In the field of quantum physics, each eigenvalue has an eigenvector, and, when the eigenvector is normalized and squared, we get the probability for the eigenvalue. The normalized eigenvector is sometimes referred to as the probability amplitude.
When all the probability amplitudes are squared and added, the total should be 1. We can represent this with the Pythagorean theorem and the right triangle below:
The above diagram consists of two probability amplitudes: 'a' and 'b.' One is a wave function cos(theta) and the other is a wave function sin(theta).
Now, suppose there are more than two eigenvalues/eigenvectors? The diagram below shows that a and b can be broken up into smaller pieces or smaller and more numerous probability amplitudes. As before, when they are all squared and summed, they give us a total of 1.
It is possible to break up 'a' and 'b' into as many pieces as we like. Below we focus on amplitude 'a':
We can imagine breaking up amplitude 'a' into as many as an infinite number of sub-amplitudes. This can be done in both Euclidean and curved space. Equation 10 below shows how amplitude 'a' and its sub-amplitudes are invariant within flat or curved space.
With a little algebra, we can derive equation 14:
Equation 14 shows amplitude 'a' consists of an infinite number of eigenvalues (eta), each with its own probability (P(eta)). Without the probabilities, the etas would add up to infinity, and that would necessitate some sort of re-normalization technique. If we assume, however, that all quantum numbers have a probability, we will not get infinity; rather, we get the expectation value, i.e., the value actually observed.
What kind of probability values yield a finite result when eta increases linearly to infinity? Probability values that decrease exponentially. Below we derive such a probability function by using the natural-log function and converting eta to 'n':
At 16.2 we have a probability function that will reduce the probability exponentially. It gives us a number between 0 and 1, but we can derive a better function that gives us a number between 0 and 1, and, we can make a substitution. The end game is equation 16.9:
Equation 16.9 claims that if n = Q, the probability of Q (P(Q)) equals the definite integral of the probability function over a range from Q-1 to Q. We can further justify this claim with the diagram below which shows the relation between discrete values (in red) with continuous values (blue line).
Note how the area under the blue line, say, from Q-1 to Q is the same as the area of the red squares from Q-1 to Q. Equation 17 models the fact the the area under the blue line is the same as the area of the red squares over the entire range.
Now, to get a finite expectation value (amplitude 'a') we could combine equations 16.9 and 17, but the math would be more complicated than need be. To simply the math we will encounter later, let's first stretch the above diagram vertically:
Next, we draw a yellow line from zero N+1. This new line is going to make our lives easier and has the same area beneath it as the red line. Wouldn't it be nice if we could nix the red and substitute the yellow? Sure! But first we have to rotate the diagram:
Ah ... now we're in business! Below is the adjusted diagram and equation 18 with a new slope of N/(N+1):
The integral has a new range of zero to N+1, so we give the probability integral the same range:
Let's combine equations 18 and 19 to get 20:
If the limit of N is infinity, equation 20 will always give us the finite probability amplitude 'a.' No re-normalization required.
Using the diagram below and equations 21 and 22, we can derive a formula that finds probability densities:
What we've covered so far allows to find probabilities for integer values. This works fine if the value is, for example, the number of vertices in a Feynman diagram. Albeit, energy can have values of n+.5. Below is the math for that circumstance:
Notice if we divide both sides of equation 26 by Q+.5, and use summation signs, we arrive at equation 23, the formula for finding probability densities.
Now that we have the math the way we want it, let's put it to a test. Let's say we want to add up an infinite number of quantum numbers to get a finite value. Let's assume that the principle of least action applies: the most probable value will be the least action (e.g. least energy, least time, least distance, least resources required, etc.). The least probable value will be the action or event that requires the most resources, time, energy, etc. So we expect the probability to drop exponentially as the value of 'n' increases linearly--this will ensure a finite result.
Let's also assume that experiments confirm that probabilities change according to equation 27:
OK, now we only have to do some complicated math to find the expectation value 'a,' right? Wrong! At 28 and 29 below we convert the right side of 27 to a natural exponent function. If we look at equation 20, it becomes obvious that we can solve this problem by mere inspection. Looking at the exponent, everything to the right of -n is 1/a. Thus equation 31 is our final result.
Here's another test: What is the probability that a particle will travel a distance 'Q' along a pathway 'omega'? Equation 34 below can answer that. At 32 we assume that each pathway has the same probability if the distance traveled is constant, since the action is the same along each pathway (except for the direction, angles, curves, twists, turns, etc.).
Equation 35 gives us a definite answer if we want to know the probability density of a range of distances and pathways the particle can travel:
As you can see, stochastic trigonometry simplifies mathematics that can turn into a complicated, ugly, and infinite mess. It can also improve statistical mechanic's coarse-graining techinique:
Why use squares when you can use triangles?
Update: The following math shows both a convergent series and divergent series can yield a finite number 'q.' First, we start with a divergent series and make it convergent by using the probability function we derived above.Next, we take a divergent series and assume the coefficients (the c's) don't add up to 1. Each could be any finite size; they could be a random series. The strategy is to factor out 'c' from the coefficients and use one of Ramanujan's techniques:
Another update: The following math generalizes the idea that a finite value can result from any arbitrary convergent or divergent series:
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