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Consider a three-particle model (
$ ANN $ system), with the$ A $ particle having a mass that is about 10 times that of the$ N $ particle, which has mass$ m $ . The$ A $ particle thus mimics the nuclear core. To be specific, let us calculate$ AN\to AN $ scattering. For simplicity, we use a separable potential of the form$\begin{aligned}[b] V^{AN}_H(p, p') = &\dfrac{1}{\sqrt{2\omega_{A}(p)2\omega_{N}(p)}} g f(p,\Lambda)f(p',\Lambda)\\ & \times \dfrac{1}{\sqrt{2\omega_{A}(p)2\omega_{N}(p)}}\; , \end{aligned}$
(1) with the regulator function
$ f(p,\Lambda) = \dfrac{\Lambda^2}{p^2+\Lambda^2}\; , $
(2) where
$ \omega_i(q) = \sqrt{m_i^2+q^2} $ , and$ g $ is the coupling constant. The normalization is explained in the Appendix. In what follows, we set$ m_A = 10\, $ GeV, and$ m = \Lambda = 1\, $ GeV. We are interested in the case in which the$ AN $ system has a very weakly bound state$ B $ with binding energy in the keV range, so the coupling$ g $ is tuned accordingly.We can then construct the Hamiltonian in the finite volume and find its eigenvalues [19]. The Hamiltonian matrix is defined as follows:
$ \begin{aligned}[b] H = &H_0+H_I,\\ \left(H_{0}\right)_{ij} =& \delta_{ij}\left(\omega_A(k_i)+\omega_N(k_i)\right), \end{aligned} $
(3) $ \left(H_I\right)_{ij} = \dfrac{\sqrt{C_3(i)C_3(j)}}{4\pi} \left(\frac{2\pi}{L}\right)^3 V_H(E,k_i,k_j), $
(4) where
$ k_i = \sqrt{i}2\pi/L $ and$ C_3(i) $ represents the number of ways to sum the square of the three integers to equal$ i $ . Further, the factor$ (\sqrt{C_3(i)C_3(j)} \; )/(4\pi) (2\pi/L)^3 $ is due to the quantization in a finite box of size$ L $ , as explained in Refs. [18, 19].For the full
$ ANN $ system with a fixed total momentum, we have two free momenta. This leads to a Hamiltonian matrix with a huge dimension in the finite volume. For simplification, we thus consider the$ BN $ system instead of the$ ANN $ system; this means that we use a version of the dimer approximation, see, e.g., [25], reminiscent of the so-called Faddeev fixed center approximation, see, e.g., Refs. [26, 27]. We thus consider the scattering process$ BN\to BN $ .The left diagram in Fig. 1 shows the attractive interaction between
$ B $ and$ N $ since the$ AN $ system has a weak attractive interaction. To prepare this diagram, we need to obtain the coupling of the$ B\to AN $ process. Since$ B $ is a loosely bound state of$ AN $ , one can obtain the coupling from the amplitude of$ AN \to AN $ around the pole position of$ B $ as follows:$\begin{aligned}[b] T^{AN}_H\Big(E\sim m_B, q = &q_0(E), q' = q_0(E) \Big) \\ =& \dfrac{1}{2m_{A}}\dfrac{1}{2m_{N}}4\pi \dfrac{\tilde{g}^2}{2m_B(E-m_B)}, \end{aligned}$
(5) where
$ q_0(E) $ is the on-shell momentum with energy$ E $ , and$ T^{AN}_H $ is obtained from Eq. (22) with the potential$ V^{AN}_H $ . The factor$ 1/(2m_{A}) \cdot 1/(2m_{N}) $ originates from the difference between$ V_H $ and$ V_L $ (see the Appendix). The momentum is on-shell, so it is close to the mass of the particle, and the factor$ 4\pi $ is from the angular integration since we only consider the s-wave. Further, the coupling$ \tilde{g} $ has dimension energy. Given this, the potential of$ BN \to BN $ from$ A $ -exchange takes the form$ \begin{aligned}[b] V^{BN1}_H(p, p') = & \dfrac{2\pi}{\sqrt{2\omega_B(p)2\omega_N(p)}} \\ &\times \int {\rm d} \cos\theta \dfrac{\tilde{g}^2} {(E_A^2)-(\vec{p}-\vec{p}\,')^2-m^2_A}\\ &\times\dfrac{1}{\sqrt{2\omega_B(p')2\omega_N(p')}}, \end{aligned} $
(6) $ E_A^2 = \frac{1}{2}\left((\omega_B(p)-\omega_N(p'))^2+\left(\omega_N(p)-\omega_B(p')\right)^2\right)\; . $
(7) Since our potential should be independent of the total energy, we take the average of the two processes
$ B\to AN $ and$ NA\to B $ . Next, we need to extract the s-wave contribution from this diagram, so we perform the angular integration between$ \vec{p} $ and$ \vec{p}\,' $ . Finally, the equation for the potential takes the form$ &V^{BN1}_H (p, p') = \dfrac{\tilde{g}^2}{\sqrt{2\omega_B(p)2\omega_N(p)2\omega_B(p')2\omega_N(p')}} \dfrac{\pi}{pp'} \ln\left(\dfrac{m_B^2+m_N^2-m_A^2-(\omega_B(p)\omega_N(p')+\omega_N(p)\omega_B(p'))+2pp'} {m_B^2+m_N^2-m_A^2-(\omega_B(p)\omega_N(p')+\omega_N(p)\omega_B(p'))-2pp'}\right). $
(8) Note that this potential should be negative because in Eq. (6), the propagator of the exchanged
$ A $ particle is negative.Now we consider the contribution from the right diagram in Fig. 1. This includes a triangle loop, and the main interaction is the
$ NN \to NN $ interaction. The$ NN \to NN $ interaction can be written as$\begin{aligned}[b] V^{NN}_H(p, p') =& \dfrac{1}{\sqrt{2\omega_{N}(p)2\omega_{N}(p)}} g_{NN} f(p,\Lambda)f(p',\Lambda)\\ &\times \dfrac{1}{\sqrt{2\omega_{A}(p)2\omega_{N}(p)}}\; , \end{aligned}$
(9) where the regulator function is chosen to be the same as that for the
$ AN $ interaction. In this model, we want to describe the situation in which the$ BN $ system cannot form a bound state. The coupling$ g_{NN} $ is the only parameter that allows this to be achieved.Next, we determine the potential based on the diagram on the right side of Fig. 1:
$\begin{aligned}[b] V^{BN2}_H(p, p') = & \dfrac{2\pi}{\sqrt{2\omega_{B}(p)2\omega_{N}(p)}} \int {\rm d}\cos\theta\, V^{BN2}_L(\vec{p},\vec{p}') \\ & \times \dfrac{1}{\sqrt{2\omega_{B}(p')2\omega_{N}(p')}}, \end{aligned}$
(10) where
$ \begin{aligned}[b] V^{BN2}_L(\vec{p},\vec{p}') = & \int {\rm d}^4q \; \tilde{g}^2 \dfrac{1}{q^2_0-\vec{q}^2-m^2_A} \dfrac{1}{(\omega_B(p)-q_0)^2-(\vec{p}-\vec{q})^2-m^2_N)} \dfrac{1}{(\omega_B(p')-q_0)^2-(\vec{p}'-\vec{q})^2-m^2_N} T^{L}_{NN} \\ \!=& \! \!\int\! {\rm d}^3\vec{q} \; \tilde{g}^2 \dfrac{1}{2\omega_A(q)} \dfrac{1}{(\omega_B(p)\!-\!\omega_A(q))^2\!-\!(\vec{p}-\vec{q})^2\!-\!m^2_N} \dfrac{1}{(\omega_B(p')-\omega_A(q))^2-(\vec{p}'-\vec{q})^2-m^2_N)} T^{NN}_{L}\; , \end{aligned} $
(11) where
$ T_{L}^{NN} $ is the amplitude of$ NN \to NN $ . In the calculation of$ T_{L}^{NN} $ , we make some further assumptions. First, we assume$ T_{L}^{NN} \sim V_{L}^{NN} $ , which should be acceptable since we are not interested in the detailed structure of the$ NN $ scattering amplitude. Moreover, we require this interaction in a boosted frame. Although the form of$ V_L $ is not Lorentz invariant, we can rewrite the potential in a special form and define all the inputs to the center-of-mass (CM) system. This means that we write$ V^{NN}_L $ as$ T^{NN}_{L} \sim V^{NN}_{L} = g_{NN} f(k^*,\Lambda)f(k'^*,\Lambda) , $
(12) $ k^{*\,2} = E_{NN}^2/4-m_N^2, $
(13) $ E_{NN}^2 = \left(\sqrt{(\vec{q}-\vec{p})^2+m_N^2}+\sqrt{\vec{p}^2+m_N^2}\;\right)^2-q^2, $
(14) $ k^{'*\,2} = E_{NN}^{'2}/4-m_N^2, $
(15) $ E_{NN}^{'2} = \left(\sqrt{(\vec{q}-\vec{p}^{\,'})^2+m_N^2}+\sqrt{\vec{p}^{\,'2}+m_N^2}\;\right)^2-q^2. $
(16) We then obtain
$ V_H^{BN2} $ as defined in Eqs. (10-16) as follows:$\begin{aligned}[b] V^{BN2}_H(p, p') = & \dfrac{4\pi^2}{\sqrt{2\omega_{B}(p)2\omega_{N}(p)2\omega_{B}(p')2\omega_{N}(p')}} \\ & \times \int q^2{\rm d}q \dfrac{\tilde{g}^2 g_{NN}}{2\omega_A(q)} H(p,q)H(p',q), \end{aligned}$
(17) where
$\begin{aligned}[b] H(p,q) = \int {\rm d}\cos\theta \dfrac{1}{m^2_B+m^2_A-m^2_N-2\omega_B(p)\omega_A(q)+2pq\cos\theta} \dfrac{4\Lambda^2}{\!\left(\sqrt{q^2+p^2-2pq\cos\theta+m_N^2}+\!\!\sqrt{p^2+m_N^2}\right)^2\!\!-\!\!q^2-4m_N^2+4\Lambda^2}\; , \end{aligned}$
(18) which can easily be evaluated numerically.
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First, we need to fix the coupling constant
$ g $ . In the left panel of Fig. 2, we show the binding energy of the two-particle system as a function of the coupling$ g $ . The latter is chosen in a suitable range such that the binding is weak, and indeed, at$ g = -30.65 $ , there is no more bound state. In what follows, we choose$ g = -31.0 $ , for which one finds a loosely bound state at$ |E_B| = 11.15\, $ keV. In the right panel of Fig. 2, the corresponding scattering phase shift in the close-to-threshold region is shown; it exhibits the typical features of a weakly bound state close to the threshold.Figure 2. (left panel) Binding energy of the
$ AN $ system as a function of the coupling constant$ g $ . (right panel) Phase shift for$ AN\to AN $ scattering for$ g = -31.0 $ . The black points are calculated from the corresponding energy levels depicted in Fig. 3 using the Lüscher equation.The corresponding energy levels in the finite volume are shown in Fig. 3. The bound state level is clearly visible: its bending downwards for smaller lattice sizes is an expected finite volume effect. For sufficiently large
$ L $ , these finite volume effects are visibly absent.Figure 3. Energy levels for the weakly bound
$ AN $ system in a finite volume$ L^3 $ . The meaning of the black squares at$ L = 10 $ fm is explained in text; see also Fig. 2.It is also instructive to compare our formalism with the Lüscher equation [16, 17]. For that, we select 17 energy levels at
$ L = 10 $ fm, and then, we use the following Lüscher equation to calculate the phase shifts from the corresponding energy levels:$ \delta(q_E) = \tan^{-1}\left(\frac{q_E L \sqrt{\pi}}{2 {\cal{Z}}(1; (\dfrac{q_EL}{2\pi})^2)} \right)+n\pi, $
(19) where
$ q_E $ is the on-shell momentum corresponding to the energy$ E $ :$ q_E = \frac{E}{2} \sqrt{\left(1-\left(\frac{m_N+m_A}{E}\right)^2\right)\left(1-\left(\frac{m_N-m_A}{E}\right)^2\right)}, $
(20) and
$ {\cal{Z}}(1;q^2) $ is the well known Zeta-function. After regularization, it can be calculated as follows:$\begin{aligned}[b] {\cal Z}(1; q^2) = &\sum\limits_{\vec{n}\in \mathbb{Z}^3}\dfrac{1}{\vec{n}^2-q^2} \\ =& -\dfrac{1}{q^2}-8.91363292+16.53231596 q^2 \\ &+ \sum\limits_{\vec{n}\in \mathbb{Z}^3}\dfrac{q^4}{\vec{n}^4\left(\vec{n}^2-q^2\right)}\; . \end{aligned}$
(21) We find that the phase shifts calculated in this way are all on the phase shift curve calculated directly from the scattering function, see Fig. 2. This shows that our calculation is consistent with the Lüscher equation.
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Here, we briefly discuss the normalization of our scattering T-mat-rix. This normalization is similar to the formalism used in Ref. [28]. Consider the s-wave of the process
$ AN\to AN $ , with the scattering function given by$ \begin{aligned}[b] T_H(E,|\vec{k}|, |\vec{k}'|) = &V_H(|\vec{k}|, |\vec{k}'|) + \int q^2 {\rm d}q V_H(|\vec{k}|,q)\\& \times \frac{1}{E-\omega_A(q)-\omega_N(q)+{\rm i}\epsilon} T_H(E, q, |\vec{k}'|)\; , \end{aligned} \tag{A1}$
(A1) where
$ \omega_i(q) = \sqrt{m_i^2+q^2} $ . Correspondingly, the Bethe-Salpter function, where$ k,k' $ are four-momenta, takes the form$ \begin{aligned}[b] T_L(P, k, k') = & V_L(P, k, k') + \int {\rm d}^4q V_L(P,k,q) \frac{1}{q^2-m_A^2+{\rm i}\epsilon} \\ &\times \dfrac{1}{(P-q)^2-m_N^2+{\rm i}\epsilon} T_L(P, q, k') . \end{aligned} \tag{A2}\!\!\!$
(A2) Actually, Eq. (A1) can be recognized as the three-dimensional reduction of Eq. (A2) by using
$ \int {\rm d}^4q \frac{1}{q^2-m_A^2+{\rm i}\epsilon} \frac{1}{(P-q)^2-m_N^2+{\rm i}\epsilon} \sim \int q^2 {\rm d}q \frac{1}{E-\omega_A(q)-\omega_N(q)+{\rm i}\epsilon}\frac{1}{2\omega_A(q)2\omega_N(q)}\; . \tag{A3}$
(A3) Therefore, we obtain the relationship between
$ V_H $ and$ V_L $ :$ V_H(E,|\vec{k}|,|\vec{k}'|) = \dfrac{2\pi}{\sqrt{2\omega_A(k)2\omega_N(k)}} \int {\rm d} \cos\theta V_L(P, k, k') \times \dfrac{1}{\sqrt{2\omega_A(k')2\omega_N(k')}}\; . \tag{A4}$
(A4) Note that we have been cavalier with some factors such as
$ (2\pi)^n $ since these are absorbed into the coupling in$ V_L $ . This equation is also the simple form of Eqs. (23,24) in Ref. [28].
Towards the continuum coupling in nuclear lattice effective field theory I: A three-particle model
- Received Date: 2020-07-21
- Available Online: 2020-12-01
Abstract: Weakly bound states often occur in nuclear physics. To precisely understand their properties, the coupling to the continuum should be worked out explicitly. As the first step, we use a simple nuclear model in the continuum and on a lattice to investigate the influence of a third particle on a loosely bound state of a particle and a heavy core. Our approach is consistent with the Lüscher formalism.