The hydrogen atom contains two particles, the proton and the electron. For a system of two particles 1 and 2 with coordinates $\left(x{1}, y{1}, z{1}\right)$ and $\left(x{2}, y{2}, z{2}\right)$, the potential energy of interaction between the particles is usually a function of only the relative coordinates $x{2}-x{1}, y{2}-y{1}$, and $z{2}-z{1}$ of the particles. In this case the two-particle problem can be simplified to two separate one-particle problems, as we now prove.
Consider the classical-mechanical treatment of two interacting particles of masses $m{1}$ and $m{2}$. We specify their positions by the radius vectors $\mathbf{r}{1}$ and $\mathbf{r}{2}$ drawn from the origin
FIGURE 6.1 A two-particle system with center of mass at $C$.
of a Cartesian coordinate system (Fig. 6.1). Particles 1 and 2 have coordinates $\left(x{1}, y{1}, z{1}\right)$ and $\left(x{2}, y{2}, z{2}\right)$. We draw the vector $\mathbf{r}=\mathbf{r}{2}-\mathbf{r}{1}$ from particle 1 to 2 and denote the components of $\mathbf{r}$ by $x, y$, and $z$ :
\(
\begin{equation}
x=x{2}-x{1}, \quad y=y{2}-y{1}, \quad z=z{2}-z{1} \tag{6.28}
\end{equation}
\)
The coordinates $x, y$, and $z$ are called the relative or internal coordinates.
We now draw the vector $\mathbf{R}$ from the origin to the system's center of mass, point $C$, and denote the coordinates of $C$ by $X, Y$, and $Z$ :
\(
\begin{equation}
\mathbf{R}=\mathbf{i} X+\mathbf{j} Y+\mathbf{k} Z \tag{6.29}
\end{equation}
\)
The definition of the center of mass of this two-particle system gives
\(
\begin{equation}
X=\frac{m{1} x{1}+m{2} x{2}}{m{1}+m{2}}, \quad Y=\frac{m{1} y{1}+m{2} y{2}}{m{1}+m{2}}, \quad Z=\frac{m{1} z{1}+m{2} z{2}}{m{1}+m{2}} \tag{6.30}
\end{equation}
\)
These three equations are equivalent to the vector equation
\(
\begin{equation}
\mathbf{R}=\frac{m{1} \mathbf{r}{1}+m{2} \mathbf{r}{2}}{m{1}+m{2}} \tag{6.31}
\end{equation}
\)
We also have
\(
\begin{equation}
\mathbf{r}=\mathbf{r}{2}-\mathbf{r}{1} \tag{6.32}
\end{equation}
\)
We regard (6.31) and (6.32) as simultaneous equations in the two unknowns $\mathbf{r}{1}$ and $\mathbf{r}{2}$ and solve for them to get
\(
\begin{equation}
\mathbf{r}{1}=\mathbf{R}-\frac{m{2}}{m{1}+m{2}} \mathbf{r}, \quad \mathbf{r}{2}=\mathbf{R}+\frac{m{1}}{m{1}+m{2}} \mathbf{r} \tag{6.33}
\end{equation}
\)
Equations (6.31) and (6.32) represent a transformation of coordinates from $x{1}, y{1}, z{1}, x{2}, y{2}, z{2}$ to $X, Y, Z, x, y, z$. Consider what happens to the Hamiltonian under this transformation. Let an overhead dot indicate differentiation with respect to time. The velocity of particle 1 is [Eq. (5.34)] $\mathbf{v}{1}=d \mathbf{r}{1} / d t=\dot{\mathbf{r}}_{1}$. The kinetic energy is the sum of the kinetic energies of the two particles:
\(
\begin{equation}
T=\frac{1}{2} m{1}\left|\dot{\mathbf{r}}{1}\right|^{2}+\frac{1}{2} m{2}\left|\dot{\mathbf{r}}{2}\right|^{2} \tag{6.34}
\end{equation}
\)
Introducing the time derivatives of Eqs. (6.33) into (6.34), we have
\(
\begin{aligned}
T= & \frac{1}{2} m{1}\left(\dot{\mathbf{R}}-\frac{m{2}}{m{1}+m{2}} \dot{\mathbf{r}}\right) \cdot\left(\dot{\mathbf{R}}-\frac{m{2}}{m{1}+m{2}} \dot{\mathbf{r}}\right) \
& +\frac{1}{2} m{2}\left(\dot{\mathbf{R}}+\frac{m{1}}{m{1}+m{2}} \dot{\mathbf{r}}\right) \cdot\left(\dot{\mathbf{R}}+\frac{m{1}}{m{1}+m{2}} \dot{\mathbf{r}}\right)
\end{aligned}
\)
where $|\mathbf{A}|^{2}=\mathbf{A} \cdot \mathbf{A}[$ Eq. (5.24)] has been used. Using the distributive law for the dot products, we find, after simplifying,
\(
\begin{equation}
T=\frac{1}{2}\left(m{1}+m{2}\right)|\dot{\mathbf{R}}|^{2}+\frac{1}{2} \frac{m{1} m{2}}{m{1}+m{2}}|\dot{\mathbf{r}}|^{2} \tag{6.35}
\end{equation}
\)
Let $M$ be the total mass of the system:
\(
\begin{equation}
M \equiv m{1}+m{2} \tag{6.36}
\end{equation}
\)
We define the reduced mass $\mu$ of the two-particle system as
\(
\begin{equation}
\mu \equiv \frac{m{1} m{2}}{m{1}+m{2}} \tag{6.37}
\end{equation}
\)
Then
\(
\begin{equation}
T=\frac{1}{2} M|\dot{\mathbf{R}}|^{2}+\frac{1}{2} \mu|\dot{\mathbf{r}}|^{2} \tag{6.38}
\end{equation}
\)
The first term in (6.38) is the kinetic energy due to translational motion of the whole system of mass $M$. Translational motion is motion in which each particle undergoes the same displacement. The quantity $\frac{1}{2} M|\dot{\mathbf{R}}|^{2}$ would be the kinetic energy of a hypothetical particle of mass $M$ located at the center of mass. The second term in (6.38) is the kinetic energy of internal (relative) motion of the two particles. This internal motion is of two types. The distance $r$ between the two particles can change (vibration), and the direction of the $\mathbf{r}$ vector can change (rotation). Note that $|\dot{\mathbf{r}}|=|d \mathbf{r} / d t| \neq d|\mathbf{r}| / d t$.
Corresponding to the original coordinates $x{1}, y{1}, z{1}, x{2}, y{2}, z{2}$, we have six linear momenta:
\(
\begin{equation}
p{x{1}}=m{1} \dot{x}{1}, \quad \ldots, \quad p{z{2}}=m{2} \dot{z}{2} \tag{6.39}
\end{equation}
\)
Comparing Eqs. (6.34) and (6.38), we define the six linear momenta for the new coordinates $X, Y, Z, x, y, z$ as
\(
\begin{array}{rlrl}
p{X} & \equiv M \dot{X}, & p{Y} \equiv M \dot{Y}, & p{Z} \equiv M \dot{Z} \
p{x} \equiv \mu \dot{x}, & p{y} \equiv \mu \dot{y}, & p{z} \equiv \mu \dot{z}
\end{array}
\)
We define two new momentum vectors as
\(
\mathbf{p}{M} \equiv \mathbf{i} M \dot{X}+\mathbf{j} M \dot{Y}+\mathbf{k} M \dot{Z} \quad \text { and } \quad \mathbf{p}{\mu} \equiv \mathbf{i} \mu \dot{x}+\mathbf{j} \mu \dot{y}+\mathbf{k} \mu \dot{z}
\)
Introducing these momenta into (6.38), we have
\(
\begin{equation}
T=\frac{\left|\mathbf{p}{M}\right|^{2}}{2 M}+\frac{\left|\mathbf{p}{\mu}\right|^{2}}{2 \mu} \tag{6.40}
\end{equation}
\)
Now consider the potential energy. We make the restriction that $V$ is a function only of the relative coordinates $x, y$, and $z$ of the two particles:
\(
\begin{equation}
V=V(x, y, z) \tag{6.41}
\end{equation}
\)
An example of (6.41) is two charged particles interacting according to Coulomb's law [see Eq. (3.53)]. With this restriction on $V$, the Hamiltonian function is
\(
\begin{equation}
H=\frac{p{M}^{2}}{2 M}+\left[\frac{p{\mu}^{2}}{2 \mu}+V(x, y, z)\right] \tag{6.42}
\end{equation}
\)
Now suppose we had a system composed of a particle of mass $M$ subject to no forces and a particle of mass $\mu$ subject to the potential-energy function $V(x, y, z)$. Further suppose that there was no interaction between these particles. If $(X, Y, Z)$ are the coordinates of the particle of mass $M$, and $(x, y, z)$ are the coordinates of the particle of mass $\mu$, what is the Hamiltonian of this hypothetical system? Clearly, it is identical with (6.42).
The Hamiltonian (6.42) can be viewed as the sum of the Hamiltonians $p{M}^{2} / 2 M$ and $p{\mu}^{2} / 2 \mu+V(x, y, z)$ of two hypothetical noninteracting particles with masses $M$ and $\mu$. Therefore, the results of Section 6.2 show that the system's quantummechanical energy is the sum of energies of the two hypothetical particles [Eq. (6.23)]: $E=E{M}+E{\mu}$. From Eqs. (6.24) and (6.42), the translational energy $E{M}$ is found by solving the Schrödinger equation $\left(\hat{p}{M}^{2} / 2 M\right) \psi{M}=E{M} \psi{M}$. This is the Schrödinger equation for a free particle of mass $M$, so its possible eigenvalues are all nonnegative numbers: $E{M} \geq 0$ [Eq. (2.31)]. From (6.24) and (6.42), the energy $E_{\mu}$ is found by solving the Schrödinger equation
\(
\begin{equation}
\left[\frac{\hat{p}{\mu}^{2}}{2 \mu}+V(x, y, z)\right] \psi{\mu}(x, y, z)=E{\mu} \psi{\mu}(x, y, z) \tag{6.43}
\end{equation}
\)
We have thus separated the problem of two particles interacting according to a potential-energy function $V(x, y, z)$ that depends on only the relative coordinates $x, y, z$ into two separate one-particle problems: (1) the translational motion of the entire system of mass $M$, which simply adds a nonnegative constant energy $E_{M}$ to the system's energy, and (2) the relative or internal motion, which is dealt with by solving the Schrödinger equation (6.43) for a hypothetical particle of mass $\mu$ whose coordinates are the relative coordinates $x, y, z$ and that moves subject to the potential energy $V(x, y, z)$.
For example, for the hydrogen atom, which is composed of an electron (e) and a proton $(p)$, the atom's total energy is $E=E{M}+E{\mu}$, where $E{M}$ is the translational energy of motion through space of the entire atom of mass $M=m{e}+m{p}$, and where $E{\mu}$ is found by solving (6.43) with $\mu=m{e} m{p} /\left(m{e}+m{p}\right)$ and $V$ being the Coulomb's law potential energy of interaction of the electron and proton; see Section 6.5.