Up to this point, we have solved only one-particle quantum-mechanical problems. The hydrogen atom is a two-particle system, and as a preliminary to dealing with the H atom, we first consider a simpler case, that of two noninteracting particles.
Suppose that a system is composed of the noninteracting particles 1 and 2. Let $q{1}$ and $q{2}$ symbolize the coordinates $\left(x{1}, y{1}, z{1}\right)$ and $\left(x{2}, y{2}, z{2}\right)$ of particles 1 and 2 . Because the particles exert no forces on each other, the classical-mechanical energy of the system is the sum of the energies of the two particles: $E=E{1}+E{2}=T{1}+V{1}+T{2}+V{2}$, and the classical Hamiltonian is the sum of Hamiltonians for each particle: $H=H{1}+H{2}$. Therefore, the Hamiltonian operator is
\(
\hat{H}=\hat{H}{1}+\hat{H}{2}
\)
where $\hat{H}{1}$ involves only the coordinates $q{1}$ and the momentum operators $\hat{p}{1}$ that correspond to $q{1}$. The Schrödinger equation for the system is
\(
\begin{equation}
\left(\hat{H}{1}+\hat{H}{2}\right) \psi\left(q{1}, q{2}\right)=E \psi\left(q{1}, q{2}\right) \tag{6.18}
\end{equation}
\)
We try a solution of (6.18) by separation of variables, setting
\(
\begin{equation}
\psi\left(q{1}, q{2}\right)=G{1}\left(q{1}\right) G{2}\left(q{2}\right) \tag{6.19}
\end{equation}
\)
We have
\(
\begin{equation}
\hat{H}{1} G{1}\left(q{1}\right) G{2}\left(q{2}\right)+\hat{H}{2} G{1}\left(q{1}\right) G{2}\left(q{2}\right)=E G{1}\left(q{1}\right) G{2}\left(q{2}\right) \tag{6.20}
\end{equation}
\)
Since $\hat{H}{1}$ involves only the coordinate and momentum operators of particle 1 , we have $\hat{H}{1}\left[G{1}\left(q{1}\right) G{2}\left(q{2}\right)\right]=G{2}\left(q{2}\right) \hat{H}{1} G{1}\left(q{1}\right)$, since, as far as $\hat{H}{1}$ is concerned, $G{2}$ is a constant. Using this equation and a similar equation for $\hat{H}{2}$, we find that (6.20) becomes
\(
\begin{equation}
G{2}\left(q{2}\right) \hat{H}{1} G{1}\left(q{1}\right)+G{1}\left(q{1}\right) \hat{H}{2} G{2}\left(q{2}\right)=E G{1}\left(q{1}\right) G{2}\left(q{2}\right) \tag{6.21}
\end{equation}
\)
\(
\begin{equation}
\frac{\hat{H}{1} G{1}\left(q{1}\right)}{G{1}\left(q{1}\right)}+\frac{\hat{H}{2} G{2}\left(q{2}\right)}{G{2}\left(q{2}\right)}=E \tag{6.22}
\end{equation}
\)
Now, by the same arguments used in connection with Eq. (3.65), we conclude that each term on the left in (6.22) must be a constant. Using $E{1}$ and $E{2}$ to denote these constants, we have
\(
\begin{gather}
\frac{\hat{H}{1} G{1}\left(q{1}\right)}{G{1}\left(q{1}\right)}=E{1}, \quad \frac{\hat{H}{2} G{2}\left(q{2}\right)}{G{2}\left(q{2}\right)}=E{2} \
E=E{1}+E{2} \tag{6.23}
\end{gather}
\)
Thus, when the system is composed of two noninteracting particles, we can reduce the two-particle problem to two separate one-particle problems by solving
\(
\begin{equation}
\hat{H}{1} G{1}\left(q{1}\right)=E{1} G{1}\left(q{1}\right), \quad \hat{H}{2} G{2}\left(q{2}\right)=E{2} G{2}\left(q{2}\right) \tag{6.24}
\end{equation}
\)
which are separate Schrödinger equations for each particle.
Generalizing this result to $n$ noninteracting particles, we have
\(
\begin{gather}
\hat{H}=\hat{H}{1}+\hat{H}{2}+\cdots+\hat{H}{n} \
\psi\left(q{1}, q{2}, \ldots, q{n}\right)=G{1}\left(q{1}\right) G{2}\left(q{2}\right) \cdots G{n}\left(q{n}\right) \tag{6.25}\
E=E{1}+E{2}+\cdots+E{n} \tag{6.26}\
\hat{H}{i} G{i}=E{i} G_{i}, \quad i=1,2, \ldots, n \tag{6.27}
\end{gather}
\)
For a system of noninteracting particles, the energy is the sum of the individual energies of each particle and the wave function is the product of wave functions for each particle. The wave function $G{i}$ of particle $i$ is found by solving a Schrödinger equation for particle $i$ using the Hamiltonian $\hat{H}{i}$.
These results also apply to a single particle whose Hamiltonian is the sum of separate terms for each coordinate:
\(
\hat{H}=\hat{H}{x}\left(\hat{x}, \hat{p}{x}\right)+\hat{H}{y}\left(\hat{y}, \hat{p}{y}\right)+\hat{H}{z}\left(\hat{z}, \hat{p}{z}\right)
\)
In this case, we conclude that the wave functions and energies are
\(
\begin{gathered}
\psi(x, y, z)=F(x) G(y) K(z), \quad E=E{x}+E{y}+E{z} \
\hat{H}{x} F(x)=E{x} F(x), \quad \hat{H}{y} G(y)=E{y} G(y), \quad \hat{H}{z} K(z)=E_{z} K(z)
\end{gathered}
\)
Examples include the particle in a three-dimensional box (Section 3.5), the three-dimensional free particle (Prob. 3.42), and the three-dimensional harmonic oscillator (Prob. 4.20).