Chapter 9 The Two-Body Problem

Chapter 9
1. The Two-Body Problem
In classical mechanics, a system composed of two particles interacting by a potential which depends only on the relative coordinate of the two particles is separable and can be solved and the solution is given by a definite integral [1]. This separability is carried over to quantum mechanics where the problem is reduced to that of the motion of a particle in a field of force. As we will see the angular part is completely solvable and the solution for the radial part of some potentials can be found by factorization method.
We start by writing the Hamiltonian for the two-body problem as

where and are the masses of the two particles, and represent their momenta and and their coordinates.
We can separate the translational part of the Hamiltonian by introducing the center of mass and relative coordinates;

and the center of mass and relative momenta

Substituting for and from (9.2) and (9.3) in (9.1) and simplifying the result we find

Here is the total mass, , and is the reduced mass

Equation (9.4) which is identical to its classical counterpart shows that is a cyclic operator since it commutes with and with , and the Hamiltonian is independent of the center of mass coordinate . Thus

and is a constant of motion. For this reason in the Hamiltonian (9.4) which has positive eigenvalues can be replaced by a -number multiplied by a unit operator.
The second term in , i.e. , is the kinetic energy of the internal degrees of freedom and can be split further into rotational and vibrational kinetic energies. To this end we define the radial momentum conjugate to by

where is added to to make Hermitian [2]. Next we observe that

Thus from (9.7) and (9.8) we have

Now taking the Hermitian conjugate of the operator relation (9.7) we get

Here we have used

and

where both and are assumed to be Hermitian operators.
The fact that does not prove that that is self-adjoint. Thus if we choose and to be two square integrable functions, then for the self-adjointness of we must have

We note that from (9.7) that we can write as

Using the differential form of as given by (9.14) and integrating by parts we find that [3]

Here we have used the integral

noting that is the end point of integration [4].
A different way of investigating the self-adjointness of is to find its deficiency indices. In Sec. 3.9 we have seen that the deficiency indices of the operator given by (9.14) is obtained by solving the differential equation [5]

with the result that

From this result it follows that while is square integrable for the range , and the function is not, and thus the deficiency indices are . We conclude that is not an observable, but as we will see it is a useful operator for calculating the eigenvalues .
Our aim is to express in terms of and for this we decompose into components parallel and perpendicular to . This can be done with the well-known formula from vector calculus (see Sec. 3.3)

Next we find the scalar product of in (9.19) with multiplying from the left;

We also have

Thus from (9.20) and (9.21) we find

Using the definition of

we can write as

In addition we have

Finally from (9.24) and (9.25) we obtain

where

is the angular momentum about the center of mass. The reason that we can write the second term in (9.26) without specifying the order of and is that these two operators commute. In fact commutes with and with any function of (see Sec. 3.3).