e^+ е^-  -> µ^+ µ^- annihilation. Processes e^+e^-\to\bar q q and e^+e^-\to hadrons at high energies. Process e \mu\to e \mu and cross symmetry
Annihilation of electrons and positrons into a muonic pair. It was described in this diagram: a positron collides with an electron and a pair of muons is produced
Let me remind you that the electron is considered to be the particle e, so the particle is μ^-. If here the momenta are -p_1 and p_2, then here they are -p_3 and p_4
The Mandelstam variables...
We’ve studied out what the differential cross section of this  process looks like. In the center-of-mass system it will be a quantity proportional to (1 + cos ^{2}θ),
which corresponds to --- this is worth remembering… for high energies, when s is much larger than the squared quadrupled muon mass. … corresponds to... α^2/(3s)
Under the same conditions, the cross section itself is (4/3)(α^2/s), which  looks like 87 nb divided by s, which in this case is expressed in (GeV)^2
Such is the picture we studied out in quite detail last time. This picture will give us an approach to a variety of other interesting things
As always, I’ll also try to make our calculations within the abstract quantum field theory, in its lower orders too, related to certain processes, including those that are investigated at BINP
The first today’s theme is the processes that are relatives to this one. First of all, processes with production of quarks
and processes with production of hadrons on colliding high-energy beams
Let's start with the first process, which corresponds to annihilation of a lepton pair to a pair of quarks. As we know, there are several types of quarks, so here we point out that this is annihilation to a concrete quark-antiquark pair
If relevant diagrams are drawn, the beginning will be exactly the same: the positron and electron are annihilated into a virtual photon with the momentum p_1 + p_2, and then a pair of quarks is produced
Here we have a quark of a sort a, and here we have an antiquark of the sort a. Let's see the difference from this reaction
Two things, actually. Firstly, another vertex, i.e. here we have the particle charge, whereas here we have the quark charge
This is the first major difference where Q_a is a dimensionless quantity corresponding to the charge of the quarks in the elementary charge units.
The second, also significant, difference is that unlike muons, quarks have an additional quantum number of color, which does not show here, except for the fact that there may be three such colors
Therefore, if we compare the cross section of the process e+e- to a quark-antiquark pair with that cross section, which is a standard one,
the difference is due primarily to the fact that an extra factor --- the quark charge --- will occur in the amplitude, and an extra factor --- the squared quark charge (Q_a)^2 --- will occur in the cross section
Secondly, for these final particles we will have to perform summation --- in addition to summing over the spins --- over the colors, too
The three colors will give an additional factor of 3. This is the main difference between this process and this process. If we are talking of production of hadrons,
it is a complex process, and in circumstances of production of hadrons from an electron-positron pair of low energy, we have, in principle, the same picture
Here we have some block of the transition into hadrons, and here we have a virtual photon, which has the momentum p_1 + p_2, and thus, with a virtuality that corresponds to the value q^2, produces this hadron system
The characteristic distances that are included in the problem will be of the order of ћ/sqrt(q^2) -- again, from the ratio of uncertainties. In our case, ћ/sqrt(s)
If we are talking of the birth of hadrons near the threshold… Where does the threshold begin? When a pair of pions is produced; that is about 280 MeV
It is easy to figure out that the corresponding distance is at the level of 10^{-13} cm. This distance is characteristic to strong interactions
Q: Is it the distance the beams have to be drawn together? 
A: These are distances at which hadrons are produced, which govern this block of production of hadrons
On the one hand, we believe that all this must be described by quantum chromodynamics. On the other hand, this offers not much use, because we cannot solve exactly the equations of quantum chromodynamics,
and, as classics wrote, only in a state of deep despair can one use the perturbation theory with the corresponding constant not small
Therefore, in these conditions, while we are near the threshold, we have to use a variety of models, and often the answer looks very tricky
Namely, when talking of the birth of charged particles π^+π^-, at first everything looks very similar to quantum electrodynamics,
i. e. in some approximation, π^+π^- can be considered as point particles, but this is only an approximation
And if we go a little bit further, there arise catastrophic differences from quantum electrodynamics (in which a pion is regarded as a point particle): rise of resonances and so on
Experimenters, including those working with colliding e^+e^- beams, are trying to sort out this mess, this spectroscopic area, with utmost precision
Models are used and are constantly refined and so on. But when it comes to high energies, we can start using quantum chromodynamics, because everything happens at short distances
At short distances we can use the perturbation theory of quantum chromodynamics
What will happen in the first order, or rather, in the lowest order of the perturbation theory of quantum chromodynamics? Quarks production, which we can calculate
And we can try to do the calculation not only in the lowest approximation, but also considering further, subsequent terms of the perturbation theory of quantum chromodynamics.
The subsequent terms will correspond to the possibility of exchanges between the produced quarks via gluons; there may be emitted additional gluons
All this will not be easy, but we have a guiding thread of the perturbation theory, and hence all these complications will be in the next order and will be corrections, at least in some sense, and the main truth will take place here
Then, from this main truth, we can calculate this cross section. This is good; everything takes place at short distances. However, there appear hadrons,
and the transformation of quarks into a set of hadrons again occurs at large distances, and thus we can say nothing special, basing on the first principles
So, what could be the use of this quantum-chromodynamics situation? Well, we knew how quarks are produced at small distances
How can we traverse a bridge to production of hadrons at high energies? Due to a simple assertion that the quarks will certainly turn into hadrons, with absolute certainty
Therefore, we calculate this section, and with absolute certainty it will be the cross section of this process
Therefore, we can say that this complex process can be described by this one, which is relatively simple, with corrections
Q: Is the reverse is true? When hadrons are produced, do they necessarily pass the quark stage?
 It may be the quark stage or the gluon one. Do you understand? In this sense, not always the reverse is true. This is particularly clearly seen in collisions of strongly interacting particles
E.g. at the LHC, where protons collide, the occurring elementary processes can be collisions of not only quarks, but gluons too, and often these processes dominate
But here the picture is, in this sense, simpler, and therefore it is said that even if some particles are discovered on colliding proton-proton beams,
their detailed study will still be done on colliding electron-positron beams, where the picture is much clearer
Q: In general, is it not one type of quarks, one flavor, that makes contribution to the cross section into hadrons?
 Absolutely right. Let's see how the energy increase may change this picture
If we take the ratio of the cross section of hadron production to the cross section of muon production (this value is referred to as just "ratio" R), how could we express it, according to this result?
That would be three quark charges squared, and summed over all those quarks ... which quarks? Quarks allowed by this statement; this answer was true in the region of large masses,
and this answer will be true if s is much larger than the squared four quark masses
Let's try to see what it will equal. If the energies are such that we go a little bit beyond the resonance region, where we cannot directly derive anything in terms of the fundamental theory,
then beyond the region, e.g. of 1.5-2 GeV in the total energy, a set of "up" and "down" quarks of the first generation may participate, as well as a strange quark
What does this yield? I take out the common factor of 3; the "up" quark has a charge of 2/3; we take the square of 2/3; the "down" and "strange" quarks have a charge of 1/3 each; so two summands are added here
What do we finally get?
Four ... five, six ... and total  "two". This is a kind of prediction for R. In which area? In an area where the energy is for sure considerably larger than twice the mass of the quarks
If the energy is such, this set of quarks is complemented with a charmed quark c, then, in addition to the things that were here, i.e. in addition to this "two", a contribution corresponding to the c quark will be added
And that will be the same "three", multiplied by the squared charge of the c quark, which will give us 2 plus 4/3
If, as a result, we will be able to watch ... and when can that be? When we move beyond the threshold of production of c quarks, which is about 3 GeV in energy
If we go beyond the threshold of production of b quark, we’ll have, in addition to this standard set, the b quark contribution, the same as the contribution of d or s quark. I.e. that will be 1/9 …by 3 ... plus 1/3
Let's try to see --- pure theory ---- what are relevant experimental results? It is quite interesting to compare. Of course, this is a simplest reasoning,
because, apart from this process in the first order, there are another processes where gluon exchange occurs, where gluons can be emitted additionally, and so further
All this is taken into account in accurate calculations, but the resulting picture is very impressive
Let’s look at it
Look at the first picture, which shows just the cross section of hadron production in e+e- collisions. The cross section in millibarns is laid in the vertical scale
Let me remind that the millibarn is 10^(-27)cm^ 2, because the barn is 10^(-24) cm^2. This cross this section is considered huge as a barn
What can we see from this picture? The value √s is laid in the bottom. If this cross section behaves as the muon cross section does, we definitely see that this quantity is decreasing quadratically with increasing s
Then, in a logarithmic scale, which is used both vertically and horizontally, it must be a straight line
This is clearly seen. As soon as we cross a border of 1.5 or 2 GeV… What do we see in the beginning?
First, there is a ρ meson peak; then a narrow ω peak, even more narrow φ peak, and a ρ' peak. As soon as we pass beyond ρ', everything is arranged in roughly straight lines, within a huge interval
We see production of the J/ψ family, and then, before the Z boson, there is the Υ (Upsilon) family. Everything is in an almost straight line, which corresponds to this relationship 1/s
And look how good the range is: from about 2-3 GeV to about 80 GeV; and near 90 GeV there is a gift of nature, Z boson, which exceeds the background level by 3 orders
Meanwhile, the cross section decreases from 10^(-4) mbarn to 10^(-7), i.e. by 3 orders. But this picture shows what the cross section looks like. The lower curve shows the ratio R
Again (in the same scale) these are almost straight lines. After ρ', there goes one straight line to ψ, then to Υ, and a small jump after Υ
In fact, it looks as if we pass from 2 to 2+4/3 and to 2+4/3+1/3. Of course, the scale is logarithmic, and it's not seen very well, but what will you see in the next page?
The same picture, but in a smaller scale. Look at the very top: this is the area where, according to our ideas, R should be defined by this value, i.e. about 2. And what do we see?
Experimental points in the range of 1.5 to 3 GeV: a lot of them near ρ', and then only separate points. By the way, VEPP-4M used to work well in this area
All of them are very well fitted by a straight horizontal line
The dotted line below corresponds to this prediction, which is called the "naive quark model", in which a pair of quarks and nothing more is taken
The straight line above represents the refined quantum chromodynamics with corrections and so on
It is seen that quantum chromodynamics is in better agreement with experiment. The next piece includes the interval from 3 to 5 GeV
We see some “song and dance” because of the rise of resonances associated with c quarks. But after that, again good agreement with the quantum chromodynamics predictions
But the dotted line is not constant, because of attempts to take the threshold effect into account. First such a value and then we go to this one. Production of c and anti-c quarks begins
Finally, the next piece, approximately from 9 to 11 GeV. It is the area where production of ͞bb quarks begins
We can see that except for the narrow resonances the experiment is in remarkably and amazingly good agreement with the quantum chromodynamics predictions
Thus this science really has a very rational kernel, confirmed by experiment. That was about the structure of hadron production in e^+e^- collisions
Further, I said that if we considered the angular distribution of muons at high energies, it would be described by this simple relationship
By the way, this simple relationship is strongly connected with the assertion that the muon has spin ½
If a scalar particle were produced here, the relationship would be substantially different
On the one hand, all this is very good and useful, because the distribution is broad and detectors that cannot operate with small angles cover a significant part of this cross section
Next, when it comes to production of quarks in the lowest order of quantum chromodynamics, the picture must be about the same: the angular distribution of quarks must be described by this formula
Of course, there will be complications because of corrections, but the main relationship will be the same: a general dependence of the type 1+cos^{2}θ
If quarks had another spin, they would have a different distribution, but quarks are objects we cannot see directly
On the other hand, these quarks for certain turn into hadrons. At high energies, the hadrons look like jets
Indeed, at high energies, generally speaking, the hadron distribution is broad, but this broad distribution often has clearly notable jets
Examination of the angular distribution of these jets confirms this relationship, and thus they came from particles with spin ½
This is one of the first experimental evidence on quarks. Of course, what we are watching is a prediction based on the lowest order of the perturbation theory
Generally speaking, there could be production of gluons, which must also turn into jets
Such three-jet events were also observed with good energy of the e+e- collision. Moreover, in a sense, experimental research on the hadronization revealed differences between quark and gluon jets
In these pictures, the gluon jet is slightly wider and has less energy than the quark jet. So, from this point of view, study of the processes of the lower orders allows us to touch the mysteries of hadron physics
In conclusion of this section, let me mention that in addition to this kind of processes that are directly studied in the annihilation of electrons and positrons it would be possible, with appropriate accelerators at hand,
to watch processes in which muons are produced in collision of photons, if we had colliding beams of photons of sufficient energy
The answer looks approximately like this, but with some significant modifications because of a different factor and the dependence being not only 1/s, but also ln(s)
But it does not matter; it is important that at high energies this cross section is the same as the cross section of annihilation of electrons and positrons, and even larger
But now, if we proceed to the production of hadrons in gamma-gamma collisions and start considering quarks, the answer will include not one but two quark apexes,
and then the cross section of this process will be proportional to… the amplitude will be proportional to the squared quark charge and the cross section will be proportional to the quark charge to the power of four
We can introduce our own R for γγ collisions --- as opposed to this one ---, which will be equal to the ratio of the cross section of "γγ into hadrons" to that of "γγ into μ^+μ^-"
That will be a substantially different function: it will be proportional to the fourth power of quark charges
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And that function will be much more sensitive to the charge of quarks than here, because of the fourth degree: if the quark charges were not standard, not 2/3 or 1/3, that would have been felt
You will study a process of this kind at the seminars, but there will be nothing new except for some complexification
Therefore, the ideas used in quantum chromodynamics in calculation of such processes are the same as there
Now a question: this thing has been well known for a long time and has been studied right here, and what can we say about such opportunities?
Where will they come from? Where can one take photon beams similar to electron and positron beams?
With a lot of high energy photons in the beams? Of course, we know beams with a lot of photons --- laser flashes. I'm going to talk of them
A laser flash can comprise many times more photons than electrons in VEPP-4M bunches
However, the laser physics is applied in the visible light; the energies are of the order of eV, and here we need the same energies as in colliding beams, i.e. we need photons with energies of many GeV
There are no such devices so far. However, a roundabout way was found. In fact, a process of this sort can occur as a virtual one
We have talked of virtual particles, and now it is the time to use this idea and technology for such processes
Let’s consider colliding electron-positron beams in which electrons and positrons collide with each other and are not annihilated but scattered on each other, emitting photons
A positron is scattered on an electron; each of these leptons emits photons, and then the photons create this system or this system,
which at high energies turn into either hadron jets or even into a set of hadrons, hadron resonances, and so on
I told you about the benefits of the transition from the formal expansion of the perturbation theory to virtual particles and so on
Processes of this kind (with production of particles lighter than muons, electrons) were first observed on colliding beams
The problem is that when this radiation is emitted, the energy of our photons is considerably less than the energy of electrons and positrons
For the first time, the processes of this kind, the so called two-photon processes, were observed here, on VEPP-4, without the letter "M" yet, and there was observed production of electrons and positrons
Since the energy of photons was considerably less than the energy of initial beams, the energy of produced electrons and positrons was also significantly lower
Later on, processes of this kind have been observed at other major accelerators, and thus these processes but with virtual photons also were studied well,
but in less detail as compared with annihilation processes of this sort. However, they yielded a lot of useful information
Q: Like radiative corrections?
You know, this is not radiative corrections; this is a separate process
With this we conclude this topic, and I would like to talk once again of the initial process, which yielded so many useful things
Let’s consider a process that involves the same particles that were here, but in another incarnation. For example, if in this reaction we move the positron into the final state
and the muon into the initial state, then we will have a process of scattering, not annihilation: an electron is scattered on a muon with the same final states
Consideration of this process will allow us to talk of crossing symmetry
This is a very deep and, most importantly, very useful idea of the quantum field theory. It can be felt through this simple example
Break.
Let’s revert to this process and recall that when we begin to calculate such a diagram, at each vertex we have such a factor:
-i by e, coming just from the constant of the interaction Lagrangian and T exponent, and the Dirac matrix --- let's designate it as γ^α
Then here we have, accordingly, a similar quantity and another matrix. Let’s recall the matrix element corresponding to this process. I wrote it down last time, so I want to check how you remember it
I ask you to write down the matrix element of this process
First let's write these factors 
(student) Minus (ie) squared. Then we have γ^α and γ^β ...
No, we begin, as always, with the fermion line. We grasp the end. What will we have here? 
(student) Here we’ll have the factor …
No, no, no, we are writing the matrix element, the scattering amplitude, there are no exponents anymore
This will be an honest common factor, without denominators: minus (ie) squared. Great. And then the spinor describing the initial positron
Q: Why a positron? 
Positron. We are going along the line from the end. 
(student) Counterclockwise
Q: Were not we going to calculate another process? 
Now we are recalling this process. So, what is here?
(student) v.
The conjugate of v.  Let’s designate it with index 1 – with the index of "momentum and helicity." Then vertex, what is there?
(student) γ^α and --- here we have the electron in the initial state --- u with the index 2
That’s all. Then the respective summing over the spinor indices will be performed. Then goes the propagator with the factor i
Do not forget to write "i". The general rule is that both then and here you need to write i. Further, what does the propagator look like?
(student) Here we have k^2. 
What will k^2 be in this case?
 (student) It will be s
Yes. Of course, formally, we would have to add +i 0, but in this case s is a positive value, so we can omit i 0. What is there in the numerator?
 (student) In the numerator ... 
Minus ... What else? 4π and g^{..}. In Feynman gauge this will be g linking the beginning and the end ... 
(student) αβ
If here α was above, then ... 
(student) Yes, from the bottom
Further
(student) Now we need to start for this. We go ... 
Begin from here, from the end
(student) Yes, we write down the muon in the final state. Then we get u (with the index 4) the conjugate of u_4; then goes γ^{αβ}; then goes v_3
That is all. Fine. We recalled the calculation of the matrix element. We’ll do absolutely the same way for this process. What will this process look like? We have a heavy particle
Let’s draw a picture for comparison: the heavy particle muon (we refer to μ^- as a particle) and the light particle of electron. And we can write down the same things that were further
Then we made the following procedure. Of course, we throw this i away. We squared the matrix element. So what? What did we do?
The electrons and positrons are non-polarized, and we do not distinguish the polarization of the muons, and what did we do?
We performed summing over the polarizations of the final muons and averaging over the polarizations of the initial electrons,
i.e. we performed summing over σ_1 with a factor of 1/2 (this is averaging) and over σ_2 also with a factor of ½
And then we got the following picture: ¼; this factor of 4π; e^2 = α, all squared; this factor s squared. Further were spurs
How did we arrange them? I ask you to recall that. And then we could immediately say that since there was such a factor, we could eliminate it and make the indexes identical, for the picture to be simpler. So, please
Here we’ll have two spurs. How will we write them?
S:I’m trying to recall ... 
Yes, please. Again we catch over this line, and what in the spur will correspond to this external positron?
After summing up what? You can take v by the conjugate of v, perform summing over its helicities, and there will arise a nice factor, which includes the momentum of this lepton multiplied by the gamma matrix;
p_1, struck-through. We introduced such a thing…
(student) Yes, the Feynman slash
... p ^μ by γ_μ is the Feynman slash. And plus or minus the mass of the particle; minus for the antiparticle. Minus the mass of the electron. Further was γ^α; then this was matched by a similar structure
(student) Plus ...
…plus the mass of the electron. Then all this was multiplied by the conjugate value; so there arose an identical vertex, but the summing was different: β, γ ...
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No, no, we had to multiply this piece, with summation over α, by the conjugate value ...
... and the conjugate value gives a similar thing, but over another summation index, over β. Therefore here we must have such a quantity. How will we write down this piece?
(student) also with a spur
There again, we clutch at the end; it will be matched by ... 
(student) p_4
p_4, slash and --- this is a particle, which implies the plus sign --- the mass of the muon. Further.. 
(student) γ^α
γ^α. This antiparticle was matched by ... 
(student) p_3 minus the mass of the muon
And when we multiplied that by the conjugate value, then we had summing over the other vertex, γ^β. To make it good, if here α was above, then here it is below, and, respectively, the same with β. Fine
Then we put this square of the matrix element in the expression for the cross section and obtained what we needed. Sit down please
What will change if we start considering this picture? Purposely I’ll take very different momenta: let this be the momentum p; this p'; here we’ll have the momentum Ƥ and here the momentum Ƥ'
All these will be particles. What will the matrix element look like? Everybody is to work at the blackboard
(student) The propagator will be the same. 
Then write γ^α and γ_α here too
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 (student) Will we directly write γ^α?
Yes, let's write it directly, i.e. we’ll perform summation over the indices in mind. Well, what will yield? Look upwards; do not forget this thing. Then, what will we take from here?
(student) the electron in the final state, i.e. \bar u ... 
\bar u of p'. Further ...
(student) γ^α, and here we have the electron part. 
Yes, again, u of p. And this all is a self-closed reflection. What do we take here?
(student)? (…) 
Let's write the propagator
(student) How is the momentum transfer written? 
You do write it. Draw a line as you like, e.g. downwards.
(student) OK
What will the momentum be equal to? 
(student) p-p '. 
OK. Below will be p-p' squared
(student) And here we need also to write i, minus, 4π, and g was taken into account ... 
g^{αβ} was taken into account. Now, this thing ...
(student) By the way, perhaps we need to add the index (e) 
It may be so
(student) And here we’ll have the muon ... 
Conjugate?
(student) Yes, in the final state, and it has the momentum Ƥ; γ_α and U ...
Ƥ, of the muon. Fine. If we compare one thing with the other, what will happen? Let's try to understand
The initial antiparticle, the positron, turned into the electron in the end. The law of conservation for this process will be written as follows: p + Ƥ = p' + Ƥ'
The law of conservation here was as follows: p_1+p_2 = p_3+p_4. So, judging by the conservation laws, what happened? Let's think
This particle, the electron, remained as it was in the beginning. Hence, in the transition from this process to this one, which we call the crossing process, there were such changes: p_2 was replaced with…what?
This particle remained as it was; p_2 was replaced with p. The final muon remained as it was: Ƥ'. The other particles underwent radical changes
The initial positron p_1 turned into the final electron. If we pull p_1 this way, it will go with the minus sign, so the transition of particle to antiparticle is accompanied by a sign change
In this picture, what did p_1 turn into?
 (student) -p '? 
-p'. And p_3, which was the antimuon in the end, turned into the initial muon
What are the changes? What will p_3 turn into? Into -Ƥ. Let's see what we got in the end?
Instead of s, which was equal to (p_1+p_2) squared, in strict reasoning we got such a quantity that is fully described by these rules
Similarly, everything will be the same here: antiparticle ͞v_1 turned into ... this, yes?
No, the initial antiparticle turned into the particle at the end ... No, no, no, not so: p_2 turned into p. Sorry, the picture will be like this
And the initial antiparticle becomes the particle; the antiparticle was in the beginning and the particle was in the end. A similar story happens to the muons
If now I try to unwind the square of the matrix element, which is,
again, summed over all the final and averaged over all the initial polarizations, what will I get?
As before, this factor will appear and will now be divided, not by (p_1+p_2)^2, but, as we said, by (p-p')^2
And then again, two spurs. It’s easy to see that these spurs will also obey the same rule of permutation. Let's do this without hurry, to be sure in all this
Q: Over there, in the square of the matrix element, should not there be the fourth power of the momentum in the denominator?
I’d ask you ... How will we write the spur? 
(student) Going from the end ...
What will match it? 
(student) p' plus the mass of the electron m_e
Strikethrough  p'. Further... 
(student) γ^α
Further...  
(student) p plus the electron mass
Is p also strikethrough? 
(student) Yes
and when I multiply this expression by the conjugate one… here we have the summing over γ^α and γ^α, and in the conjugate expression, for example, the summation over γ^β and γ^β… So, it will be so
 (student) We take the lower line, and again go from the end. It will be ... 
Ƥ', yes. Further... 
 (student) Here is the subscript. 
Yes ... right
If the transition in the matrix element is quite tricky and complicated ...
Look, the conjugate bispinors of the antiparticle in the beginning were replaced with the bispinors of the final particle, and the same was the case with v_3, which was replaced with u of Ƥ
All that regards the spurs is in a pure form. Yes? Do you have a question? What is the question? 
Q: This expression, ¼ M squared…
This is the matrix element squared and summed over the spins of the final particles and averaged over the spins of the initial particles. Here and there, we took exactly the same; namely this thing must enter the cross section
Note that no tricks here: no spinors; just replacement of momenta and nothing more. This opportunity to describe such processes, related or so-to-say crossing, with simple replacements of 4-momenta
is one of the biggest advantages of our covariant perturbation theory
No need in a new calculation; it is sufficient to take the expression for the square of the matrix element of the initial process and perform a replacement of the 4-momenta in it,
and we get the square of the matrix element of the crossing process
Then we meet a difference, because the cross section is expressed not only via the square of the matrix element,
but it also includes the momenta of the initial and final particles: they are a little bit different in these two reactions. But that will be a simple, purely kinematic change
As regards the most difficult part, i.e. the calculation of the square of the matrix element, it is done in such a simple way
After that (as you remember, this square of the matrix element has been calculated) we replace the 4-momenta it in, then express the 4-momenta via quantities in whatever system and get the final answer
In particular, very interesting ... so, we assume that we’ve done all this in mind and I’ll write only the part that will be
of interest to this piece. Let’s consider an option, when the following process occurs:
the muon is at rest and the electron is scattered. The energy of the electron is much smaller than the mass of the muon. But this electron, since the muon mass is about 105 MeV, may be relativistic as well
It is possible, yes? In this case the answer is as follows
In these circumstances, the muon experiences very small recoil, and then this can be considered as scattering of an electron (including, perhaps, relativistic one) on an immovable center,
which is the Coulomb field of the muon. For scattering on an immovable center of the Coulomb type the answer is known --- it is the Rutherford cross section. Let me recall that the Rutherford cross section
looks approximately the same for the relativistic and non-relativistic cases
and has such a form: α^2, 4v^2 by p^2 and by this remarkable factor, sin^4(θ/2), which appeared in Rutherford's experiments
The difference between the relativistic and non-relativistic pieces is as follows: in the non-relativistic case, the momentum was mv, and in the relativistic case the Lorentz factor was added;
in this case, v is the electron velocity
Such is the Rutherford cross section. Our answer, obtained from the exact result of this  will look like this: it will contain the Rutherford cross section,
with such an additional multiplier. I remind you that, in fact, this v should be written as v/c. In the non-relativistic case, this thing is not interesting at all; it can be discarded.
Whereas in the relativistic case, also included in this consideration, this factor may be significant. How does this factor behave in dependence on the angle θ?
In forward scattering, this is exactly 1, and our quantum electrodynamic cross section does not differ from the classical Rutherford one
With the increasing angle, what value does the sine reach? When θ is equal to π, here we’ll have π/2 and the sine is equal to 1
And this thing is cut off; the result is (1-v^2), and for a relativistic particle this may be a value very close to zero. So, for a relativistic particle, this additional factor behaves somewhat like this
I.e. the Rutherford cross section is already rapidly decreasing with increasing angle. In particular, for the electron, there occurs additional suppression due to this factor
When you study the quantum electrodynamics in more detail, the root causes of the appearance of this factor will be disclosed: it is associated with the conservation of the helicity of the electron at high energies
This concludes our study of relatives of this simple process. Let’s move on to the penultimate process that we consider today in this course
it's the Compton scattering, i.e. the process γe→ γe
It is convenient to compare this process with its analogue as concerns scalar particles. We had the process π^0π^-→π^0π^-. I recall that it was described with two diagrams
So, π^0 and π^- could go into virtual π-, and then π^0 and π^- appeared again
That was a process in which exchange of virtual π^- occurred in the s channel. And another option we also considered …
... It looks like this: π^0, π^-, ... π^0, π^-. In this case, in comparison with this process,
that will be exchange of virtual π^- in the u channel. If we now take the Compton process γe→ γe, then everything will be very similar...
... and the Feynman diagrams will be just the same: the electron, photon, photon, and electron ---exchange of virtual electrons in the s channel
And the second diagram, with exchange of the virtual electron in the t channel. If you recall the respective contributions to the matrix element,
e.g. to this one ... I’ll write it in full: i multiplied by the scattering amplitude includes -ig squared ... The vertexes are here
No spinors or polarization vectors here, and thus only this propagator is included. The propagator for the exchange in the s channel will contain a momentum equal to ...
Now we need to distinguish between… If indications here are as follows: k_1, k_2, p_1, and p_2, then I'll stick to the same notation here: k_1, k_2, p_1, and p_2
The propagator will depend on the sum of the momenta, (k_1 + p_1). The second diagram will contain the propagator depending on this sum of momenta: this is k_1; this is p_2,
and thus this momentum will be (p_2 - k_1). A similar thing can be written here, and we'll do that later, but the following is important
As regards the external vertexes, we know what to do to them. However, as regards the propagator that is found here, it is a new object and we have to learn more about it
So let's talk a little bit more of the propagator of the electron
Here we can do without big guns, which will be applied starting from the expression of the operator of the S matrix, S operator in the second order, and so on. Let’s just use the known things: we know how these propagators were found
Let me recall. There we had the propagator of the scalar particle (π^-), which looked as follows: it was a matrix element over the vacuum ... of what?
of the T product of the operator φ(x) by φ†(x'). No need in writing this: we had all this in the coordinate representation
Then we rewrite this all in the momentum representation, and analyzing all this in detail, ultimately, we found this value D(p) ...
That was such a simple thing ... We can certainly say that in relation to the electron, everything will be the same: here will be the matrix element of the T product of ... of what?
Instead of the operator φ we’ll have the operator Ψ(x) and the Dirac conjugate of Ψ. So far, only the first complication: Dirac conjugation instead of the usual conjugation
The second, not so pleasant complexification is because of the fact that these are spinors; they have their own indexes, e.g. this one has the index j, and this one the index k
And thus the result should be referred to as the propagator (often designated with the letter G),
but now that will be a matrix with its indexes, which can be expanded to a sum of plane waves and so on.. But we can do a little bit trickier
We saw that D(x) is none other than Green's function for the Klein-Fock-Gordon equation. We checked that if the operator that entered the Klein-Fock-Gordon equation
acted on the D(x), the result will be... I'll have to ... what is this operator? This operator is i∂^μ; this operator is i∂_μ, ...
This operator in turn was a denominator we already met, and the Fourier transform of it was included
If this operator acts on the exponential, the denominator, of course, will remain, as well as the exponential...
but if this thing acts on this exponential, an additional factor (-ip) will pop up, and this i will give just p. In turn, the covariant operator will also give p but with a subscript; m^2 will remain. And what do we get?
The numerator and the denominator will cancel each other, and the integral of such exponential is well known.
(student) This will yield a delta function
A delta function of x. So if we look at the beginning, this is the operator from the Klein-Fock -Gordon equation, but the right side includes the delta function instead of zero
We can try to apply the same technology here: this propagator must be Green's function for the Dirac equation. If we take use of this, we’ll get the answer very fast. Let’s do this next time, without hurry
