }_{\Lambda, {\rm{p}}}-{\psi }_{{\rm{RP}}})$ and$\mathit{\boldsymbol{p}}_{\Lambda, {\rm{p}}}^{y}=|\mathit{\boldsymbol{p}}_{\Lambda, {\rm{p}}}^{T}|\sin ({\phi }_{\Lambda, {\rm{p}}}-{\psi }_{{\rm{RP}}})$ , where ϕΛ, p are the azimuthal angles of the Λ hyperon and proton, respectively.Figure 1. (color online) In the coordinate system (x, y, z), the beam direction is along the negative z-direction, the impact parameter vector is in the x-direction, and the orbital angular momentum is in the y-direction. The direction of the proton momentum can be described by the polar angle θp and the azimuthal angle ϕp. The coordinate system (x′, y′, z′) is used in experiment. The z′-axis is just the z-axis. The azimuthal angle of the impact parameter vector in the (x′, y′, z′) system is ψRP.
We can further simplify Eq. (16) by using the elliptic flow coefficients. The distribution of pΛ is not isotropic but satisfies
$ \begin{eqnarray}\begin{array}{lll}{\langle |\mathit{\boldsymbol{p}}_{\Lambda }^{x}{|}^{2}\rangle }_{{\rm{ev}}} & \approx & {\langle |\mathit{\boldsymbol{p}}_{\Lambda }^{T}{|}^{2}\rangle }_{{\rm{ev}}}{\langle {\cos }^{2}({\phi }_{\Lambda }-{\psi }_{{\rm{RP}}})\rangle }_{{\rm{ev}}}\\ & \approx & {\langle |\mathit{\boldsymbol{p}}_{\Lambda }^{T}{|}^{2}\rangle }_{{\rm{ev}}}\frac{1}{2}(1+{v}_{2}^{\Lambda }), \\ {\langle |\mathit{\boldsymbol{p}}_{\Lambda }^{y}{|}^{2}\rangle }_{{\rm{ev}}} & \approx & {\langle |\mathit{\boldsymbol{p}}_{\Lambda }^{T}{|}^{2}\rangle }_{{\rm{ev}}}{\langle {\sin }^{2}({\phi }_{\Lambda }-{\psi }_{{\rm{RP}}})\rangle }_{{\rm{ev}}}\\ & \approx & {\langle |\mathit{\boldsymbol{p}}_{\Lambda }^{T}{|}^{2}\rangle }_{{\rm{ev}}}\frac{1}{2}(1-{v}_{2}^{\Lambda }), \end{array}\end{eqnarray} $
(17) where
${v}_{2}^{\Lambda }$ is the elliptic flow of the Λ hyperon. Since the global angular momentum is along the y-axis, we have 〈px〉ev = 〈pz〉ev = 0, and the only non-vanishing component is$ \begin{eqnarray}\begin{array}{lll}{\langle {\mathit{\boldsymbol{\hat p}}}_{y}^{* }\rangle }_{{\rm{ev}}} & \approx & \frac{1}{|\mathit{\boldsymbol{p}}^{* }|}{\left(1+{\left\langle \frac{|\mathit{\boldsymbol{p}}_{\Lambda }^{T}{|}^{2}{\sin }^{2}({\phi }_{\Lambda }-{\psi }_{{\rm{RP}}})}{({E}_{\Lambda }+{m}_{\Lambda }){m}_{\Lambda }}\right\rangle }_{{\rm{ev}}}\right)}^{-1}\\ & & \times {\langle |\mathit{\boldsymbol{p}}_{T}|\sin ({\phi }_{{\rm{p}}}-{\psi }_{{\rm{RP}}})\rangle }_{{\rm{ev}}}\\ & \approx & \frac{1}{|\mathit{\boldsymbol{p}}^{* }|}{\left[1+\frac{1}{2}(1-{v}_{2}^{\Lambda }){\left\langle \frac{|\mathit{\boldsymbol{p}}_{\Lambda }^{T}{|}^{2}}{({E}_{\Lambda }+{m}_{\Lambda }){m}_{\Lambda }}\right\rangle }_{{\rm{ev}}}\right]}^{-1}\\ & & \times {\langle |\mathit{\boldsymbol{p}}_{T}|\sin ({\phi }_{{\rm{p}}}-{\psi }_{{\rm{RP}}})\rangle }_{{\rm{ev}}}\end{array}\end{eqnarray} $
(18) In the central rapidity region
$|\mathit{\boldsymbol{p}}_{\Lambda }^{z}|\ll |\mathit{\boldsymbol{p}}_{\Lambda }^{T}|$ and$|\mathit{\boldsymbol{p}}_{\Lambda }^{T}|\approx |\mathit{\boldsymbol{p}}_{\Lambda }|$ , so that Eq. (18) becomes$ \begin{eqnarray}\begin{array}{ll}{\langle {\mathit{\boldsymbol{\hat p}}}_{y}^{* }\rangle }_{{\rm{ev}}}\approx & \frac{1}{|\mathit{\boldsymbol{p}}^{* }|}{\left[1+\frac{1}{2}(1-{v}_{2}^{\Lambda })({\langle {\gamma }_{\Lambda }\rangle }_{{\rm{ev}}}-1)\right]}^{-1}\\ & \times {\langle |\mathit{\boldsymbol{p}}_{T}|\sin ({\phi }_{{\rm{p}}}-{\psi }_{{\rm{RP}}})\rangle }_{{\rm{ev}}}\end{array}\end{eqnarray} $
(19) In the non-relativistic limit, γΛ ≈ 1 and |vΛ| ≈ 0, we obtain
$ \begin{eqnarray}{\langle {\mathit{\boldsymbol{\hat p}}}_{y}^{* }\rangle }_{{\rm{ev}}}\approx \frac{1}{|\mathit{\boldsymbol{p}}^{* }|}{\langle |\mathit{\boldsymbol{p}}_{T}|\sin ({\phi }_{{\rm{p}}}-{\psi }_{{\rm{RP}}})\rangle }_{{\rm{ev}}}\end{eqnarray} $
(20) The difference with respect to the STAR method is that now we are taking the event average using the proton momenta in the lab frame.
Another method is to use the Lorentz transformation of the energy associated with Eq. (14)
$ \begin{eqnarray}{E}_{{\rm{p}}}={\gamma }_{\Lambda }{E}_{{\rm{p}}}^{* }+\frac{\mathit{\boldsymbol{p}}^{* }\cdot \mathit{\boldsymbol{p}}_{\Lambda }}{{m}_{\Lambda }}\end{eqnarray} $
(21) to replace (p*·pΛ)/mΛ with
${E}_{{\rm{p}}}-{\gamma }_{\Lambda }{E}_{{\rm{p}}}^{* }$ . Eq. (14) then becomes$ \begin{eqnarray}\begin{array}{ll}\mathit{\boldsymbol{p}} & =\mathit{\boldsymbol{p}}^{* }+({E}_{{\rm{p}}}-{\gamma }_{\Lambda }{E}_{{\rm{p}}}^{* })\frac{\mathit{\boldsymbol{p}}_{\Lambda }}{{E}_{\Lambda }+{m}_{\Lambda }}+\frac{{E}_{{\rm{p}}}^{* }}{{m}_{\Lambda }}\mathit{\boldsymbol{p}}_{\Lambda }\\ & =\mathit{\boldsymbol{p}}^{* }+\frac{{E}_{{\rm{p}}}}{{E}_{\Lambda }+{m}_{\Lambda }}\mathit{\boldsymbol{p}}_{\Lambda }+\frac{{E}_{{\rm{p}}}^{* }}{{E}_{\Lambda }+{m}_{\Lambda }}\mathit{\boldsymbol{p}}_{\Lambda }.\end{array}\end{eqnarray} $
(22) Taking the event average and using 〈pΛ/(EΛ+mΛ)〉 ≈ 0, we obtain
$ \begin{eqnarray}\begin{array}{ll}{\langle \mathit{\boldsymbol{p}}^{* }\rangle }_{{\rm{ev}}} & ={\langle \mathit{\boldsymbol{p}}\rangle }_{{\rm{ev}}}-{\left\langle \frac{{E}_{\Lambda }}{{E}_{\Lambda }+{m}_{\Lambda }}{E}_{{\rm{p}}}{v}_{\Lambda }\right\rangle }_{{\rm{ev}}}\\ & ={m}_{{\rm{p}}}{\left\langle {\gamma }_{{\rm{p}}}\left({v}_{{\rm{p}}}-\frac{{\gamma }_{\Lambda }}{{\gamma }_{\Lambda }+1}{v}_{\Lambda }\right)\right\rangle }_{{\rm{ev}}}, \end{array}\end{eqnarray} $
(23) where γp and γΛ are Lorentz contraction factors for the proton and Λ hyperon, respectively. The right-hand-side of the above equation involves only momenta in the lab frame. We can project Eq. (23) onto the y-direction (the direction of the orbital angular momentum) to obtain
${\langle {\mathit{\boldsymbol{p}}}_{y}^{* }\rangle }_{{\rm{ev}}}$ .With
${\langle {\hat{\mathit{\boldsymbol{p}}}}_{y}^{* }\rangle }_{{\rm{ev}}}$ given by one of Eqs. (18, 19, 23), we can obtain the global polarization of Λ from Eq. (13). In the next section we compare these methods by simulations.
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The UrQMD model [31, 32] has been used to produce an ensemble of Λ hyperon four-momenta (EΛ, pΛ) from Au+Au collisions with an impact parameter of 6 fm and collision energies listed in Table 1. In each event there are a few Λ hyperons produced. All these hyperons are collected. Each hyperon is allowed to decay into a proton and a pion, whose angular distribution in the Λ rest frame is given by
energy/GeV method 1 Eq. (8) method 2 Eq. (11) method 3 Eq. (18) method 4 Eq. (23) method 5 Eq. (19) number of Λs (full rapidity) 200 0.33581 0.335851 0.3324 0.33014 0.308495 1304795 180 0.330877 0.33141 0.326565 0.329057 0.306966 927717 140 0.338745 0.337673 0.338942 0.335862 0.351934 892533 120 0.333962 0.333688 0.329696 0.334152 0.318965 995522 100 0.336686 0.334685 0.34669 0.34522 0.360992 971596 62.4 0.331964 0.33118 0.324133 0.333466 0.353216 918787 40 0.330536 0.330302 0.332092 0.331782 0.323459 795837 39 0.337252 0.337516 0.332983 0.331683 0.312195 847367 19.6 0.328531 0.328434 0.339587 0.328939 0.31276 707868 7.7 0.341257 0.3417 0.364069 0.34862 0.302301 434697 Table 1. Simulation results for the global polarization of Λ hyperons. We set
${{\mathscr{P}}}_{\Lambda }=1/3$ , i.e. the Λ hyperons are completely polarized. By analyzing the momentum distribution of daughter protons in the lab frame, we determine the Λ polarization. The results of five methods are presented: methods 1 and 2, Eqs. (8, 11) are used in the STAR experiment [21]; methods 3-5, given by Eqs. (18, 23, 19) are proposed in this paper. The number of events collected are 4 × 104 at 200 GeV and 2.5 × 104 at other energies. The results of method 1-4 are from events in the full rapidity range, while those of method 5 are in the rapidity range [−0.5, 0.5]$ \begin{eqnarray}\frac{{\rm{d}}N}{{\rm{d}}{\Omega }^{* }}=\frac{1}{4\pi }\left(1+{\alpha }_{H}{{\mathscr{P}}}_{\Lambda }\frac{{\mathit{\boldsymbol{n}}}^{* }\cdot \mathit{\boldsymbol{p}}^{* }}{|\mathit{\boldsymbol{p}}^{* }|}\right), \end{eqnarray} $
(24) where
${{\mathscr{P}}}_{\Lambda }$ denotes the Λ polarization. By taking a specific value of${{\mathscr{P}}}_{\Lambda }$ , we sample proton momenta in Λ rest frames. For each Λ hyperon, the proton momentum in its rest frame is then boosted back to the lab frame. In this way we create an ensemble of proton momenta in the lab frame. With the ensemble of momenta for protons and Λ hyperons, we obtain${\langle \mathit{\boldsymbol{p}}_{y}^{* }\rangle }_{{\rm{ev}}}$ . Here we choose the direction of the global angular momentum along the y-direction. Finally, we obtain${\langle {p}_{y}^{* }\rangle }_{{\rm{ev}}}$ from Eq. (13). Simulation results for the global polarization of Λ hyperons using the methods given by Eqs. (8, 11, 18, 23, 19) are shown in Table 1. We see that all proposed methods give equivalent results to the STAR method. Fig. 2 shows the dependence of the simulation results on the rapidity ranges. We conclude that all methods work well for the chosen rapidity ranges, except method Eq. (19) when used in the full rapidity range, or in ranges [−1.5, 1.5] and [−1, 1]. This is understandable since Eq. (19) is only valid for central rapidity. When applied in the rapidity range [−0.5, 0.5], this method also works well.Figure 2. (color online) The dependence of simulation results on rapidity ranges for the global polarization of the Λ hyperon. The same parameters and number of events are used as in Table 1.