D.11 Extension to three-dimensional solutions

Maybe you have some doubt whether you really can just multiply one-di­men­sion­al eigenfunctions together, and add one-di­men­sion­al energy values to get the three-di­men­sion­al ones. Would a book that you find for free on the Internet lie? OK, let’s look at the details then. First, the three-di­men­sion­al Hamiltonian, (really just the kinetic energy operator), is the sum of the one-di­men­sion­al ones:

\begin{displaymath}
H = H_x + H_y + H_z
\end{displaymath}

where the one-di­men­sion­al Hamiltonians are:

\begin{displaymath}
H_x = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}...
...
H_z = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial z^2}
\end{displaymath}

To check that any product $\psi_{n_x}(x)\psi_{n_y}(y)\psi_{n_z}(z)$ of one-di­men­sion­al eigenfunctions is an eigenfunction of the combined Hamiltonian $H$, note that the partial Hamiltonians only act on their own eigenfunction, multiplying it by the corresponding eigenvalue:

\begin{displaymath}
\begin{array}{l}
(H_x+H_y+H_z)\psi_{n_x}(x)\psi_{n_y}(y)...
...+
E_z \psi_{n_x}(x)\psi_{n_y}(y)\psi_{n_z}(z)
\end{array}
\end{displaymath}

or

\begin{displaymath}
H \psi_{n_x}(x)\psi_{n_y}(y)\psi_{n_z}(z)
=
(E_x + E_y + E_z) \psi_{n_x}(x)\psi_{n_y}(y)\psi_{n_z}(z).
\end{displaymath}

Therefore, by definition $\psi_{n_x}(x)\psi_{n_y}(y)\psi_{n_z}(z)$ is an eigenfunction of the three-di­men­sion­al Hamiltonian, with an eigenvalue that is the sum of the three one-di­men­sion­al ones. But there is still the question of completeness. Maybe the above eigenfunctions are not complete, which would mean a need for additional eigenfunctions that are not products of one-di­men­sion­al ones.

The one-di­men­sion­al eigenfunctions $\psi_{n_x}(x)$ are complete, see [40, p. 141] and earlier exercises in this book. So, you can write any wave function $\Psi(x,y,z)$ at given values of $y$ and $z$ as a combination of $x$-​eigenfunctions:

\begin{displaymath}
\Psi(x,y,z)=\sum_{n_x} c_{n_x} \psi_{n_x}(x),
\end{displaymath}

but the coefficients $c_{n_x}$ will be different for different values of $y$ and $z$; in other words they will be functions of $y$ and $z$: $c_{n_x}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $c_{n_x}(y,z)$. So, more precisely, you have

\begin{displaymath}
\Psi(x,y,z)=\sum_{n_x} c_{n_x}(y,z) \psi_{n_x}(x),
\end{displaymath}

But since the $y$-​eigenfunctions are also complete, at any given value of $z$, you can write each $c_{n_x}(y,z)$ as a sum of $y$-​eigenfunctions:

\begin{displaymath}
\Psi(x,y,z)=
\sum_{n_x}
\left(\sum_{n_y} c_{n_xn_y} \psi_{n_y}(y)\right)
\psi_{n_x}(x),
\end{displaymath}

where the coefficients $c_{n_xn_y}$ will be different for different values of $z$, $c_{n_xn_y}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $c_{n_xn_y}(z)$. So, more precisely,

\begin{displaymath}
\Psi(x,y,z)=
\sum_{n_x}
\left(\sum_{n_y} c_{n_xn_y}(z) \psi_{n_y}(y)\right)
\psi_{n_x}(x),
\end{displaymath}

But since the $z$-​eigenfunctions are also complete, you can write $c_{n_xn_y}(z)$ as a sum of $z$-​eigenfunctions:

\begin{displaymath}
\Psi(x,y,z)=
\sum_{n_x}
\left(
\sum_{n_y}\left(\sum_...
...}\psi_{n_z}(z)\right)\psi_{n_y}(y)
\right)
\psi_{n_x}(x).
\end{displaymath}

Since the order of doing the summation does not make a difference,

\begin{displaymath}
\Psi(x,y,z)=
\sum_{n_x} \sum_{n_y} \sum_{n_z}
c_{n_xn_yn_z} \psi_{n_x}(x) \psi_{n_y}(y) \psi_{n_z}(z).
\end{displaymath}

So, any wave function $\Psi(x,y,z)$ can be written as a sum of products of one-di­men­sion­al eigenfunctions; these products are complete.