# Decoupling the two-particle Hamiltonian

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercise 10.1.3.

Shankar shows that, for a two-particle system, the state vector ${\left|\psi\right\rangle }$ is an element of the direct product space ${\mathbb{V}_{1\otimes2}}$. Its evolution in time is determined by the Schrödinger equation, as usual, so that

$\displaystyle i\hbar\left|\dot{\psi}\right\rangle =H\left|\psi\right\rangle =\left[\frac{P_{1}^{2}}{2m_{1}}+\frac{P_{2}^{2}}{2m_{2}}+V\left(X_{1},X_{2}\right)\right]\left|\psi\right\rangle \ \ \ \ \ (1)$

The method by which this equation can be solved (if it can be solved, that is) depends on the form of the potential ${V}$. If the two particles interact only with some external potential, and not with each other, then ${V}$ is composed of a sum of terms, each of which depends only on ${X_{1}}$ or ${X_{2}}$, but not on both. In such cases, we can split ${H}$ into two parts, one of which (${H_{1}}$) depends only on operators pertaining to particle 1 and the other (${H_{2}}$) on operators pertaining to particle 2. If the eigenvalues (allowed energies) of particle ${i}$ are given by ${E_{i}}$, then the stationary states are direct products of the corresponding single particle eigenstates. That is, in general

$\displaystyle H\left|E\right\rangle =\left(H_{1}+H_{2}\right)\left|E_{1}\right\rangle \otimes\left|E_{2}\right\rangle =\left(E_{1}+E_{2}\right)\left|E_{1}\right\rangle \otimes\left|E_{2}\right\rangle =E\left|E\right\rangle \ \ \ \ \ (2)$

Thus the two-particle state ${\left|E\right\rangle =\left|E_{1}\right\rangle \otimes\left|E_{2}\right\rangle }$. Since a stationary state ${\left|E_{i}\right\rangle }$ evolves in time according to

$\displaystyle \left|\psi_{i}\left(t\right)\right\rangle =\left|E_{i}\right\rangle e^{-iE_{i}t/\hbar} \ \ \ \ \ (3)$

the compound two-particle state evolves according to

 $\displaystyle \left|\psi\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle e^{-iE_{1}t/\hbar}\left|E_{1}\right\rangle \otimes e^{-iE_{2}t/\hbar}\left|E_{2}\right\rangle \ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-i\left(E_{1}+E_{2}\right)t/\hbar}\left|E\right\rangle \ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-iEt/\hbar}\left|E\right\rangle \ \ \ \ \ (6)$

In this case, the two particles are essentially independent of each other, and the compound state is just the product of the two separate one-particle states.

If ${H}$ is not separable, which will occur if ${V}$ contains terms involving both ${X_{1}}$ and ${X_{2}}$ in the same term, we cannot, in general, reduce the system to the product of two one-particle systems. There are a couple of instances, however, where such a reduction can be done.

The first instance is if the potential is a function of ${x_{2}-x_{1}}$ only, in other words, that the interaction between the particles depends only on the distance between them. Shankar shows that in this case we can transform the system to that of a reduced mass ${\mu=m_{1}m_{2}/\left(m_{1}+m_{2}\right)}$ and a centre of mass ${M=m_{1}+m_{2}}$. We’ve already seen this problem solved by means of separation of variables. The result is that the state vector is the product of a vector for a free particle of mass ${M}$ and of a vector of a particle with reduced mass ${\mu}$ moving in the potential ${V}$.

Another case where we can decouple the Hamiltonian is in a system of harmonic oscillators. We’ve already seen this system solved for two masses in classical mechanics using diagonalization of the matrix describing the equations of motion. The classical Hamiltonian is

$\displaystyle H=\frac{p_{1}^{2}}{2m}+\frac{p_{2}^{2}}{2m}+\frac{m\omega^{2}}{2}\left[x_{1}^{2}+x_{2}^{2}+\left(x_{1}-x_{2}\right)^{2}\right] \ \ \ \ \ (7)$

The earlier solution involved introducing normal coordinates

 $\displaystyle x_{I}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(x_{1}+x_{2}\right)\ \ \ \ \ (8)$ $\displaystyle x_{II}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(x_{1}-x_{2}\right) \ \ \ \ \ (9)$

and corresponding momenta

 $\displaystyle p_{I}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(p_{1}+p_{2}\right)\ \ \ \ \ (10)$ $\displaystyle p_{II}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(p_{1}-p_{2}\right) \ \ \ \ \ (11)$

These normal coordinates are canonical as we can verify by calculating the Poisson brackets. For example

 $\displaystyle \left\{ x_{I},p_{I}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial x_{I}}{\partial x_{i}}\frac{\partial p_{I}}{\partial p_{i}}-\frac{\partial x_{I}}{\partial p_{i}}\frac{\partial p_{I}}{\partial x_{i}}\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (13)$ $\displaystyle \left\{ x_{I},x_{II}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial x_{I}}{\partial x_{i}}\frac{\partial x_{II}}{\partial p_{i}}-\frac{\partial x_{I}}{\partial p_{i}}\frac{\partial x_{II}}{\partial x_{i}}\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (15)$

and so on, with the general result

 $\displaystyle \left\{ x_{i},p_{j}\right\}$ $\displaystyle =$ $\displaystyle \delta_{ij}\ \ \ \ \ (16)$ $\displaystyle \left\{ x_{i},x_{j}\right\}$ $\displaystyle =$ $\displaystyle \left\{ p_{i},p_{j}\right\} =0 \ \ \ \ \ (17)$

We can invert the transformation to get

 $\displaystyle x_{1}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(x_{I}+x_{II}\right)\ \ \ \ \ (18)$ $\displaystyle x_{2}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(x_{I}-x_{II}\right) \ \ \ \ \ (19)$

and

 $\displaystyle p_{1}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(p_{I}+p_{II}\right)\ \ \ \ \ (20)$ $\displaystyle p_{2}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(p_{I}-p_{II}\right) \ \ \ \ \ (21)$

Inserting these into 7 we get

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \frac{1}{4m}\left[\left(p_{I}+p_{II}\right)^{2}+\left(p_{I}-p_{II}\right)^{2}\right]+\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{m\omega^{2}}{4}\left[\left(x_{I}+x_{II}\right)^{2}+\left(x_{I}-x_{II}\right)^{2}+x_{II}^{2}\right]\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{p_{I}^{2}}{2m}+\frac{p_{II}^{2}}{2m}+\frac{m\omega^{2}}{2}\left(x_{I}^{2}+2x_{II}^{2}\right) \ \ \ \ \ (24)$

We can now subsitute the usual quantum mechanical operators to get the quantum Hamiltonian:

$\displaystyle H=-\frac{\hbar^{2}}{2m}\left(P_{I}^{2}+P_{II}^{2}\right)+\frac{m\omega^{2}}{2}\left(X_{I}^{2}+2X_{II}^{2}\right) \ \ \ \ \ (25)$

In the coordinate basis, this is

$\displaystyle H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial x_{I}^{2}}+\frac{\partial^{2}}{\partial x_{II}^{2}}\right)+\frac{m\omega^{2}}{2}\left(x_{I}^{2}+2x_{II}^{2}\right) \ \ \ \ \ (26)$

The Hamiltonian is now decoupled and can be solved by separation of variables.

We could have arrived at this result by starting with 7 and promoting ${x_{i}}$ and ${p_{i}}$ to quantum operators directly, then made the substitution to normal coordinates. We would then start with

$\displaystyle H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial x_{1}^{2}}+\frac{\partial^{2}}{\partial x_{2}^{2}}\right)+\frac{m\omega^{2}}{2}\left[x_{1}^{2}+x_{2}^{2}+\left(x_{1}-x_{2}\right)^{2}\right] \ \ \ \ \ (27)$

The potential term on the right transforms the same way as before, so we get

$\displaystyle \frac{m\omega^{2}}{2}\left[x_{1}^{2}+x_{2}^{2}+\left(x_{1}-x_{2}\right)^{2}\right]\rightarrow\frac{m\omega^{2}}{2}\left(x_{I}^{2}+2x_{II}^{2}\right) \ \ \ \ \ (28)$

To transform the two derivatives, we need to use the chain rule a couple of times. To get the first derivatives:

 $\displaystyle \frac{\partial\psi}{\partial x_{1}}$ $\displaystyle =$ $\displaystyle \frac{\partial\psi}{\partial x_{I}}\frac{\partial x_{I}}{\partial x_{1}}+\frac{\partial\psi}{\partial x_{II}}\frac{\partial x_{II}}{\partial x_{1}}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\frac{\partial\psi}{\partial x_{I}}+\frac{\partial\psi}{\partial x_{II}}\right)\ \ \ \ \ (30)$ $\displaystyle \frac{\partial\psi}{\partial x_{2}}$ $\displaystyle =$ $\displaystyle \frac{\partial\psi}{\partial x_{I}}\frac{\partial x_{I}}{\partial x_{2}}+\frac{\partial\psi}{\partial x_{II}}\frac{\partial x_{II}}{\partial x_{2}}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\frac{\partial\psi}{\partial x_{I}}-\frac{\partial\psi}{\partial x_{II}}\right) \ \ \ \ \ (32)$

Now the second derivatives:

 $\displaystyle \frac{\partial^{2}\psi}{\partial x_{1}^{2}}$ $\displaystyle =$ $\displaystyle \frac{\partial}{\partial x_{I}}\left(\frac{\partial\psi}{\partial x_{1}}\right)\frac{\partial x_{I}}{\partial x_{1}}+\frac{\partial}{\partial x_{II}}\left(\frac{\partial\psi}{\partial x_{1}}\right)\frac{\partial x_{II}}{\partial x_{1}}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\frac{\partial}{\partial x_{I}}\left(\frac{\partial\psi}{\partial x_{I}}+\frac{\partial\psi}{\partial x_{II}}\right)+\frac{\partial}{\partial x_{II}}\left(\frac{\partial\psi}{\partial x_{I}}+\frac{\partial\psi}{\partial x_{II}}\right)\right]\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\frac{\partial^{2}\psi}{\partial x_{I}^{2}}+2\frac{\partial^{2}\psi}{\partial x_{I}\partial x_{II}}+\frac{\partial^{2}\psi}{\partial x_{II}^{2}}\right]\ \ \ \ \ (35)$ $\displaystyle \frac{\partial^{2}\psi}{\partial x_{2}^{2}}$ $\displaystyle =$ $\displaystyle \frac{\partial}{\partial x_{I}}\left(\frac{\partial\psi}{\partial x_{2}}\right)\frac{\partial x_{I}}{\partial x_{1}}+\frac{\partial}{\partial x_{II}}\left(\frac{\partial\psi}{\partial x_{2}}\right)\frac{\partial x_{II}}{\partial x_{1}}\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\frac{\partial}{\partial x_{I}}\left(\frac{\partial\psi}{\partial x_{I}}-\frac{\partial\psi}{\partial x_{II}}\right)-\frac{\partial}{\partial x_{II}}\left(\frac{\partial\psi}{\partial x_{I}}-\frac{\partial\psi}{\partial x_{II}}\right)\right]\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\frac{\partial^{2}\psi}{\partial x_{I}^{2}}-2\frac{\partial^{2}\psi}{\partial x_{I}\partial x_{II}}+\frac{\partial^{2}\psi}{\partial x_{II}^{2}}\right] \ \ \ \ \ (38)$

Combining the two derivatives, we get

$\displaystyle \frac{\partial^{2}\psi}{\partial x_{1}^{2}}+\frac{\partial^{2}\psi}{\partial x_{2}^{2}}=\frac{\partial^{2}\psi}{\partial x_{I}^{2}}+\frac{\partial^{2}\psi}{\partial x_{II}^{2}} \ \ \ \ \ (39)$

Inserting this, together with 28, into 27 we get 26 again.

# Hamiltonian for the two-body problem

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.5; Exercise 2.5.4.

Here we derive the equations of motion of the two-body problem using the Hamiltonian formalism.

The Hamiltonian is given by

$\displaystyle H\left(q,p\right)=\sum_{i}p_{i}\dot{q}_{i}-L\left(q,\dot{q}\right) \ \ \ \ \ (1)$

where the velocities ${\dot{q}_{i}}$ are expressed in terms of the positions ${q_{i}}$ and momenta ${p_{i}}$. In this case, we start with the Lagrangian in terms of the centre of mass position ${\mathbf{r}_{CM}}$ and the relative position ${\mathbf{r}}$ of mass 2 to mass 1.

 $\displaystyle L$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(m_{1}+m_{2}\right)\left|\dot{\mathbf{r}}_{CM}\right|^{2}+\frac{1}{2}\frac{m_{1}m_{2}}{m_{1}+m_{2}}\left|\dot{\mathbf{r}}\right|^{2}-V\left(\mathbf{r}\right)\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{M}{2}\left|\dot{\mathbf{r}}_{CM}\right|^{2}+\frac{\mu}{2}\left|\dot{\mathbf{r}}\right|^{2}-V\left(\mathbf{r}\right) \ \ \ \ \ (3)$

where ${M=m_{1}+m_{2}}$ is the total mass and ${\mu=\frac{m_{1}m_{2}}{m_{1}+m_{2}}}$ is the reduced mass.

There are potentially 6 velocity components and 6 coordinate components in the Lagrangian, but the 3 components of ${\mathbf{r}_{CM}}$ do not appear, which simplifies things a bit. To convert to a Hamiltonian, we need the momenta

$\displaystyle p_{i}=\frac{\partial L}{\partial\dot{q}_{i}} \ \ \ \ \ (4)$

The ${x}$ component of momentum of the centre of mass is

$\displaystyle p_{CM,x}=\frac{\partial L}{\partial\dot{r}_{CM,x}}=M\dot{r}_{CM,x} \ \ \ \ \ (5)$

The other two components of the centre of mass velocity, and of the relative velocity, have a similar form, and in general we can write

 $\displaystyle p_{CM,i}$ $\displaystyle =$ $\displaystyle M\dot{r}_{CM,i}\ \ \ \ \ (6)$ $\displaystyle p_{i}$ $\displaystyle =$ $\displaystyle \mu\dot{r}_{i} \ \ \ \ \ (7)$

In vector notation, this becomes

 $\displaystyle \dot{\mathbf{r}}_{CM}$ $\displaystyle =$ $\displaystyle \frac{\mathbf{p}_{CM}}{M}\ \ \ \ \ (8)$ $\displaystyle \dot{\mathbf{r}}$ $\displaystyle =$ $\displaystyle \frac{\mathbf{p}}{\mu}\ \ \ \ \ (9)$ $\displaystyle \left|\dot{\mathbf{r}}_{CM}\right|^{2}$ $\displaystyle =$ $\displaystyle \frac{\left|\mathbf{p}_{CM}\right|^{2}}{M^{2}}\ \ \ \ \ (10)$ $\displaystyle \left|\dot{\mathbf{r}}\right|^{2}$ $\displaystyle =$ $\displaystyle \frac{\left|\mathbf{p}\right|^{2}}{\mu^{2}} \ \ \ \ \ (11)$

The Lagrangian thus becomes

$\displaystyle L=\frac{\left|\mathbf{p}_{CM}\right|^{2}}{2M}+\frac{\left|\mathbf{p}\right|^{2}}{2\mu}-V\left(\mathbf{r}\right) \ \ \ \ \ (12)$

The Hamiltonian is

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \mathbf{p}\cdot\dot{\mathbf{r}}+\mathbf{p}_{CM}\cdot\dot{\mathbf{r}}_{CM}-L\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left|\mathbf{p}\right|^{2}}{\mu}+\frac{\left|\mathbf{p}_{CM}\right|^{2}}{M}-\left[\frac{\left|\mathbf{p}_{CM}\right|^{2}}{2M}+\frac{\left|\mathbf{p}\right|^{2}}{2\mu}-V\left(\mathbf{r}\right)\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left|\mathbf{p}_{CM}\right|^{2}}{2M}+\frac{\left|\mathbf{p}\right|^{2}}{2\mu}+V\left(\mathbf{r}\right) \ \ \ \ \ (15)$

Once we’ve got the Hamiltonian, we can apply Hamilton’s canonical equations to get the equations of motion.

 $\displaystyle \frac{\partial H}{\partial p_{i}}$ $\displaystyle =$ $\displaystyle \dot{r}_{i}\ \ \ \ \ (16)$ $\displaystyle -\frac{\partial H}{\partial r_{i}}$ $\displaystyle =$ $\displaystyle \dot{p}_{i} \ \ \ \ \ (17)$

Since ${\mathbf{r}_{CM}}$ does not appear in the Hamiltonian, we have

 $\displaystyle \dot{\mathbf{p}}_{CM}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (18)$ $\displaystyle \mathbf{p}_{CM}$ $\displaystyle =$ $\displaystyle \mbox{constant} \ \ \ \ \ (19)$

so the momentum of the centre of mass does not change, as expected.

For ${\mathbf{r}}$, we have

 $\displaystyle \frac{\partial H}{\partial p_{i}}$ $\displaystyle =$ $\displaystyle \frac{p_{i}}{\mu}=\dot{r}_{i}\ \ \ \ \ (20)$ $\displaystyle \frac{\partial H}{\partial r_{i}}$ $\displaystyle =$ $\displaystyle \frac{\partial V}{\partial r_{i}}=-\dot{p}_{i} \ \ \ \ \ (21)$

The first equation tells us nothing new, while the second is just Newton’s law for a central force: ${\mathbf{\dot{p}}=-\nabla V}$.

# Lagrangian for the two-body problem

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.3; Exercise 2.3.1.

A fundamental problem in classical physics is the two-body problem, in which two masses interact via a potential ${V\left(\mathbf{r}_{1}-\mathbf{r}_{2}\right)}$ that depends only on the relative positions of the two masses. In such a case, the Lagrangian can be decoupled so that the problem gets reduced to a one-body problem.

The Euler-Lagrange equations are

$\displaystyle \frac{d}{dt}\frac{\partial L}{\partial\dot{q}_{i}}-\frac{\partial L}{\partial q_{i}}=0 \ \ \ \ \ (1)$

where ${q_{i}}$ and ${\dot{q_{i}}}$ are the generalized coordinates and velocities, respectively. For systems where the potential energy ${V\left(q_{i}\right)}$ is independent of the velocities ${\dot{q}_{i}}$, the Lagrangian can be written as

$\displaystyle L=T-V \ \ \ \ \ (2)$

where ${T}$ is the kinetic energy. In terms of the absolute positions and velocities, we have

$\displaystyle L=\frac{1}{2}m_{1}\left|\dot{\mathbf{r}}_{1}\right|^{2}+\frac{1}{2}m_{2}\left|\dot{\mathbf{r}}_{2}\right|^{2}-V\left(\mathbf{r}_{1}-\mathbf{r}_{2}\right) \ \ \ \ \ (3)$

To decouple this equation, we define two new position vectors:

 $\displaystyle \mathbf{r}$ $\displaystyle \equiv$ $\displaystyle \mathbf{r}_{1}-\mathbf{r}_{2}\ \ \ \ \ (4)$ $\displaystyle \mathbf{r}_{CM}$ $\displaystyle \equiv$ $\displaystyle \frac{m_{1}\mathbf{r}_{1}+m_{2}\mathbf{r}_{2}}{m_{1}+m_{2}} \ \ \ \ \ (5)$

Here ${\mathbf{r}}$ is the relative position, and ${\mathbf{r}_{CM}}$ is the position of the centre of mass.

We can invert these equations to get

 $\displaystyle \mathbf{r}_{1}$ $\displaystyle =$ $\displaystyle \mathbf{r}+\mathbf{r}_{2}\ \ \ \ \ (6)$ $\displaystyle \left(m_{1}+m_{2}\right)\mathbf{r}_{CM}$ $\displaystyle =$ $\displaystyle m_{1}\mathbf{r}+\left(m_{1}+m_{2}\right)\mathbf{r}_{2}\ \ \ \ \ (7)$ $\displaystyle \mathbf{r}_{2}$ $\displaystyle =$ $\displaystyle \mathbf{r}_{CM}-\frac{m_{1}}{m_{1}+m_{2}}\mathbf{r}\ \ \ \ \ (8)$ $\displaystyle \mathbf{r}_{1}$ $\displaystyle =$ $\displaystyle \mathbf{r}_{CM}-\frac{m_{2}}{m_{1}+m_{2}}\mathbf{r} \ \ \ \ \ (9)$

To decouple the Lagrangian, we insert these last two equations into 3.

 $\displaystyle m_{1}\left|\dot{\mathbf{r}}_{1}\right|^{2}$ $\displaystyle =$ $\displaystyle m_{1}\left[\dot{\mathbf{r}}_{CM}-\frac{m_{2}}{m_{1}+m_{2}}\dot{\mathbf{r}}\right]\cdot\left[\dot{\mathbf{r}}_{CM}-\frac{m_{2}}{m_{1}+m_{2}}\dot{\mathbf{r}}\right]\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle m_{1}\left|\dot{\mathbf{r}}_{CM}\right|^{2}-2\frac{m_{1}m_{2}}{m_{1}+m_{2}}\dot{\mathbf{r}}_{CM}\cdot\dot{\mathbf{r}}+m_{1}\left(\frac{m_{2}}{m_{1}+m_{2}}\right)^{2}\left|\dot{\mathbf{r}}\right|^{2}\ \ \ \ \ (11)$ $\displaystyle m_{2}\left|\dot{\mathbf{r}}_{2}\right|^{2}$ $\displaystyle =$ $\displaystyle m_{2}\left[\dot{\mathbf{r}}_{CM}+\frac{m_{1}}{m_{1}+m_{2}}\dot{\mathbf{r}}\right]\cdot\left[\dot{\mathbf{r}}_{CM}+\frac{m_{1}}{m_{1}+m_{2}}\dot{\mathbf{r}}\right]\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle m_{2}\left|\dot{\mathbf{r}}_{CM}\right|^{2}+2\frac{m_{1}m_{2}}{m_{1}+m_{2}}\dot{\mathbf{r}}_{CM}\cdot\dot{\mathbf{r}}+m_{2}\left(\frac{m_{1}}{m_{1}+m_{2}}\right)^{2}\left|\dot{\mathbf{r}}\right|^{2}\ \ \ \ \ (13)$ $\displaystyle \frac{1}{2}m_{1}\left|\dot{\mathbf{r}}_{1}\right|^{2}+\frac{1}{2}m_{2}\left|\dot{\mathbf{r}}_{2}\right|^{2}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(m_{1}+m_{2}\right)\left|\dot{\mathbf{r}}_{CM}\right|^{2}+\frac{1}{2}\frac{m_{1}m_{2}^{2}+m_{2}m_{1}^{2}}{\left(m_{1}+m_{2}\right)^{2}}\left|\dot{\mathbf{r}}\right|^{2}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(m_{1}+m_{2}\right)\left|\dot{\mathbf{r}}_{CM}\right|^{2}+\frac{1}{2}\frac{m_{1}m_{2}}{m_{1}+m_{2}}\left|\dot{\mathbf{r}}\right|^{2} \ \ \ \ \ (15)$

The Lagrangian 3 thus becomes

 $\displaystyle L$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(m_{1}+m_{2}\right)\left|\dot{\mathbf{r}}_{CM}\right|^{2}+\frac{1}{2}\frac{m_{1}m_{2}}{m_{1}+m_{2}}\left|\dot{\mathbf{r}}\right|^{2}-V\left(\mathbf{r}\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle L_{CM}+L_{r} \ \ \ \ \ (17)$

with

 $\displaystyle L_{CM}$ $\displaystyle \equiv$ $\displaystyle \frac{1}{2}\left(m_{1}+m_{2}\right)\left|\dot{\mathbf{r}}_{CM}\right|^{2}\ \ \ \ \ (18)$ $\displaystyle L_{r}$ $\displaystyle \equiv$ $\displaystyle \frac{1}{2}\frac{m_{1}m_{2}}{m_{1}+m_{2}}\left|\dot{\mathbf{r}}\right|^{2}-V\left(\mathbf{r}\right) \ \ \ \ \ (19)$

Thus ${L}$ decouples into two Lagrangians, one of which depends only on ${\dot{\mathbf{r}}_{CM}}$ and the other of which depends only on ${\mathbf{r}}$ and ${\dot{\mathbf{r}}}$. The absence of ${\mathbf{r}_{CM}}$ means that, from 1

 $\displaystyle \frac{d}{dt}\frac{\partial L}{\partial\dot{r}_{i,CM}}$ $\displaystyle =$ $\displaystyle \frac{d}{dt}\frac{\partial L_{CM}}{\partial\dot{r}_{i,CM}}=\frac{m_{1}+m_{2}}{2}\frac{d\dot{r}_{i,CM}}{dt}=0\ \ \ \ \ (20)$ $\displaystyle \dot{r}_{i,CM}$ $\displaystyle =$ $\displaystyle \mbox{constant} \ \ \ \ \ (21)$

which is separately true for each component of ${\dot{\mathbf{r}}_{CM}}$, which shows that the velocity of the centre of mass is a constant, as we’d expect for an isolated two-body system with no external force.

From the other Lagrangian, we get

$\displaystyle \frac{m_{1}m_{2}}{m_{1}+m_{2}}\ddot{\mathbf{r}}=-\nabla V\left(\mathbf{r}\right) \ \ \ \ \ (22)$

which is the equation of motion of a single particle of mass ${\frac{m_{1}m_{2}}{m_{1}+m_{2}}}$, called the reduced mass. Viewed from the centre of mass frame, where ${\dot{\mathbf{r}}_{CM}=0}$, ${\mathbf{r}}$ becomes the absolute position of the reduced mass. We can transform the result back to the ‘absolute’ frame by using 4.