CritPt / Challenge_52_main / The Efimov effect is one of the few examples of a three-body problem that can be…

# Problem setup:

The Efimov effect is one of the few examples of a three-body problem that can be solved exactly.  Let's calculate the the angular portion of the three-body wave function and its normalization constant for the Efimov effect.  

 Consider three identical bosons with spatial coordinates (r)→₁, (r)→₂, and (r)→₃.  Define Jacobi vectors (r)→_(ij)=(r)→_(i)−(r)→_(j) and (ρ)→_(k,ij)=(2(r)→_(k)−(r)→_(i)−(r)→_(j))/√(3), where (i,j,k) is a permutation of (1,2,3).  The hyperradius is defined as R²=(r_(ij)²+ρ_(k,ij)²)/2.  We use notation r_(ij)=|(r)→_(ij)| and ρ_(ij)=|(ρ)→_(ij)|. The hyperangle α is defined as r_(ij)=2Rsinα_(k) and ρ_(k,ij)=√(2)Rcosα_(k) such that α_(k)=arctan(r_(ij)/ρ_(k,ij)).  The range of the hyperangle is restricted between 0 and π/2 such that r_(ij) and ρ_(k,ij) remain always positive.  We define the shorthand α≡α₃ as well as permutation operators (P)^₁₃ and (P)^₂₃, where (P)^_(ij) is the permutation operator that swaps particle indices i and j.

Consider the hyperangular part of the three-body wave function for an Efimov state ϕ(s,α)=(1+(Q)^)F(s,α)/√(N(s)), where F(s,α)=φ(s,α)/sin(2α), φ(s,α)=sin(s(π/2−α)), (Q)^=(P)^₁₃+(P)^₂₃, and N(s) is the normalization factor.

The values s are obtained from solving
(dφ(s,0))/(dα)+(8)/(√(3))φ(s,π/3)=0.
Assume s is a real number.  We are concerned with the first non-integer value of s that solves this equation.  Calculate your answer to three significant decimal places and record the value in the variable s₁.


Next, calculate the overlap of the hyperangular parts of the non-interacting and Efimov wave functions for s=s₁.  This overlap is P(s)=G(s)²/(N(s)H) for s=s₁.  

Here N(s), G(s), and F(s) are overlap integrals given by:

N(s)=∫₀^(π/2)dαsin(2α)²ϕ(s,α)²,

G(s)=∫₀^(π/2)dαsin(2α)²ϕ(s,α),

H=∫₀^(π/2)dαsin(2α)².


# Main problem:

Calculate N(s), H, and G(s) and use these results to obtain P(s₁) to three decimal places.  

# Problem setup:

The Efimov effect is one of the few examples of a three-body problem that can be solved exactly.  Let's calculate the the angular portion of the three-body wave function and its normalization constant for the Efimov effect.  

 Consider three identical bosons with spatial coordinates $\vec{r}_1$, $\vec{r}_2$, and $\vec{r}_3$.  Define Jacobi vectors $\vec{r}_{ij} = \vec{r}_i - \vec{r}_j$ and $\vec{\rho}_{k,ij} = (2\vec{r}_k - \vec{r}_i - \vec{r}_j)/\sqrt{3}$, where $(i,j,k)$ is a permutation of $(1,2,3)$.  The hyperradius is defined as $R^2 = (r_{ij}^2 + \rho_{k,ij}^2)/2$.  We use notation $r_{ij}= |\vec{r}_{ij}|$ and $\rho_{ij}= |\vec{\rho}_{ij}|$. The hyperangle $\alpha$ is defined as $r_{ij}=2R\sin\alpha_k$ and $\rho_{k,ij}=\sqrt{2}R\cos\alpha_k$ such that $\alpha_k = \arctan(r_{ij}/\rho_{k,ij})$.  The range of the hyperangle is restricted between $0$ and $\pi/2$ such that $r_{ij}$ and $\rho_{k,ij}$ remain always positive.  We define the shorthand $\alpha\equiv \alpha_3$ as well as permutation operators $\hat{P}_{13}$ and $\hat{P}_{23}$, where $\hat{P}_{ij}$ is the permutation operator that swaps particle indices $i$ and $j$.

Consider the hyperangular part of the three-body wave function for an Efimov state $\phi(s,\alpha) = (1 + \hat{Q})F(s,\alpha)/\sqrt{N(s)}$, where $F(s,\alpha) = \varphi(s,\alpha)/ \sin(2\alpha)$, $\varphi(s,\alpha) = \sin(s(\pi/ 2 - \alpha))$, $\hat{Q} = \hat{P}_{13} + \hat{P}_{23}$, and $N(s)$ is the normalization factor.

The values $s$ are obtained from solving
$$\frac{d\varphi(s,0)}{d\alpha} + \frac{8}{\sqrt{3}}\varphi(s,\pi/3)=0.$$
Assume $s$ is a real number.  We are concerned with the first non-integer value of $s$ that solves this equation.  Calculate your answer to three significant decimal places and record the value in the variable $s_1$.


Next, calculate the overlap of the hyperangular parts of the non-interacting and Efimov wave functions for $s=s_1$.  This overlap is $P(s) = G(s)^2/(N(s)H)$ for $s=s_1$.  

Here $N(s)$, $G(s)$, and $F(s)$ are overlap integrals given by:

$N(s) = \int_0^{\pi/2} d\alpha \sin(2\alpha)^2\phi(s,\alpha)^2$,

$G(s) = \int_0^{\pi/2} d\alpha \sin(2\alpha)^2\phi(s,\alpha)$,

$H = \int_0^{\pi/2} d\alpha \sin(2\alpha)^2$.


# Main problem:

Calculate $N(s)$, $H$, and $G(s)$ and use these results to obtain $P(s_1)$ to three decimal places.