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:

$$H = H_x + H_y + H_z$$

where the one-dimensional Hamiltonians are:

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

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:

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

or

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

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.

Well, the one-dimensional eigenfunctions $\psi_{n_x}(x)$ are complete, see [39, 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:

$$\Psi(x,y,z)=\sum_{n_x} c_{n_x} \psi_{n_x}(x),$$

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 $z$: $c_{n_x}=c_{n_x}(y,z)$. So, more precisely, you have

$$\Psi(x,y,z)=\sum_{n_x} c_{n_x}(y,z) \psi_{n_x}(x),$$

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:

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

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

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

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

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

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

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

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