2170 - Notes, Syllabus Section IV, lecture 30 (4/01/96)

You might think of a marble rolling back and forth. Classically, it rolls out to the point where the total energy crosses the potential, and there it stops and turns around.

Here is a rough sketch of the ground state wave function, using the ideas developed in the last lectures

I drew the potential again, on top of the wave function, to help guide the eye. Where I say it is "sine-like", I really mean cosine like, but since V is constantly changing, so is wavelength, so it really isn't quite a cos function at all. But at least, it should be concave down throughout the "classically allowed" region. The forbidden region, like before, can now have some wave in it, but it has to be exponentially damped. Since V-E is constantly changing, the exponent isn't a constant any more, so it isn't an ordinary exponential either! But at least, it should be concave up and head asymptotically towards zero. The potential is symmetric, so the wave function is too, and the ground state has zero nodes.

The next higher level, E1, should be odd. It looks like this:

It has one node, and is vaguely sine-like in the allowed region, and (just like above) vaguely exponential-like outside.

In all the wave functions, the sign of the second derivative flips right at the dividing line between classically allowed and classically forbidden. (Can you see why?)

Notice that, because E1>E0, the classically allowed region is slightly larger in the first excited wave function.

In a certain sense, almost any system in the world that has an equilibrium position can be modeled or approximated as a harmonic oscillator. That's why its such an incredibly important case to consider.

I can prove this as follows: Classically, force is , so for a harmonic oscillator, , the old familiar "spring force", or "Hookes Law". Now consider some more arbitrary, physical force F(x). Taylor's theorem says that F can be expanded near x=0 as

(There's a mistake in that previous equation: I left out the factor of "x" in the dF/dx term! Sorry.)

If we define x=0 as the equilibrium position, then F(0)=0. (that's what you mean by equilibrium - at that point, there's no net force on the object) So this gives

For sufficiently small x, the quadratic term is guaranteed to be less important than the leading one, which means

If F' were positive, then we would have unstable equilibrium, because the force would get bigger as you moved away from the origin, forcing you still further away. For any stable equilibrium, the derivative is some negative number, which I define as C.

So, F(x)=-Cx is the approximate force law for almost any real force, no matter what the detailed form is, for small vibrations. It's pretty easy to imagine that most small (quantum) systems look a lot like harmonic oscillators!

In general, we should already expect a lot of features based on our previous qualitative studies. I expect:

Quantized, discrete energies En.

Alternating even and odd wave functions.

A non-zero minimum energy .

A dying off of the wave function past the classical turning point, which is where E=(1/2)Cx^2.

Since V(x) keeps growing up to infinity, there is no maximum energy.

Since V(x) is a single, smooth function (unlike our boxes, which had square edges), we can expect the wave function to be a single, smooth function. We won't have to separate our problem into "allowed" and "forbidden" regions, and match at the boundary, like we did before. The only boundary is at infinity, where the wave function must vanish.

So we're going to solve the time independent Schrodinger equation,

F+T (sect 4-3) go through the solution. I'm going to follow their method (there are various types of approach possible) but with one change. I want to change variables, so I have dimensionless equations. This is a standard trick - we've done it already once for the finite well. It just makes the math a lot cleaner. So we write

Then we notice that has units of kg (N/m) / (J s)^2

=kg (kg/s^2) / (kg m^2/s)^2 = 1/meter^4, which means that

is unitless. So, let's define a new variable, y, which is "just like x", but rescaled so it's unitless (unitless distance!)

.

Similarly, we rewrite energy E as

The factor of 2 downstairs is purely arbitrary and conventional (you'll see why it's convenient shortly), and the classical frequency is just that - the oscillation frequency of a classical harmonic oscillator with the same spring constant as we have,

.

is then our new variable, it is the "dimensionless energy", since it's linearly proportional to energy, but manifestly a pure number. We now rewrite our Schrodinger equation, but change variables from y to x. (It's a simple proportionality shift):

(I canceled a common term of everywhere to get to the last line)

I really didn't have to do any of this - I could just as well solve

which is mathematically exactly the same. It's just messier because of all those coefficients you have to keep dragging around.

Now we just need to solve the rather innocuous looking equation

Looks simple enough - but how to solve it? You can try a few obvious guesses (like simple polynomials), but none of them work. There is a dependable way we may recall from Physics 2140 - power series solution!

Write

plug this back in the diff eq., and see if you can get relations between the a's. This will work!! Unfortunately, the answer you get is pretty non-intuitive. So, I'm going to use a trick which is often a good start with differential equations - find out what the asymptotic limit is. (i.e., what happens as y -> infinity?)

(for large y)

Let's just mess around. If , then we know that works as a solution. So, if , we might naively guess that (??) This is surely nonsense, but let's go with it....

If

(the last step being true only for large y)

This is almost what we want! But, that factor of 4 is a pain. How about

,

just as desired.

What has this messing around taught us? Only that, for large y,

is going to work. But it appears to fail in general for finite y (? See below!)

Here's the trick then: Let's guess a solution of the form

When solving diff eq's, you can always try things like this! By finding the "asymptotic" behaviour, , and factoring it out, you are hoping that what's left behind (f(y)) will be simpler.

Let's now think of a power series solution for f! But wait - f better not end up being an infinite power series. Because if it is, then our result probably won't be behaving like at large y any more. So let's hope that f is a finite polynomial,

Sidetrack: You might say to yourself, in what sense is behaving "like" , for large y?! F+T (p. 166) have a nice argument - think of the change in , or , as y goes from y to y+1 at large y.

while

For large y, there's no difference in the "change factor", because1/y goes to zero. These 2 functions, fractionally speaking, change the same as y -> y+1. This is true for any power of y, as long as it's finite.

Here's another way to see it (this is my argument).

Suppose rather than being , your wave function is instead . Then

For large y, the 3 is negligible, and thus , which is what we wanted. The polynomial out front didn't ruin or change this relation.

We haven't derived anything yet, we've just been investigating our differential equation, coming up with a plausible form for the solution, which has the right behaviour at large y. It could all be in vain, if we try it and it doesn't work! But at this point, we're guessing (hoping) that perhaps a wave function of the general form

might work. I have no idea where to truncate this polynomial, as long is it doesn't get infinitely long, which could mess up our asymptotic behaviour.

Let's just start trying this out; We'll begin with n=0, and see what happens. That is, try

(Later, we can try n=1, i.e. , etc.)

You should remember our T.I.S.E, which was

, where .

Taking the derivatives, I get

Hey, this works! But ONLY if !

We've discovered an eigenvalue, a magic value of energy (or alpha), namely , and a corresponding eigenfunction, namely

(We can find a_0 later by normalizing the wave function. You'll do this in your homework, don't worry about it now)

How about when n=1? You might be tempted to try

, but this function is neither even nor odd, and I know it must be one or the other. The only n=1 polynomial which is either even or odd is simply a_1 y, i.e.

(There's a mistake in the first line for the second derivative above. The very last term should have a factor of "y" in it! Sorry)

Again we have a solution of the TISE, but ONLY iff ,

and then we have the eigenfunction

(Again, the norm will fix a_1)

You might be worried here. Sure, I'm finding some valid solutions, but who says there aren't any other ones, that simply can't be expressed as f(y) Exp(-y^2/2)? The best answer I can give to that is, go back and try a general power series solution. You will find the exact same solutions, and no others. And now you see why I went to all the troubles to "play around" with asymptotic solutions - the power series for Exp(-y^2/2) doesn't exactly pop out at you if you don't know what you're looking for, so you might never have noticed the elegant, simple, analytic form of these solutions. (I still haven't proven that there are no non-analytic, ugly log type solutions with no power series solution, but there really aren't!) (Also, I can study the solutions I've generated, and notice that the number of nodes increases one by one, another clue that I'm not missing any possible solutions!)

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