This is the conclusion of the Ritz variation discussion we’ve been doing. Tomorrow we’ll get back to less arcane and more entertaining physics, I promise! We’re capping things off today by demonstrating the actual technique in detail for an example where we already know the answer. What we’re doing is finding the ground state of the 1-d square well.
Before we even start, we can look up the actual answer and use it to give us an idea of how accurate our Ritz approximation ends up being. The ground state energy of a particle in a 1-d box is:
Now that we know what the answer is, we can go ahead and try our method and see how close we get. When I last talked about this technique, I said that the first step was to recognize that since we didn’t know the actual ground state wavefunction, we would just make one up. It could be anything at all, though if for some reason we had an a priori suspicion that the actual wavefunction might look a certain way it would be a good idea for our made-up function to be similar. But that’s not entirely true. We can’t quite just make up anything. Our trial wavefunction has to be expressible in terms of the eigenstates of the actual solution to our problem. Because these states are mathematically complete, the set of possible functions is pretty much anything we care to name… so long as we’re careful to make sure our function satisfies the boundary conditions of the problem. It would be a mistake and cause a wrong answer if for example we picked a trial function that was nonzero outside of the box, for instance.
For our particle in a box, the only boundary condition is that our chosen function be zero at and beyond the two walls of the box, located at x = 0 and x = L. So just defining our function to be zero outside the box, it only remains to pick a function that’s zero at the walls themselves. The simplest thing I can think of is the polynomial x(x – L), with of course the provision that x must be between 0 and L. Our polynomial is zero at the right places, so let’s give it a shot.
We’re looking to find an upper limit on the ground state energy:
Where the upper case Greek letter psi is our trial wavefunction, x(x-L). Let’s evaluate the top first:
Ok, part one done. Not too bad. Time for the bottom half of our original fraction. It’s easier:
Now we can place those into our original fraction and get the final answer:
Thus in the natural units of the problem, our approximate answer is 5. The real answer which we calculated above is about 4.93. And we picked the absolute simplest, most trivially lazy possible guess function in the world. We actually skipped the entire “variation” process that’s the whole point of this approximation and we still got a percent error of only around 1.3%. That’s an amazingly good result considering the lack of effort.
Now we won’t actually go through the math, but imagine we had picked as our trial wavefunction almost the same thing, but with an extra Gaussian profile centered at the middle of our well:
Here alpha is a parameter that controls the width of the Gaussian function. We would go through the exact same procedure, with this function substituted for the unmodified x(x – L). The final result would include that parameter alpha. We would then use the methods of calculus to find what value of alpha minimizes the ground state energy. With any luck, this will get us even closer to a right answer. Unfortunately the algebra and calculus is fairly ugly and when you differentiate the result and set equal to zero you’ll end up having to solve for alpha numerically anyway. But that’s trivial on a computer. The final answer is 4.93491. Very, very close despite what’s still a somewhat crude trial function.
Ok, now that I’ve driven off all but the most intrepid of my readers I can only hope that at least a few people will find this interesting or helpful. But hey, I think it’s fun stuff.