Suppose you want to solve Schrodinger's equation for the hydrogen atom. It's the simplest atom in the universe: only one proton and one electron, so finding the solution should be as easy as blowing up a hydrogen balloon, right? Here's an operation Schrodinger's equation requires you to perform:
Three partial derivatives with respect to spherical coordinates of the function psi. That doesn't sound too bad, but take a look at the function psi:
Obviously it's going to take you more than a minute to find a solution. Looking at the function above, it is no surprise many QM books never take you past the hydrogen atom. To find the ionization energies of all the atoms in the periodic table seems like a daunting task, requiring a supercomputer. But there is a quick-and-dirty way to achieve the task with no computer help. However, it requires that we simplify the wave function.
Often, the basis chosen for the wave function involves space coordinates, but the basis we will choose is going to be time or space-time.
Equations 1 and 2 show how a single dimension (ct or x4) subsumes all three space dimensions and time. This will allow us to do one double derivative of the function instead of the more complicated Laplacian operation.
At 3 below we express the Schrodinger equation using the space-time variable (x4).
Here is the wave function:
At 5 we choose a value for the wave number (k). At 6 we solve the kinetic-energy term and convert the units to electron volts.
At 7 through 11 we solve the potential-energy term (V):
We put all the terms together and do the math. At the other end the solution comes out (see equation 22):
Finding the ground-state ionization energy for hydrogen is now just a plug-n-chug operation:
With our new Schrodinger solution, we can easily find the ionization energy for more complicated atoms such as carbon:
The main trick is finding the values for constants A, B, and C. Here they are for atoms with up to 12 electrons:
OK, so how do we find A, B, and C? Let's take a real-world example. We want to find the values of A, B, and C for atoms with at least 13 electrons. The smallest of which is Aluminum (Al). We look at some data (click here to see the data) and choose a sample size of five elements (Z >= 13). We then make a pretty little table:
Notice as the table progresses downward it represents the equivalent of taking a double derivative of the energies. Notice at the last row, the values are pretty similar, around 3.5. Let's figure out the average:
Let's take the integral of the average with respect to Z, and then find B:
Notice we didn't use the integer Z values to find B. We used 13.5, 14.5, etc. That's because we used the numbers in the second to the last table row. If you follow the broken lines leading to the top of the table, they fall half way between the Z values. We are dealing with energy differences and not the energies themselves. Each energy difference corresponds to a half-way point.
Now that we know B, let's do another integral to find A:
Let's not forget C. This time we are dealing with the energies and not energy differences, so we plug in the integer Z values:
And here's our final answer:
We should check this against the database (click here for data) for every 13th electron listed. We may have to tweak A, B and C a little to get a near-perfect fit.
