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.
Imagine a line x along a plane. How many such lines can the plane hold? An infinite number:
Imagine a plane within a cube. How many such planes can the cube hold? An infinite number:
Now this is harder: imagine a cube within a 4D space. How many cubes can the 4D space hold? An infinite number:
Thus there are an infinite amount of cubes or 3D spaces inside 4D. According to the equation above, this is true no matter how small wxyz is. We can infer that space that is 4D or higher can accommodate an infinite amount of 3D space. The above video shows Brian Greene talking about a little ant crawling around in a tiny, curled-up higher dimension. Here is an illustration:
According to Greene, the ant is able to enter and exit the higher dimension with ease. However, according to the math above, the ant would have to travel an infinite distance. It would never make the trip within its puny lifetime. Below, the red arrows represent the ant's journey. The vertical line represents 3D and the circle represents a higher curled-up dimension. Note the lines within the circle. They represent 3D spaces--an infinite number of them.
Now where there's infinite 3D space, there is bound to be infinite ground-state energy and mass:
So within a higher dimension (no matter how small) there should be infinite energy and mass! If these higher dimensions exist, every volume of stuff we measure should have infinite energy and mass! But they don't. That suggests strongly that higher dimensions are imaginary. There is also the Heisenberg Uncertainty principle which tells us that something very small with infinite energy should exist for zero amount of time:
Then again, what about the many many dimensions of Hilbert space? Where would quantum physics be without those extra dimensions? We should take a close look at Hilbert space. Let's start by examining a familiar 3D space, then we will analyze a 6D object I recently discovered.
Consider vector A below. It is composed of unit vectors i, j, and k:
Let's take the dot product of A with itself.
Taking the dot product of A with itself leads to the identity matrix or Kronecker delta. Because space dimensions are orthogonal (90 degrees to each other), the products of the cross terms equal zero. "Orthogonal" is a good thing--it suggests the 3D space is legit. If the dimensions are orthogonal, the dimensions should also be linearly independent. Let's check this. First we define the eigenvectors:
Next, we multiply each eigenvector by a coefficient (ci,cj,ck), then add them to get a column vector with all zeros.
It is clear that all the coefficients equal zero. This spells linear independence. Now, below is the math for the 6D object I mentioned earlier. As you can see, its dimensions are also orthogonal and linearly independent.
This 6D object does not have infinite mass or energy or infinite 3D space within. The 6D object is none other than a single die:
Notice that the unit vectors i and j are parallel lines; i.e., ijcos(0). Their product does not equal zero, so they are not orthogonal. When we test for linear independence, equation 19 reveals that ci does not necessarily equal zero, nor does cj. So the die's spacial dimensions are not 6D, but 3D.
So what are the die's six dimensions and why are they orthogonal and linearly independent? We can think of equation 20 below as the wave function of the die. At equation 21 we take the dot or inner product of the wave function. Notice at 22 the cross-term products are all zeros within the matrix. How is that possible? (See equations 23 to 25.)
Equation 25 reveals the 6D has nothing to do with space or its angles. It has to do with probabilities! The cross-term products are zero because there is a zero probability the die will yield any numbers that aren't 1, 2, 3, 4, 5, 6. Equations 26 and 27 provide an example of a cross-term product and its probability:
Thus the die's dimensions are orthogonal because they are each statistically independent--not because of right angles. However, we can say that p (the variable subsuming probabilities) is equivalent to cosine-phi for Cartesian unit vectors--at least for the cross-term products, since they are all zero.
This equivalence might lead one to believe there are six spacial dimensions, but what we really have are six statistically independent states within Hilbert space. We can determine the expectation value of these states as follows:
As you can see, having extra dimensions, parameters or states can be useful--and they can reside within 3D space.
Perhaps you heard the story (that's code for unsubstantiated rumor) where Paul Dirac, a legend in the field of quantum physics, woke up one morning, put on his trousers, put on his shirt, put on his socks and shoes. He then stepped into the shower, forgetting he had already dressed. His mind was elsewhere, but the brisk chill of cascading water drenching his clothes brought him back. "That's it!" he said. "Square the amplitude and you get the probability!"
He was, of course, referring to the fact that you can take the inner product of an eigenvector (or ket) with its complex conjugate (bra) and get a probability of a particle's position, momentum, or whatever. It's a pretty cool trick and it works. The question is why? Why does squaring a wave amplitude yield a probability? Below is a mathematical equation that I thought up while I was taking a shower (with my clothes on):
In the numerator you may recognize Euler's identity. Basically, the numerator is the sum of all possible wave amplitudes. The denominator is a normalization factor: it ensures that when you calculate the inner product of the equation's right-side expression with its complex conjugate you get one. One is a good number to get, since it is the sum of all probabilities. Most importantly, the equation shows the connection between wave amplitudes and probabilities.
Below is a visual aid that I hope will clarify the connection. It is a well-known fact in the subject of trigonometry that sine squared plus cosine squared always equals one. This fact can be used to model not only waves, but probabilities as well, since the sum of all probabilities also equals one. However, to reach a total of one, one must divide the wave amplitude by X1 in the case below where n is equal to one, i.e., where there is only the sum of one Euler's identity multiplied by a coefficient X1.
The lower part of the visual aid shows the inner product between the bra and ket vectors.
As you can see in the diagram below, the sum of squared wave amplitudes equals one, and probabilities P(A), P(B) add up to one.
This system works well if you have only two probabilities: cosine squared can represent one probability and sine squared can represent the other, but what if there are three or more probabilities? Well, that's why I put together the equation we started with. Let's say there are three probabilities, and they must add up to one. In such a case we need to raise n to 2:
There are now three wave amplitudes. When you square them, add them, and divide by the normalization factor (x1^2 + x2^2), you get one. You also get one when you calculate the inner product of the bra-ket vectors:
And, of course, the probabilities also add up to one:
