It’s vitally important that we listen to other points of view. No, you may not reply to that.
Inter faith buffets still continue, but they’re fewer and more strained. There are plenty of people who want to exploit division.
Fortunately, we have the Invisible Magic Friend to tell us what absolute truth is. And the great thing about the absolute truths from the Invisible Magic Friend is that there are so many to choose from.
Most of the interesting stuff is either too low, or behind trees, or too faint for my telescope. T Corona Borealis still hasn’t gone bang.
My usual way to take photos is to try and get a few hours sleep and then get up about midnight, but this doesn’t work for T Cor Bor any more. By 1am it’s behind my neighbour’s apple tree and out of sight. So I’ll have to start staying up late, or at least trying to, if I’m going to catch it going nova.
Moon was up a couple of nights ago.
And on the plus side, Andromeda is beginning to rise in the east. In a couple of months time it will be in the southern sky and I might get some decent pictures of it. At the moment, this is the best I can do.
Got bored and decided to stack a couple of images of the beautiful orange-blue double star, Albireo in Cygnus.
We did this by calculating the overlap of the |bc> state with the |a> state after the interaction potential, gABC, had been applied to it, and found that the result was just g.
< bc | gABC | a > = g
We now examine the case of a B and C particle approaching one another, interacting, and then moving off into the distance.
The initial state is |bc>, with one B particle and one C particle.
When the particles interact this means that their state is acted upon by the interaction potential, gABC. The result will be:
gABC | bc >
We then want the probability that this will create a final state which is also |bc>. We find this by taking the overlap.
< bc | gABC | bc >.
But this isn’t going to work. Remember that ABC creates all the permutations of the A, B and C creation and annihilation operators.
Any of the permutations that includes A-, will give
A-|bc>.
But there are no A particles in the |bc> state, so this will give us zero.
A-|bc> = 0.
However, any of the permutations with an A+ will give us A+|bc>. This creates an A particle
A+ |bc> = |abc>
But this does not correspond to our final state which only has B and C particles in it. The overlap between the two states is zero.
< bc | A+ | bc > = < bc | abc> = 0
In other words, all permutations result in zero.
< bc | gABC | bc > = 0
The calculation is saying that there is no possibility of a B and C particle interacting and giving us a B and C particle back! However, if you read the comment on last week’s post, you’ll already know that the interaction potential is just the first in a whole series of terms that need to be applied to the initial state. The first term, gABC, gives us zero, so we move to the next term, the interaction squared term.
< bc | (gABC) (gABC) | bc >
Notice that, whereas gABC by itself created 8 permutations of three operators, acting with the interaction potential twice produces 64 permutations of six operators. Despite the extraordinary simplicity of our model, where we’ve ignored almost every physical attribute of the particles, the number of calculations quickly becomes exceedingly large. Every physical attribute that we take into consideration: momentum, mass, charge, spin etc. adds a new level of complexity to the calculation. Even a simple scattering problem like this can take pages and pages of maths.
If you remember, a couple of weeks ago, when I first introduced creation and annihilation operators, I mentioned that QFT tries to get all the annihilation operators on the right. This is why. We quickly end up with very large products of fields. Unless we work to get rid of most of them then the calculations become unmanageable.
So that you know which term I’m referring to, I’m going to call the gABC on the right hand side (gABC)R and the one on the left hand side (gABC)L. These are just labels. So our desired overlap now looks like the following.
< bc | (gABC)L (gABC)R | bc >
Although (gABC)R contains eight terms, we can get rid of all the ones with an A- in them as there are no A particles in the initial state.
A- | bc > = 0
So only the terms in (gABC)R with A+ in them will count. But this means that we’ll end up with a state with an “A” particle created. Our final state also has no “A” particles in it, so (gABC)L must have an A- in it to remove the “A” particle created by (gABC)R. This reduces the total number of terms we must consider from 64 to 16, 4 with A+ in (gABC)R and 4 with A- in (gABC)L.
We can reduce the count even further. Since the initial and final states both have only a B and C particle in them, whatever (gABC)R does, (gABC)L must do the opposite. So our choice of (gABC)R completely determines what (gABC)L must be. We are now down to only four possibilities.
< bc | (gABC)L (gABC)R | bc >
= < bc | (gA-B-C-)L (gA+B+C+)R | bc > + < bc | (gA-B-C+)L (gA+B+C-)R | bc > +< bc | (gA-B+C-)L (gA+B-C+)R | bc > +< bc | (gA-B+C+)L (gA+B-C-)R | bc >
Let’s look at one of them.
(gA-B-C-)(gA+B+C+) | bc >
For simplicity, we’ll only look at the “B” particles.
B- B+ |b>
It may not seem obvious, but this is actually the same as.
B+ B- |b>
I made a point of explaining that B-B+ is quite definitely NOT the same as B+B-, and that is true. What happens in this case is that, when we swap them round, we get extra, constant, terms. And that is what happens here.
In general, switching from “time ordered” to “normal ordered” introduces a function called the “Feynman Propogator” which generally depends on the type of particle involved. As we’ve ignored all particle properties, in our case the Feynman Propogator is just a number. As all it does is produce a scale factor, I’m simply going to ignore it. The same holds true for the C particles.
So, in the end, we really just need a single term out of the original 64 combinations.
< bc | (gABC)L (gABC)R | bc >
= < bc | (gA-B+C+)L (gA+B-C-)R | bc >
The first copy of the interaction will destroy the B and C particles and create an A particle, leaving this
= < bc | (gA-B+C+)L g | a >
The second copy of the interaction will do the opposite, destroy the A particle and create a B and C particle.
= < bc | g g | bc >
As g is just a number and < bc | bc > = 1, our final answer is.
< bc | (gABC)L (gABC)R | bc > = g²
Now this calculation doesn’t really make sense. I’ve ignored everything about the particles, including their initial and final momenta. When these are included, what we get is a probability, given the initial momenta, of finding the final momenta. This in turn can be used to calculate the probability of all possible scattering angles. And this in turn can be used to calculated the classical form of the force law. In the case of electromagnetism, g is the charge of the electron and it can be shown that the scattering amplitude recovers the usual Coulomb’s law.
What I really wanted to show here was the intermediate “A” particle, that doesn’t appear in either the initial or final states. This is called a “virtual” particle, which I’ll discuss a bit more next week.
In real life systems, dealing with the large number of combinations of operators can get very messy. (If you really want to mess with your brains, have a look at Wick’s Theorem, https://en.wikipedia.org/wiki/Wick%27s_theorem .) But this was the way QFT was done until that nice Mr Feynman came along and taught everyone how to draw pictures instead.
In our final post on QFT, we’ll take a look at Feynman diagrams.
Platitude of the Day 