Why Does Squaring a Wave Amplitude Yield a Probability?

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:

String Theory's Graviton and Spin-2 Particle Controversy

According to the current physics dogma, the graviton is a spin-2 particle. One argument for this involves the fact that Einstein's field equations are rank-2 tensors. Rank-2 tensor corresponds with spin-2 particle? If that's the case, then a 1/2-spin electron should correspond with a 1/2-rank tensor--but there is no such tensor! Just when you've given up hope that the graviton is a spin-2 particle, another argument is thrown against the wall to see if it will stick. In the following video, Leonard Susskind presents that argument:

There is a rule that says that left-moving energy must equal right-moving energy along a closed string. Each point along the string (represented by sigma) could serve as a starting point and the end point (equal zero and 2pi); i.e., the closed string and its points are invariant. Because right-moving and left-moving energy adds to zero, and because the invariant string and its points are equivalent, the closed string must have right-moving energy and left moving energy, and those two energies must be equal. Clear as mud, right? But that's essentially the argument in favor of spin-2 particles. Here is how the graviton is represented:

The a's and b's in parenthesis are the creation operators for the x and y axis, respectively. Complex numbers are used to show angular momentum. Notice there are two angular-momentum creation terms; one for left motion and one for right motion. There are two possible states: the two angular momenta going clockwise or counter-clockwise. Now compare this setup to the photon's:

The photon has only one momentum or one spin. There is no right-left motion business. And why should there be? When angular momentum completes a cycle, the total displacement is zero. If there is a vector moving left, there is an equal, opposite vector moving right. No need for an extra spin. An extra spin is redundant. This is why the left-right-motion argument fails to justify a spin-2 particle. Nature only requires one spin to complete a cycle that adds to zero. Here is the math:

Take the integral of the entire cycle of the spin (360 degrees) and you get the equivalent of the right-left-motion business--which is zero.

Bearing all this in mind, does the graviton (assuming it exists) need to be a spin-2 particle?

Why A Quantum Gravity Theory May Be Impractical

Finding that elusive graviton would be a boon to the Standard Model--but is a quantum theory of gravity practical?

According to the Wilkinson Microwave Anisotropy Probe, the ground-state energy density of the vacuum of space is E-10 joules per cubic meter.  This is essentially the energy of nothing.  Energy less than this is less than nothing.  For all intents and purposes, energies less than the vacuum don't exist.  For the graviton to exist, it should have an energy density greater than E-10.

Suppose we take two protons and place them a Bohr radius apart.  How much is the gravitational energy?  What is the gravitational energy density?

The formula used to measure gravitational energy is E = Gmm/r.  Where E is energy; G is Newton's constant (E-11); m is the proton mass (E-27) and r is the bohr radius (E-11).

E = (E-11)(E-27)(E-27)/E-11 = E-54 joules.

To determine the energy density, divide by the the volume, the cube of the Bohr radius:

E-54/E-33 = E-21 joules per cubic meter!

This is approximately 100 billion times less than the vacuum energy density!  No wonder the graviton has not been found.

Our gravitational energy formula suggests that we could reduce the radius to an infinitesimal size.  Surely we would get an astronomical amount of gravitational energy and gravitons would be as common as sand on on the beach.

There are a couple of problems with this strategy:  1.  Gmm/r can't be greater than (m^2c^4 + p^2c^2)^.5--the total energy of the protons.  2.  Most of that energy is due to the strong force and the quarks that make up the protons.  For quarks and protons to exist, they need the lion share of the available energy.

To get an energy density greater than the vacuum's there needs to be a lot more particles than just two protons.  According to the gravitational-energy formula, when you double the mass, you increase the gravitational energy four times.  The chances of finding a graviton increase exponentially as you add more mass.  Or does it?

When you add more mass, you add a haystack of particles for the graviton to hide in.  You might say that the graviton has its own uncertainty principle: the closer you get (quantum scale), the harder it is to find.  If you step back and look at the big picture (the mass of a double-star system) you can detect a very faint gravitational wave.  Such a wave has little or nothing to do with quantum physics, since you need huge masses just to get started.

These are some of the reasons why a quantum gravity theory may be impractical.

Update:  An elementary particle such as an electron at rest needs E-14 joules of energy to exist.  The gravitational energy of a proton, using the proton's classical radius, is (E-11)(E-54)/E-15 = E-50 joules.  This is not nearly enough energy to create an elementary particle such as the electron--so it appears the graviton does not exist at the quantum level, unless its energy requirement is a thousand billion trillion trillion times lower than the electron's.