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.

An Amplitude Squared Equals a Probability--a Mathematical Proof

Start with a wave:

Add more waves. Waves are in and out of phase with each other (constructive and destructive interference):

The range is from zero degrees out of phase to 180 degrees out of phase. On average, any pair of waves is 90 degrees out of phase, making a sine wave and a cosine wave:

Each has an amplitude (Ao). We can add the waves together, making a complex number (equation 1). Since our intention is to isolate and square the amplitude, we need a complex conjugate (equation 2). Using Euler's identity we write the following equations:

When we square the the left sides and square the right sides of equations 1 and 2, here's what we get:

It's entirely possible we don't get a probability. Instead, we could get a value greater than one. Plus we have distance units--so we need to normalize the amplitude:

Now, let's express our oscillating wave(s) in terms of Hooke's law and derive an energy (E):

Below we make a couple of substitutions to derive equation 10:

At 11 we define the probability of E. It could be from zero to one:

Multiply both sides of equation 10 by the probability P(E) then add a pinch of algebra to get equation 16:

Equation 16 shows the probability is equal to the square of the reduced normalized amplitude. Of course we are not limited to just energy eigenvalues. We can show that any type of eigenvalue has a probability equal to the square of a normalized amplitude:

In fact, if we express energy E in terms of momentum (p), mass (m), position (x), time (t) and wave number (k), we discover that the momentum, mass, position, time and wave number each have the same probability as the energy: the square of the amplitude A'. This is due to the energy being dependent on a specific value of each of these other eigenvalues. Each term below contains one eigenvalue and constants (which don't change). So the energy eigenvalue correlates with each of these other eigenvalues, and, likewise, the probabilities correlate.

In conclusion it is safe to say a wave amplitude squared does not guarantee a probability; however, it looks as though the square of a normalized amplitude does.

Proving the Schwartz Inequality and Heisenberg's Uncertainty Principle

In this post we once again derive the Heisenberg uncertainty principle, but this time we make use of the Schwartz inequality and the position-momentum commutator. We begin our proof by defining the variables:

Below we have the Schwartz inequality:

Is it true? Let's prove it. At lines 2 and 3 we map inner products with multiple (n) dimensions to simpler 2D Pythagorean expressions. At 4 through 7 we define the bras and kets and their inner products in terms of a, b, c, d.

Next, we take the products of the inner products and derive 11 below:

At 12 and 13 we convert the variables ad and bc to x and x+h. You may recognize h from calculus texts. In this case, it is just any arbitrary number. At 14 and 15 we do a little algebra to get 16:

At 16 it is obvious the absolute value of h^2 is greater than or equal to zero. Thus, the Schwartz inequality is true.

Before we put it to work, we need to define energy (E) and time (t). E is the lowest possible energy (ground-state) and t is the reciprocal of frequency (f). We sandwich these between the normalized bras and kets at 20. That brings us to the energy-time uncertainty at 21.

At 22 and 23 we do a quick-and-dirty derivation of the momentum-position uncertainty:

We can also find the momentum-position uncertainty by making use of its commutator and a wave function (psi). At 24 we define the commutator; at 25 we define momentum (p). After making a substitution for p at 26, we do some more algebra until we get the desired outcome at 30.

Equation 30 is looking good, but there is a slight problem: it's an equation! We want an inequality, so we make use of the Schwartz inequality one more time:

Ah, that's better.

Proving the Dirac Delta Function, Etc.

In this post we prove the Dirac Delta function and its sampling property. Here are the variables we need:

So we ask, is equation 1 below true?

To prove equation 1, we first define the Dirac Delta function:

And it helps if we draw a diagram:

In the diagram above, we have a line segment along the x axis ranging from minus infinity to infinity. If we let the value of epsilon go to infinity (or assume an infinite number of points along any epsilon distance) we can make a substitution and create the diagram below:

At line 3 we create a new integral that is equivalent to the one we started with. From there to line 5 we show that the integral does indeed equal 1.

Now let's prove the sampling property. Suppose a particle is located at position 'a' instead of zero. Is the value of the integral f(a)?

We assume the following are true:

We make some changes to equation 6 to get 7:

We draw a new diagram to account for the fact that x equals 'a' instead of zero:

We can now derive equation 8. From there we derive the calculus difference quotient or derivative formula at line 14.

Line 15 above confirms the Dirac Delta function's sampling property.

Deriving the Fourier Transform

In the field of quantum mechanics the Fourier transform shows the relationship between momentum/wave-number space, phi(k), and position space psi(x). Equations 1 and 2 below are a typical example of this relationship:

Let's see if we can derive equations 1 and 2 from scratch. We start with perhaps the simplest transforms:

At equation 3 above, we make the right side a function of x by multiplying by e^ikx. If we multiply both sides by e^-ikx, we get equation 4 and equation 4's right side transforms from a function of x to a function of k.

Next, we take equation 3 and find the integral of both sides with respect to k:

We solve the integral on the left side first.

It looks like we are going to get infinity. Darn! We want something finite. Here's what we are dealing with: Imagine a line segment with point zero at the center. The furthest point to the right is an infinite number of points from point zero. Going to the left, there are a minus infinite number of points.

Suppose we bend the line segment into a half circle like this:

We deduce that pi/2 radians is equivalent to an infinite number of points.

We take the square root of both sides of 7 to get 8:

It is most convenient that the square root of infinity is still infinity and the square root of pi/2 radians is still equivalent to an infinite number of points from zero to (pi/2)^.5. To make the math less cluttered we do the following:

Using some high-school algebra we derive equation 14 below:

By repeating the steps above we derive equation 2 (aka: equation 24):