## Tuesday, May 30, 2017

### How To Derive the Prime Number Theorem

I read the article on Wikipedia about prime number distribution and I was surprised to find that the inverse logarithimc function was a very recent (late-nineteenth century) innovation, and that the proof was very deep. I have my own derivation which seems fairly simple, and I wonder if it is correct; and if it is, whether it has been previously written up.

It’s based on the Sieve of Eratosthenes, and it goes like this. You know that when checking for primes, you only have to go up to the square root of the number you’re testing. So if you’re looking for primes in the vicinity of 1,000,000 you only have to work the seive up to 1000.

Now, let’s say you’re checking for primes in the vicinity of 1,200,000. Now you have to work the sieve a little farther – in fact, up to about 1100. So the sieving process is going to be more thorough. That’s why the density of primes goes down as you move upwards.

How much more thorough is the sieve going to be? It will be more thorough by exactly the number of primes between 1000 and 1100. In short, the rate of change of the density function in the vicinity of 1000000 depends on the value of the density function in the vicinity of 1000. Can we construct a differential equation from these constraints?

Yes we can, and you will see that when we solve it we get 1/lnx, the desired result which eluded mathematicians for so long. Yes, there are more accurate functions for the distribution, but this is ground zero, the starting point.

- - - - - - - - -

Would you like to see how I construct the differential equation? You might not like how I do it, but it works for me. I find it’s not so hard if you throw some numbers into the game. Let’s suppose that the density of primes in the vicinity of 1000 is 1/8. So there are approximately 12.5 primes between 1000 and 1100. (The numbers don’t matter…we’re going to throw them away in the end and convert them into functions, variables, and differentials. But I like having numbers in front of me.)

Now let’s assume that the density of primes in the vicinity of 1 million is 1/40. So there are 25 primes between 1,000,000 and 1.001, 000? Then we ask: how many primes are there between 1,2000,000 and 1,201, 000

Well, if you applied the same Sieve of Eratosthenes to the higher interval, you would get the same number of primes – 25 primes. But in fact…the Sieve of Eratosthenes is more severe on the higher interval. You have an additional 12.5 primes between 1000 and 1100, and each of them strikes the new interval exactly once. (I’ve sized the interval so it works out that way.)

Let us call a number a prime candidate in the new interval if it survived the Sieve up to 1000. There are of course 25 prime candidates. How many will be struck by the sieve over the interval 1000 to 1100?

At first glance you might think we will lose half (12.5) of the prime candidates. But of course it’s not so. Most of the new sieving numbers fall on those 975 composite numbers that have already been eliminated. A little careful thought shows that if the sieving is truly random, the chance that a given number will be struck by the additional impact is 12.5/1000. So of the 25 prime candidates, we stand to lose only .375, leaving 24.625 prime numbers in the interval from 1.200 million to 1.201 million.

Shall we now convert the preceding narrative into a differential equation? It’s not so hard. Let’s consider n as being 1000 and dn = 100.  then  (using Greek rho for density) ρ(n) = 1/8.

Now let’s look at the density in the neighborhood of 1,000,000…that is,

We’re going to ask how fast ρ is changing as you go from 1 million to 1.2 million…in other words from   .

We’ve already done this with numbers. The number of primes between 1 million and 1.001 million is

The number of primes between 1.2 million and 1.201 million is, formally

But we’ve already figured out how to calculate this. It’s just the same as the number in the vicinity of 1 million, subtract a correction which we’ve figured out as:
We are very close to having our differential equation. The density of prime numbers in the vicintiy of 1.2 million is just (25 - .375)/1000, or converting this all into symbols:

This is a differential equation and it solves. Multiplying and regrouping, we get:

From here we get:

I’m not gonna pretend I can solve this diffrerential equation, but I know how to substitute in. And if I put 1/ln(n) in for the density, I get a true equation. Try it yourself if you don’t believe me.

## Saturday, February 27, 2016

### There Are No Pea-Shooters For Photons

I wrote an article several years ago titled "There Are No Peashooters For Photons". The link will take you to the where the article is posted. In the article, I take apart the various arguments which claim to debunk the wave theory of light in favor of the particle theory. You know the arguments: the black-body spectrum, the photo-electric effect, the Compton effect and and so forth. None of these arguments, which were so persuasive in the early days of quantum mechanics, remained valid after the discovery of the Schroedinger equation in 1926. All those arguments depended on the difficulties of having a wave (light) interacting with a particle (electron). None of those arguments remained valid once we understood that the electron was also a wave. Yet  by then the phton picture, with its magical "quantum leaps", was so well entrenched that the dominant physicists refused to abandon it. Instead, they reinvented the old quantum leap in the guise of Born's "collapse of the wave function", and particles ruled. Waves were out.

Even Schroedinger was cowed into acceptance of the new Copenhagen orthodoxy, and he did not seriously attempt to challenge it again until the 1950's when he was no longer taken seriously by people at the "cutting edge". But over the years, this or that group (notably the Jaynes group in the 1960's) has worked out the alternative formulation of quantum mechanics in terms of waves rather than photons; and by now it is recognized by most people at the highest level that at the very least, the photo-electric effect has a straightforward interpretation in terms of the wave theory of light. But even people who should know better continue to claim to be troubled by the apparent discrepancy between the power density of the e-m wave and the physical cross-section of the atom. People who are trouble by this should read up a little more on classical antenna theory, especial my blogpost on the Crystal Radio. But that's another story.

Even the people who accept the wave explanation of the photo-electric effect are quick to stipulate that the Compton effect is impossible to explain by waves. Schroedinger's own beautiful explanation, published in 1927, languishes in such obscurity that when I independently rediscovered it on my own ten years ago, I thought I would be a shoo-in for the Nobel Prize. The Compton effect remains a huge debating point in favor of the particle theory of light, but in fact it is no more than a debating point.

Feynmann weighs in on the wave-versus particle argument by citing the photo-multiplier tube as incontrovertible experimental evidence of the particle nature of light: you can actually hear individual photons being counted by the detector. But Feynmann is simply wrong. There is a 1200-volt power supply hooked up to the tube, and there is no way you can argue from logic that it must be a particle rather than a wave which triggers the process whereby that 1200-volt supply generates a pulse of current. I explain the wave mechanism in my article on Quantum Siphoning. The article actually deals with the mechanism for generating flecks of silver on a photographic plate, but the principle is the same.

But the proponents of the particle theory have one final impenetrable refuge, which they have relied upon since probably the mid-1980's or perhaps a little earlier. There are now experiments, so they say, where you can actually shoot a single photon at a beam splitter and measure that it goes either one way or the other; it is never split in two as you would expect if light were a wave.

This is a very weighty argument, and I talked about this in my article, "There Are No Pea-Shooters For Photons". It turns out that the experiment is a little more complicated than shooting a single photon at a target. You have to use something called parametric down-conversion, which creates entangled pairs, and by measuring one member of the pair, you can be sure that you have shot the other member at your desired target. As they tell us, "it comes to exactly the same thing as shooting single photons at a target."

Or does it? In my original article, I expressed my doubts. Hence the title of the article. But that was just me. Would any serious physicists back me up?

Maybe they have now. I came upon this post in the Physics Stack Exchange site the other day. It seems that some guys have analyzed the statistics of parametric down-conversion, and it is after all not identical to what you would get from a true single-photon source. The original Physical Review article by Bashansky, Vurgaftman, Pipino and Reintes is here, behind a paywall. From the abstract, it looks pretty significant to me.

Almost all of the cosmic paradoxes of Quantum Mechanics originate with the hypothetical, "suppose you shoot a single photon at...". But what if you can't? That would represent a huge paradigm shift in the way we have to look at the microscopic universe.

I've been saying it for years, and it turns out not that maybe I was right. Bashansky et al have my back on this: there really are no pea-shooters for photons.

At least not yet.

## Thursday, June 25, 2015

### Polish Guy Singing "Crazy"

Jacek (that's YAH-tsek, but you can call him Jack) is a friend of mine who takes part in a Thursday evening rehearsal session downtown that I lead at a program called ArtBeat. I think this is a phenomenal version of Crazy, the Patsy Cline signature song. (Not everyone knows that Willie Nelson wrote it.) I helped a little with the arrangement on this, but despite our best efforts the two patch changes on the instrumental solo were kind of surprising to us (as you might be able to tell from Jacek's facial expression.)

If you like this video, pass it on to your friends.

## Saturday, April 4, 2015

### I. L. Peretz 100th Yohrzeit

Today is the 100th anniversary of the death of Yiddish writer I. L. Peretz. They say that a hundred thousand people thronged the streets of Warsaw for his funeral.

Peretz was, in his best moments, a genius without parallel in the Yiddish world. I've had a personal role in trying to keep Peretz's work alive. It's brilliant work and it's actually important if you care at all about who you are and where you came from. Of course, not everyone sees it that way. In fact, my work hasn't been much appreciated in the Jewish world. But at least I've done something.

My biggest project was my musical setting of Peretz's epic poem "The Ballad of Monsich". You can see it on Youtube here:

This was taken from the Winnipeg Fringe Festival in 2005, and there's actually eleven videos. This is the first one, but if you want the whole playlist, it should link from here.

But I almost forgot one other project I did, which was an incredibly chilling poem about power and hypocrisy called Solomon's Throne, which still rings true today. It was never published anywhere, and so I thought it would be fitting on this, the 100th anniversary of Peretz's death, to post it on the internet. You can link to the PDF here: it includes the transcribed Yiddish (using German spelling) side-by-side with the English translation.  I'm going to say without modesty that my translation is pretty f#%\$ing brilliant.

## Tuesday, February 3, 2015

### Penetrating the Barrier: Some Calculations

Yesterday I calculated the difference between the two ground states of the double well (the infinte well with a finite barrier in the middle.) The two states were of course the symmetric and the antisymmetric cases, with the antisymmetric having a slightly higher energy:

I picked a wavelength of 44 angstroms (7 angstroms = 1 radian) for the sine wave, and a decay length of 2 angstroms for the exponential region, so the ration of the two parameters was 7:2. And then I picked the dimensions of the box so that the waves would just fit inside. And that's pretty much all you need to do the calculation. The Wikipedia formula asks you to calculate the transmission coefficient in terms of E and V, the energies in the two zones respectively, but the formula they give is a bit redundant...you can re-express it just in terms of the ratio (7:2 in this case) of the two characteristic lengths:
You can see I've got r = 7/2, and ka = 4, so plugging in the numbers, we get a transmission coefficient of 1/2483, or close to 0.04%.   Is this the same as I got using the steady state solutions?

Yesterday I calculated that the wavelengths of the symmetric and antisymmetric modes were different by 0.4% (about one part in 250). But in quantum mechanics wavelength is momentum, and energy is frequency (momentum squared). So the frequencies are different by one part in 125.

Here is where it gets interesting. You start out with the electron all on one side of the barrier. How do you do that? By having the symmetric and antisymmetric modes in phase so they re-inforce on the left and cancel on the right. After 125 cycles, they will be back in phase again. But after 62.5 cycles, the relative phases will be reversed...so all the wave function will be on the right hand side.

Is this the same result as we got from the Wikipedia formula? It's hard to say, because the Wikipedia formula is expressed in terms of the Copenhagen interpretation as the "probability" of a particle getting through the barrier. Where is my "particle" in terms of standing waves?

Well, one way to interpret it would be to imagine the particle bouncing back and forth in the potential well. If it bounces once on each cycle, that means after hitting the wall 62 times, it gets completely through to the other side. That's like a 1.5% penetration on each cycle...more like 3% or 4%, actually, because there's the competing probability of it returning from whence it came. That's a lot more than the Wikipedia calculation. Have we done something wrong?

It's actually not quite that bad. Remember in quantum mechanics there's a discrepancy between the phase velocity and the group velocity. For electrons, the group velocity is twice the phase velocity. We've been treating the electron as though it travels with the phase velocity of the wave function...it's actually twice that, so where we thought it was hitting the wall 62.5 times, it was actually 125 times. So our nominal penetration ration goes down by half, to below one percent. That's a little better, but still a long way off. What gives?

There is a fascinating answer to this question, and it gives us a very deep insight into the whole subject of quantum resonance. We said that each time an electron strikes the barrier, it has a 1/2500 chance of getting through. But what does this mean in terms of the wave function? It means the amplitude of the transmitted function is one fiftieth. The power goes as the amplitude squared.

In the coupled well system, the transmitted wave is in phase each time it re-strikes. So the amplitude on the transmitted side goes as one fiftieth, two fiftieths, three fiftieths, etc. After only fifty strikes, it is 100 percent through!

Not exactly....because once the amplitude starts building up on the right hand side, there is the probability that it will come back through the other way. But early in the game, that probability is negligible. So while the amplitude is growing as 1/50, 2/50, 3/50....   the probability is growing as 1/2500, 4/2500, 9/2500... in other words, it is growing parabolically.

How close does this parabola fit to the sine wave which represents the oscillating probability? Pretty close, as it turns out. I'm not going to do it in detail, but if you look at the Taylor Expansion for the cosine function, the x-squared term projects to -1 at when x = 1.41 radians (the square root of two. On our sine wave, where 125 strikes (the electron striking the wall) makes is a half-cycle, that comes to 56 strikes.

It's pretty close.

### Quantum Tunneling: A Different KInd of Calculation

I did some interesting physics yesterday so I thought I should write it up. The question was about quantum tunneling. It came from my friend's 3rd year Physical Electronics course in Electrical Engineering. You had a MosFET with a silicon dioxide insulating layer between the gate and the channel. They gave you the the barrier potential in the insulator, and told you to assume an electron with a certain energy (less than the barrier potential) in the gate. What was the "probability" or rate of tunneling?

This is a pretty standard problem which you can look up on Wikipedia. You calculate the form of the free-electron solutions in all three regions and match up the boundary condition. It's a little messy but it works.

I thought I could do better. The problem is messy because the conditions are different on the left and right hand sides. So I came up with the idea of working with symmetries. Instead of taking a freely propagating wave from the left and following it through the barrier, I put the whole thing inside a bigger potential well and considered the steady state solutions. Of course there are two minimum-energy solutions: the symmetric and the anti-symmetric, separated by a very tiny energy difference.

I don't like to carry too many letter symbols so I picked numbers that would come out nicely. From the terms of the problem, it's easy to calculate the free-propagation constants in both regions. So I allowed myself to tweak the problem parameters so the propagation constants come out to nice integers. Assuming pi = 22/7, here is the problem as I set it up:

You see I gave the well a width of 20 (you can call it 20 Angstroms if you like or 20 millimeters, it won't matter in the end...it's all about the geometry.) Can you see why this fits the sine wave perfectly? It's because the wave penetrates into the barrier. I've assumed that the penetration length is 2, so 20+2=22 and everything fits. The anti-symmetric solution is the same basic idea:

You can see I set it up so that there is a wavelength of 44, which fits perfectly into the box. But actually that's a bit of a cheat. If I were matching up a sine wave to an exponential at the boundary, then I would indeed get a wavelength of exactly 44, because of the well-known property of an exponential whereby the projection of the slope intersects the x-axis at exactly the decay length (I'm also assuming the sine wave is in the small-angle regime where sin(x)=x):

I don't really care about the constant in front of the sine wave, but I just threw it in here to show that it's easy. I used the property of sine waves that the projection of the tangent from the zero-crossing reaches the altitude of the sine wave after one radian. The point is that if I'm matching my sine wave (period = 44) to an exponential (decay length = 2) then this is how they line up. Everything fits.

Except I'm not exactly matching the sine wave to an exponential. I'm matching it to cosh (in the symmetric case) and sinh (in the antisymmetric case. Cosh is just a little higher and a little less steep...so it matches the sine wave at a slightly different position. Actually...the only way to match it up is to make the sine wave just a little bit longer. Instead of a half-length of 22 angstroms, the sine wave stretches to around 22.04....close to 0.2%, if you like. Similarly, in the anti-symmetric solution, the sine wave has to be shortened by the same amount. So the anti-symmetric solution has a slightly higher frequency than the symmetric solution.

We'll see what the implications of this are when we come back.

Oh...how did I calculate the 0.04 angstroms? That's the beauty of this method...it's pure geometry, just looking at the ration of the function with its derivative on both sides of the boundary, and using the small-angle approximation tan(x) = x. It falls right out.

## Thursday, December 4, 2014

### Wave Function Collapse Explained by Quantum Siphoning

It's almost five years since I posted my original article on Quantum Siphoning and I think it's time for a second look. The idea is to explain the collapse of the wave function by means of normal time evolution. Specifically, how does the very weak "wave function" of the light from a distant star induce the reduction of silver bromide ("collapse of the wavefunction") on a photographic plate?

The traditional explanation is that this phenomenon proves the existence of photons, because the energy of the wave has to be concentrated in a point in order to provide:

a) the positive energy needed to account for the energy difference between silver bromide and metallic silver; and,

b) the additional energy needed to overcome the "bump" of promoting an electron into the conduction band.

The traditional picture looks like this:

And they say you can't explain this with classical electromagnetism because the classical wave is far to weak to concentrate enough energy in that one little silver atom. Hence the photon.

I don't buy it. The first problem with this argument is that the thermodynamics is flawed. The photographic plate is not a device which captures energy and converts it to chemical form. The energy is already present in the plate in chemical form. Yes, it's true that the standard enthalpy of metallic silver is greater than that of silver bromide. But to calculate the spontaneity of the reaction you need to take into account the concentrations. The silver concentrations in an exposed photographic plate are parts per trillion. At those concentrations, the free energy of the reactions actually tilts the other way! I've done the calculation here. You don't need the energy of the "photon" to drive the reaction. The energy is already available in the chemistry of the plate.

But what about the "activation energy"...the bump of energy needed to get the electron into the conduction band, the intermediate stage of the process? That's where Quantum Siphoning comes in. The energy released when the target silver atom gets reduced is pumped back into the crystal to break up the silver bromide bonds. But how can it do that?

I couldn't figure this out for the longest time because I was trapped in the paradigm of a single electron getting transferred from one site in the crystal to another. But that's not what electrons are. They aren't particles with their own distinct identities. They are a collective wave function with multiple excitations. Anyone who knows about quantum field theory knows this is true.You simply can't describe an atom with two electrons by saying "this electron is here and that one is there". They are both excitations of a single wave function.

And the funny thing is, the same is true on some level for the x-trillion silver atoms which are part of a single silver bromide crystal. You can't say that a photon comes along and knocks an electron out of a single silver atom. What you can say is that a wave passes through the crystal and disturbs the wave function of the whole crystal so it is driven, ever so slightly, into a superposition of states where there is some amplitude that "an electron" is in the conduction band.

This is where people have trouble understanding where I'm going next. And the reason they have trouble is the way quantum mechanics is taught, from a strictly Copenhagen perspective of particles and quantum leaps. Nobody tells them about the equally-valid Schroedinger picture of wave functions and time evolution, which gives exactly the same results for all kinds of ordinary things, including the black body spectrum, the photo-electric effect, and even the Compton effect. I explain the connection between the two pictures in a series of blogposts starting here.

What people don't understand is that everything an atom does in its interactions with ordinary thermal light can be understood by looking at the superpositions of states calculated by the Schroedinger equation, and applying the charge density interpretation to the resulting wave functions, instead of Copenhagen's "probability density". People don't know this!

But it's much worse than that. They certainly don't know the full implications of the wave function picture, as I've listed two paragraphs back. That is not so surprising. What is horrifying to me is that they don't even know the basic and obvious fact that is you take the superposition of the s and p states of a hydrogen atom, you get an oscillating charge density...in other words, and antenna.

They don't know this...and when I tell them, they don't believe it. Even though its an obvious consequence of the well-known solutions of the Schroedinger equation.

And if they don't believe that the Schroedinger equations gives you a hydrogen atom with an oscillating charge distribution, then how are they going to believe that the very same tiny oscillating charge behaves exactly like a classical antenna? That's the calculation I did in those articles I pointed out above, comparing the Copenhagen picture to the Schroedinger interpretation.

And yet there is nothing that should be terribly controversial about anything I have said so far in this article. It's certainly unfashionable to talk about charge densities instead of probabilities, but there is nothing objectively wrong with it. What is appalling to me is that it is so very unfashionable that educated physicists scoff at the very notion that applying Maxwell's equations to the charge density picture gives you correct quantum-mechanical results. But it most certainly does.

So how does all this apply to the photographic plate? It's really very simple. The trillion-or-so silver atoms in the silver bromide crystal are little receiving antennas. The presence of a very week electromagnetic wave drives them ever so slightly into the excited state. What state is that? It's a state where the conduction band is ever-so-slightly excited.

Now there is one particular silver atom which is the target site, the site which we hope to convert to metallic silver. The presence of charge density in the conduction band will naturally couple to this target atom. And this coupling turns that special site into...an antenna. Just like the trillion little silver bromide antennas...but with a difference. Those silver bromide sites were receiving antennas. This target silver atom is...a transmitting antenna. Just as an atom being driven from a lower energy state to a higher energy state functions as a receiving antenna, so does an atom capturing an electron from a higher energy state into a lower energy state function as a transmitting antenna. I explain how these things work in this blogpost about the Crystal Radio.

The very weak electromagnetic wave is gone. It disturbed the silver bromide crystal to a very tiny extent, leaving a small amount of energy distributed among the ground state (silver bromide) and the two other states...the conduction band and the reduced silver site. And the combination of those two states creates a tiny classical antenna...a transmitting antenna buried in the middle of that micro-crystal. And now Quantum Siphoning takes over.

I always knew that the target silver atom would re-transmit energy at the optical frequency. I just never knew how the silver bromide "molecule" could re-capture that energy, which was going out in all directions. Until I realized it's not a single silver-bromide site which is involved...it's the whole silver-bromide crystal. Each silver-bromide site within that crystal is a receiving antenna, and those receiving antennas surround the target silver atom, which is transmitting. It's a perfect classical siphon. As the electron amplitude flows into the target silver atom, the conduction band is being depleted. But the energy released at the target site, in the form of re-radiated e-m waves, is captured by the surrounding silver-bromide sites...thereby replenishing the conduction band.

There is no photon. There is no collapse of the wave function. There is nothing but the ordinary, natural time-evolution of the Schroedinger function, working together with Maxwell's equations.

It's a perfect classical siphon.