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.
Reality says, "It's finite," but your math says,"It's infinite." Why? And, is there a simple way to solve the problem? Today we present an incredibly simple formula for removing unwanted infinities. First, let's define the variables we will use:
Now that we got that out of the way, let's examine why infinities occur. One way to get infinity is to add up an infinite number of numbers as the following integral-summation shows:
You may have noticed something missing in equation 1 above. The thing that's missing makes the infinity finite:
The missing ingredient is none other than dx -- a really tiny number. When the infinity is multiplied by dx it magically becomes a finite number.
So one way to rid ourselves of an unwanted infinity is to multiply it by dx or something equivalent. Equation 5 below is a general formula that always provides a finite value by canceling infinity should one arise.
The first factor is equivalent to dx. The second factor could be equivalent to an infinite sum. Let's apply equation 5 to a classical example. Suppose we have a lattice with N cells. Each lattice cell contains some fraction (epsilon) of the total energy (E):
We could just add up all the parts to get the total energy (E):
We could calculate the average energy per cell (bar-epsilon*E) and multiply that value by the number of cells (N) to get the total energy:
But let's see what happens when we apply our new formula:
At equations 13 and 14, the dx equivalent is simply 1 which is often true in a classical case. It's a bit redundant but yields the right answer as do the more familiar methods above. In this example, it's like counting bricks to get the total. Very simple but effective. It's effective because the average energy per cell is simply the total energy divided by N--the number of cells. Thus, the bar-epsilon in our formula is the reciprocal of N--and the two variables cancel, leaving the total energy E. But look what happens when bar-epsilon doesn't equal the reciprocal of N:
In the above example we have infinite or infinitely uncertain energy in each cell, but the bar-epsilons and N's cancel, the uncertainty and/or infinity is nullified and we end up with the finite energy E.
The following example shows how infinity can rear its ugly head in a seemingly classical lattice.
Here's a data table and chart for different values of n:
Notice how the density increases as n decreases. As n increases the density levels off. The same is true for a 3D lattice:
When n is increased to infinity, the density is virtually constant and behaves like bricks. Increasing and decreasing the number of bricks does not change their mass density.
When n is reduced to zero, the density blows up to infinity! No more brick-density behavior:
We can translate the lattice equation into something resembling perturbation theory:
At equation 29 above we take epsilon to infinity (which is equivalent to taking n to zero) and the answer is 1 + infinity. We can use a famous ad hoc method to rid ourselves of the infinity: simply subtract it or ignore it. By pretending it isn't there, we get a legitimate finite energy value. On the other hand, it would make more mathematical sense to reduce epsilon to zero:
Let's apply our new formula to this problem. We begin with the original lattice equation and reduce n to zero to get the infinite result. At equation 33 we multiply the infinite density by the lattice volume to get the total energy--which is also infinite.
At 33b we restate the formula for convenience. At 34 we plug in the values we get from equation 33. At 35 we see the total energy is no-longer infinite, but the correct finite value.
Now, let's consider an even more classical setup. Here's the lattice diagram:
Surely the density should always behave like bricks, since we have one dot per cell. Sixteen dots per sixteen cells has the same density as one dot per one cell:
But even bricks have a limit. If you divide a brick up into smaller and smaller pieces, eventually you end up with something that no-longer resembles a brick. We can simulate this by reducing n to 1, and k to 0 (see 39 and 40). Once again we get the dreaded infinity--but we can fix that using the new formula (see 41 through 44).
How about a real-world example? At 45 below we set the total energy of two electrons equal to their Coulomb energy at a fixed radius (ro). You will note the total energy (2Ee) is quantized at a fixed integer value (no).
If the radius were to shrink to zero, one would think the energy would explode to infinity. To solve this problem we do some algebra to derive equation 52 below:
At 51 and 52 we see the radius cancelling itself. The total energy of the electrons is determined by the quantum number n. Apparently it is possible to have zero to infinite Coulomb energy (or force) without impacting the overall energy? The crude diagrams below sheds some light on this:
The diagrams show two Coulomb energies: E1 and E2 with radii of r1 and r2, respectively. There is a unit area A. Over a large surface area there are more units of A and vice versa. There are two circles; each representing two different surface areas. Using this information we derive equation 58:
Equation 58 tells us that a large Coulomb energy over a small surface area is equal to a small Coulomb energy over a large surface area, and the total energy is the Coulomb energy over unit area A times the surface area. The total energy is, of course, the total energy of the two electrons--and that energy is finite and conserved! This result is consistent with our formula:
If we plug in the following values and do the math, we get the finite energy of the two electrons:
Thus, we now have an incredibly simple and redundant formula that shows the connection between the parts and the whole at the classical and quantum levels.
There once was a physics student named Richard Feynman who asked his professor (re: the double-slit experiment), "What happens if you increase the number of slits to infinity?" This question led to summing the infinite number of paths a particle can take from point A to point B. If each path has a finite energy and we add up all those energies, we should get infinite energy! But that can't be right, since we started with one particle with a finite energy. Energy should be conserved, so what's wrong here? That's what we shall address in this post. First, let's define the variables we will use:
To tame this infinity problem we borrow a great idea from calculus: the integral. The integral is used to find the area beneath a curve (see diagram below). Equations 1 and 2 define the integral in terms of a summation.
At equation 2, notice how we can take N (the number of y's) to infinity and end up with a finite area beneath the curve. A finite number is definitely what we are after. Another way to get the finite area under the curve is to determine the average y (see equation 3), then map our curve to a simple rectangle (see diagram below) that has the same area as the curve shape. Equation 4 shows that the integral is equal to 'a' (the x value) times the average y value.
Now, let's apply what we know to the slit experiment. At the bottom of the next diagram, a gun fires an electron with energy Ee. Let's assume that all lights and detectors are off and the electron goes through all the slits simultaneously. The energy at each slit is some multiple or fraction of Ee. These energies can vary due to constructive and destructive interference. As with our y's above, we determine the average epsilon or coefficient for each Ee.
We don't know the average energy per slit. We do know it must be greater than zero according to the Heisenberg uncertainty principle and the Compton wavelength formula (see equations 5 and 5b). To have zero energy requires infinite time or an infinite wavelength. Obviously the space we use for the experiment is less than infinity, and, the longest time on record is the age of the universe--also less than infinity, so yes, the energy per slit must be greater than zero.
If N equals infinity, we might naively multiply N by the average energy at each slit to get infinite energy:
But here's a better approach. First let's reset our dimension variables:
Let's express x and y in terms of the electron's initial energy:
Next, let's express x and y in terms of the average energy per slit:
To get dx we divide by the number of slits (N).
At 12 we sum all the y's. At 13 we convert that sum to its energy equivalent:
We solve the integral at 14. At 16 the infinities and epsilons cancel. That brings us to 17, the area under the curve, but expressed in terms of energy.
Multiply both sides by E^2/xy, find the square root which gives a finite energy for the electron.
Thus energy is conserved. Our final energy is the same as our initial energy instead of being absurdly infinite.
