A Perturbation Theory Proof

To prove the validity of the perturbation expansion, let's start with a simple and obvious statement:

It seems intuitive that a variable x can be the sum of two terms containing two new variables. In fact, the same can be said of variable a1:

And, a2 can be the sum of two terms containing two new variables. We can repeat this exercise as many times as we like until we reach a(sub-n):

Using equations 2 through 5 we make a series of substitutions which transform equation 1 into the familiar perturbation expansion:

However, equation 6 is an exact solution which can't be had unless we know the value of a(sub-n). Let's assume we don't. Let's eliminate a(sub-n). If we do, we get an approximate solution:

Now, how do we apply this approximate series known as a perturbation series? Consider a function of x:

Let's assume this function of x is too hard to solve. So we re-write f(x) above to show the easy, solvable part and the seemingly unsolvable part:

Let's suppose the hard part is g(x). Wouldn't it be great if we could just get rid of it? Not permanently. Just for the time being. We do away with g(x) by multiplying it by epsilon, and epsilon has a zero limit.

We end up with equation 12 which is easy to solve. The solution can be found at 13. However, we didn't solve variable x--we solved what is called an unperturbed x or x0. We can now plug that solution into equation 7 and equation 8. Because x is approximately equal to x0 plus the rest of the series, we can make a substitution at equation 11 to get the following:

At 15 we set the equation to zero. The idea here is to solve the xi coefficients. This is often done by grouping terms on the basis of their epsilon powers. The equation is broken up into smaller, simpler equations that are easy to solve. At 18 below, the solved coefficients are plugged into the series and epsilon is set back to 1 (giving us back the equivalent to the missing hard part of the original equation). At 19 and 20 we simply add up the coefficients to get a close approximation of the true value of x.

Since the intent of this post was to do a general proof of perturbation theory, no specific problem-solving examples were provided. Click here to see an example of a specific problem solved using perturbation theory.

How to Solve Complex Equations When Perturbation Theory Fails

Perturbation theory is useful to any mathematician or physicist who needs to solve a seemingly unsolvable equation such as the one below:

Equation 1 may be hard to solve, but its structure looks simple. Perturbation theory works fine in its case. Later, we will try to solve a much more complex equation, where perturbation theory ceases to be practical, and we will use an alternative method. But for now, let's explore the benefits and limitations of perturbation theory. To solve equation 1 we first reformat it as follows:

Then we expand x and x^2:

Next, we make substitutions:

Let's assume the following terms equal zero:

Using a little algebra and more substitutions we solve for x0, x1, and x2. When we add these x's up we get the value of the original x:

Not bad! But check out this monster:

OK, let's set some values for y and z and solve for x. We derive equation 19:

Should we solve it with perturbation theory? Well, let's see. We first have to expand x^5:

I don't know about you, dear reader, but I don't even want to think about expanding x^5! In our previous example we only had to expand x^2 to only the second power of epsilon. Can you imagine the can of worms we will open if we expand x^5 to two or more powers of epsilon? There's got to be an easier way ... and there is.

To solve equation 19 we use a systematic guess-and-check method. The idea is to choose a value for x that will give us an answer that is close to zero if not exactly zero. We first test -1, 0, and 1. We see that 1 is closer to zero. We try 10. If 10 is better, we try 100, and higher powers of 10 as we get closer to zero. After all, the value for x could be any real number from negative infinity to positive infinity.

Now, after checking 10, we see it's no good. So we now know x is between 1 and 10. We try 2. Whoops, 1 was better. So we try 1.1 and see an improvement. Then we try 1.2--even better. We increment by tenths to 1.3. At 1.4, we overshoot. So the answer is around 1.3, so we try 1.29 and 1.31, i.e., we increment in hundredths. Our final answer is 1.29. If we want more accuracy, we increment in thousandths, and ever smaller powers of 10. Below is a table of our results:

The chief advantage of this method is it is possible to write a computer program that will yield a quick answer.

Update: Click here for a proof of perturbation theory

Which Way Will the Pencil Fall?--Playing with Perturbation Theory

On a smooth, flat table you have a pencil balancing on its tip. Which way will it fall? Or will it just stay balanced on its tip? Variable L is the displacement. If the pencil stays balanced, L equals zero. If the cat jumps on the table ... L is greater than zero. In fact, L is the length of the pencil laying on its side, thanks to Felix the feline pencil flopper.

We can mathematically represent this feline faux pas with the following equation straight out of perturbation theory:

Whether the displacement (L) is zero or greater depends on the value of epsilon. If epsilon is zero, then L just equals Lo which is zero. In that instance, the pencil is perfectly balanced on that smooth, shiny tabletop. Epsilon represents a small disturbance (or a big one). When Felix jumps on the table, epsilon is greater than zero, and that causes L1 and L2 to come into play, which causes L to be greater than zero--so the pencil falls on its side.

The above equation, however, does not tell us which way the pencil fell (or the color of Felix's fur). It does not tell us much about the forces that make up epsilon. We want to know which way the pencil fell and the magnitude and direction of the forces involved. We could represent the direction and angle of displacement (L) if we use a complex number: a + bi. We can also use angle phi:

The maximum displacement (L) (i.e., the length of the pencil laying on its side) is equal to the square root of the complex number times its complex conjugate:

Epsilon can be divided into three dimensions of force: epsilon(a) is the force(s) that causes the pencil to fall in the "a" direction or "x" direction. Epsilon(b) is the force(s) pertaining to the "bi" direction or "y" direction. Since the pencil does not rise vertically along the "z" axis, or drill into the table, we only need to consider two dimensions.

The equation below tells us what epsilon is along directions "a" and "b." You will note that the complex number is divided by itself. This yields a one or a zero. The value of N equals one if the pencil falls, and zero if it does not.

We can find the values of a and b as follows:

And let's not forget the values of epsilon(a) and epsilon(b):

Last, but not least, we can find angle phi:

Click here for a totally excellent tutorial on perturbation theory.