Eckart frame Hamiltonians in the three-body problem

Abstract

The Eckart frame is used to separate out the collective rotations in the quantum three-body problem. Explicit expressions for the corresponding rotational and vibro-rotational (i.e. Coriolis) Hamiltonians are derived. Special attention is paid to the situation when two principal moments of inertia are equal in the equilibrium configuration.

Appendices

Appendix A: Relations between Eckart and Jacobi-bond coordinates

Below the connections between the set of Eckart and Jacobi internal coordinates are presented. Such connections follow from the definition of Eckart vectors (9) and the properties of the principal inertia axes. Omitting details of somewhat cumbersome computations we present here only the final results,

$$\begin{aligned} f_1^{2} = \frac{I_2}{I_1 - I_2} \left( r_1^{2} (I_1 - \rho _1^{2}) + r_2^{2} (I_1 - \rho _2^{2}) - 2 (\mathbf{r}_1 \cdot \mathbf{r}_2 )\, ( \varvec{\rho }_1 \cdot \varvec{\rho }_2 ) \right). \end{aligned}$$

(34)

For the vector $\mathbf{f}_2$ one has

$$\begin{aligned} f_2^{2} = \frac{-I_1}{I_1 - I_2} \left( r_1^{2} (I_2 - \rho _1^{2}) + r_2^{2} (I_2 - \rho _2^{2}) - 2 (\mathbf{r}_1 \cdot \mathbf{r}_2 )\, ( \varvec{\rho }_1 \cdot \varvec{\rho }_2 ) \right), \end{aligned}$$

(35)

The scalar product of Eckart vectors expresses as

$$\begin{aligned} (\mathbf{f}_1 \cdot \mathbf{f}_2) = \frac{| \varvec{\rho }_1 \times \varvec{\rho }_2 |}{I_1 - I_2} \left( (\varvec{\rho }_1 \cdot \varvec{\rho }_2)\, (r_1^{2}-r_2^{2}) - (\mathbf{r}_1 \cdot \mathbf{r}_2)\, (\rho _1^{2} - \rho _2^{2}) \right). \end{aligned}$$

(36)

As is seen, the scalar product $(\mathbf{f}_1 \cdot \mathbf{f}_2)$ vanishes in the rigid-body limit when the vibration amplitudes tend to zero. The analysis of the rigid body limit of parameters $f_1$ and $f_2$ is slightly more complicated. Namely, in the rigid-body limit we have $\mathbf{r}_1 \rightarrow \varvec{\rho }_1$ and $\mathbf{r}_2 \rightarrow \varvec{\rho }_2$ which, noting the definitions (9), leads to

$$\begin{aligned} \left(f_1^{2} \right)_{eq} = x_1^{2} \rho _1^{2} + x_2^{2} \rho _2^{2} + 2 x_1 x_2 (\varvec{\rho }_1 \cdot \varvec{\rho }_2). \end{aligned}$$

(37)

Now we re-write $\rho$’s in terms of their coordinates according to (8). This yields

$$\begin{aligned} \left(f_1^{2} \right)_{eq} = \left( x_1^{2} + x_2^{2} \right)^{2} + (x_1 y_1 + x_2 y_2)^{2} = I_2^{2}. \end{aligned}$$

(38)

Here, in the course of derivations we have utilized Eqs. (17) and (16).

Thus, in the limit of zero vibrations one has

$$\begin{aligned} (f_1)_{eq}&= I_2, \quad (f_2)_{eq} = I_1, \nonumber \\ \phi _{eq}&= \frac{\pi }{2}. \end{aligned}$$

(39)

Clearly, the above Eqs. (34)–(36) are invalid when the equilibrium principal moments of inertia are equal. According to Eq. (44) of the next Appendix the condition $I_1 = I_2$ is met only if

$$\begin{aligned} \rho _1 = \rho _2, \quad \theta _e = \pi /2, \end{aligned}$$

(40)

where $\theta _e$ is the mutual angle in the equilibrium configuration, see Fig. 1. Thus, the vectors $\varvec{\rho }_1$ and $\varvec{\rho }_2$ are perpendicular and, hence, we can choose them to define the Cartesian basis, i.e.

$$\begin{aligned} \mathbf{e}_1 = \frac{\varvec{\rho }_1}{\rho _1}, \quad \mathbf{e}_2 = \frac{\varvec{\rho }_2}{\rho _2}. \end{aligned}$$

(41)

As a consequence we have that the coordinates in Eq. (8) become

$$\begin{aligned} x_1 = \rho _1, \quad y_2 = \rho _2, \quad x_2 = y_1 =0. \end{aligned}$$

From these identities and from (9) we obtain that at $I_1 = I_2$ the Eckart vectors can be chosen as

$$\begin{aligned} \mathbf{f}_1 = \rho \mathbf{r}_1, \quad \mathbf{f}_2 = \rho \mathbf{r}_2, \end{aligned}$$

(42)

where $\rho = \rho _1 = \rho _2$. The corresponding Eckart parameter (12) expresses as

$$\begin{aligned} \mathcal{F }^{2} = \rho ^{2} \, \bigl ( r_1^{2} + r_2^{2} + 2 r_1 r_2 \sin \theta \bigr ). \end{aligned}$$

(43)

Appendix B: The equilibrium inertia moments

The straightforward calculation of the eigenvalues of the inertia tensor leads to the following relations for the principal inertia moments of the equilibrium configuration

$$\begin{aligned} I_1&= \frac{\rho _1^{2}+\rho _2^{2}}{2} + \frac{1}{2}\sqrt{ (\rho _1^{2} - \rho _2^{2})^{2} + (2 \rho _1 \rho _2 \cos \theta _e)^{2}}, \nonumber \\ I_2&= \frac{\rho _1^{2}+\rho _2^{2}}{2} - \frac{1}{2}\sqrt{ (\rho _1^{2}-\rho _2^{2})^{2} + (2 \rho _1 \rho _2 \cos \theta _e )^{2}}. \end{aligned}$$

(44)

We remind that $\rho _{1,2}$ are mass-scaled Jacobi vectors. To obtain expressions in terms of conventional Jacobi vectors one should apply the replacements $\rho _{1,2} \rightarrow \rho _{1,2} \, \sqrt{\mu }_{1,2}$ to the above Eq. (44).

As is seen from (44) the principal inertia moments are equal if Jacobi vectors are perpendicular (i.e. when $\theta _e = \pi /2$) and their lengths satisfy the equation

$$\begin{aligned} \mu _1 \rho _1^{2} = \mu _2 \rho _2^{2}. \end{aligned}$$

(45)