This article takes a look at a 3D potential that forms a box, as you see in the following figure. You want to get the wave functions and the energy levels here.
![A box potential in 3D.](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd22_397258.image0.jpeg)
Inside the box, say that V(x, y, z) = 0, and outside the box, say that
![image1.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd05_397259.image1.png)
So you have the following:
![image2.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd1b_397260.image2.png)
Dividing V(x, y, z) into Vx(x), Vy(y), and Vz(z) gives you
![image3.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd38_397261.image3.png)
Okay, because the potential goes to infinity at the walls of the box, the wave function,
![image4.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd01_397262.image4.png)
must go to zero at the walls, so that's your constraint. In 3D, the Schrödinger equation looks like this in three dimensions:
![image5.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd1f_397263.image5.png)
Writing this out gives you the following:
![image6.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd35_397264.image6.png)
Take this dimension by dimension. Because the potential is separable, you can write
![image7.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd11_397265.image7.png)
Inside the box, the potential equals zero, so the Schrödinger equation looks like this for x, y, and z:
![image8.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd28_397266.image8.png)
The next step is to rewrite these equations in terms of the wave number, k. Because
![image9.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd31_397267.image9.png)
you can write the Schrödinger equations for x, y, and z as the following equations:
![image10.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd25_397268.image10.png)
Start by taking a look at the equation for x. Now you have something to work with — a second order differential equation,
![image11.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd14_397271.image13.png)
Here are the two independent solutions to this equation, where A and B are yet to be determined:
![image12.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd18_397270.image12.png)
So the general solution of
![image13.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd14_397271.image13.png)
is the sum of the last two equations:
![image14.png](https://cdn.prod.website-files.com/6634a8f8dd9b2a63c9e6be83/669d4a41be209577ccdccd3d_397272.image14.png)