In my next-to-last post on this subject, I left the question hanging in the air: how does a reaction proceed when the Gibb's Free Energy is positive? I am now going to try to answer this.

The reaction in question is the decomposition of methane in the presence of CO2:

CH4 + CO2 ===> 2CO + 2H2

As I recall I determined that at STP conditions (25 C and 1 atm) the Free Energy was definitely unfavorable. I think it worked out to something on the order of 90 kiloJoules. There was a complication however: the reaction takes place in a nuclear reactor at a temperature of 300 C. The effect of this is to bring the Free Energy down quite a bit (because the entropy term is favored by the smaller mole fragments on the product side of the equation) but not nearly enough to make it go negative. Let's say it came to around 65 kJ.

The funny thing is that these numbers lead to all kinds of consequences in terms of measuring the actual gas concentrations. To see what happens we have to convert the Gibbs Free Energy to the corresponding Equilibrium Constant. The equilibrium constant k is given by the formula:

k = exp(-(Gibbs Free Energy) /RT)

where RT is the gas constant product from the formula PV=nRT. It's easy to remember that RT comes to exactly 2.2 kJ because one litre of air at STP occupies just 22 litres.

Now it gets interesting. The equilibrium constant in the first case (room temperature) works out, on the ratio of 90/2.2, to around e^-40, or dividing by 2.3 you get 10^-17. At the higher temperature, however, the numbers come out a little less extreme, at around 10^-12. These are all still very tiny numbers. How could they make much difference?

It turns out that the difference is critical in the case of interest. In a previous blog I showed how to use the equilibrium constant to calculate the concentrations of a reaction. Now lets apply it to the trace gasses in the nuclear reactor. We assume that reactor coolant is leaking into the CO2 purge gas stream through a tiny crack in a pressure tube. The coolant, a complex hydrocarbon, will typically break down into methane-like fragments. Let us suppose the initial concentration of methane in CH4 is 20 ppm, or 2 * 10^-5.

How can we balance the equilibrium equation in this case? We will write the equation as

k = [CO]^2 * [H2]^2 / [CH4] * [CO2]

For a reactor temperature of 300 degrees, we have k = 10^-13 and we need to make the concentrations in the above equation work out to that value. By toying with numbers, we can see that if the initial concentration of methane is 90% dissociated, so that only 2 ppm remains, that the CO and H2 concentrations must each be close to 40 ppm: we then get for the numerator a value of 256 *10^-20, which requires the denominator to be 256 * 10^-8, or just over 2 ppm. That's where the equation becomes balanced.

The situation is quite different, however, when the gas returns to room temperature. Now the equilibrium constant is 10^-17. The math is a little hard to write out in ASCII, but it turns out if only 10% of the methane dissociates, you get a concentration of 4 ppm for both the CO and the H2. Multiply out the terms and the equation balances at these concentrations. You have 18 ppm methane.

These numbers describe exactly what was seen to be happening in the leak detection system! When gas was initially sampled at the detector, it showed definite methane readings in the tens of parts per million. But after drawing for several minutes on a sample line, the reading "settled down" to zero, reassuring everyone that there was no cause for concern. What was actually happening was that the initial reading came from the room temperature gasses that had been sitting in the tubes: the methane that had initially decomposed at high temperatures had recombined into its original form. But as you kept drawing on the sample line, you then got to the fresh gasses which had just decomposed at reactor temperatures. Even if the sample point was at room temperature, the gasses had not yet had enough time to recombine, so the methane reading was basically zero.

I started off telling this story because almost thirty years later, it would have critical consequences in my development of the theory of Quantum Siphoning. I was involved in an online discussion of the quantum nature of the photographic process, and my opponents had thrown in my face the fact that the photo-reduction of silver bromide to silver was thermodynamically non-spontaneous. It was only because of my previous experience in analyzing the decomposition of trace methane in CO2 that I was able to see the possibility of doing the same thing for trace concentrations of silver in a silver bromide crystal. This insight was to have critical consequences for my ability to see the way to Quantum siphoning.

There was one other physics problem from my younger days that gave me an unusual perspective on processes involve waves and the interchange of energy. That was the problem of the crystal radio. But that's a story for another day.

There is a final postscript to the story of the nuclear reactor that needs to be told. Having figured out what was the problem with the leak monitoring system, I naturally wanted to tell people all about it; in no small measure, to show people how smart I was. I called a meeting of the interested parties. The department manager was the chair of the meeting. I began my presentation but almost immediately the manager interrupted me. "Before we go too far, I want to ask one question". He then turned to the reactor superintendant and said: "Even if what Marty says is right, would it make any difference to how you operate the reactor?".

"Absolutely not," replied the superintendant. The manager then turned to me and said, "Well, I guess there's no need to continue with this meeting". And that was it.

I didn't object because at the time I was actually impressed at what a good job the manager had done: he had gotten right down to the essentials without wasting a lot of time over theoretical issues. I was young and I wanted to be a team player, so I accepted the result.

It wasn't until I started writing this series of articles several weeks ago that it struck me: I had been set up! The manager and the superintendant had gotten together and figured out how to shut me up. It's dirty trick that people use when they want to dismiss what you're saying without letting you make your point: they start off with "even if what you say is true...". Of course, if what I was saying were true, it should have made a difference. They just didn't want to admit it.

## Sunday, August 22, 2010

## Monday, July 19, 2010

### Why do solids absorb light?

I used to participate in an online discussion group at physicsforums.com until I was banned for life earlier this year. I still sometimes notice some topics that I am motivated to comment on, and today was one such occasion. Since I can't post to the newsgroup, I will use my blogsite to put in my two cents worth.

A "newbie" using the name of "Infrasound" posted a question about absorption of light by solids. It's a very good question. Basically he notes that the traditional explanation of exitation of electron orbitals fails to explain where the light actually goes. Every electron that is excited to a higher energy level must sooner or later decay back to the ground state. It's a non-dissipative process which cannot absorb energy. So where does the light go?

Infrasound was quickly put in his place by a several veteran posters including the cryptically named cthugha and alxm, and of course the omniscient ZapperZ. Basically they tell Infrasound to get a life before wasting their time with such simple questions whose answer is to be found in any first-year text. It is not, they say, the atomic energy levels which are excited: those are too high to be susceptible to optical or infrared; it is rather the collective energy levels, which can be much closer spaced.

Infrasound rightly points out that this answer merely evades the question without answering the fundamental point about absorption. As long as the modes are driven by a specific frequency of light, they will give off the same frequency when they decay back to the ground state. So the colors are scattered but never absorbed. This objection is met by scoffing on the part of the above-mentioned smug guardians of truth who dominate the forum.

I'm going to propose an answer to this question, which I don't believe I've seen anywhere else. The mechanism I'm going to propose is most closely related to Compton scattering; and recall that in Compton scattering, the scattered light has a lower frequency than the incident light. So there is actual transfer of energy from light to matter.

When people hear Compton scattering they thing of free electrons being impacted by photons. That's not what I'm talking about. Most people thing that the Compton effect is conclusive proof of the particle nature of light, but in fact back in 1927 Schroedinger showed that the Compton effect can be explained purerly as a wave-on-wave interaction between classical e-m radiation and standing waves of electrons. The feature that characterises this interaction is that the light waves and matter waves interact when they have exactly the same wavelength. This is very different from the semi-classical explanation for the photoelectric effect, which is based on light and matter waves interacting at a shared frequency. Schroedinger's explanation was scoffed at and marginalized by the dominant Copenhagen group at the time, and today hardly anyone remembers it. But its mechanism is essential to a full understanding of how light interacts with matter.

In almost any solid, when the atomic lattice goes into vibration, there is a net displacement of electric charge. It is almost impossible for the positive charge lattice (the nuclei) to vibrate exactly in unison with the sea of negative charge (the electrons), so it is almost inevitable that sheets of charge density will appear along with the vibrations. It is these sheets of charge, at the exact separation of the wavelength of light, which must interact strongly with the light whose wavelength they share. The exact mechanism is difficult to explain in a few words, but the overall effect is that the light is absorbed and converted into mechanical energy of vibration.

My personal discovery of Schroedinger's wave explanation for the Compton effect is a long story which I'm going to have to save for another blogpost.

A "newbie" using the name of "Infrasound" posted a question about absorption of light by solids. It's a very good question. Basically he notes that the traditional explanation of exitation of electron orbitals fails to explain where the light actually goes. Every electron that is excited to a higher energy level must sooner or later decay back to the ground state. It's a non-dissipative process which cannot absorb energy. So where does the light go?

Infrasound was quickly put in his place by a several veteran posters including the cryptically named cthugha and alxm, and of course the omniscient ZapperZ. Basically they tell Infrasound to get a life before wasting their time with such simple questions whose answer is to be found in any first-year text. It is not, they say, the atomic energy levels which are excited: those are too high to be susceptible to optical or infrared; it is rather the collective energy levels, which can be much closer spaced.

Infrasound rightly points out that this answer merely evades the question without answering the fundamental point about absorption. As long as the modes are driven by a specific frequency of light, they will give off the same frequency when they decay back to the ground state. So the colors are scattered but never absorbed. This objection is met by scoffing on the part of the above-mentioned smug guardians of truth who dominate the forum.

I'm going to propose an answer to this question, which I don't believe I've seen anywhere else. The mechanism I'm going to propose is most closely related to Compton scattering; and recall that in Compton scattering, the scattered light has a lower frequency than the incident light. So there is actual transfer of energy from light to matter.

When people hear Compton scattering they thing of free electrons being impacted by photons. That's not what I'm talking about. Most people thing that the Compton effect is conclusive proof of the particle nature of light, but in fact back in 1927 Schroedinger showed that the Compton effect can be explained purerly as a wave-on-wave interaction between classical e-m radiation and standing waves of electrons. The feature that characterises this interaction is that the light waves and matter waves interact when they have exactly the same wavelength. This is very different from the semi-classical explanation for the photoelectric effect, which is based on light and matter waves interacting at a shared frequency. Schroedinger's explanation was scoffed at and marginalized by the dominant Copenhagen group at the time, and today hardly anyone remembers it. But its mechanism is essential to a full understanding of how light interacts with matter.

In almost any solid, when the atomic lattice goes into vibration, there is a net displacement of electric charge. It is almost impossible for the positive charge lattice (the nuclei) to vibrate exactly in unison with the sea of negative charge (the electrons), so it is almost inevitable that sheets of charge density will appear along with the vibrations. It is these sheets of charge, at the exact separation of the wavelength of light, which must interact strongly with the light whose wavelength they share. The exact mechanism is difficult to explain in a few words, but the overall effect is that the light is absorbed and converted into mechanical energy of vibration.

My personal discovery of Schroedinger's wave explanation for the Compton effect is a long story which I'm going to have to save for another blogpost.

## Tuesday, June 22, 2010

### Where did the entropy go?

I ended my last post rather abruptly when I got confused about the equilibrium equation for the reaction we were discussing. We had:

CH4 + CO2 ===> 2CO + 2 H2

The Gibbs Free Energy for this reaction is positive, so conventionally it "shouldn't go". I had considered what would actually happen in the case of a reaction whose Free Energy was exactly zero, and I said it should be in equilibrium just as it was written. That's where things got messed up.

For a Free Energy of zero, you get k=1 for the equilibrium constant. I plugged in the concentrations as written above into the equilibrium formula

k = [CO]^2 * [H2]^2 / [CH4] * [CO2]

and it comes to k = 16, so something is wrong.

Here's what I've figured out so far. Free energy is calculated for gasses at STP conditions: Standard Temperature and Pressure, or 1 atmosphere at 25 degrees Celsius. If you think about it, you'll see that it is physically impossible to combine these four gasses, in their stoichiometric ratios, in a single container at STP conditions. Just think about it. There is twice as much H2 as there is CH4, so the partial pressure of hydrogen must be twice the partial pressure of methane. They can't both be at STP.

By confining the gasses in a cylinder with a piston, we can vary the total pressure and it is not surprising that the equilibrium concentrations will change. This is just an application of Le Chatelier's principle. The value of k remains constant but the equilibrium point moves to the left or right: as the piston is compressed, the heavier molecules are favored because they take up less space, and vice versa. At some arbitray position, the gasses will inevitably be present in their stoichiometric proportions: but it is physically impossible for them to be present in the precise conditions specified in the balanced chemical equation: namely, all of them simultaneoulsy at STP.

By fiddling with the numbers, you can in fact verify that the equilibrium equation is satisfied for the following concentrations:

1/4 CH4 + 1/4 CO2 ===> 1/2 CO + 1/2 H2

It's just the balanced equation divided through by 4. So whatever enthalpies and entropies we had for the balanced equation, they are all altered in proportion for this modified equation. The free energy change is clearly zero.

What is still bothering the hell out of me is the fact that you still cannot put these four components into a single container at the exact conditions represented in the equation. The equation is written for STP conditions, and when you combine these gasses the partial pressures of the product side are still goind to be double the pressures of the reactant side. So it still needs to be explained: why does the equilibrium hold?

Let's look closely at what it means to combine these gasses.You have in the reaction as written 1/4 litre of methane, 1/4 litre of CO2, 1/2 litre of CO, etc; all of them at one atmosphere. You put them into a 1 litre container, and they fill the container: now, the partial pressures are respectively 1/4 atmosphere, 1/4 atmosphere, 1/2 atmosphere, etc. (You may notice that the total pressure inside the container is now 1.5 atmospheres but that is neither here nor there.) The point is that none of the gasses are at STP any more, so their enthalpies and entropies are all different. Why then does the equilibrium hold?

The problem is not so much with the enthalpies as with the entropies. In fact, the enthalpy of an ideal gas at a given temperature does not depend on the pressure or size of the containment vessel. (This fact in itself ought to be surprising but that is a story I'm not going to open up at this point.) But the entropies certainly change with expansion. And the gasses on the left hand of the equation expand by a factor of four, while those on the right hand side (the products) expand only by a factor of two. So the entropy changes ought to be different.

Most troubling of all is the fact that the entropy change tends to be logarithmic with the change in volume. So if we are looking to balance out the competing energy changes, it seems odd to me that the physical solution should come out in terms of nice fractional ratios of the original chemical equation. Clearly I haven't fully understood the situation yet. Maybe someone else will have a better explanation. But for now I'm going to put this little side issue to rest and continue with my story about Atomic Energy of Canada and the hydrocarbon detection system.

CH4 + CO2 ===> 2CO + 2 H2

The Gibbs Free Energy for this reaction is positive, so conventionally it "shouldn't go". I had considered what would actually happen in the case of a reaction whose Free Energy was exactly zero, and I said it should be in equilibrium just as it was written. That's where things got messed up.

For a Free Energy of zero, you get k=1 for the equilibrium constant. I plugged in the concentrations as written above into the equilibrium formula

k = [CO]^2 * [H2]^2 / [CH4] * [CO2]

and it comes to k = 16, so something is wrong.

Here's what I've figured out so far. Free energy is calculated for gasses at STP conditions: Standard Temperature and Pressure, or 1 atmosphere at 25 degrees Celsius. If you think about it, you'll see that it is physically impossible to combine these four gasses, in their stoichiometric ratios, in a single container at STP conditions. Just think about it. There is twice as much H2 as there is CH4, so the partial pressure of hydrogen must be twice the partial pressure of methane. They can't both be at STP.

By confining the gasses in a cylinder with a piston, we can vary the total pressure and it is not surprising that the equilibrium concentrations will change. This is just an application of Le Chatelier's principle. The value of k remains constant but the equilibrium point moves to the left or right: as the piston is compressed, the heavier molecules are favored because they take up less space, and vice versa. At some arbitray position, the gasses will inevitably be present in their stoichiometric proportions: but it is physically impossible for them to be present in the precise conditions specified in the balanced chemical equation: namely, all of them simultaneoulsy at STP.

By fiddling with the numbers, you can in fact verify that the equilibrium equation is satisfied for the following concentrations:

1/4 CH4 + 1/4 CO2 ===> 1/2 CO + 1/2 H2

It's just the balanced equation divided through by 4. So whatever enthalpies and entropies we had for the balanced equation, they are all altered in proportion for this modified equation. The free energy change is clearly zero.

What is still bothering the hell out of me is the fact that you still cannot put these four components into a single container at the exact conditions represented in the equation. The equation is written for STP conditions, and when you combine these gasses the partial pressures of the product side are still goind to be double the pressures of the reactant side. So it still needs to be explained: why does the equilibrium hold?

Let's look closely at what it means to combine these gasses.You have in the reaction as written 1/4 litre of methane, 1/4 litre of CO2, 1/2 litre of CO, etc; all of them at one atmosphere. You put them into a 1 litre container, and they fill the container: now, the partial pressures are respectively 1/4 atmosphere, 1/4 atmosphere, 1/2 atmosphere, etc. (You may notice that the total pressure inside the container is now 1.5 atmospheres but that is neither here nor there.) The point is that none of the gasses are at STP any more, so their enthalpies and entropies are all different. Why then does the equilibrium hold?

The problem is not so much with the enthalpies as with the entropies. In fact, the enthalpy of an ideal gas at a given temperature does not depend on the pressure or size of the containment vessel. (This fact in itself ought to be surprising but that is a story I'm not going to open up at this point.) But the entropies certainly change with expansion. And the gasses on the left hand of the equation expand by a factor of four, while those on the right hand side (the products) expand only by a factor of two. So the entropy changes ought to be different.

Most troubling of all is the fact that the entropy change tends to be logarithmic with the change in volume. So if we are looking to balance out the competing energy changes, it seems odd to me that the physical solution should come out in terms of nice fractional ratios of the original chemical equation. Clearly I haven't fully understood the situation yet. Maybe someone else will have a better explanation. But for now I'm going to put this little side issue to rest and continue with my story about Atomic Energy of Canada and the hydrocarbon detection system.

## Monday, June 14, 2010

### Point of Equilibrium

It is a well-known fact that a chemical reaction will not procede if the Gibbs Free Energy is positive. But as with so many well-known facts, there's more to the story than that.

Last week I wrote out the chemical reaction

CH4 + CO2 ===> 2CO + 2 H2

and noted that even at 300 degrees C, the Free Energy was positive. The obvious conclusion is that the hydrocarbons leaking into our leak detection system should not decompose, and if they exist they should be measurable at the hydrocarbon detector.

Anyone who has taken first year chemistry will be able to follow this straightforward logic. But it is an oversimplification of the truth. Now I'm going to explain why.

The much-repeated claim that the Gibbs Free Energy describes the spontaneity of a reaction is strictly true only when the components of the reaction are present in their stoichiometric proportions: which is to say, the proportions as given when the chemical reation is written out in its balanced form, as it is here. So if you mix one mole of methate, one mole of carbon dioxide

But in our leak detection system, the situation is very different. We are far from having the gasses present in their stoichiometric proportions. In fact we begin with pure CO2, and then introduce a tiny amount of methane, measured in the parts per million. What happens then?

It's a question of equilibrium. There comes a point where the rate of decomposition on the part of the heavier species equals the rate of recombination of the lighter species. The proportions of the mixture then stabilize. And not everyone remembers this, but in fact it is also part of the first-year chemistry curriculum (because otherwise I would have had no way of knowing it myself, not having any other education in the subject): you can use the Gibbs Free Energy to determine where exactly that point of equilibrium lies!

At this point I'm not going to explain why it works but I'm just going to write out the formula for equilibrium: it should look something like this:

k = [CO]^2 * [H2]^2 / [CH4] * [CO2]

The quantities in brackets are just the quantities of chemicals expressed in mole fractions. Reactants on the bottom, products on top. Because there are two moles of carbon monoxide in the balanced formula, you have to take the square of the concentration. Etcetera. Remember, I'm not explaining why this works, I'm just saying it's how the formula goes.

There's one more formula we need to make this work. The quantity k in the expression above is called the "equilibrium constant". To get the equilibrium constant you use the Gibbs Free Energy. It's an exponential formula: you take the ratio of the Gibbs Free Energy to the Ideal Gas Law constant for the temperature in question, and that ratio becomes the argument of the exponential function. If the ratio is positive, then the constant is greater than one; if negative, it is less than one.

If the Gibbs free energy is zero, then the equilibrium constant is equal to one. What does this mean? Just plug in the numbers. It means the numerator and the denominator of the fraction have to be equal. There are many ways you can do this. You can have one mole of each component. Or you can have, for example, two moles of methane, two moles of CO2, two moles of CO and one mole of H2...then your fraction comes to four on top and four on the bottom. Or whatever.

Even as a write these words, I can see that something is wrong. When the Gibbs Free Energy is zero, with k=1, the reaction ought to be balanced exactly as it is written. But if I try to plug those numbers into the equilibrium formula, I get 2^2 * 2^2 on top (because there are two moles each of hydrogen and carbon monoxide) and 1*1 on the bottom, so my equilibrium constant is way off. I'm going to take a pause here while I figure out what's wrong.

Last week I wrote out the chemical reaction

CH4 + CO2 ===> 2CO + 2 H2

and noted that even at 300 degrees C, the Free Energy was positive. The obvious conclusion is that the hydrocarbons leaking into our leak detection system should not decompose, and if they exist they should be measurable at the hydrocarbon detector.

Anyone who has taken first year chemistry will be able to follow this straightforward logic. But it is an oversimplification of the truth. Now I'm going to explain why.

The much-repeated claim that the Gibbs Free Energy describes the spontaneity of a reaction is strictly true only when the components of the reaction are present in their stoichiometric proportions: which is to say, the proportions as given when the chemical reation is written out in its balanced form, as it is here. So if you mix one mole of methate, one mole of carbon dioxide

*, two*moles of carbon monoxide and two moles of hydrogen, it is true that there will be no further production of the lighter species; in fact, given the right stimulus (e.g. a spark) the tendency would be for the reaction to go the other way, namely recombination into methane and CO2.But in our leak detection system, the situation is very different. We are far from having the gasses present in their stoichiometric proportions. In fact we begin with pure CO2, and then introduce a tiny amount of methane, measured in the parts per million. What happens then?

It's a question of equilibrium. There comes a point where the rate of decomposition on the part of the heavier species equals the rate of recombination of the lighter species. The proportions of the mixture then stabilize. And not everyone remembers this, but in fact it is also part of the first-year chemistry curriculum (because otherwise I would have had no way of knowing it myself, not having any other education in the subject): you can use the Gibbs Free Energy to determine where exactly that point of equilibrium lies!

At this point I'm not going to explain why it works but I'm just going to write out the formula for equilibrium: it should look something like this:

k = [CO]^2 * [H2]^2 / [CH4] * [CO2]

The quantities in brackets are just the quantities of chemicals expressed in mole fractions. Reactants on the bottom, products on top. Because there are two moles of carbon monoxide in the balanced formula, you have to take the square of the concentration. Etcetera. Remember, I'm not explaining why this works, I'm just saying it's how the formula goes.

There's one more formula we need to make this work. The quantity k in the expression above is called the "equilibrium constant". To get the equilibrium constant you use the Gibbs Free Energy. It's an exponential formula: you take the ratio of the Gibbs Free Energy to the Ideal Gas Law constant for the temperature in question, and that ratio becomes the argument of the exponential function. If the ratio is positive, then the constant is greater than one; if negative, it is less than one.

If the Gibbs free energy is zero, then the equilibrium constant is equal to one. What does this mean? Just plug in the numbers. It means the numerator and the denominator of the fraction have to be equal. There are many ways you can do this. You can have one mole of each component. Or you can have, for example, two moles of methane, two moles of CO2, two moles of CO and one mole of H2...then your fraction comes to four on top and four on the bottom. Or whatever.

Even as a write these words, I can see that something is wrong. When the Gibbs Free Energy is zero, with k=1, the reaction ought to be balanced exactly as it is written. But if I try to plug those numbers into the equilibrium formula, I get 2^2 * 2^2 on top (because there are two moles each of hydrogen and carbon monoxide) and 1*1 on the bottom, so my equilibrium constant is way off. I'm going to take a pause here while I figure out what's wrong.

## Tuesday, June 8, 2010

### Where did the methane go?

Getting back to the question of the oil-cooled nuclear reactor in Pinawa, Manitoba: my assignment was to get a new monitoring system installed on the leak detection system. The fifty-seven pressure tubes of the reactor were each surrounded by a larger tube, and these outer tubes were continuously purged with CO2 gas. The leak detection system was based on monitoring the purge gas for trace hydrocarbons, which would indicate that a pressure tube was leaking. To pinpoint the location of any possible leaks, the 57 tubes were separately routed through a system of solenoid valves to a monitoring station.

To the great chagrin of the reactor operators, the detection system indicated all kinds of leaks! At least a quarter of the channels showed hydrocarbon readings well into the tens of parts per million. This was a very serious matter.

Until someone got the bright idea of just letting the multiplexer valve sit on a single channel for a while. It turned out that after ten minutes the reading would go back down to zero. You could clearly see it on the strip chart recorder which left a telltale line of ink, one after another, for each channel in sequence. The high readings were obviously some kind of instrumentation glitch, a "surge" they called it. The true reading was the number showing after ten minutes purging a single channel. Problem solved.

And this was where things sat when I was given the assignment. I was most definitely not expected to get into the question of explaining the "surges": the system was working just fine. All we needed was some more modern equipment. Strip chart recorders were after all very very 1960's: this was the 80's and we were converting to computers for all our data monitoring. And one of the major benefits of computer data logging would be to get rid of those annoying surge readings. You would just program the computer to switch channels, wait ten minutes, and then log the reading only when it had had a chance to "settle down".

I still don't know what made me think of it, but it occurred to me that I might be able to explain the surges. What if there was some chemical reaction taking place in the sampling lines whereby hydrocarbons were being broken down to some other lighter species which were then going through the detector without showing their presence? Then, when the valve switched to the next sampling line, some unsampled gas would still be left sitting in the last tube. With 57 channels at 10 minutes each, it would be almost six hours before that tube would be sampled again. Maybe that was enough time for the reaction to reverse and the products be converted back into methane. That would explain the surge, and it would explain why the surge disappeared after ten minutes: once the fresh gas reached the detector, the remaining hydrocarbons would have been washed out of the system.

But what kind of reaction could be responsible for this wierd behavior. Taking methane as an example of a typical hydrocarbon, I wrote out:

CH4 + CO2 => ???

I did some trial and error and came up a couple of possible reactions. The one that seemed most interesting was:

CH4 + CO2 => 2CO + 2H2

If you haven't done chemistry for a while, you might want to note that the left and right hand sides of this equation each have two carbons, two oxygens, and four hydrogens. So it is indeed a balanced chemical equation. The question is: does this reaction actually take place?

There's a way to tell if a reaction is expected to take place or not, and it's something you learn in first year chemistry. It's called the Gibbs Free Energy and its a formula that combines the enthalpy and entropy of a reaction into a combined measure of spontaneity. In short, all things being equal, if the Gibbs Free Energy is negative, the reaction should go. If it's positive, then it won't.

Let's ignore for a moment just what is the physical meaning of enthalpy and entropy: the fact is you can look them up in the chemical handbooks and add them up. It's not that hard and what you find is that the Gibbs Free Energy for the reaction in question is decidedly positive. So my theory was wrong: thermodynamically speaking, the reaction shouldn't occur.

Was that correct? Maybe I'd made a mistake in the calculation. I went over it again and got the same result. Then, I noticed something: in the formula for Gibbs Free Energy, the entropy term is multiplied by the temperature. Out of habit I had used STP (Standard Temperature and Pressure) conditions, but of course the reactor ran at a coolant temperature of 300 degrees Celsius. Maybe this would make a difference?

Sure enough, it did. For this reaction, the internal energy of the molecules (the enthalpy) was definitely higher on the right hand side, which inhibited the reaction: but the entropy contribution was in the other direction; since it was multiplied by temperature, the reaction became more favorable the hotter it ran. This makes sense because entropy is a measure of disorder, and the products consist of four molecules while the reactants are only two. So at sufficiently high temperatures the entropy should dominate and the reaction procede.

I quickly redid the figures, and once again I was disappointed. Even with the corrected temperature, the additional contribution of the entropy term was still not sufficient to tilt the balance from positive to negative. The reaction was still a "no go."

And yet: the reaction does take place, and I was able to prove it by directly measuring carbon monoxide in a freely running sample line! The explanation of this mystery will follow in my next blog post.

To the great chagrin of the reactor operators, the detection system indicated all kinds of leaks! At least a quarter of the channels showed hydrocarbon readings well into the tens of parts per million. This was a very serious matter.

Until someone got the bright idea of just letting the multiplexer valve sit on a single channel for a while. It turned out that after ten minutes the reading would go back down to zero. You could clearly see it on the strip chart recorder which left a telltale line of ink, one after another, for each channel in sequence. The high readings were obviously some kind of instrumentation glitch, a "surge" they called it. The true reading was the number showing after ten minutes purging a single channel. Problem solved.

And this was where things sat when I was given the assignment. I was most definitely not expected to get into the question of explaining the "surges": the system was working just fine. All we needed was some more modern equipment. Strip chart recorders were after all very very 1960's: this was the 80's and we were converting to computers for all our data monitoring. And one of the major benefits of computer data logging would be to get rid of those annoying surge readings. You would just program the computer to switch channels, wait ten minutes, and then log the reading only when it had had a chance to "settle down".

I still don't know what made me think of it, but it occurred to me that I might be able to explain the surges. What if there was some chemical reaction taking place in the sampling lines whereby hydrocarbons were being broken down to some other lighter species which were then going through the detector without showing their presence? Then, when the valve switched to the next sampling line, some unsampled gas would still be left sitting in the last tube. With 57 channels at 10 minutes each, it would be almost six hours before that tube would be sampled again. Maybe that was enough time for the reaction to reverse and the products be converted back into methane. That would explain the surge, and it would explain why the surge disappeared after ten minutes: once the fresh gas reached the detector, the remaining hydrocarbons would have been washed out of the system.

But what kind of reaction could be responsible for this wierd behavior. Taking methane as an example of a typical hydrocarbon, I wrote out:

CH4 + CO2 => ???

I did some trial and error and came up a couple of possible reactions. The one that seemed most interesting was:

CH4 + CO2 => 2CO + 2H2

If you haven't done chemistry for a while, you might want to note that the left and right hand sides of this equation each have two carbons, two oxygens, and four hydrogens. So it is indeed a balanced chemical equation. The question is: does this reaction actually take place?

There's a way to tell if a reaction is expected to take place or not, and it's something you learn in first year chemistry. It's called the Gibbs Free Energy and its a formula that combines the enthalpy and entropy of a reaction into a combined measure of spontaneity. In short, all things being equal, if the Gibbs Free Energy is negative, the reaction should go. If it's positive, then it won't.

Let's ignore for a moment just what is the physical meaning of enthalpy and entropy: the fact is you can look them up in the chemical handbooks and add them up. It's not that hard and what you find is that the Gibbs Free Energy for the reaction in question is decidedly positive. So my theory was wrong: thermodynamically speaking, the reaction shouldn't occur.

Was that correct? Maybe I'd made a mistake in the calculation. I went over it again and got the same result. Then, I noticed something: in the formula for Gibbs Free Energy, the entropy term is multiplied by the temperature. Out of habit I had used STP (Standard Temperature and Pressure) conditions, but of course the reactor ran at a coolant temperature of 300 degrees Celsius. Maybe this would make a difference?

Sure enough, it did. For this reaction, the internal energy of the molecules (the enthalpy) was definitely higher on the right hand side, which inhibited the reaction: but the entropy contribution was in the other direction; since it was multiplied by temperature, the reaction became more favorable the hotter it ran. This makes sense because entropy is a measure of disorder, and the products consist of four molecules while the reactants are only two. So at sufficiently high temperatures the entropy should dominate and the reaction procede.

I quickly redid the figures, and once again I was disappointed. Even with the corrected temperature, the additional contribution of the entropy term was still not sufficient to tilt the balance from positive to negative. The reaction was still a "no go."

And yet: the reaction does take place, and I was able to prove it by directly measuring carbon monoxide in a freely running sample line! The explanation of this mystery will follow in my next blog post.

## Sunday, May 30, 2010

### Karma and Carbon Monoxide

Earlier this spring I got into quite a heated debate on physicsforums.com about the nature of wave function collapse. In the course of that debate I leaned pretty heavily on the assumption that the basic chemistry of the photographic process involved a thermodynamically spontaneous transition from the undeveloped film to the exposed film. To be fair I never said I knew this to be true for a fact; I just said it seemed like a pretty good assumption, partly based on the fact that I never heard of anyone being able to "regenerate" used photographic film by, say for example, gently heating it to convert the exposed crystals back to their unexposed state.

I was ridiculed pretty soundly for trying to inject thermodynamics into the argument. Not knowing the chemistry, I kept pressing my opponents (who claimed some expertise in that field) to write out the reactions so we could evaluate it. Eventually it came out: at least in a simplified form, it appeared that we had to account for the reduction of silver bromide to elemental silver. Ignoring what happens to the bromine, we are nevertheless faced with a reaction enthalpy of 99 kJoules/mole in the positive direction: in other words, the reaction, far from being thermodynamically spontaneous, requires a significant input of energy. If we convert this to atomic terms, it comes to a near infrared "photon" for each atom of silver. It appeared that I was completely wrong.

And yet within a few days I had posted a counterargument which unexpectedly, and quite effectively rescued my argument from the ashes. It was a clever argument, backed up with sound mathematics; and one which I was uniquely predisposed to be able to make. Because thirty years previously, I had done a thermodynamic analysis of a chemical system in a completely different context which turned out to have the same essential features as the present case. I'm going to tell you about that now.

When I graduated from engineering in 1984, my first job was with Atomic Energy of Canada at their nuclear research station in Pinawa, Manitoba. You know that there are several different reactor designs out there, with the coolant system being one of the design variables. Some reactors are cooled with ordinary "light" water, and others with heavy water. The Pinawa reator was unique in being oil-cooled, which meant it ran at higher temperature with relatively lower coolant pressure.

Nevertheless one or the reactor's protective systems involved checking for cracks in the pressure tubes containing the coolant. This was done via a tube-within-a-tube geometry whereby the pressure tubes were surrounded by a containment tube full of CO2 gas, which was constantly purged and sampled at a monitoring station. Potential cracks in the pressure tubes could theoretically be detected by monitoring for trace hydrocarbon contamination in the CO2 purge gas.

The system seemed to work reasonably well for about twenty years, although it had a few quirks that no one worried to much about. Nevertheless, for one reason or another a decision had been made to upgrade and modernize the instrumentation, and as a new junior engineer I was given this relatively straightforward assignment.

That's where the story gets interesting. I'll continue with my next blogpost.

I was ridiculed pretty soundly for trying to inject thermodynamics into the argument. Not knowing the chemistry, I kept pressing my opponents (who claimed some expertise in that field) to write out the reactions so we could evaluate it. Eventually it came out: at least in a simplified form, it appeared that we had to account for the reduction of silver bromide to elemental silver. Ignoring what happens to the bromine, we are nevertheless faced with a reaction enthalpy of 99 kJoules/mole in the positive direction: in other words, the reaction, far from being thermodynamically spontaneous, requires a significant input of energy. If we convert this to atomic terms, it comes to a near infrared "photon" for each atom of silver. It appeared that I was completely wrong.

And yet within a few days I had posted a counterargument which unexpectedly, and quite effectively rescued my argument from the ashes. It was a clever argument, backed up with sound mathematics; and one which I was uniquely predisposed to be able to make. Because thirty years previously, I had done a thermodynamic analysis of a chemical system in a completely different context which turned out to have the same essential features as the present case. I'm going to tell you about that now.

When I graduated from engineering in 1984, my first job was with Atomic Energy of Canada at their nuclear research station in Pinawa, Manitoba. You know that there are several different reactor designs out there, with the coolant system being one of the design variables. Some reactors are cooled with ordinary "light" water, and others with heavy water. The Pinawa reator was unique in being oil-cooled, which meant it ran at higher temperature with relatively lower coolant pressure.

Nevertheless one or the reactor's protective systems involved checking for cracks in the pressure tubes containing the coolant. This was done via a tube-within-a-tube geometry whereby the pressure tubes were surrounded by a containment tube full of CO2 gas, which was constantly purged and sampled at a monitoring station. Potential cracks in the pressure tubes could theoretically be detected by monitoring for trace hydrocarbon contamination in the CO2 purge gas.

The system seemed to work reasonably well for about twenty years, although it had a few quirks that no one worried to much about. Nevertheless, for one reason or another a decision had been made to upgrade and modernize the instrumentation, and as a new junior engineer I was given this relatively straightforward assignment.

That's where the story gets interesting. I'll continue with my next blogpost.

## Wednesday, May 12, 2010

### Do atoms behave as waves?

As you may recall, I was recently banned for life from the discussion group physicsforums.com. However, from time to time discussions take place in which I would like to put in my two cents worth. Today I am going to comment on the thread "Do atoms behave as waves?".

The most compelling evidence of atoms behaving as waves is the anomalous specific heat of diatomic gasses, which was known in the nineteenth century and caused great concern for people like Maxwell and Boltzmann. Here is the problem: the thermal energy of the typical diatomic gas such oxygen, nitrogen, or hydrogen is classically accounted for by counting the five modes: three translational and two rotational. (The third rotational mode contributes no energy because it is oriented along the axis of the dumbbell.) The theory works well in practise, except at low temperatures hydrogen begins to diverge from the expected value. It's as though the rotational modes stop contributing and only the translational modes remain in effect.

This is a clear instance of the wave nature of atoms. In order to properly drive the rotational modes, the dumbbell has to be cleanly struck by another atom. With classical billiard balls connected by pegs, this is no problem. But since atoms behave as waves, it's not so simple. At lower temperatures, the atoms move slower, so the wavelength gets longer. At a certain point, the wavelenght of the molecule is longer than the length of the dumbell, so it is impossible to strike one atom without also striking the other one as well. Therefore it is impossible to set the molecule in rotation to the full extent which is possible with simple billiard balls on pegs.

If you do the simple calculation of de Broglie wavelengths at the mean molecular speed and set it equal to bond length of the molecule, you get (within an order of magnitude) the well-known classical Wien formula for peak radiation frequency at a given temperature. It is interesting that you can get this result without recourse to the assumption that energy is quantized in lumps.

The most compelling evidence of atoms behaving as waves is the anomalous specific heat of diatomic gasses, which was known in the nineteenth century and caused great concern for people like Maxwell and Boltzmann. Here is the problem: the thermal energy of the typical diatomic gas such oxygen, nitrogen, or hydrogen is classically accounted for by counting the five modes: three translational and two rotational. (The third rotational mode contributes no energy because it is oriented along the axis of the dumbbell.) The theory works well in practise, except at low temperatures hydrogen begins to diverge from the expected value. It's as though the rotational modes stop contributing and only the translational modes remain in effect.

This is a clear instance of the wave nature of atoms. In order to properly drive the rotational modes, the dumbbell has to be cleanly struck by another atom. With classical billiard balls connected by pegs, this is no problem. But since atoms behave as waves, it's not so simple. At lower temperatures, the atoms move slower, so the wavelength gets longer. At a certain point, the wavelenght of the molecule is longer than the length of the dumbell, so it is impossible to strike one atom without also striking the other one as well. Therefore it is impossible to set the molecule in rotation to the full extent which is possible with simple billiard balls on pegs.

If you do the simple calculation of de Broglie wavelengths at the mean molecular speed and set it equal to bond length of the molecule, you get (within an order of magnitude) the well-known classical Wien formula for peak radiation frequency at a given temperature. It is interesting that you can get this result without recourse to the assumption that energy is quantized in lumps.

## Tuesday, May 11, 2010

### A Tale of Two Strikes

Some people might remember me from the 2004 (?) professor's strike at the University of Manitoba. I'm the guy who got arrested for taking over a physics class from the absent professor. A good story on its own, but not what I'm remembering right now about that strike.

I remember driving in to the campus and having to cross a picket line. The professors were milling about in the middle of the road while police stood by and made sure the cars waited until the profs were ready to let them through. Every group of cars had to pay its symbolic obeisance to the sanctity of the picket line.

I got up and started giving bloody hell to one of the officers on duty. "These people have no right to block traffic! Your job is to keep the streets clear, not to support the strikers!"

To be fair, it should be noted that these were campus police, not city police. Still, they wore uniforms and gave orders to civilians and called themselves police. I found it revolting that officers of the law would take sides in a strike.

Flash forwart to last weekend when I was on the other side of a work action. There was the company, the police, and the striker. I was the striker.

As for my union, I was alas a union of one. But surely this entitled me to no fewer civil rights than the powerful professor's union. Discrimination on the basis of union membership is surely prohibited by the human rights code. Shouldn't discrimination based of

*lack*of union membership be just as wrong?And if the laws protect the rights of people with good, high-paying jobs who think they deserve even more than they're getting, shouldn't it also protect the rights of low-life bottom-feeders like me with minimum wage jobs who are only asking to be paid the money they actually agreed to work for in the first place? If the professors are allowed to block traffic to ask for more money, shouldn't I be allowed to at least stand by the curb holding up a sign asking to be paid what I'm owed?

For what's about to come, let's assume for the sake of argument I was a terrible employee and the boss had every right to fire me. I can argue the opposite all I want and it doesn't make a hairs worth of difference, because the story I'm going to tell isn't about my dispute with the employer, it's about my treatment at the hands of the police. And as far as that is concerned, the facts of my dispute with the employer are irrelevant. So what if I was fired because I slept in the hayloft for four weeks instead of working, did I have less justice on my side than those professors with their six-figure salaries, their pensions and their sabbaticals, who struck in mid-term so as to hold the students hostage? But that's another argument. Suffice it to say that after numerous unsuccessful attempts to collect, I finally drove out to the hotel where I had been working and set up my sign beside the road. Accordian slung over my shoulders, I paced back and forth in front of the entrance. The question is: if the professors have the right to march in front of their place of employement carrying signs, what about me?

Within half an hour the Gimli RCMP was there, flashers blazing. An officer got out while his young female partner waited in the car. "What's going on here?" I pointed to my sign in reply: "Mike Bruno owes me four weeks wages AND WON'T PAY UP. The officer was unimpressed. "You're blocking traffic. You need to move your car."

I looked at my car. There was basically no shoulder on that stretch of road and little room before the precipice to the ditch. I pointed to my wheels about 18 inches inside the cement line. "Is it OK if I can get it off the pavement?"

This is where the run-around started. The officer wasn't about to give me the definition of a legal parking spot on the edge of the highway. To every location that I pointed at, he just said "you can't park there. Just get off the highway. Go find a side road somewhere and maybe you'll find a spot."

That was a specific as he was willing to get. He was no more cooperative on telling me what would be an acceptable placement for my sign, which had up til then been leaning on the back of my car. He basically just wanted me out of there and as much as told me so.

I got in my car and drove over to a nearby hiking trail and pulled into the entry area. "Is that OK?" All he would say is "if I get any complaints about it, you're getting towed". As to whose "complaints" would be sufficient to justify my eviction, he was silent. I leaned my sign on a hydro pole and he drove into the hotel compound with his young female partner.

Fifteen minutes later he was back. He pulled up and rolled down his window, asking me to come over. "How much money do they owe you?"

I didn't like where this was going. "Why do you want to know?" I didn't want to say "it's none of your business", even though that's what I was thinking.

He pressed me further. It finally came to the dreaded response. "Yes it is my business", he said, "because my business is keeping the peace and maybe I have a solution for your problem".

Up til then there had been no problem with "peace", and I was skeptical about his ability to negotiate a solution on my behalf, especially given his actions "on my behalf" up to this point. It seems that the hotel manager had given him a long story about what a bad employee I had been and how I should be glad to settle for a "reasonable" fraction of the hours I had put in for. Although I knew it would make me seem ungrateful, I simply had to tell the officer I wasn't interested in his mediation efforts. I turned to walk over to my sign.

He got out of the car and followed me. "Don't walk away when I'm talkiung to you. I'm trying to help you".

I turned back and said, "I'm sorry but I don't want your help. I want to do this my way." And then I walked away again.

"I told you not to walk away when I'm talking to you. Get back here right now". This was the man with the gun talking, and I can tell you it was very intimidating. Mustering my faltering resolve, I repeated: "I'm sorry but I'm not interested in talking to you any more".

"Well then I'm taking your sign". This was too much. "Take your hand off my sign. That's my property and you have no right to confiscate it". After a short tussle he changed tack. "All right, then I'm towing your car".

He had found nothing wrong with my parking spot fifteen minutes earlier, and now it was about to be towed. "Where do you want me to move it?"

"Why don't you find a paking lot?" he sneered. Obviously the only parking lot for miles around was off limits to me.

I got in my car and started driving, and almost immediately pulled into a private yard. The homeowner was outside and readily agreed to my offer of $5 for permission to park for the evening. "But I don't wan't your money. Park for free".

Elated that my problem had been so unexpectedly solved, I once again saddled up my accordian and headed back towards my picket station. But before I could leave the property, the RCMP car pulled into the driveway. To my shock and amazement, the officer went over to the homeowner and began trying to convince him to revoke my permission to park!

This was too much. Furious, I strode towards the officer. "Get the hell out of this property", I snarled. "I have a private arrangement with this homeowner and you have no right to interfere with it". At the very last word I approached within inches of his face and at that moment, he cried out jubilantly: "Assaulting an officer!" I found myself whipped around and thrust against the back of the cruiser. "You just make a big mistake," he gloated as the aghast homeowner watched in horror.

After locking me in the back of the cruiser, the officer planned one more treat for me. "Now", he said, "your'e car is getting towed". He went over to the homeowner but to the man's enormous credit, he would not be pressured by the police. "Let the man leave his car here. As long as he wants. And make sure it is locked before you drive away".

Denied this last bit of satisfaction, the officer contented himself with booking me for assault and throwing me in a cell with a metal toilet while he processed my paperwork. An hour later I was released, trudging on foot the three miles back to my picket.

Oddly enough, it felt pretty good. I had taken the worst they could give me and was still standing. My sign had been confiscated, but I still had two hunks of drywall in my car and a felt marker. After setting up my new signs, I pulled out the accordian and started playing.

I had one more surprise in store. Within half an hour a cruiser pulled up, this time with two different officers. "We've had complaints that you're blocking traffic."

Here we go again. "Is there anything wrong with my car?" No. (I hardly need mention that it was at that moment in a spot where the first officer had said he would have it towed.) "Is there anything wrong with my sign?" No. "Well in that case I've had just about enough of you people for tonight", I said, looking square at him. "If you see me breaking any laws, why don't you arrest me? Otherwise, get the hell out of my business!"

It felt pretty good watching them drive away, knowing that I had established my right to...well, to my rights as a citizen, as a human being. I'm not talking about so-called "human rights", which I despise. Those aren't rights. Rights aren't just for minority groups and trade unions and civil servants and lesbians...they're also for regular trailer trash white people like me. If I went to the human rights commission to complain about what the police did to me, they'd tell me to get lost. I'm not a protected group.

The rights I stood up for were the thing that actually makes freedom worthwhile. You just don't realize how fragile they are until you find yourself being stripped of them. And how easy it would have been for me to pack up and go home, to give up in the face of intimidation. But I stood my ground and faced them down. Because I knew that if I didn't make a stand then and there, that I might as well be living in Russia. Or North Korea. Or whatever country is the current poster boy for dictatorships.

The next day the police were back, with two more trumped-up citations for trespassing and blocking traffic. One hundred eleven dollars each. Not to mention the small matter of assaulting an officer. Never mind. I'll have my day in court. Today I am a free man.

## Tuesday, April 6, 2010

### Banned for life

I keep getting farther behind. There are so many things that I have to post and new things keep getting in the way. I was talking about the isoelectronic series of helium last week and never quite finished. When I left off I had worked out a formula for the ground state energies of all the two-electron ions. It turns out the data for the first four ions is available. It starts off out of whack but rapidly converges: I'm out by ten percent on the first one (H-), a couple percent on Helium, less than one percent on Li+ and half that on Be++. I have a table somewhere that I'll post someday when I get around to it, but it's a very cool outcome. The error on the light ionsmakes total sense because the calculation uses a direct product for the wave function of two electrons, which is an oversimplification. It works for the heavy ions but not for the light ones, and the reason goes back to the same thing I talked about in my blog entry on the two-electron potential well: the small box limit corresponds, "counter-intuitively", to the case of no repulsion between the electrons.

I put the word "counter-intuitively" in quotes because I used it ironically: it's the sense in which the world at large uses the term, as though to suggest that we shouldn't trust our intuition, we should only believe what the math tells us. My belief is the opposite: there is nothing more important in physics than developing your intuition. A "counter-intuitive" result is in fact an opportunity for us to develop and refine our intuition. It means we guessed wrong and now have to figure out why, so next time we'll guess right. It doesn't mean we shouldn't guess at all!

But that's not what I started to blog about today. The big news in the world of physics happened yesterday: the evil Conway was banned for life from the on-line forum physicsforums.com. This forum is run by a small clique of, shall we say tin-pot dictators, whose authority to expel members is not subject to appeal. The unusual thing about this particular expulsion has to be the wave of jubilation that ensued in its aftermath; you'd think that someone had just announced the discovery of the elusive Higgs Boson or whatever it is they spent all those billions of dollars looking for in that big hadron collider. But no, it was just me; yes, I was the evil Conway and I'm sure physicsforums.com has never seen such a display of high-fiving and backslapping as you'll find if you check the last page of the thread on "decoherence". (The username was chosen in honor of country music legend Conway Twitty in case you wondered.) So the question has to be asked: why was I expelled?

Technically, I was expelled for technical reasons that really don't really matter. In fact I was expelled because people hated me passionately. It's interesting because I took special care to stay away from the kind of personal sparring that often characterises these online discussions. It's almost as though I drew hatred strictly for the physics. There is a certain personality type that you find on internet forums which consists of professional physicists who take it upon themselves to be guardians of the truth. It's an attitude I don't really understand. People post nonsense on these forums all the time and I ignore it. I'm interested in engaging in discussions with people who know what they're talking about, whether or not I think they are the most upstanding gentlement or the most obnoxious assholes. Because either way I stand to learn physics from the exchange. But it seems that the people who chose to argue with me are the opposite: they are driven to seek out those who spout incorrect heresies, so they can argue those heretics down to the ground and thereby defend and protect their beloved edifice of revealed truth. I don't think the above characterisation particularly misrepresents either my psychology or theirs.

Anyhow, what's done is done. If anyone's interested I think this link

http://www.physicsforums.com/showthread.php?t=385341

should get you to where all the fireworks happened. It was fun while it lasted but now I have to get back to doing physics.

Oh, one more thing. I have the impression that the physicists running the newsgroup are mostly from England, so they may not see the irony in the fact that the system administrator who actually pulled the plug on Conway Twitty was someone by the name of George Jones.

I put the word "counter-intuitively" in quotes because I used it ironically: it's the sense in which the world at large uses the term, as though to suggest that we shouldn't trust our intuition, we should only believe what the math tells us. My belief is the opposite: there is nothing more important in physics than developing your intuition. A "counter-intuitive" result is in fact an opportunity for us to develop and refine our intuition. It means we guessed wrong and now have to figure out why, so next time we'll guess right. It doesn't mean we shouldn't guess at all!

But that's not what I started to blog about today. The big news in the world of physics happened yesterday: the evil Conway was banned for life from the on-line forum physicsforums.com. This forum is run by a small clique of, shall we say tin-pot dictators, whose authority to expel members is not subject to appeal. The unusual thing about this particular expulsion has to be the wave of jubilation that ensued in its aftermath; you'd think that someone had just announced the discovery of the elusive Higgs Boson or whatever it is they spent all those billions of dollars looking for in that big hadron collider. But no, it was just me; yes, I was the evil Conway and I'm sure physicsforums.com has never seen such a display of high-fiving and backslapping as you'll find if you check the last page of the thread on "decoherence". (The username was chosen in honor of country music legend Conway Twitty in case you wondered.) So the question has to be asked: why was I expelled?

Technically, I was expelled for technical reasons that really don't really matter. In fact I was expelled because people hated me passionately. It's interesting because I took special care to stay away from the kind of personal sparring that often characterises these online discussions. It's almost as though I drew hatred strictly for the physics. There is a certain personality type that you find on internet forums which consists of professional physicists who take it upon themselves to be guardians of the truth. It's an attitude I don't really understand. People post nonsense on these forums all the time and I ignore it. I'm interested in engaging in discussions with people who know what they're talking about, whether or not I think they are the most upstanding gentlement or the most obnoxious assholes. Because either way I stand to learn physics from the exchange. But it seems that the people who chose to argue with me are the opposite: they are driven to seek out those who spout incorrect heresies, so they can argue those heretics down to the ground and thereby defend and protect their beloved edifice of revealed truth. I don't think the above characterisation particularly misrepresents either my psychology or theirs.

Anyhow, what's done is done. If anyone's interested I think this link

http://www.physicsforums.com/showthread.php?t=385341

should get you to where all the fireworks happened. It was fun while it lasted but now I have to get back to doing physics.

Oh, one more thing. I have the impression that the physicists running the newsgroup are mostly from England, so they may not see the irony in the fact that the system administrator who actually pulled the plug on Conway Twitty was someone by the name of George Jones.

## Sunday, March 28, 2010

### Sorting out the helium atom

I said in my last post that I was going to generalize the helium atom so I could calculate the ground state energy for the whole isoelectonic series. The first step is to work out the helium atom.

The key to working out the energy is pretty simple (in retrospect! it was hard for me to do it the first time). There are three pieces: the kinetic energy, the potential energy, and the interaction energy (which comes from the electron-electron repulsion.) The starting point will be what I called the "naive model", where we take the wave functions to be hydrogen like, we drop both electrons into the ground state, and add the interaction energy as an afterthought. Here is how it goes.

With just one electron, helium scales from the hydrogen atom by a factor of two on linear dimension, which gives a factor of four on energy. So where hydrogen is -1 Ry, He+ (singly ionized) is -4 Ry. This is true theoretically and experimentally; it's pretty much an exact result.

We can look inside this quantity of energy and break it down further into kinetic and potential. This is the same thing you can do with a planet orbiting the sun. The kinetic energy is positive and the potential energy is negative. There is something called the virial theorem which tells you to expect these quantities to be in the ration of 2:1; it works for planets, and without getting into the reasons why, suffice to say it works for atoms too. The kinetic energy of He+ is +4 Ry and the potential is -8 Ry. Total: -4 Ry.

Now we add the second electron and assume it goes into exactly the same energy state as the first. Total energy of the atom is now -8 Ry. But that's neglecting the mutual repulsion of the electrons. As I mentioned in my last post, that's an exact integral that comes to 2.5 Ry, giving a grand total of -5.5 Ry. Let's write it out:

(eqn. 1) -16 Ry potential + 8 Ry kinetic + 2.5 Ry interactive = -5.5 Ry

That's what I've called the "naive model". Now we're ready to have some fun.

The first thing to do is to optimize the wave function to try and minimize the total energy. The calculation above assumes that the wave function is just the same as the hydrogen wave function, scaled by a factor of two on linear dimension. In fact, because the two electrons are crowding each other, we might expect the wave function to spread out a bit to try and compensate. The wave function is essentially of the form exp(-r), and the easiest way to spread it out is to put a variable parameter k into the exponential decay. The nice thing is that with this simple change, the three terms in our energy equation are altered in a very simple way.

(The traditional explanation is a little different, as you can see on the Wikipedia page. They talk about mutual screening and "effective nuclear charge Z". This is confusing for a couple of reasons, mostly because Z is normally reserved for nuclear charge number and later I'm going to need to vary just that parameter. So I stick with what I'm calling a relaxation parameter, k, which just relaxes the shape of the wave function without any particular justification.)

The question is, how does my relaxation factor change the energy? Looking at it term by term, it's pretty simple. The kinetic energy is quadratic in k and the other terms are linear. Why? The kinetic is quadratic because in quantum mechanics you differentiate twice to get KE, and each differentiation gives a factor of k. The potential is linear because of the 1/r term in the equation for potential energy. Change the average distance by a factor of k, and the energy goes by the same factor. So we can write, without any particular difficulty, the revised energy equation as follows:

(eqn. 2) -16k (potential) + 8k^2 (kinetic) + 2.5k (interactive) = Energy

You have got to check out Wikipedia to see how much work they went through to get from (eqn. 1) to (eqn. 2). I did it with virtually no work at all by using a simple scaling argument. Once you're here, it's a very simple first-year calculus exercise to idenfiy the minium energy as occuring when k= 27/32. (On Wikipedia they get 27/16, but they define their optimization parameter a little differently.) You get Energy = 729/128 Ry, or just about 77.5 eV. It's about 1.5 eV short of the true energy.

This is where they basically leave off, letting you think that if you keep tinkering with the shape of the wave function, you'll get closer and closer to the true energy. You won't. You have to do something really radical to improve on this calculation much beyond the point where we've already taken it. You have to come up with a wave function that includes the coordinates of both electrons independently. But that's a story for another day.

What we're going to do next is take the optimized function for helium and generalize it to the rest of the isoelectronic series. We're going to need to introduce a parameter Z for nuclear charge. And we're going to think about how the energy terms in (eqn. 2) scale with Z.

It's going to be pretty easy, but it's just a little different than when we scaled to k. This time, the potential and kinetic terms are both going to be quadratic, and only the interaction will be linear. It works this way because a factor of Z on nuclear charge gives you a factor of 1/Z on linear dimension. For kinetic energy, this gives you Z-squared because of the double differentiation. For potential energy, you get a factor of Z for dimension and another factor of Z for the increased charge: hence, Z-squared. For the interaction however, although you get a factor of Z on dimension, the charges are the same two electrons you started out with, so there's no additional factor. It's just linear in Z.

So we take eqn. 2 and just put Z or Z-squared in front of the terms as appropriate:

(eqn. 3) -16k(Z^2) (potential) + 8(k^2)(Z^2) (kinetic) + 2.5kZ (interactive)

Except that we have to fix it up so that when Z=2, we get back (eqn. 2) unaltered. So we just divide the quadratic terms by 4 and the linear terms by 2:

(eqn. 4) -4k(Z^2) (potential) + 2(k^2)(Z^2) (kinetic) + 1.25kZ (interactive) = Energy

And that's our great result. We're going to look at the implication of this in our next post.

The key to working out the energy is pretty simple (in retrospect! it was hard for me to do it the first time). There are three pieces: the kinetic energy, the potential energy, and the interaction energy (which comes from the electron-electron repulsion.) The starting point will be what I called the "naive model", where we take the wave functions to be hydrogen like, we drop both electrons into the ground state, and add the interaction energy as an afterthought. Here is how it goes.

With just one electron, helium scales from the hydrogen atom by a factor of two on linear dimension, which gives a factor of four on energy. So where hydrogen is -1 Ry, He+ (singly ionized) is -4 Ry. This is true theoretically and experimentally; it's pretty much an exact result.

We can look inside this quantity of energy and break it down further into kinetic and potential. This is the same thing you can do with a planet orbiting the sun. The kinetic energy is positive and the potential energy is negative. There is something called the virial theorem which tells you to expect these quantities to be in the ration of 2:1; it works for planets, and without getting into the reasons why, suffice to say it works for atoms too. The kinetic energy of He+ is +4 Ry and the potential is -8 Ry. Total: -4 Ry.

Now we add the second electron and assume it goes into exactly the same energy state as the first. Total energy of the atom is now -8 Ry. But that's neglecting the mutual repulsion of the electrons. As I mentioned in my last post, that's an exact integral that comes to 2.5 Ry, giving a grand total of -5.5 Ry. Let's write it out:

(eqn. 1) -16 Ry potential + 8 Ry kinetic + 2.5 Ry interactive = -5.5 Ry

That's what I've called the "naive model". Now we're ready to have some fun.

The first thing to do is to optimize the wave function to try and minimize the total energy. The calculation above assumes that the wave function is just the same as the hydrogen wave function, scaled by a factor of two on linear dimension. In fact, because the two electrons are crowding each other, we might expect the wave function to spread out a bit to try and compensate. The wave function is essentially of the form exp(-r), and the easiest way to spread it out is to put a variable parameter k into the exponential decay. The nice thing is that with this simple change, the three terms in our energy equation are altered in a very simple way.

(The traditional explanation is a little different, as you can see on the Wikipedia page. They talk about mutual screening and "effective nuclear charge Z". This is confusing for a couple of reasons, mostly because Z is normally reserved for nuclear charge number and later I'm going to need to vary just that parameter. So I stick with what I'm calling a relaxation parameter, k, which just relaxes the shape of the wave function without any particular justification.)

The question is, how does my relaxation factor change the energy? Looking at it term by term, it's pretty simple. The kinetic energy is quadratic in k and the other terms are linear. Why? The kinetic is quadratic because in quantum mechanics you differentiate twice to get KE, and each differentiation gives a factor of k. The potential is linear because of the 1/r term in the equation for potential energy. Change the average distance by a factor of k, and the energy goes by the same factor. So we can write, without any particular difficulty, the revised energy equation as follows:

(eqn. 2) -16k (potential) + 8k^2 (kinetic) + 2.5k (interactive) = Energy

You have got to check out Wikipedia to see how much work they went through to get from (eqn. 1) to (eqn. 2). I did it with virtually no work at all by using a simple scaling argument. Once you're here, it's a very simple first-year calculus exercise to idenfiy the minium energy as occuring when k= 27/32. (On Wikipedia they get 27/16, but they define their optimization parameter a little differently.) You get Energy = 729/128 Ry, or just about 77.5 eV. It's about 1.5 eV short of the true energy.

This is where they basically leave off, letting you think that if you keep tinkering with the shape of the wave function, you'll get closer and closer to the true energy. You won't. You have to do something really radical to improve on this calculation much beyond the point where we've already taken it. You have to come up with a wave function that includes the coordinates of both electrons independently. But that's a story for another day.

What we're going to do next is take the optimized function for helium and generalize it to the rest of the isoelectronic series. We're going to need to introduce a parameter Z for nuclear charge. And we're going to think about how the energy terms in (eqn. 2) scale with Z.

It's going to be pretty easy, but it's just a little different than when we scaled to k. This time, the potential and kinetic terms are both going to be quadratic, and only the interaction will be linear. It works this way because a factor of Z on nuclear charge gives you a factor of 1/Z on linear dimension. For kinetic energy, this gives you Z-squared because of the double differentiation. For potential energy, you get a factor of Z for dimension and another factor of Z for the increased charge: hence, Z-squared. For the interaction however, although you get a factor of Z on dimension, the charges are the same two electrons you started out with, so there's no additional factor. It's just linear in Z.

So we take eqn. 2 and just put Z or Z-squared in front of the terms as appropriate:

(eqn. 3) -16k(Z^2) (potential) + 8(k^2)(Z^2) (kinetic) + 2.5kZ (interactive)

Except that we have to fix it up so that when Z=2, we get back (eqn. 2) unaltered. So we just divide the quadratic terms by 4 and the linear terms by 2:

(eqn. 4) -4k(Z^2) (potential) + 2(k^2)(Z^2) (kinetic) + 1.25kZ (interactive) = Energy

And that's our great result. We're going to look at the implication of this in our next post.

## Saturday, March 27, 2010

### The Isoelectronic Series of Helium (Part II)

I ended up writing a paper about what I figured out for helium. The paper starts off following pretty much the analysis of the helium atom that you'll find in the Wikipedia article. (There's actually a better version of this analysis on a University of Texas website http://farside.ph.utexas.edu/teaching/qmech/lectures/node128.html

but I'm just going to refer to Wikipedia for convenience.) There are basically three stages in the refinement of the analysis:

1. First case, you treat it as a hydrogen atom with two electrons in the same ground orbital. Scaling for the charge and reduced dimension, you get a ground state energy or -108.8 eV, or exactly -8 Rydbergs (where hydrogen is -1 Ry).

2. First improvement: you keep the same wave function as in case (1), but you calculate the repulsion energy of the two electrons. It turns out to be an exact integral, and it comes to 2.5 Ry. (It's not hard to do numberically either because of the spherical symmetry.) So the refined estimate of the energy is -5.5 Ry, or -74.8 eV. It's actually pretty close to the experimental value of -79.0 eV. I call this the "naive model".

3. Second improvement. You tweak the wave function by putting a variable parameter in front of the exponential decay. The nice thing is because you're still basically using the same wave function except for a scaling factor, you don't have to redo any of the hard 3-dimensional integrals. You just scale the ones you already have.

When you do case (3), you get an optimization parameter that lets you bring the energy down to -77.5 eV. Now you're within 1.5 eV of the true value. The funny thing is, why didn't you get even closer? The "naive model" was already within 4 eV of the true value, so this optimization trick has really only closed the gap by 60%.

That's why I find it odd what the guy at U of Texas says at the end of his article. He says: "Obviously, we could get even closer to the correct value of the helium ground-state energy by using a more complicated trial wave-function with more adjustable parameters. ". It's a funny statement because how obvious is it? Based on the numbers so far it doesn't appear to me that we're closing in fast enough.

In hindsight, I know the reason. You can make the wave function as complicated as you want and it will only get you so close and no closer. To get the correct ground state energy you have to do something quite tricky and unexpected, at least based on the analysis so far I'd have to say it's unexpected. But I'll get to that a little later.

What I did in my analysis was to generalize the equations in the Wikipedia article so they work on any size nucleus, as long as it has two electrons. So it works for the hydrogen negative ion (H-); it works for neutral helium; it works for singly ionized Lithium (Li+); and it works for doubly ionized beryllium (Be++). It works for all the atoms of course, but the first four are the ones I was able to look up and check the actual energies.

But before we get to my results, I'm going to brag about how clever I was in deriving these equations. I did a bunch of stuff using basic scaling principles that let me go from A to B in a single line when Wikipedia (and Texas) use a whole page of hard math to do the same thing. I think I'll tell you about that in my next post.

but I'm just going to refer to Wikipedia for convenience.) There are basically three stages in the refinement of the analysis:

1. First case, you treat it as a hydrogen atom with two electrons in the same ground orbital. Scaling for the charge and reduced dimension, you get a ground state energy or -108.8 eV, or exactly -8 Rydbergs (where hydrogen is -1 Ry).

2. First improvement: you keep the same wave function as in case (1), but you calculate the repulsion energy of the two electrons. It turns out to be an exact integral, and it comes to 2.5 Ry. (It's not hard to do numberically either because of the spherical symmetry.) So the refined estimate of the energy is -5.5 Ry, or -74.8 eV. It's actually pretty close to the experimental value of -79.0 eV. I call this the "naive model".

3. Second improvement. You tweak the wave function by putting a variable parameter in front of the exponential decay. The nice thing is because you're still basically using the same wave function except for a scaling factor, you don't have to redo any of the hard 3-dimensional integrals. You just scale the ones you already have.

When you do case (3), you get an optimization parameter that lets you bring the energy down to -77.5 eV. Now you're within 1.5 eV of the true value. The funny thing is, why didn't you get even closer? The "naive model" was already within 4 eV of the true value, so this optimization trick has really only closed the gap by 60%.

That's why I find it odd what the guy at U of Texas says at the end of his article. He says: "Obviously, we could get even closer to the correct value of the helium ground-state energy by using a more complicated trial wave-function with more adjustable parameters. ". It's a funny statement because how obvious is it? Based on the numbers so far it doesn't appear to me that we're closing in fast enough.

In hindsight, I know the reason. You can make the wave function as complicated as you want and it will only get you so close and no closer. To get the correct ground state energy you have to do something quite tricky and unexpected, at least based on the analysis so far I'd have to say it's unexpected. But I'll get to that a little later.

What I did in my analysis was to generalize the equations in the Wikipedia article so they work on any size nucleus, as long as it has two electrons. So it works for the hydrogen negative ion (H-); it works for neutral helium; it works for singly ionized Lithium (Li+); and it works for doubly ionized beryllium (Be++). It works for all the atoms of course, but the first four are the ones I was able to look up and check the actual energies.

But before we get to my results, I'm going to brag about how clever I was in deriving these equations. I did a bunch of stuff using basic scaling principles that let me go from A to B in a single line when Wikipedia (and Texas) use a whole page of hard math to do the same thing. I think I'll tell you about that in my next post.

## Friday, March 26, 2010

### The isoelectronic series of helium: Part I

In my last post, I talked about how the solution of the two-electron well becomes simple in the two extreme cases: the very large box, and the very small box. And how these cases correspond to having a fixed size of box while varying the strength of the electric repulsion. The funny thing is, the very small box corresponds to the case of weak repulsion, and the very big box is the case of strong repulsion. Why is this?

The answer turns out to be almost obvious: The repulsive energy is a 1/r force, but the kinetic energy goes as 1/r-squared. You know about the 1/r from basic electrostatics; you see it in the fomula for the potential of a spherical charge. It comes from the 1/r-squared law of forces. Since energy is the integral of force along a path, it always comes to something in 1/r for spherical geometry. The kinetic energy, however, is a purely quantum-mechanical thing. For the particle confined in a box, the kinetic energy is the square of the momentum operator. The momentum operator is differention (well, it's the del operator in three dimension). If you make the box twice as small, the derivatives are twice as big, so when applied twice, you get a factor of 4. In other words, it's an inverse square relationship on the dimension of the containment. So for a very small box the kinetic term dominates the potential term.

The interesting thing is that you can see this in a physical case: it's called the isoelectronic series of helium, and it goes all the way back to my second post about why the helium atom doesn't have a miniature replica in the hydrogen atom. It has to do with the scaling effect being different for the potential and kinetic energy terms, and I did some really cool calculations to show how this works for certain atomic energy levels: hence, "the isoelectronic series of helium".

You may know that when you solve the equation for the hydrogen atom, you automatically get by analogy the solution for a whole series of other atoms in their ionized states, when they have only one electron. For example, singly ionized helium is exactly the same as hydrogen except the wave functions are compressed by a factor of two and all the energy levels are four times greater; ditto for doubly-ionized lithium, except it's three times on size and nine times on energy. Etcetera. It's called the isoelectronic series of hydrogen.

The series for helium is a little different. You don't get the same geometric scaling effect, and the reason is because you're basically putting two eletrons in a box, and it makes a difference how big the box is. You can see it pretty clearly if you look up (as I did) the energy levels for the atomic states belonging to this sequence.

In my next posting I'm going to explain some more about this.

The answer turns out to be almost obvious: The repulsive energy is a 1/r force, but the kinetic energy goes as 1/r-squared. You know about the 1/r from basic electrostatics; you see it in the fomula for the potential of a spherical charge. It comes from the 1/r-squared law of forces. Since energy is the integral of force along a path, it always comes to something in 1/r for spherical geometry. The kinetic energy, however, is a purely quantum-mechanical thing. For the particle confined in a box, the kinetic energy is the square of the momentum operator. The momentum operator is differention (well, it's the del operator in three dimension). If you make the box twice as small, the derivatives are twice as big, so when applied twice, you get a factor of 4. In other words, it's an inverse square relationship on the dimension of the containment. So for a very small box the kinetic term dominates the potential term.

The interesting thing is that you can see this in a physical case: it's called the isoelectronic series of helium, and it goes all the way back to my second post about why the helium atom doesn't have a miniature replica in the hydrogen atom. It has to do with the scaling effect being different for the potential and kinetic energy terms, and I did some really cool calculations to show how this works for certain atomic energy levels: hence, "the isoelectronic series of helium".

You may know that when you solve the equation for the hydrogen atom, you automatically get by analogy the solution for a whole series of other atoms in their ionized states, when they have only one electron. For example, singly ionized helium is exactly the same as hydrogen except the wave functions are compressed by a factor of two and all the energy levels are four times greater; ditto for doubly-ionized lithium, except it's three times on size and nine times on energy. Etcetera. It's called the isoelectronic series of hydrogen.

The series for helium is a little different. You don't get the same geometric scaling effect, and the reason is because you're basically putting two eletrons in a box, and it makes a difference how big the box is. You can see it pretty clearly if you look up (as I did) the energy levels for the atomic states belonging to this sequence.

In my next posting I'm going to explain some more about this.

## Wednesday, March 24, 2010

### The two-electron well: part 2

When I left off yesterday, I had sketched out the proposed symmetric solution of the two-electron potential well. I got it by taking two simple solutions where the electrons crowded themselves to opposite sides of the box, and adding those two solutions together.

The peculiar thing is there is another possible solution that you have to also consider: it's called the anti-symmetric solution, and you get it by taking the difference instead of the sum. The wave function sketched in the coordinates of the two electrons looks like this:

And the question is: which one has lower energy - the symmetric or the antisymmetric combination? And it's not totally obvious to me what the answer should be.

So I posted this question on an interned forum, and one guy gave the very helpful suggestion of "turning off" the repulsion of the two electrons, and then turning it on again very slowly. When you turn it off, the two blobs come together (it's the product of simple sine waves); then, when you allow the electrons to repel just a little bit, this blob can only deform in one way: it has to stretch towards the two corners just like the symmetric case. So that's what the solution has to look like. And in fact the energy of the anti-symmetric case turns out to be quite a bit higher.

I have to say it was a shock for me to contemplate that the shape of the solution might depend on how strong the repulsion was between the electrons. This really wasn't what I expected. And then it occured to me that if the solution depended on the strength of the repulsion, then it also depended on the size of the box. For real electrons, whose strength is constant, you get different waveshapes as you vary the size of the containment.

This is amazing. For the case of weak repulsion, the solution approaches the simple product function of two sine waves. In other words, both electrons fill the lowest energy state with little regard for what the other one is doing. Obviously for real electrons this must correspond to the case of the very large box.

But it doesn't! It turns out that the case of weak repulsion corresponds to the very tiny box. The repulsion only gets to be important as you make the box bigger and bigger. And a funny thing happens in the limit of the very big box: the symmetric and anti-symmetric solutions converge to the same energy! (In technical terms they become "degenerate".)

That's what happens, and I'm going to explain why in my next post.

The peculiar thing is there is another possible solution that you have to also consider: it's called the anti-symmetric solution, and you get it by taking the difference instead of the sum. The wave function sketched in the coordinates of the two electrons looks like this:

And the question is: which one has lower energy - the symmetric or the antisymmetric combination? And it's not totally obvious to me what the answer should be.

So I posted this question on an interned forum, and one guy gave the very helpful suggestion of "turning off" the repulsion of the two electrons, and then turning it on again very slowly. When you turn it off, the two blobs come together (it's the product of simple sine waves); then, when you allow the electrons to repel just a little bit, this blob can only deform in one way: it has to stretch towards the two corners just like the symmetric case. So that's what the solution has to look like. And in fact the energy of the anti-symmetric case turns out to be quite a bit higher.

I have to say it was a shock for me to contemplate that the shape of the solution might depend on how strong the repulsion was between the electrons. This really wasn't what I expected. And then it occured to me that if the solution depended on the strength of the repulsion, then it also depended on the size of the box. For real electrons, whose strength is constant, you get different waveshapes as you vary the size of the containment.

This is amazing. For the case of weak repulsion, the solution approaches the simple product function of two sine waves. In other words, both electrons fill the lowest energy state with little regard for what the other one is doing. Obviously for real electrons this must correspond to the case of the very large box.

But it doesn't! It turns out that the case of weak repulsion corresponds to the very tiny box. The repulsion only gets to be important as you make the box bigger and bigger. And a funny thing happens in the limit of the very big box: the symmetric and anti-symmetric solutions converge to the same energy! (In technical terms they become "degenerate".)

That's what happens, and I'm going to explain why in my next post.

## Tuesday, March 23, 2010

### The Double-electron Potential Well

I promised to catch up on some topics I skipped over, so here goes. This problem arose when I was floundering on the helium atom; no, not the helium atom exactly, but the case of two hydrogen atoms far apart. I couldn't figure out why I kept getting "mini-helium" as a solution. So I backtracked to the simpler case of two potential wells. I thought that would help.

Then I realized I didn't know how to do even ONE potential well if it had a second electron in it. So I started playing around.

First I said, what if the two electrons make some minimal effort to stat away from each other. You know that the ground state is a sine wave, so I added a little “second harmonic” to each one. Something like so:

Then I realized I didn't know how to do even ONE potential well if it had a second electron in it. So I started playing around.

First I said, what if the two electrons make some minimal effort to stat away from each other. You know that the ground state is a sine wave, so I added a little “second harmonic” to each one. Something like so:

Now you see electron A is crowding towards the left, and electron B towards the right. So far so good. But the thing about two electrons, is they really want you to analyze them in terms of what they call the six-dimensional wave function in the vector coordinates of both electrons. At least for 3-d space, you get a six-dimensional wave function. For the one-dimensional well, you just get a two-dimensional plot, where the two axes are the coordinate for each electron. It looks something like this:

I've taken the wave function to be a direct product of A and B, which is the simplest thing you can do. It means other than the general crowding over, there is no particular detailed correlation between A and B. You can see from the graph that the wave function density is the highest when A is towards the left and B is towards the right.

The next step is a little interesting. In quantum mechanics you're not allowed to have a wave function that distinguishes in any meaningful way between two electrons. If you have A here and B there, you have to say that it's just as likely that B is here and A is there. So the function I've drawn is technically illegal. There is a way to make it right however, and it's called symmetrization. You just reverse A and B to get a new function. (It will look the same except the blob will be in the lower left corner instead of the upper right. Then you just add the two functions together. That way, the roles of A and B are automatically interchangeable:

And that's it: that's the hypothetical wave function for the two-electron well, chosen so that the electrons make some effort to stay away from each other, and symmetrized to comply with the requirement that the electrons are fundamentally indistinguishable.

It turns out there is another different way to symmetrize our wave function, but I'm going to have to wait til my next post to talk about that...

## Thursday, March 18, 2010

### I'm getting way behind

It's hard for me to understand but I've done an awful lot of physics since I started this blog. Two blogs ago I said something about having some more stuff to post, and the next thing I posted was Quantum Siphoning. That's not what I was planning. It just came up all of a sudden, and it might actually be the solution to the long standing measurement problem, the so-called "collapse of the wave function". So it had to be posted.

But meanwhile a couple of cool problems got left aside. They were the fallout from my flirtation with mini-helium. One problem was the case of two electrons in a potential well. It has a surprising outcome. The other problem is the iso-electronic series of helium. It's also a two-electron problem, and the solution to the simple potential well pointed me in the right direction for the more complicated case. I actually wrote up the isoelectronic series and submitted it to a journal. But I'm still going to post it.

The one other very recent calculation that I haven't put up yet is my thermodynamic argument on how the reaction AgBr ---> Ag + 1/2 Br2 becomes spontaneous in the conditions of photographic exposure. That result led to my discovery of Quantum Siphoning, so I really should post the calculation. I'm getting around to it...

But meanwhile a couple of cool problems got left aside. They were the fallout from my flirtation with mini-helium. One problem was the case of two electrons in a potential well. It has a surprising outcome. The other problem is the iso-electronic series of helium. It's also a two-electron problem, and the solution to the simple potential well pointed me in the right direction for the more complicated case. I actually wrote up the isoelectronic series and submitted it to a journal. But I'm still going to post it.

The one other very recent calculation that I haven't put up yet is my thermodynamic argument on how the reaction AgBr ---> Ag + 1/2 Br2 becomes spontaneous in the conditions of photographic exposure. That result led to my discovery of Quantum Siphoning, so I really should post the calculation. I'm getting around to it...

### Ramanujan and the Casimir Effect

Damn, there's a lot of stuff I need to post. This is a calculation I did last year. The writeup it's pretty much barebones, but it's a very cool calculation.

EDIT: Right now (Jan 19 2014) this post is getting more hits than usual because of an upsurge of interest in the Ramanujan series. So I thought I'd revisit the whole thing and see if I couldn't explain it any better. You'll find my updated version here.)

The bizarre thing is how this weird mathematical series actually plays a role in modern physics. It can be used to calculate the Casimir Effect. Here's how the calculation goes.

1 + 2 + 3 .... - (4 + 8 + 12...)

which is...1 - 2 + 3 - 4 + 5 .... = 0.25!

If you take the pressure of the lowest mode in the smaller box to be 4, you see that the pressure difference between the small box and the big box is one sixteenth of the lowest-mode pressure. You can then take a chain of boxes, each one double the next one, to get the pressure relative to infinity: 1/16 + 1/64 + 1/256...= 1/12

And that's the calculatation. It's not really right for the three dimensional case, but I think if you apply it to a small cube and take the pressure defect to be 1/12 of the lowest mode pressure, you get something close to the ballpark for the parallel plate if you fill the gap with those small cubes. Or something like that. It's certainly true that the actual Casimir effect pressure is a definite fraction of the lowest mode pressure.

EDIT: Right now (Jan 19 2014) this post is getting more hits than usual because of an upsurge of interest in the Ramanujan series. So I thought I'd revisit the whole thing and see if I couldn't explain it any better. You'll find my updated version here.)

Once in a long while I actually calculate something. I think I figured out how to do the Casimir effect. And I used that series 1 + 2 - 3 + 4....

First the series: it's one of those bizarre and counterintuitive results credited to the tragic Indian prodigy Ramanujan. For reasons that no one seems to know, he could intuitively recognize that certain alternating series which would appear to anyone else to be divergent actually had a finite sum. His conjectures were tested and almost always found to be correct.

I don't know how the mathematicians do it, but I can also make these series add up using my own little tricks. For example, the alternating arithmetic progression written above: you put it in an Excel spredsheet and then spread an envelope over it, a very gentle Gaussian. It quickly adds up to 0.25 which is the right answer. You can put the same Gaussian over the alternating squares, and it adds up to zero which is also correct. Of course you need a wider envelope to swallow up the much bigger terms.

I don't know how the mathematicians do it, but I can also make these series add up using my own little tricks. For example, the alternating arithmetic progression written above: you put it in an Excel spredsheet and then spread an envelope over it, a very gentle Gaussian. It quickly adds up to 0.25 which is the right answer. You can put the same Gaussian over the alternating squares, and it adds up to zero which is also correct. Of course you need a wider envelope to swallow up the much bigger terms.

The bizarre thing is how this weird mathematical series actually plays a role in modern physics. It can be used to calculate the Casimir Effect. Here's how the calculation goes.

We simplify things a little by putting everything in a one-dimensional box. So the energy modes are 1,2,3, etc. We can choose our units so that the pressure is numerically equal to the energy.

Now make the box twice small. The energy modes are 2,4,6...etc. But the PRESSURE is energy per unit volume (length, since it's one-dimensional)...so the pressures are (get this:) 4,8,12...

What's the difference in pressure between the big box and the small box?

1 + 2 + 3 .... - (4 + 8 + 12...)

which is...1 - 2 + 3 - 4 + 5 .... = 0.25!

If you take the pressure of the lowest mode in the smaller box to be 4, you see that the pressure difference between the small box and the big box is one sixteenth of the lowest-mode pressure. You can then take a chain of boxes, each one double the next one, to get the pressure relative to infinity: 1/16 + 1/64 + 1/256...= 1/12

And that's the calculatation. It's not really right for the three dimensional case, but I think if you apply it to a small cube and take the pressure defect to be 1/12 of the lowest mode pressure, you get something close to the ballpark for the parallel plate if you fill the gap with those small cubes. Or something like that. It's certainly true that the actual Casimir effect pressure is a definite fraction of the lowest mode pressure.

## Wednesday, March 17, 2010

### Quantum Siphoning

“The collapse of the wave function” has been with us for eighty-some years (almost a hundred if we include its immediate precursor, the “quantum leap”). It has to be considered one of the great philosophical bugaboos (?) of quantum mechanics and modern physics in general. But what exactly is it? If we could possibly agree on the definition of wave function collapse, I imagine it would have to be something close to the following: that in a physical process, governed by laws of nature in the form of differential equations, where the state of a system (the wave function) evolves continously through time, there comes a moment when something happens to the wave function at one place, and simultaneously, the wave function everywhere else simply ceases to exist. It’s a paradox.

Perhaps there are people who would find fault with the above characterisation of wave function collapse. No matter; there are surely countless examples which we could draw on to demonstrate the phenomenon. Unfortunately, I have not been able to find anywhere a published list of the 10 or 100 most compelling examples of wave function collapse. Quite the contrary; the Wikipedia article for example contains not a single concrete instance (although it does link to an article on Schroedingers Cat). So I have no choice but to go out on a limb and choose my own example.

I choose to focus on the specks of silver which appear on a photographic plate when it is exposed to the weak light of a distant star. Few would dispute that at least in the mass culture, this example would be near the top of almost anybody’s list. But in precisely what manner does it so convincingly epitomize the concept? We will find that question is not so easy to answer.

Superficially, the argument probably goes something like this. The reduction of metallic silver from silver bromide takes a certain amount of energy. This energy can only come the light of the distant star. We can consider light in the form of electromagnetic waves, and easily calculate its power density at the location of the silver atom. It is easy to show that this density, calculated over the cross section of the atom, is far too low to account for the chemical reduction of silver in any realistic time frame. Ipso facto, collapse of the wave function.

This argument is shaky on the grounds that it uses classical electromagnetic light as the driving force. Classical wave functions don’t collapse; only quantum mechanical ones do. In quantum mechanics, light is made of point particles called photons. There is no philosophical problem with a photon striking a silver atom and driving the conversion; it’s just a matter of probability. No need to collapse a wave function.

The resolution is of course to combine the two arguments. Particle or not, the photon is still governed by a quantum mechanical wave function which is essentially the same as the classical wave function. Until a speck of silver appears on the plate, the wave function of the photon is spread out everywhere. Then, at the moment the speck appears, the wave function of the photon vanishes everywhere. It has to vanish everywhere because all of its energy was required at that one point in space in order to drive the conversion, of which the visible evidence is the residual fleck of silver.

----------------------------

But there is another way in which one can challenge the idea of wave function collapse in the case of the photographic plate. Is it really true that the reduction reduction of silver from silver bromide requires the full measure of energy from a single photon? At first glance, the question seems almost absurd in its naivety. The enthalpy of the chemical reaction

AgBr ------------> Ag + ½ Br2

is 99 kJ/mol, or nearly one electron volt per atom of silver, and it is in the positive direction, which means it requires the input of energy.

However, it turns out that the detailed physical chemistry of the process is somewhat more complicated than the simple one-line reaction written above, and even today it is probably not fully understood to the last detail. It is said that the role of crystal defects and trace impurities is critical in the efficient functioning of photographic film.

Furthermore, in calculating the free energy of the reaction from the enthalpy above, we must not forget to include the effect of concentration. A silver bromide crystal containing trillions of atoms may be developable after exposure even if it contains a mere handful of reduced silver atoms; and there is a fascinating thermodynamic argument based the on the entropy of mixing which shows that at such concentrations, the point of equilibrium shifts so far to the right as to make it at least arguably plausible (plausibly arguable?) that there may be locations within the crystal where the reduction of silver is in fact thermodynamically favored.

All this must surely seem highly speculative. But it is upon this faint hope which the argument to follow hinges. I am suggesting the possibility that the appearance of the silver fleck represents the transition of the crystal from a less stable to a more stable state energetically. And therefore it did not require the full energy of the photon to procede.

Even this radical assumption does not solve the problem of “collapse of the wave function”! Let us see why not.

The problem has to do with the energy barrier associated with the transition. In accordance with our proposed suspension of disbelief, we are going to suppose that the exposed AgBr crystal with its reduced silver atom is in a lower energy state than the undisturbed crystal. If this were the whole story, we would not need the energy of a photon to drive the transition. But the problem is in getting from A to B. According to our best understanding, the incident photon liberates a valence band electron by promoting it to the conduction band. I would like to say definitively how much energy this takes, but I have not been able to find this data. It is certainly a known fact that photographic film can be safely exposed to the deep red light of a dark room, but I haven’t found the actual cutoff frequency. Let’s assume it is 2 eV for the sake of argument.

The problem is that once we assume the photon is absorbed in promoting the electron to the valence band, we have implicitly assumed the collapse of the photon’s wave function. And that’s just what I’m trying to avoid.

The goal is to follow the process through the time evolution of the Schroedinger wave function to see if we can bring it to completion without at any stage invoking the collapse of the wave function. To facilitate this analysis, I have simplified the system to what I hope will be most easily manageable while retaining the essential features of the actual process.

What I have analyzed is the case of two potential wells inside a box. One well (A) is at a slightly higher potential than the other (B), and it is at (A) that the electron is trapped. Both wells are located inside a larger box (C) so the problem is confined to a finite volume.

So far, so good. But in what follows, it is a little tedious to have to remember which is A, which is B, and which is C. So I’m going to rename them to make things a little easier to follow. Remember, these are just names, and they aren’t to be taken as meaning any more than that.

We will refer to the well at A as “the silver halide”. The well at B, at a slightly lower potential , will be called “the silver atom”. And the big box, C, will be called “the conduction band”. Remember, despite what we call them, they are just two potential wells inside a box, and the similarities to any real chemical processes are, if not purely coincidence, at least incomplete. Oh, one more thing: the potential required to promote an electron from the “silver halide” to the “conduction band” (you see how this is going to work?) will be called “the band gap”.

We will assume that for one reason or another, the electron cannot get from the silver halide to the silver atom by tunneling. In our model we can arrange for this by having the two wells sufficiently far apart. We will further stipulate that the quantum of energy required to get into the conduction band is 2 eV.

What I proposed was that we consider what happens if instead of absorbing all the energy from a discrete photon, silver halide absorbs from a continuous electromagnetic field, only a fraction of that quantum of energy before the field disappears. Then the electron is then left in a superposition of states; say, 90% silver halide and 10% conduction band.. Since the conduction band is also coupled to the silver atom, the wave function may evolve further so that the electron finds itself in a superposition of 90% silver halide, 10% conduction band and 1% silver atom. (Due to rounding, percentages may not add to exactly 100.) And now the question: where does the system go from here? In particular, is it possible for the system to end up with the electron entirely in the silver atom?

Using the Born rule, standard QM would seem to tell us that there is a 90% probability of finding it in the silver halide, a 10% chance of finding in the conduction band, and a 1% chance of finding it in the silver atom. But that is not especially helpful. For one thing, as long as the conduction band is partially filled, there will be radiating energy which we are going to assume can escape from the box. And as the radiant energy is lost from the system, the conduction band will be depleted to the benefit of both the silver halide and the silver. We might then expect the system to stabilize in the proportions of 90/10 silver halide to silver. The exact percentages don’t matter; the point is, there is a probability of finding the electron in the silver atom.

But the Born rule which we used to get those probabilities essentially demands the collapse of the wave function, and that’s just the thing we’re trying to avoid. We avoided it at the point the photon was absorbed by saying we wouldn’t take a whole photon, we’d just take a portion of a continuous e-m wave. It defeats our whole purpose to now invoke the Born Rule to force it into the silver atom. We return to the question: can the electron get from the mixed state of silver/silver halide/condutcion band to a state where it is entirely in the silver atom, and can it get there by means of a process which evolves naturally in space and time?!!

There is enough energy available to drive the process; that’s not the problem. When the conduction band couples to the silver atom, it is true that the probability drains toward the silver which is what we want. And energy is released which we would like to use in order to replenish the silver halide, to keep the process moving. Very much like an ordinary siphon. The probelm is that the energy at the silver atom is released in the form of radiating electromagnetic energy, and it was not clear to me how could I recapture it with good efficiency at the silver halide, especially if the two are relatively far apart.

I though about this for a long time and couldn’t get around it. Then the answer came to me. You can’t do it. You can’t recapture the energy lost radiated away from the silver atom, receive it at the silver halide, and pump it back up to the conduction band.

But you can do it if you have millions and millions of silver halide sites! We will call such a collection of potential wells (remember, that’s all they are) a “crystal”. In the very middle of the crystal is one special well, the “silver atom”, just a little deeper than all the other wells, and it is empty.

Now pass a wave of light through the “crystal”. Not a very strong wave, but at a frequency sufficient to couple the silver halides to the conduction band. When the wave is gone, the conduction band is excited to the extent of 10%, and all the silver halides ground state wave functions are filled to the extent of only 99.99999%.

Notice carefully that according to my picture, you haven’t captured enough energy to put a whole electron into the conduction band. Just 10% of one. And yet now I’m going to show how the system will evolve through time so that the electron ends up in the silver atom. The process is what I call quantum siphoning.

It’s not hard to get a little bit of wave function excitation happening in the silver atom. There’s plenty available in the conduction band to excite it to the level of, say 1%. But as it starts to fills up, the conduction band is emptying out, and it is not being replenished. The energy which was available is simply being radiated away from the site of the silver atom.

But the silver atom is surrounded by millions of receiving antennas which are tuned to its exact frequency. These are none other than all the surrounding silver halides. As the electromagnetic wave radiates outwards, it can’t help but excite the millions of potential wells in its path. And as it does, it leaves a little of its energy behind at each one, getting weaker and weaker as it goes. At the same time a wee liitle portion of all the electron wave functions in those millions of silver halides is is promoted to the conduction band. In fact, the penetration depth of the em wave emanating from the silver atom decays exponentially, until essentially one hundred percent of it is absorbed. And every bit of that absorbed energy goes towards replenishing the conduction band. There is literally nowhere else for it to go. It is in fact the quantum mechanical version of a siphon. The replenished conduction band can now continue to refill the silver atom ground state until it’s completely full. Full of exactly one electron.

Let us recap: A weak, diffuse electromagnetic wave passed through the crystal. A small portion of a quantum of energy was absorbed in the passage, and the effect of this absorption was recorded by the appearance of an electron in the silver atom. Every step of the process occured according to the ordinary time evolution laws of the Schroedinger and Maxwell equations, and at no point was it necessary for a wave to “collapse to an eigenfunction of the state being measured.”

And finally, not to argue the point to strongly, although it is just a though experiment with potential wells scattered in a big box, it does seem to shares some of the characteristic features of the traditional photographic plate, with its flecks of silver appearing one by one under the influence of very weak starlight. Is light made of photons? Does the wave function collapse? Or can we not perhaps explain it all by wave-on-wave interactions according to the well-known laws of Schroedinger and Maxwell?

## Sunday, March 14, 2010

### I found my mistake

It's been a while, but I finally got to the bottom of my problem with the helium atoms. You remember that I had two isolated protons and I tried to solve the Schroedinger equation by sharing two electrons between them so each atom looked like a miniature helium. Now I know what I did wrong.

You can account for the energy of the system by adding up five terms. They are:

(1) the kinetic energy of electron A

(2) the kinetic energy of electron B

(3) the potential energy of electron A

(4) the potential energy of electron B

(5) the repulsion energy of electron A versus electron B

If you have a solution to the Shroedinger equation, and you make a new wave function where all these terms are exact multiples of your old solution, then the new wave function will also be a solution. That's what I was trying to do.

I took the helium atom solution and spread it out in space so it was twice as wide. Then I cloned it and put one replica at proton A and one replica at proton B. Looking at the five components of system energy, it appeared to me that each one was exactly one quarter of the original, giving me a valid solution. That was my mistake.

The kinetic energy of electron A is indeed one quarter of the original, and so is the kinetic energy of electron B. It works because the del-squared operator automatically gives you one-quarter the result when you double the linear dimension.

The potential energy of electron A is also one quarter of the original, as is the potential energy of electron B. It works because at each atom you have one-eighth the energy: half the nuclear charge, half the electron charge, and twice the distance. At first glance you might think there ought to be extra terms in the potential energy on account of the attraction of proton A for electron B and vice versa, but I can reduce these terms arbitrarily close to zero by putting the atoms far apart. No, the potential energy works out OK. It is the repulsion energy which is messed up.

The repuslion energy of the two electrons appears at first glance to work out exactly the same as the potential energy. At each atom you have half an electron repelling half an electron at twice the distance: one-eighth the energy. Double it for the second atom and you are back to one quarter, so everything seems proportional. But it isn't.

I am not a fan of the probability density interpretation of the wave function but in this instance I don't have a better explanation. The interpretation that works is not that you have half an electron repelling half an electron. It is that you have a 50% probability of a whole electron

repelling a whole electron. This gives you twice the energy as what I calculated, so this term goes out of whack with the other four terms.

It has to work this way because otherwise, you could apply this technique in the opposite direction and solve the doubly ionized beryllium atom (Be++) as a squeezed-down replica of the helium atom. All the energy levels would be exactly four times as big. In fact you do just this when going from the hydrogen atom to the He+ ion. It works in that case because with only one

electron there is no repulsion term. The isoelectronic series of hydrogen consists scaled copies of the identical wave function. But the isoelectronic series of helium doesn't work that way.

So I can't create mini-helium by sharing two electrons between two isolated protons. But that doesn't mean my problem doesn't have a solution. It just means that the wave function I chose does not minimize the energy of the system. There is a solution, and it is in the shape of the hydrogen negative ion, a little-known form of hydrogen with an extra electron. It seems that there is just enough attraction between a hydrogen atom and a free electron to make this a stable species.

So it means you can take the wave function of H- and clone it so each proton gets a copy. Then you share the two electrons between the two protons. It looks strange but it's a solution of the Schroedinger equation. Each proton has two "half-electrons" bound to it.

There's a more conventional solution where both electrons go to one proton and the other proton sits there all alone. That's the familiar solution. Is there a relationship between the two solutions?

Yes, and it's called symmetrization. It's something you actually do all the time in quantum mechanics. You observe that there is nothing special about A or B, so any solution which distinguishes them must have a counterpart where the roles are reversed. You take these two complementary solutions and make a new solution by adding them together. That's called symmetrization and it gives you my distributed mini-ions. Using the hydrogen negative ion of course rather than helium.

There is another side to symmetrization: if you can take the sum of two wave functions you can also take the difference. It's called, not without some logic, "anti-symmetrization". In quantum mechanics you can always do that with states: instead of working with states A and B, you work with the sums and differences: A+B (symmetric) and A-B (antisymmetric). What if you're really interested in what happened to state A? Easy: just apply the same trick to your new states. You can see that if you add the symmetric and antisymmetric states together, it just returns you to state A. The difference gives you back state B.

When I started this blog a month ago I said I was fed up with physics because I couldn't solve a single problem with two electrons in it. The funny thing is maybe I have now. Not just this problem but a couple more, which I'm going to talk about in my next post.

You can account for the energy of the system by adding up five terms. They are:

(1) the kinetic energy of electron A

(2) the kinetic energy of electron B

(3) the potential energy of electron A

(4) the potential energy of electron B

(5) the repulsion energy of electron A versus electron B

If you have a solution to the Shroedinger equation, and you make a new wave function where all these terms are exact multiples of your old solution, then the new wave function will also be a solution. That's what I was trying to do.

I took the helium atom solution and spread it out in space so it was twice as wide. Then I cloned it and put one replica at proton A and one replica at proton B. Looking at the five components of system energy, it appeared to me that each one was exactly one quarter of the original, giving me a valid solution. That was my mistake.

The kinetic energy of electron A is indeed one quarter of the original, and so is the kinetic energy of electron B. It works because the del-squared operator automatically gives you one-quarter the result when you double the linear dimension.

The potential energy of electron A is also one quarter of the original, as is the potential energy of electron B. It works because at each atom you have one-eighth the energy: half the nuclear charge, half the electron charge, and twice the distance. At first glance you might think there ought to be extra terms in the potential energy on account of the attraction of proton A for electron B and vice versa, but I can reduce these terms arbitrarily close to zero by putting the atoms far apart. No, the potential energy works out OK. It is the repulsion energy which is messed up.

The repuslion energy of the two electrons appears at first glance to work out exactly the same as the potential energy. At each atom you have half an electron repelling half an electron at twice the distance: one-eighth the energy. Double it for the second atom and you are back to one quarter, so everything seems proportional. But it isn't.

I am not a fan of the probability density interpretation of the wave function but in this instance I don't have a better explanation. The interpretation that works is not that you have half an electron repelling half an electron. It is that you have a 50% probability of a whole electron

repelling a whole electron. This gives you twice the energy as what I calculated, so this term goes out of whack with the other four terms.

It has to work this way because otherwise, you could apply this technique in the opposite direction and solve the doubly ionized beryllium atom (Be++) as a squeezed-down replica of the helium atom. All the energy levels would be exactly four times as big. In fact you do just this when going from the hydrogen atom to the He+ ion. It works in that case because with only one

electron there is no repulsion term. The isoelectronic series of hydrogen consists scaled copies of the identical wave function. But the isoelectronic series of helium doesn't work that way.

So I can't create mini-helium by sharing two electrons between two isolated protons. But that doesn't mean my problem doesn't have a solution. It just means that the wave function I chose does not minimize the energy of the system. There is a solution, and it is in the shape of the hydrogen negative ion, a little-known form of hydrogen with an extra electron. It seems that there is just enough attraction between a hydrogen atom and a free electron to make this a stable species.

So it means you can take the wave function of H- and clone it so each proton gets a copy. Then you share the two electrons between the two protons. It looks strange but it's a solution of the Schroedinger equation. Each proton has two "half-electrons" bound to it.

There's a more conventional solution where both electrons go to one proton and the other proton sits there all alone. That's the familiar solution. Is there a relationship between the two solutions?

Yes, and it's called symmetrization. It's something you actually do all the time in quantum mechanics. You observe that there is nothing special about A or B, so any solution which distinguishes them must have a counterpart where the roles are reversed. You take these two complementary solutions and make a new solution by adding them together. That's called symmetrization and it gives you my distributed mini-ions. Using the hydrogen negative ion of course rather than helium.

There is another side to symmetrization: if you can take the sum of two wave functions you can also take the difference. It's called, not without some logic, "anti-symmetrization". In quantum mechanics you can always do that with states: instead of working with states A and B, you work with the sums and differences: A+B (symmetric) and A-B (antisymmetric). What if you're really interested in what happened to state A? Easy: just apply the same trick to your new states. You can see that if you add the symmetric and antisymmetric states together, it just returns you to state A. The difference gives you back state B.

When I started this blog a month ago I said I was fed up with physics because I couldn't solve a single problem with two electrons in it. The funny thing is maybe I have now. Not just this problem but a couple more, which I'm going to talk about in my next post.

## Sunday, February 14, 2010

### The clicking detectors

The detection of clicks or spots on a photographic plate has long been held to be clear evidence for the particle nature of light. But attempts to make a mathematical case have always fallen just short. The reason is that the statistics are ambiguous. It turns out the classical theory supports the facts just as well as the quantum theory.

By "classical" I'm talking about the theory where the waves are classical but the atoms are quantum. You have to make one assumption: that the detection probability, say of a silver halide crystal changing state, is proportional to the intensity of the incident wave. Why should a crystal behave this way? That's another question, but at least it's a reasonable working theory. It is of course the only theory which makes the classical wave agree with the photon picture, and it happens to be a reasonably plausible theory.

It works fine when you have lots of photons, but gets tricky when you go down to "single photons". You do this by attenuating your source so the photons only strike once every ten minutes. Wouldn't the waves be too weak to concentrate their power into a detection event? Not according to our working assumption. Remember, by the way, the appearance of a dot on a photographic plate is what we call thermodynamically spontaneous - you don't need a net energy input to drive the transition. The wave theory still works even down to the very lowest levels.

But now let's add a twist. We're going to split the beam with a half-silvered mirror. Now the wave and particle theories make different predictions. The wave procedes at half strength to each detector, so there's some chance that both will click. But the particle goes one way or the other, so both detectors can never click.

Or can they? The problem is you're never quite sure that you have only one photon, no matter how weak the beam. Let's do the statistics. Take a beam with one photon every ten seconds. Use a one-second "coincidence" window. The beam splits in two so each detector fires every twenty seconds (100% efficiency). The wave theory says that there is a certain, constant wave intensity at each detector, that they fire on average every twenty seconds, and it doesn't matter if one or the other fires. In any given second, there is a one-in-twenty chance of either detector firing, and a one-in-four-hundred chance of both.

What do photons do? Photons are different, because if Detector A fires, it "uses up" the energy that "might have" also fired Detector B. That's the theory anyways. What do the statistics say?

Let the photons obey Poisson statistics (typical for laser). You can see where this is going. In one second there is a one-in-ten chance of a photon. What are the chances of two photons within the same "window"? One in...get this...

*two*hundred. (Write the formula for Poisson if you don't believe me). But wait...just because there are two photons doesn't mean that you get two clicks. They might both go to the same detector. Four ways...AA, AB, BA, and BB. Only half the options give you coincidence counts. The odds? One in four hundred.You can see that we can make the beam weaker and the window tighter, and it still doesn't hlep. The photon theory and the wave theory will always predict exactly the same number of clicks.

Now...IF you could fire one photon at a time, there is a very simple experiment which would totally go against the wave theory. You just put up a beam splitter and wait. With single photons, you can NEVER get two clicks. And the wave theory predicts that whatever your detection efficiency, call it n, you should get two clicks with a probability of n-squared/4. (25% of the time for 100% efficiency). It's a very simple experiment.

Interestingly enough, the much-referenced paper by Thorne et al http://physics.princeton.edu/~mcdonald/examples/QM/thorn_ajp_72_1210_04.pdf

proposes an experiment which is generally similar to what I've described except it is decidedly more complicated. There is a third beam used for gating, and there are electronic coincidence detectors. Then you have to calculate something called "second-order coherence". It's just overly complicated enough that I might plausibly claim to not be convinced by the results. But of course that would be unreasonable of me.

Yet I have to wonder. The authors of this experiment make it a major selling point that it clearly demonstrates the particle nature of light at a level suitable for an undergraduate lab. Yet their version is far more complicated than my bare-bones version. If the case is so clear-cut, why the complications?

Yet I have to wonder. The authors of this experiment make it a major selling point that it clearly demonstrates the particle nature of light at a level suitable for an undergraduate lab. Yet their version is far more complicated than my bare-bones version. If the case is so clear-cut, why the complications?

## Wednesday, February 10, 2010

### Something I just figured out yesterday

People go on and on about the deep mysteries of quantum mechanics. It seems like the double slit experiment is a huge mind-bender for people. Not for me. I can do the double-slit calculations and they make sense. Basically I can handle stuff in quantum mechanics when there is one electron at a time. It's when you add the second electron that it gets messed up.

I finally decided that it should be possible to work some things out by taking the very simplest two-electron case. No, it's not the Helium atom, although that's also an important one. No, it's not the Hydrogen molecule either. No, it's not the scattering of one electron by another. And it's not even the infinite potential well with two electrons in it. I can't do any of these problems.

I finally decided that it should be possible to work some things out by taking the very simplest two-electron case. No, it's not the Helium atom, although that's also an important one. No, it's not the Hydrogen molecule either. No, it's not the scattering of one electron by another. And it's not even the infinite potential well with two electrons in it. I can't do any of these problems.

What I've tried to do is the very simplest possible case. It's the case of two isolated hydrogen atoms, treated as a single system with two electrons. Now why would you do it that way? You can solve for a single hydrogen atom, and then if you have a second one, you've already solved it. It's not exactly easy to solve for one electron in the sense that you've got to differentiate stuff in polar coordinates, but it's at least straighforward.

But that's why I chose to do it as a two-electron problem. You just get the familiar solution centered around each atom, and then combine them into symmetric and antisymmetric combinations. The symmetric state is the lowest energy.

Now bring in the electrons. You can put the first electron in that ground state and it is then shared between the two atoms. If you want to localize the electron at one atom, you have to use the combination of symmetric plus antisymmetric. But since the energies of these two states are slightly different, they gradually go out of phase with each other and eventually you find the electron at the other atom, even if that other atom is quite far away. That's what they call tunneling.

But that's why I chose to do it as a two-electron problem. You just get the familiar solution centered around each atom, and then combine them into symmetric and antisymmetric combinations. The symmetric state is the lowest energy.

Now bring in the electrons. You can put the first electron in that ground state and it is then shared between the two atoms. If you want to localize the electron at one atom, you have to use the combination of symmetric plus antisymmetric. But since the energies of these two states are slightly different, they gradually go out of phase with each other and eventually you find the electron at the other atom, even if that other atom is quite far away. That's what they call tunneling.

What gets interesting is the second electron. Forget everything you heard about two electrons being able to occupy the same state if they have the opposite spin: technically that's correct, but it fools people into thinking they can just drop two electrons into the states they've already calculated. While that would give you two nice hydrogen atoms at first glance, it's not really correct. It ignores the interaction energy between the two electrons.

If you've ever glanced at the solution for the helium atom it starts by writing the six-dimensional Schroedinger equation which basically has terms for the potential and kinetic energy similar to the hydrogen atom, plus a new term in 1/(r1-r2) which is the interaction potential. Now, what occurs to me in the case of two hydrogen atoms, is that it's exactly the same equation. The way nature solves this equation in practise is that it anti-correlates the two electrons by putting one here whenever the other one is there, so the interaction term basically gives you zero and you get the familiar solutions. But it occurs to me: why can't you let the hydrogen atoms solve this problem exactly the same way that helium does? In other words, take the "known" solution for helium - well, you can't easily write it down but somehow nature "knows" a solution" - and scale it by a factor of 2 on length so it fits the hydrogen atom. How can this not possibly be a correct solution of the Schroedinger equation? True, it's not the lowest energy solution, but it's still a solution.

So each hydrogen atom becomes a mini-helium, with two half-electrons vying for the same space. The energy is exaclty one-quarter of the helium ground state energy of -76 eV, which comes to something like -9.5 eV per electron. Of course, you can't ionize it with 9.5 eV, you need to use 19 eV because its a coupled system and you have to ionize both at the same time. But regardless, you've got a mathematical state which it seems to me should have some physical consequences.

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