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:

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

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:

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

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.

The one-di­men­sion­al eigen­func­tions $\psi_{n_x}(x)$ are com­plete, see [41, 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 $z$: $c_{n_x}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $c_{n_x}(y,z)$. 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 $z$, $c_{n_xn_y}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $c_{n_xn_y}(z)$. 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:

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

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.