# Sunday Function

“So,” Herr Schrodinger says to us, “I’m looking for a function of x. It needs to be equal to the negative of its second derivative, up to a constant factor. When x is zero, the function itself should be zero. And when x equals L for some constant L greater than zero, the function also needs to be zero. Can you find me such a function?”

Sure we can. As you know, the derivative of a function is its rate of change. Its second derivative is the rate of change of the rate of change. There are functions which are equal to their derivative (and second derivative), e^x is one of them. But we’re looking for a function which is equal to the negative of its own second derivative. There’s two of them, the sine and the cosine function. And since the sum of a derivative is the derivative of the sum, we know the answer to Schrodinger’s problem is:

Differentiate that sucker twice, and you’ll sure enough get the negative of the very same function back, times an overall constant which happens to be the square of k. We have to have the constants A, B, and k so that we don’t miss any of the possibilities. As an example if we wanted a function equal to its derivative, e^x works, but so does any constant times e^x. Those are perfectly legitimate functions in their own right, and we can’t ignore them.

Now we have the answer to the first part of Schrodinger’s question. But our function isn’t zero at x = 0, and it isn’t zero at x = L. Probably some of the ways we might select the constants A, B, and k might work, but certainly not all of them. We have to begin chopping away the ones that don’t work.

First, we know that at x = 0, f(x) must equal zero too. But while sin(0) = 0 automatically, cos(0) is never zero no matter what k we select. The only way we can force f(0) = 0 is to chop that term entirely by setting B = 0. This leaves us with:

Which is promising, but not equal to 0 when x = L. At least not unless we can figure out what particular values of k cause that condition to be true. Now we know that the sine function is zero at lots of x values. The sine function is zero at x = 0, pi, 2pi, 3pi, 4pi, etc for any n times pi. But we want it to be equal to zero at x = L, so that means we have to pick k such that

Solve for k, plug into our function:

And we’re done! Our function fits every condition Schrodinger set out for us. Notice that there’s an infinite number of possible n. Assuming we set L = 1, we can graph a few (The value of A is arbitrary, I’m just going to set it equal to 1 also):

This is, in slight disguise, a very important problem in intro quantum mechanics – the wavefunction of a particle in a box. Each n corresponds to an energy level of the system. Most importantly any sum of those solutions is itself a solution, which leads us to the superposition principle and the expansion postulate. And complete sets of orthogonal functions, which are super-cool and I might just make it a topic of a future Sunday Function.

1. #1 Anonymous
April 19, 2009

awesome sause

2. #2 Uncle Al
April 19, 2009

As the frequency increases the antinodes stack more and more densely. In the classical limit.. whoa! As the frequency increases the nodes stack more and more densely. In the classical limit… there is nothing. Some say the glass is half full, others half empty. 100% full! 50% observables, 50% ullage.

http://itl.chem.ufl.edu/4412_aa/partinbox.html

3. #3 Poopy
April 19, 2009

Funky

4. #4 Chris H.
April 20, 2009

I second the talking about orthogonality and completeness for a future topic.

5. #5 CCPhysicist
April 21, 2009

Nice, except Schroedinger actually wanted the solution for a Coulomb potential in spherical coordinates. But it is a lot easier to solve and a *lot* easier when it comes time to look at orthogonality and completeness!